4. Quantum Fields and Antiparticles#

In this chapter we will construct the Hamiltonians in the form of \(H = H_0 + V\), where \(H_0\) is the Hamiltonian of free particles, and \(V = \int d^3 x~\Hscr(0, \xbf)\) is a (small) interaction term in the form of Eq.2.5.11, and the interaction density \(\Hscr(x)\) is a Lorentz scalar in the sense of Eq.2.5.12 and satisfies the cluster decomposition principle Eq.2.5.14. As a byproduct of the construction, we’ll also demystify the so-called antiparticles which have been mentioned a number of times so far without definition.

4.1. Symmetries and Quantum Fields#

Following the discussions in The Cluster Decomposition Principle, we’ll construct \(\Hscr(x)\) out of creation and annihilation operators. However, as we’ve seen in The Lorentz and CPT transformation laws, the Lorentz transformation of both \(a^{\dagger}(q)\) and \(a(q)\) involve in the coefficients a matrix element \(D_{\sigma \sigma'}(W(\Lambda, p))\) that depend on the momenta of the particles, and hence are not scalars. The idea, then, is to construct \(\Hscr(x)\) out of the so-called annihilation and creation fields defined by

(4.1.1)#\[\begin{split}\psi_{\ell}^+(x) &\coloneqq \sum_{\sigma, n} \int d^3 p~u_{\ell}(x;~\pbf, \sigma, n) a(\pbf, \sigma, n) \\ \psi_{\ell}^-(x) &\coloneqq \sum_{\sigma, n} \int d^3 p~v_{\ell}(x;~\pbf, \sigma, n) a^{\dagger}(\pbf, \sigma, n)\end{split}\]

where \(\ell\) is reserved for labeling particles later. We see, in particular, that the creation field \(\psi_{\ell}^-(x)\) is a superposition of creation operators. Applying it to the vacuum state and let \(x\) wander around the whole spacetime then creates a (quantum) field.

Note

Just like particles, fields may be either bosonic or fermionic, but not mixed. For example \(\psi^-_{\ell}(x)\) is a bosonic/fermionic if and only if all the particles created by \(a^{\dagger}\) are bosonic/fermionic, respectively.

4.1.1. Lorentz symmetries#

Now the hope is that the creation and annihilation fields transform, under (proper orthochronous) Lorentz symmetry, by a matrix that is independent of the spacetime coordinate \(x\). More precisely, we’d like the following to hold

(4.1.2)#\[\begin{split}U_0(\Lambda, b) \psi_{\ell}^+(x) U_0^{-1}(\Lambda, b) &= \sum_{\ell'} D_{\ell \ell'}(\Lambda^{-1}) \psi_{\ell'}^+ (\Lambda x + b) \\ U_0(\Lambda, b) \psi_{\ell}^-(x) U_0^{-1}(\Lambda, b) &= \sum_{\ell'} D_{\ell \ell'}(\Lambda^{-1}) \psi_{\ell'}^- (\Lambda x + b)\end{split}\]

Note that we’ve put \(\Lambda^{-1}\) inside \(D_{\ell \ell'}\) so that \(D\) furnishes a representation of the homogeneous Lorentz transformations in the sense that \(D(\Lambda_1) D(\Lambda_2) = D(\Lambda_1 \Lambda_2)\). [1] There is a priori no reason to use the same representation \(D\) for both \(\psi^{\pm}_{\ell}\), but this turns out to be possible just by calculation. Moreover, the representation \(D\) is not assumed to be irreducible. Indeed, as we’ll see later, it generally decomposes into blocks fixed by further labels.

Then we can try to construct \(\Hscr(x)\) by a formula similar to Eq.3.4.1 as follows

(4.1.3)#\[\Hscr(x) = \sum_{N,M=0}^{\infty} \sum_{\ell'_1, \cdots, \ell'_N} \sum_{\ell_1, \cdots, \ell_M} g_{\ell'_1, \cdots, \ell'_N;~\ell_1, \cdots, \ell_M} \psi_{\ell'_1}^-(x) \cdots \psi_{\ell'_N}^-(x) \psi_{\ell_1}^+(x) \cdots \psi_{\ell_M}^+(x)\]

It follows from Eq.4.1.2 that the interaction density defined by Eq.4.1.3 is a scalar if the following holds

(4.1.4)#\[\begin{split}g_{\bar{\ell}'_1, \cdots, \bar{\ell}'_N;~\bar{\ell}_1, \cdots, \bar{\ell}_M} & = \sum_{\ell'_1, \cdots, \ell'_N} \sum_{\ell_1, \cdots, \ell_M} D_{\ell'_1 \bar{\ell}'_1}(\Lambda^{-1}) \cdots D_{\ell'_N \bar{\ell}'_N}(\Lambda^{-1}) \\ &\qquad \times D_{\ell_1 \bar{\ell}_1}(\Lambda^{-1}) \cdots D_{\ell_M \bar{\ell}_M}(\Lambda^{-1}) g_{\ell'_1, \cdots, \ell'_N;~\ell_1, \cdots, \ell_M}\end{split}\]

The solution to the last problem relies on a classification of the representations of the Lorentz group, which has been discussed (in terms of the little group representations) in One-Particle States, and shall be dealt with at a later point. The main goal of this section is to pin down the conditions \(u_{\ell}\) and \(v_{\ell}\) must satisfy so that they can be explicitly solved in the following sections.

For this section, we’ll focus on the massive case. Recall the Lorentz transformation laws for the creation and annihilation of massive particles Eq.3.2.8, Eq.3.2.9 as follows

(4.1.5)#\[\begin{split}U_0(\Lambda, b) a(\pbf, \sigma, n) U_0^{-1}(\Lambda, b) &= e^{\ifrak b \cdot \Lambda p} \sqrt{\frac{(\Lambda p)_0}{p_0}} D^{(j_n)}_{\sigma \sigma'}(W^{-1}(\Lambda, p)) a(\pbf_{\Lambda}, \sigma', n) \\ U_0(\Lambda, b) a^{\dagger}(\pbf, \sigma, n) U_0^{-1}(\Lambda, b) &= e^{-\ifrak b \cdot \Lambda p} \sqrt{\frac{(\Lambda p)_0}{p_0}} D^{(j_n) \ast}_{\sigma \sigma'}(W^{-1}(\Lambda, p)) a^{\dagger}(\pbf_{\Lambda}, \sigma', n)\end{split}\]

where we’ve used the fact that \(D\) is unitary in the sense that \(D^{\dagger} = D^{-1}\) to invert \(W(\Lambda, p)\) (and flip the indexes) for later convenience – mostly because of the use of \(\Lambda^{-1}\) in Eq.4.1.2.

Using Eq.4.1.5 and Eq.4.1.1, we can compute the left-hand-side of Eq.4.1.2 as follows

\[\begin{split}& U_0(\Lambda, b) \psi_{\ell}^+(x) U_0^{-1}(\Lambda, b) \\ &\quad = \sum_{\sigma, n} \int d^3 p~u_{\ell}(x;~\pbf, \sigma, n) U_0(\Lambda, b) a(\pbf, \sigma, n) U_0^{-1}(\Lambda, b) \\ &\quad = \sum_{\sigma, \sigma', n} \int d^3 p~u_{\ell}(x;~\pbf, \sigma, n) e^{\ifrak b \cdot \Lambda p} \sqrt{\frac{(\Lambda p)_0}{p_0}} D^{(j_n)}_{\sigma \sigma'} (W^{-1}(\Lambda, p)) a(\pbf_{\Lambda}, \sigma', n) \\ &\quad = \sum_{\sigma, \sigma', n} \int d^3 (\Lambda p)~u_{\ell}(x;~\pbf, \sigma, n) e^{\ifrak b \cdot \Lambda p} \sqrt{\frac{p_0}{(\Lambda p)_0}} D^{(j_n)}_{\sigma \sigma'}(W^{-1}(\Lambda, p)) a(\pbf_{\Lambda}, \sigma', n) \\ &\quad = \sum_{\sigma', n} \int d^3 (\Lambda p) \blue{\sum_{\sigma} u_{\ell}(x;~\pbf, \sigma, n) e^{\ifrak b \cdot \Lambda p} \sqrt{\frac{p_0}{(\Lambda p)_0}} D^{(j_n)}_{\sigma \sigma'}(W^{-1}(\Lambda, p))} a(\pbf_{\Lambda}, \sigma', n)\end{split}\]

where in the last equality we’ve used that the fact from Eq.1.3.9 that \(d^3 p / p_0\) is Lorentz invariant.

Now the right-hand-side of Eq.4.1.2 can be calculated as follows

\[\begin{split}\begin{align*} & \sum_{\ell'} D_{\ell \ell'}(\Lambda^{-1}) \psi^+_{\ell'}(\Lambda x + b) \\ &\quad = \sum_{\ell'} D_{\ell \ell'}(\Lambda^{-1}) \sum_{\sigma, n} \int d^3 p~u_{\ell'}(\Lambda x + b;~\pbf, \sigma, n) a(\pbf, \sigma, n) \\ &\quad = \sum_{\sigma', n} \sum_{\ell'} D_{\ell \ell'}(\Lambda^{-1}) \int d^3 (\Lambda p)~u_{\ell'}(\Lambda x + b;~\pbf_{\Lambda}, \sigma', n) a(\pbf_{\Lambda}, \sigma', n) \\ &\quad = \sum_{\sigma', n} \int d^3 (\Lambda p)~\blue{\sum_{\ell'} D_{\ell \ell'}(\Lambda^{-1}) u_{\ell'}(\Lambda x + b;~\pbf_{\Lambda}, \sigma', n)} a(\pbf_{\Lambda}, \sigma', n) \end{align*}\end{split}\]

Equating the blue parts the two calculations, and inverting \(D^{(j_n)}_{\sigma \sigma'} (W^{-1}(\Lambda, p))\) and \(D_{\ell \ell'}(\Lambda^{-1})\), we get

(4.1.6)#\[\sqrt{\frac{p_0}{(\Lambda p)_0}} e^{\ifrak b \cdot \Lambda p} \sum_{\ell} D_{\ell' \ell}(\Lambda) u_{\ell}(x;~\pbf, \sigma, n) = \sum_{\sigma'} u_{\ell'}(\Lambda x + b;~\pbf_{\Lambda}, \sigma', n) D_{\sigma' \sigma}^{(j_n)} (W(\Lambda, p))\]

A parallel calculation for the creation field in Eq.4.1.2, which we’ll omit, gives the following

(4.1.7)#\[\sqrt{\frac{p_0}{(\Lambda p)_0}} e^{-\ifrak b \cdot \Lambda p} \sum_{\ell} D_{\ell' \ell}(\Lambda) v_{\ell}(x;~\pbf, \sigma, n) = \sum_{\sigma'} v_{\ell'}(\Lambda x + b;~\pbf_{\Lambda}, \sigma', n) D^{(j_n) \ast}_{\sigma' \sigma}(W(\Lambda, p))\]

The identities Eq.4.1.6 and Eq.4.1.7 pose the fundamental conditions on \(u_{\ell}\) and \(v_{\ell}\), respectively, which we’ll utilize to eventually solve for their solutions. Currently both \(u_{\ell}\) and \(v_{\ell}\) depend on \(x, \pbf, \sigma\) and \(n\), and the goal is to use the Lorentz symmetry to reduce the dependencies. This will be carried out in three steps, corresponding to the three types of Lorentz symmetries: translations, boosts, and rotations, as follows.

Translations

Taking \(\Lambda = 1\) in Eq.4.1.7 gives \(\exp(\ifrak b \cdot p) u_{\ell}(x;~\pbf, \sigma, n) = u_{\ell}(x + b;~\pbf, \sigma, n)\), which then implies

(4.1.8)#\[u_{\ell}(x;~\pbf, \sigma, n) = (2\pi)^{-3/2} e^{\ifrak p \cdot x} u_{\ell}(\pbf, \sigma, n)\]

and similarly

(4.1.9)#\[v_{\ell}(x;~\pbf, \sigma, n) = (2\pi)^{-3/2} e^{-\ifrak p \cdot x} v_{\ell}(\pbf, \sigma, n)\]

where we’ve slightly abused notations by keeping the names of \(u_{\ell}\) and \(v_{\ell}\), while changing their arguments. Here the seemingly redundant \((2\pi)^{-3/2}\) is inserted so that the fields

(4.1.10)#\[\begin{split}\psi^+_{\ell}(x) &= \sum_{\sigma, n} (2\pi)^{-3/2} \int d^3 p~e^{\ifrak p \cdot x} u_{\ell}(\pbf, \sigma, n) a(\pbf, \sigma, n) \\ \psi^-_{\ell}(x) &= \sum_{\sigma, n} (2\pi)^{-3/2} \int d^3 p~e^{-\ifrak p \cdot x} v_{\ell}(\pbf, \sigma, n) a^{\dagger}(\pbf, \sigma, n)\end{split}\]

look like the usual Fourier transforms.

Plugging Eq.4.1.8 and Eq.4.1.9 into Eq.4.1.6 and Eq.4.1.7, respectively, they can be simplified as follows

(4.1.11)#\[\begin{split}\sqrt{\frac{p_0}{(\Lambda p)_0}} \sum_{\ell} D_{\ell' \ell}(\Lambda) u_{\ell}(\pbf, \sigma, n) &= \sum_{\sigma'} u_{\ell'}(\pbf_{\Lambda}, \sigma', n) D^{(j_n)}_{\sigma' \sigma}(W(\Lambda, p)) \\ \sqrt{\frac{p_0}{(\Lambda p)_0}} \sum_{\ell} D_{\ell' \ell}(\Lambda) v_{\ell}(\pbf, \sigma, n) &= \sum_{\sigma'} v_{\ell'}(\pbf_{\Lambda}, \sigma', n) D^{(j_n) \ast}_{\sigma' \sigma}(W(\Lambda, p))\end{split}\]

for any homogeneous Lorentz transformation \(\Lambda\).

Boosts

Taking \(\pbf = 0\) and \(\Lambda = L(q)\) which takes a particle at rest to one with (arbitrary) momentum \(q\), we see, using Eq.1.3.5, that

\[W(\Lambda, p) = L(\Lambda p)^{-1} \Lambda L(p) = L(q)^{-1} L(q) = 1\]

In this case Eq.4.1.11 take the following form (with \(\qbf\) substituted by \(\pbf\))

(4.1.12)#\[\begin{split}\sqrt{\frac{m}{p_0}} \sum_{\ell} D_{\ell' \ell}(L(p)) u_{\ell}(0, \sigma, n) &= u_{\ell'}(\pbf, \sigma, n) \\ \sqrt{\frac{m}{p_0}} \sum_{\ell} D_{\ell' \ell}(L(p)) v_{\ell}(0, \sigma, n) &= v_{\ell'}(\pbf, \sigma, n)\end{split}\]

It follows that one can calculate \(u_{\ell}(\pbf, \sigma, n)\) for any \(\pbf\) from the special case of \(\pbf = 0\) given a representation \(D\).

Rotations

Taking \(\pbf = 0\) and \(\Lambda = \Rcal\) a \(3\)-rotation, and recalling from Eq.1.3.18 that \(W(\Lambda, p) = \Rcal\), we get special cases of Eq.4.1.11 as follows

\[\begin{split}\sum_{\ell} D_{\ell' \ell}(\Rcal) u_{\ell}(0, \sigma, n) &= \sum_{\sigma'} u_{\ell'}(0, \sigma', n) D_{\sigma' \sigma}^{(j_n)}(\Rcal) \\ \sum_{\ell} D_{\ell' \ell}(\Rcal) v_{\ell}(0, \sigma, n) &= \sum_{\sigma'} v_{\ell'}(0, \sigma', n) D_{\sigma' \sigma}^{(j_n) \ast}(\Rcal)\end{split}\]

Using Eq.1.3.11 we can further reduce it to the first order as follows

(4.1.13)#\[\begin{split}\sum_{\ell} \hat{\Jbf}_{\ell' \ell} u_{\ell}(0, \sigma, n) &= \sum_{\sigma'} u_{\ell'}(0, \sigma', n) \Jbf^{(j_n)}_{\sigma' \sigma} \\ \sum_{\ell} \hat{\Jbf}_{\ell' \ell} v_{\ell}(0, \sigma, n) &= -\sum_{\sigma'} v_{\ell'}(0, \sigma', n) \Jbf^{(j_n) \ast}_{\sigma' \sigma}\end{split}\]

where \(\hat{\Jbf}\) denotes the angular momentum vector for the representation \(D_{\ell' \ell}(\Rcal)\), in analogy with the usual angular momentum \(\Jbf^{(\jfrak)}\) for \(D^{(j)}(\Rcal)\).

4.1.2. The cluster decomposition principle#

Let’s verify that the fields defined by Eq.4.1.10, when plugged into Eq.4.1.3, indeed satisfy the cluster decomposition principle as discussed in The Cluster Decomposition Principle. It’s really just a straightforward but tedious calculation which we spell out as follows

\[\begin{split}V(t) &= \int d^3 x~\Hscr(x) \\ &= \sum_{N,M=0}^{\infty} \sum_{\ell'_1, \cdots, \ell'_N} \sum_{\ell_1, \cdots, \ell_M} g_{\ell'_1, \cdots, \ell'_N;~\ell_1, \cdots, \ell_M} \int d^3 x~\psi^-_{\ell'_1}(x) \cdots \psi^-_{\ell'_N}(x) \psi^+_{\ell_1}(x) \cdots \psi^+_{\ell_M}(x) \\ &= \sum_{N,M=0}^{\infty} \sum_{\ell'_1, \cdots, \ell'_N} \sum_{\ell_1, \cdots, \ell_M} g_{\ell'_1, \cdots, \ell'_N;~\ell_1, \cdots, \ell_M} \sum_{\sigma'_1, \cdots, \sigma'_N} \sum_{n'_1, \cdots, n'_N} \sum_{\sigma_1, \cdots, \sigma_M} \sum_{n_1, \cdots, n_M} (2\pi)^{-3(N+M)/2} \\ &\quad \times \int d^3 p'_1 \cdots d^3 p'_N d^3 p_1 \cdots d^3 p_M~\exp(\ifrak (E_1 + \cdots E_M - E'_1 - \cdots - E'_N)t) \\ &\quad \times \blue{\int d^3 x~\exp(\ifrak (\pbf_1 + \cdots \pbf_M - \pbf'_1 - \cdots - \pbf'_N) \cdot \xbf)} \\ &\quad \times v_{\ell'_1}(\pbf'_1, \sigma'_1, n'_1) \cdots v_{\ell'_N}(\pbf'_N, \sigma'_N, n'_N) u_{\ell_1}(\pbf_1, \sigma_1, n_1) \cdots u_{\ell_M}(\pbf_M, \sigma_M, n_M) \\ &\quad \times a^{\dagger}(\pbf'_1, \sigma'_1, n'_1) \cdots a^{\dagger}(\pbf'_N, \sigma'_N, n'_N) a(\pbf_1, \sigma_1, n_1) \cdots a(\pbf_M, \sigma_M, n_M) \\ &= \sum_{N,M=0}^{\infty} \sum_{\sigma'_1, \cdots, \sigma'_N} \sum_{n'_1, \cdots, n'_N} \sum_{\sigma_1, \cdots, \sigma_M} \sum_{n_1, \cdots, n_M} \int d^3 p'_1 \cdots d^3 p'_N d^3 p_1 \cdots d^3 p_M \\ &\quad \times (2\pi)^{3 - 3N/2 - 3M/2} \exp(\ifrak (E_1 + \cdots E_M - E'_1 - \cdots - E'_N)t) \\ &\quad \times a^{\dagger}(\pbf'_1, \sigma'_1, n'_1) \cdots a^{\dagger}(\pbf'_N, \sigma'_N, n'_N) a(\pbf_1, \sigma_1, n_1) \cdots a(\pbf_M, \sigma_M, n_M) \\ &\quad \times \blue{\delta^3(\pbf_1 + \cdots + \pbf_M - \pbf'_1 - \cdots - \pbf'_N)} \\ &\quad \times \Big( \sum_{\ell'_1, \cdots, \ell'_N} \sum_{\ell_1, \cdots, \ell_M} g_{\ell'_1, \cdots, \ell'_N;~\ell_1, \cdots, \ell_M} v_{\ell'_1}(\pbf'_1, \sigma'_1, n'_1) \cdots v_{\ell'_N}(\pbf'_N, \sigma'_N, n'_N) \phantom{)} \\ &\qquad \phantom{(} \times u_{\ell_1}(\pbf_1, \sigma_1, n_1) \cdots u_{\ell_M}(\pbf_M, \sigma_M, n_M) \Big)\end{split}\]

Besides re-ordering the terms, the only actual calculation is highlighted in the two blue terms, where the second one is the integral of the first. One can compare this calculation with Eq.3.4.1 and see that the cluster decomposition principle is indeed satisfied because there is a unique momentum conservation delta function in each coefficient, as long as \(g, u, v\) are reasonably smooth, i.e., it’s ok to have poles and/or branching singularities but no delta functions.

4.1.3. Causality and antiparticles#

We now turn to the other crucial condition on the Hamiltonian, namely, the causality condition Eq.2.5.14. Given the general formula Eq.4.1.3 of the interaction density, we are forced to require that \([\psi^+_{\ell}(x), \psi^-_{\ell'}(y)] = 0\) whenever \(x - y\) is space-like. However, according to Eq.4.1.10, we have

\[\begin{split}& [\psi^+_{\ell}(x), \psi^-_{\ell'}(y)]_{\pm} \\ &\quad = \sum_{\sigma, n} (2\pi)^{-3} \int d^3 p~d^3 p'~e^{\ifrak (p \cdot x - p' \cdot y)} u_{\ell}(\pbf, \sigma, n) v_{\ell'}(\pbf', \sigma, n) [a(\pbf, \sigma, n), a^{\dagger}(\pbf', \sigma, n)]_{\pm} \\ &\quad = \sum_{\sigma, n} (2\pi)^{-3} \int d^3 p~e^{\ifrak p \cdot (x - y)} u_{\ell}(\pbf, \sigma, n) v_{\ell'}(\pbf, \sigma, n)\end{split}\]

where the sign \(\pm\) is positive if the field is fermionic, and negative otherwise. This quantity is not necessarily vanishing even if \(x - y\) is space-like.

Therefore in order to construct \(\Hscr\) in the form of Eq.4.1.3 that satisfies Eq.2.5.14, we must not just use \(\psi^{\pm}(x)\) as the building blocks. It turns out that one may consider a linear combination of the two as follows

(4.1.14)#\[\psi_{\ell}(x) \coloneqq \kappa_{\ell} \psi^+_{\ell}(x) + \lambda_{\ell} \psi^-_{\ell}(x)\]

as well as its adjoint \(\psi^{\dagger}_{\ell}(x)\), and hope that they satisfy

(4.1.15)#\[[\psi_{\ell}(x), \psi_{\ell}(y)]_{\pm} = [\psi_{\ell}(x), \psi_{\ell'}^{\dagger}(y)]_{\pm} = 0\]

whenever \(x-y\) is space-like, and replace \(\psi^{\pm}_{\ell}(x)\) with \(\psi_{\ell}(x), \psi^{\dagger}_{\ell}(x)\) in Eq.4.1.3. Under these assumptions, we can then construct the interaction density \(\Hscr\) as a polynomial in \(\psi_{\ell}(x), \psi_{\ell}^{\dagger}(x)\) with an even number of fermionic fields (so that the sign in Eq.4.1.15 is negative).

There remains, however, one issue with field like Eq.4.1.15 that mixes creation and annihilation fields. Namely, special conditions must hold in order for such fields to play well with conserved quantum numbers. To be more specific, let \(Q\) be a conserved quantum number, e.g., the electric charge. Then the following hold

(4.1.16)#\[\begin{split}[Q, a(\pbf, \sigma, n)] &= -q(n) a(\pbf, \sigma, n) \\ [Q, a^{\dagger}(\pbf, \sigma, n)] &= q(n) a^{\dagger}(\pbf, \sigma, n)\end{split}\]

where \(q(n)\) denotes the quantum number of the particle species \(n\). These identities can be verified by applying both sides to \(\Psi_{\pbf, \sigma, n}\) and \(\Psi_{\VAC}\), respectively.

Now in order for \(Q\) to commute with \(\Hscr\), which is constructed as a polynomial of \(\psi_{\ell}(x)\) and \(\psi_{\ell}^{\dagger}(x)\), we better have

(4.1.17)#\[[Q, \psi_{\ell}(x)] = -q_{\ell} \psi_{\ell}(x)\]

so that each monomial (with coefficient neglected) \(\psi^{\dagger}_{\ell'_1}(x) \cdots \psi^{\dagger}_{\ell'_M}(x) \psi_{\ell_1}(x) \cdots \psi_{\ell_N}(x)\) in \(\Hscr\) will commute with \(Q\) if

\[q_{\ell_1} + \cdots + q_{\ell_N} = q_{\ell'_1} + \cdots + q_{\ell'_M}\]

Note that the negative sign in Eq.4.1.17 is a formal analogy to Eq.4.1.16, where we think of \(\psi_{\ell}(x)\) as an annihilation field even though it’s really not. Since \(\psi_{\ell}(x)\) is a linear combination of \(\psi^+_{\ell}(x)\) and \(\psi^-_{\ell}(x)\), which in turn are superpositions of annihilation and creation operators, respectively, it follows from Eq.4.1.16 that in order for Eq.4.1.17 to hold, the following conditions must be satisfied

all particles annihilated by \(\psi^+_{\ell}(x)\) must have the same charge \(q(n) = q_{\ell}\),
all particles created by \(\psi^-_{\ell}(x)\) must have the same charge \(q(n) = -q_{\ell}\), and
for any particle of species \(n\), which is annihilated by \(\psi^+_{\ell}(x)\), there exists a particle of species \(\bar{n}\), which is created by \(\psi^-_{\ell}(x)\), such that \(q(n) = -q(\bar{n})\).

The particles of species \(n\) and \(\bar{n}\) are called antiparticles of each other – they are exactly the same except for the charges which are opposite. It is the last condition that demands the existence of particle-antiparticle pairs so that one can formulate a consistent (relativistic) quantum field theory.

4.2. Scalar Fields#

We’ll start, as always, with the simplest case of scalar fields, namely, when \(\psi^+(x) = \psi^+_{\ell}(x)\) and \(\psi^-(x) = \psi^-_{\ell}(x)\) are scalar functions. We argue first that such fields can only create/annihilate spinless particles. Indeed, since \(\hat{\Jbf}\) necessarily vanishes, it follows from Eq.4.1.13 that \(u\) and \(v\) may be nonzero if and only if \(j_n = 0\). If we, for the moment, are concerned with just one particle species, then we can write \(u(\pbf, \sigma, n) = u(\pbf)\) and \(v(\pbf, \sigma, n) = v(\pbf)\). Lastly, we note that since \(D = 1\) in this case, Eq.4.1.11 become

\[\begin{split}\sqrt{p_0}~u(\pbf) &= \sqrt{(\Lambda p)_0}~u(\pbf_{\Lambda}) \\ \sqrt{p_0}~v(\pbf) &= \sqrt{(\Lambda p)_0}~v(\pbf_{\Lambda})\end{split}\]

It follows that

(4.2.1)#\[u(\pbf) = v(\pbf) = (2p_0)^{-1/2}\]

where the factor \(2\) is just conventional. In particular \(u(0) = v(0) = (2m)^{-1/2}\).

Plugging Eq.4.2.1 into Eq.4.1.10, we get

(4.2.2)#\[\begin{split}\psi^+(x) &= \int d^3 p~(2\pi)^{-3/2} e^{\ifrak p \cdot x} (2p_0)^{-1/2} a(\pbf) \\ \psi^-(x) &= \int d^3 p~(2\pi)^{-3/2} e^{-\ifrak p \cdot x} (2p_0)^{-1/2} a^{\dagger}(\pbf) = \psi^{+ \dagger}(x)\end{split}\]

In this case the interaction density \(\Hscr\), defined by Eq.4.1.3, may be constructed as any polynomial in \(\psi^{\pm}(x)\) since Eq.4.1.4 holds trivial \(D\) and scalar \(g\).

Next let’s consider the causality condition which demands that \(\left[ \psi^+(x), \psi^-(y) \right] = 0\) whenever \(x - y\) is space-like. Using the canonical commutation relation Eq.3.2.5 we calculate

(4.2.3)#\[\begin{split}\left[ \psi^+(x), \psi^-(y) \right]_{\pm} &= \int d^3 p~d^3 q~(2\pi)^{-3} e^{\ifrak (p \cdot x - q \cdot y)} (4 p_0 q_0)^{-1/2} \left[ a(\pbf), a^{\dagger}(\qbf) \right]_{\pm} \\ &= \frac{1}{(2\pi)^3} \int \frac{d^3 p}{2p_0}~e^{\ifrak p \cdot (x - y)} \eqqcolon \Delta_+(x - y)\end{split}\]

where

(4.2.4)#\[\Delta_+(x) \coloneqq \frac{1}{(2\pi)^3} \int \frac{d^3 p}{2p_0}~e^{\ifrak p \cdot x}\]

We notice that \(\Delta_+(x)\) is manifestly (proper orthochronous) Lorentz invariant – the invariance of the volume element comes from Eq.1.3.9. It is, however, not in general invariant under transformations like \(x \to -x\). But, as we’ll see, such invariance holds assuming \(x\) is space-like.

Note

The plus subscript in \(\Delta_+(x)\) is there to distinguish it from an anti-symmetrized version \(\Delta(x)\) to be introduced later.

Now we’ll restrict ourselves to the special case of a space-like \(x\) which, up to a Lorentz transformation, can be assumed to take the form \(x = (0, \xbf)\) with \(|\xbf| > 0\). In this case, we can then calculate \(\Delta_+(x)\) as follows [2]

\[\begin{split}\Delta_+(x) &= \frac{1}{(2\pi)^3} \int \frac{d^3 p}{2\sqrt{\pbf^2 + m^2}}~\exp(\ifrak \pbf \cdot \xbf) \\ &= \frac{4\pi}{(2\pi)^3} \int_0^{\infty} \frac{\pbf^2 d|\pbf|}{2\sqrt{\pbf^2 + m^2}} \int_{S^2} d^2 \hat{\pbf}~\exp(\ifrak |\pbf| |\xbf| \hat{\pbf} \cdot \hat{\xbf}) \\ &= \frac{1}{2\pi} \int_0^{\infty} \frac{\pbf^2 d|\pbf|}{2\sqrt{\pbf^2 + m^2}} \int_0^{\pi} d\theta~\exp(\ifrak |\pbf| |\xbf| \cos\theta)\end{split}\]

The last integral cannot be easily evaluated, at least without some knowledge about special functions. Nonetheless, we observe that \(\Delta_+(x) \neq 0\), which means that \(\Hscr\) cannot be just any polynomial in \(\psi^{\pm}(x)\). Moreover, we note that \(\Delta_+(x) = \Delta_+(-x)\) as promised earlier.

As already mentioned in Eq.4.1.14, let’s try

(4.2.5)#\[\psi(x) \coloneqq \kappa \psi^+(x) + \lambda \psi^-(x)\]

Using Eq.4.2.2 and Eq.4.2.3, we can then try to make Eq.4.1.15 hold by the following calculations

\[\begin{split}\left[ \psi(x), \psi(y) \right]_{\pm} &= \kappa\lambda \left(\left[ \psi^+(x), \psi^-(y) \right]_{\pm} + \left[ \psi^-(x), \psi^+(y) \right]_{\pm} \right) \\ &= \kappa\lambda (1 \pm 1) \Delta(x - y) \\ \left[ \psi(x), \psi^{\dagger}(y) \right]_{\pm} &= \left[ \kappa \psi^+(x) + \lambda \psi^-(x), \kappa^{\ast} \psi^-(y) + \lambda^{\ast} \psi^+(y) \right]_{\pm} \\ &= |\kappa|^2 \left[ \psi^+(x), \psi^-(y) \right]_{\pm} + |\lambda|^2 \left[ \psi^-(x), \psi^+(y) \right]_{\pm} \\ &= \left( |\kappa|^2 \pm |\lambda|^2 \right) \Delta(x - y)\end{split}\]

We see that Eq.4.1.15 holds for scalar fields if the fields are bosonic, i.e., the bottom sign in \(\pm\) applies, and \(|\kappa| = |\lambda|\). By adjust the phase of \(a(\pbf)\), we can actually arrange so that \(\kappa = \lambda\), in which case we have

(4.2.6)#\[\psi(x) = \psi^+(x) + \psi^-(x) = \psi^+(x) + \psi^{+ \dagger}(x) = \psi^{\dagger}(x)\]

Note

Although the arrangement of phase so that \(\kappa = \lambda\) is a mere convention, it’s a convention that needs to be applied to all scalar fields appearing in \(\Hscr\). Namely, one cannot have both \(\psi(x)\) as in Eq.4.2.6 and another

\[\psi'(x) = e^{\ifrak \theta} \psi^+(x) + e^{-\ifrak \theta} \psi^{+ \dagger}(x)\]

for some \(\theta\), because \(\psi(x)\) won’t commute with \(\psi'(y)\) even if \(x - y\) is space-like.

Now if the particle created and annihilated by \(\psi(x)\) carries a (non-vanishing) conserved quantum number \(Q\), then by the discussions on the charge conservation from the previous section, a density \(\Hscr\) made up of \(\psi(x)\) as defined by Eq.4.2.6 will not commute with \(Q\). Instead, one must assume the existence of a field \(\psi^{+ c}(x)\) that creates and annihilates the corresponding antiparticle, in the sense that

\[\begin{split}\left[ Q, \psi^+(x) \right] &= -q \psi^+(x) \\ \left[ Q, \psi^{+ c}(x) \right] &= q \psi^{+ c}(x)\end{split}\]

Here the supscript \(c\) stands for charge (conjugation). Now instead of Eq.4.2.5, let’s try

\[\psi(x) \coloneqq \kappa \psi^+(x) + \lambda \psi^{+ c \dagger}(x)\]

so that \([Q, \psi(x)] = -q \psi(x)\). We calculate the commutators, assuming the antiparticle is different from the particle, just as before as follows

\[\begin{split}\left[\psi(x), \psi(y) \right]_{\pm} &= \left[ \kappa \psi^+(x) + \lambda \psi^{+ c \dagger}(x), \kappa \psi^+(y) + \lambda \psi^{+ c \dagger}(y) \right]_{\pm} = 0 \\ \left[\psi(x), \psi^{\dagger}(y) \right]_{\pm} &= \left[ \kappa \psi^+(x) + \lambda \psi^{+ c \dagger}(x), \kappa^{\ast} \psi^{+ \dagger}(y) + \lambda^{\ast} \psi^{+ c}(y) \right]_{\pm} \\ &= |\kappa|^2 \left[ \psi^+(x), \psi^{+ \dagger}(y) \right]_{\pm} + |\lambda|^2 \left[ \psi^{+ c \dagger}(x), \psi^{+ c}(y) \right]_{\pm} \\ &= (|\kappa|^2 \pm |\lambda|^2) \Delta(x - y)\end{split}\]

where we’ve assumed, in particular that the particle and its particle share the same mass so that Eq.4.2.3 equally applies.

By the same argument as in the case where no quantum number is involved, we see that a scalar field can satisfy the causality condition if it describes a boson. Moreover, by adjusting the phase of \(a(\pbf)\), one can arrange so that \(\kappa = \lambda\) so that

(4.2.7)#\[\psi(x) = \psi^+(x) + \psi^{+ c \dagger}(x)\]

Note that this is compatible with Eq.4.2.6 in the case where the particle is its own antiparticle.

Using Eq.4.2.2, we can write \(\psi(x)\) in terms of the creation and annihilation operators as follows

(4.2.8)#\[\psi(x) = \int \frac{d^3 p}{(2\pi)^{3/2} (2p_0)^{1/2}}~\left[ e^{\ifrak p \cdot x} a(\pbf) + e^{-\ifrak p \cdot x} a^{c \dagger}(\pbf) \right]\]

with the possibility of \(a^{c \dagger}(\pbf) = a^{\dagger}(\pbf)\) in the case where the created particle is its own antiparticle.

For later use (e.g., the evaluation of Feynman diagrams), we note the following identity which holds for any, and not just space-like, \(x\) and \(y\).

(4.2.9)#\[\left[ \psi(x), \psi^{\dagger}(y) \right] = \Delta(x - y)\]

where \(\Delta(x)\) is defined as follows

(4.2.10)#\[\Delta(x) \coloneqq \Delta_+(x) - \Delta_+(-x) = \frac{1}{(2\pi)^3} \int \frac{d^3 p}{2p_0} \left( e^{\ifrak p \cdot x} - e^{-\ifrak p \cdot x} \right)\]

4.2.1. The CPT symmetries#

Let’s investigate how a scalar field transforms under spatial inversion \(\Pcal\), time inversion \(\Tcal\), and charge conjugation \(\Ccal\). This follows essentially from Eq.4.2.8 together with our knowledge about how creation/annihilation operators transform under CPT transformations in The Lorentz and CPT transformation laws. Recall that we consider the case of massive particles here, leaving the massless case to a later section.

We start with the spatial inversion \(\Pcal\) by recalling the following transformation rules

\[\begin{split}U(\Pcal) a(\pbf) U^{-1}(\Pcal) &= \eta^{\ast} a(-\pbf) \\ U(\Pcal) a^{c \dagger}(\pbf) U^{-1}(\Pcal) &= \eta^c a^{c \dagger}(-\pbf)\end{split}\]

where \(\eta\) and \(\eta^c\) are the intrinsic parities of the particle and antiparticle, respectively. In order for the scalar field Eq.4.2.8 to transform nicely with \(\Pcal\), one must have \(\eta^{\ast} = \eta^c\) (or \(\eta^{\ast} = \eta\) in the case where the particle is its own antiparticle). As a result, we have

(4.2.11)#\[U(\Pcal) \psi(x) U^{-1}(\Pcal) = \eta^{\ast} \psi(\Pcal x)\]

Next let’s consider the time inversion \(\Tcal\). We recall the transformation rules as follows

\[\begin{split}U(\Tcal) a(\pbf) U^{-1}(\Tcal) &= \zeta^{\ast} a(-\pbf) \\ U(\Tcal) a^{c \dagger}(\pbf) U^{-1}(\Tcal) &= \zeta^c a^{c \dagger}(-\pbf)\end{split}\]

Similar to the case of spatial inversions, in order for \(\psi(x)\) to transform nicely with \(U(\Tcal)\), one must have \(\zeta^{\ast} = \zeta^c\). Moreover, since \(U(\Tcal)\) is anti-unitary, we have

\[U(\Tcal) \psi(x) U^{-1}(\Tcal) = \zeta^{\ast} \psi(-\Tcal x)\]

Finally let’s consider the charge conjugation \(\Ccal\) with the following transformation laws

\[\begin{split}U(\Ccal) a(\pbf) U^{-1}(\Ccal) &= \xi^{\ast} a^c(\pbf) \\ U(\Ccal) a^{c \dagger}(\pbf) U^{-1}(\Ccal) &= \xi^c a^{\dagger}(\pbf)\end{split}\]

As before, we must have \(\xi^{\ast} = \xi^c\) and therefore

\[U(\Ccal) \psi(x) U^{-1}(\Ccal) = \xi^{\ast} \psi^{\dagger}(x)\]

4.3. Vector Fields#

The next simplest scenario after scalar field is vector field, where the representation \(D(\Lambda) = \Lambda\). Once again, let’s consider particles of one species so that we can drop the \(n\) label from, for example, \(a(\pbf, \sigma, n)\). In this case, we can rewrite Eq.4.1.10 as follows

(4.3.1)#\[\begin{split}\psi^+_{\mu}(x) &= \sum_{\sigma} (2\pi)^{-3/2} \int d^3 p~e^{\ifrak p \cdot x} u_{\mu}(\pbf, \sigma) a(\pbf, \sigma) \\ \psi^-_{\nu}(x) &= \sum_{\sigma} (2\pi)^{-3/2} \int d^3 p~e^{-\ifrak p \cdot x} v_{\nu}(\pbf, \sigma) a^{\dagger}(\pbf, \sigma)\end{split}\]

where \(\mu, \nu\) are the \(4\)-indexes. Moreover, the boost transformation formulae Eq.4.1.12 take the following form

(4.3.2)#\[\begin{split}u_{\mu}(\pbf, \sigma) &= (m / p_0)^{1/2} {L(p)_{\mu}}^{\nu} u_{\nu}(0, \sigma) \\ v_{\mu}(\pbf, \sigma) &= (m / p_0)^{1/2} {L(p)_{\mu}}^{\nu} v_{\nu}(0, \sigma)\end{split}\]

Finally the (linearized) rotation transformation formulae Eq.4.1.13 take the following form

(4.3.3)#\[\begin{split}\sum_{\sigma'} u_{\mu}(0, \sigma') \Jbf^{(\jfrak)}_{\sigma' \sigma} &= \sum_{\nu} \hat{\Jbf}_{\mu \nu} u_{\nu}(0, \sigma) \\ -\sum_{\sigma'} v_{\mu}(0, \sigma') \Jbf^{(\jfrak)}_{\sigma' \sigma} &= \sum_{\nu} \hat{\Jbf}_{\mu \nu} v_{\nu}(0, \sigma)\end{split}\]

where \(\hat{\Jbf}\) is the angular momentum vector associated with the (tautological) representation \(\Lambda\). It follows from Eq.1.2.16 and Eq.1.2.18 that

(4.3.4)#\[\begin{split}\left( \hat{\Jbf}_k \right)_{00} = \left( \hat{\Jbf}_k \right)_{0i} = \left( \hat{\Jbf}_k \right)_{i0} &= 0 \\ \left( \hat{\Jbf}_k \right)_{ij} &= -\ifrak \epsilon_{ijk}\end{split}\]

where \(\{i,j,k\} = \{1,2,3\}\). From this one can then calculate \(\hat{\Jbf}^2\) as follows

\[\begin{split}\left( \hat{\Jbf}^2 \right)_{00} &= \left( \hat{\Jbf}^2 \right)_{0i} = \left( \hat{\Jbf}^2 \right)_{i0} = 0 \\ \left( \hat{\Jbf}^2 \right)_{ij} &= \sum_{k,m=1}^3 \left( \hat{\Jbf}_k \right)_{im} \left( \hat{\Jbf}_k \right)_{mj} = \sum_{k,m=1}^3 -\epsilon_{imk} \epsilon_{mjk} = 2\delta_{ij}\end{split}\]

It follows then from Eq.4.3.3 that

(4.3.5)#\[\begin{split}\sum_{\sigma'} u_0(0, \sigma') \left( \Jbf^{(\jfrak)} \right)^2_{\sigma' \sigma} &= \sum_{\nu} \left( \hat{\Jbf}^2 \right)_{0 \nu} u_{\nu}(0, \sigma) = 0 \\ \sum_{\sigma'} u_i(0, \sigma') \left( \Jbf^{(\jfrak)} \right)^2_{\sigma' \sigma} &= \sum_{\nu} \left( \hat{\Jbf}^2 \right)_{i \nu} u_{\nu}(0, \sigma) = \sum_j 2 \delta_{i j} u_j(0, \sigma) = 2 u_i(0, \sigma) \\ \sum_{\sigma'} v_0(0, \sigma') \left( \Jbf^{(\jfrak)} \right)^2_{\sigma' \sigma} &= 0 \\ \sum_{\sigma'} v_i(0, \sigma') \left( \Jbf^{(\jfrak)} \right)^2_{\sigma' \sigma} &=2v_i(0, \sigma)\end{split}\]

where we’ve worked out the details of the calculations for \(u\), but not \(v\) because they are essentially the same.

Now recall from Eq.1.3.15 that

\[\left( \Jbf^{(\jfrak)} \right)^2_{\sigma \sigma'} = \jfrak (\jfrak + 1) \delta_{\sigma \sigma'}\]

It follows that in order for Eq.4.3.5 to have nonzero solutions, one must have either \(\jfrak = 0\), in which case only the time-components \(u_0(0)\) and \(v_0(0)\) may be nonzero, where we’ve also suppressed \(\sigma\) because spin vanishes, or \(\jfrak = 1\), in which case only the space-components \(u_i(0, \sigma)\) and \(v_i(0, \sigma)\) may be nonzero. These two cases are discussed in more details as follows.

4.3.1. Spin-\(0\) vector fields#

In this case \(\jfrak = 0\). For reasons that will become clear momentarily, let’s fix the constants \(u_0(0), v_0(0)\) as follows

\[\begin{split}u_0(0) &= \ifrak (m / 2)^{1/2} \\ v_0(0) &= -\ifrak (m / 2)^{1/2}\end{split}\]

It follows from Eq.4.3.2 (see also Eq.1.3.17) that

\[\begin{split}u_{\mu}(\pbf) &= (m / p_0)^{1/2} {L(p)_{\mu}}^0 u_0(0) \\ &= (m / p_0)^{1/2} (p_{\mu} / m) \ifrak (m / 2)^{1/2} \\ &= \ifrak p_{\mu} (2p_0)^{-1/2} \\ v_{\mu}(\pbf) &= -\ifrak p_{\mu} (2p_0)^{-1/2}\end{split}\]

where we once again have omitted the details of the calculation of \(v\) because it’s similar to that of \(u\). Plugging into Eq.4.3.1, we see that the field components take the following form

\[\begin{split}\psi^+_{\mu}(x) &= (2\pi)^{-3/2} \int d^3 p~e^{\ifrak p \cdot x} \ifrak p_{\mu} (2p_0)^{-1/2} a(\pbf) \\ \psi^-_{\mu}(x) &= (2\pi)^{-3/2} \int d^3 p~e^{-\ifrak p \cdot x} (-\ifrak p_{\mu}) (2p_0)^{-1/2} a^{\dagger}(\pbf)\end{split}\]

Comparing these with Eq.4.2.2, and thanks to the choices of \(u_0(0)\) and \(v_0(0)\) above, we see that

\[\psi^{\pm}_{\mu}(x) = \p_{\mu} \psi^{\pm}(x)\]

It follows that in fact a spinless vector field defined by \(\psi_{\mu}(x) \coloneqq \psi^+_{\mu}(x) + \psi^-_{\mu}(x)\) as usual is nothing but the gradient vector field of a (spinless) scalar field. Hence we get nothing new from spinless vector fields.

4.3.2. Spin-\(1\) vector fields#

In this case \(\jfrak = 1\). We start with the states whose spin \(z\)-component vanishes, i.e., \(u_{\mu}(0,0)\) and \(v_{\mu}(0,0)\). First we claim that they are both in the \(z\)-direction, i.e., \(u_{\mu}(0,0) = v_{\mu}(0,0) = 0\) unless \(\mu=3\). Indeed, taking the \(z\)-components of both sides of Eq.4.3.3 and recalling that \(\left( J_3^{(1)} \right)_{0 \sigma} = 0\), we have for \(\mu = 1\)

\[0 = \sum_{\nu} \left( \hat{\Jbf}_3 \right)_{1 \nu} u_{\nu}(0, 0) = -\ifrak u_2(0, 0) \implies u_2(0, 0) = 0\]

and for \(\mu = 2\)

\[0 = \sum_{\nu} \left( \hat{\Jbf}_3 \right)_{2 \nu} u_{\nu}(0, 0) = \ifrak u_1(0, 0) \implies u_1(0, 0) = 0\]

These, together with the fact that \(u_0(0, 0) = 0\) for \(\jfrak = 1\), imply that only \(u_3(0, 0)\) can be nonzero. The same conclusion can also be drawn for \(v_3(0, 0)\). Therefore up to a normalization factor, we can write

(4.3.6)#\[\begin{split}u_{\mu}(0, 0) = v_{\mu}(0, 0) = (2m)^{-1/2} \begin{bmatrix*}[r] 0 \\ 0 \\ 0 \\ 1 \end{bmatrix*}\end{split}\]

Now to calculate \(u\) and \(v\) for the other spin \(z\)-components, we’ll try to use Eq.1.3.11 as follows. First, according to Eq.4.3.3 we have the following general equality

\[\begin{split}\sum_{\nu} \left(\left( \hat{\Jbf}_1 \right)_{\mu \nu} + \ifrak \left( \hat{\Jbf}_2 \right)_{\mu \nu} \right) u_{\nu}(0, \sigma) &= \sum_{\sigma'} u_{\mu}(0, \sigma') \left( J^{(1)}_1 + \ifrak J^{(1)}_2 \right)_{\sigma \sigma'} \\ &= \sum_{\sigma'} u_{\mu}(0, \sigma') \delta_{\sigma+1, \sigma'} \sqrt{(1 - \sigma)(2 + \sigma)}\end{split}\]

Then, letting \(\sigma=0\) and \(\mu=1\), we have

\[\sqrt{2}~u_1(0, 1) = \sum_{\nu} \left(\left( \hat{\Jbf}_1 \right)_{1 \nu} + \ifrak \left( \hat{\Jbf}_2 \right)_{1 \nu} \right) u_{\nu}(0, 0) = \ifrak \left( \hat{\Jbf}_2 \right)_{13} u_3(0, 0) = -(2m)^{-1/2}\]

Then, changing to \(\mu=2\), we have

\[\sqrt{2}~u_2(0, 1) = \sum_{\nu} \left(\left( \hat{\Jbf}_1 \right)_{2 \nu} + \ifrak \left( \hat{\Jbf}_2 \right)_{2 \nu} \right) u_{\nu}(0, 0) = (\hat{\Jbf}_1)_{23} u_3(0, 0) = -\ifrak (2m)^{-1/2}\]

Finally taking \(\mu=3\), we have

\[\sqrt{2}~u_3(0, 1) = \sum_{\nu} \left(\left( \hat{\Jbf}_1 \right)_{3 \nu} + \ifrak \left( \hat{\Jbf}_2 \right)_{3 \nu} \right) u_{\nu}(0, 0) = \left( \hat{\Jbf}_1 \right)_{32} u_2(0, 0) + \ifrak \left( \hat{\Jbf}_2 \right)_{31} u_1(0, 0) = 0\]

Putting these all together, we have calculated \(u_{\mu}(0, 1)\) as follows

\[\begin{split}u_{\mu}(0, 1) = -\frac{1}{2\sqrt{m}} \begin{bmatrix*}[r] 0 \\ 1 \\ \ifrak \\ 0 \end{bmatrix*}\end{split}\]

Calculations for \(\sigma = -1\) as well as for \(v\) are similar and hence omitted. The results are listed for future reference as follows

(4.3.7)#\[\begin{split}\begin{alignat*}{2} u_{\mu}(0, 1) &= -v_{\mu}(0, -1) &&= -\frac{1}{2\sqrt{m}} \begin{bmatrix*}[r] 0 \\ 1 \\ \ifrak \\ 0 \end{bmatrix*} \\ u_{\mu}(0, -1) &= -v_{\mu}(0, 1) &&= \frac{1}{2\sqrt{m}} \begin{bmatrix*}[r] 0 \\ 1 \\ -\ifrak \\ 0 \end{bmatrix*} \end{alignat*}\end{split}\]

Applying the boosting formulae Eq.4.3.2 to Eq.4.3.6 and Eq.4.3.7, we obtain the formulae for \(u\) and \(v\) with arbitrary momentum as follows

(4.3.8)#\[u_{\mu}(\pbf, \sigma) = v_{\mu}^{\ast}(\pbf, \sigma) = (2p_0)^{-1/2} {L(p)_{\mu}}^{\nu} e_{\nu}(0, \sigma) \eqqcolon (2p_0)^{-1/2} e_{\mu}(\pbf, \sigma)\]

where

(4.3.9)#\[\begin{split}e_{\mu}(0, 0) = \begin{bmatrix*}[r] 0 \\ 0 \\ 0 \\ 1 \end{bmatrix*}, \quad \ e_{\mu}(0, 1) = -\frac{1}{\sqrt{2}} \begin{bmatrix*}[r] 0 \\ 1 \\ \ifrak \\ 0 \end{bmatrix*}, \quad \ e_{\mu}(0, -1) = \frac{1}{\sqrt{2}} \begin{bmatrix*}[r] 0 \\ 1 \\ -\ifrak \\ 0 \end{bmatrix*}\end{split}\]

Now we can rewrite the general Eq.4.3.1 more specifically as follows

(4.3.10)#\[\begin{split}\psi^+_{\mu}(x) &= \sum_{\sigma} (2\pi)^{-3/2} \int \frac{d^3 p}{\sqrt{2p_0}}~\exp(\ifrak p \cdot x) e_{\mu}(\pbf, \sigma) a(\pbf, \sigma) \\ \psi^-_{\mu}(x) &= \sum_{\sigma} (2\pi)^{-3/2} \int \frac{d^3 p}{\sqrt{2p_0}}~\exp(-\ifrak p \cdot x) e_{\mu}^{\ast}(\pbf, \sigma) a^{\dagger}(\pbf, \sigma) = \psi^{+ \dagger}_{\mu}(x)\end{split}\]

Similar to the calculation Eq.4.2.3 for scalar field, the (anti-)commutator can be calculated as follows

(4.3.11)#\[\left[ \psi^+_{\mu}(x), \psi^-_{\nu}(y) \right]_{\pm} = \int \frac{d^3 p}{(2\pi)^3 2p_0}~\exp(\ifrak p \cdot (x - y)) \Pi_{\mu \nu}(\pbf)\]

where

(4.3.12)#\[\Pi_{\mu \nu}(\pbf) \coloneqq \sum_{\sigma} e_{\mu}(\pbf, \sigma) e^{\ast}_{\nu}(\pbf, \sigma)\]

To better understand the quantity \(\Pi_{\mu \nu}(\pbf)\), let’s first evaluate it at \(\pbf = 0\) as follows

\[\begin{split}\Pi_{\mu \nu}(0) = \begin{bmatrix*}[r] 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix*} + \frac{1}{2} \begin{bmatrix*}[r] 0 & 0 & 0 & 0 \\ 0 & 1 & -\ifrak & 0 \\ 0 & \ifrak & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix*} + \frac{1}{2} \begin{bmatrix*}[r] 0 & 0 & 0 & 0 \\ 0 & 1 & \ifrak & 0 \\ 0 & -\ifrak & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix*}\end{split}\]

which is nothing but the projection to the spatial \(3\)-space, or phrased more invariantly, the orthogonal complement of the time direction. Considering the definition \(e_{\mu}(\pbf, \sigma) \coloneqq {L(p)_{\mu}}^{\nu} e_{\nu}(0, \sigma)\) as in Eq.4.3.8, we see that the general \(\Pi_{\mu \nu}(\pbf)\) is really just a projection to the orthogonal complement of \(p\), and therefore can be written as

(4.3.13)#\[\Pi_{\mu \nu}(\pbf) = \eta_{\mu \nu} + \frac{p_{\mu} p_{\nu}}{m^2}\]

because of the mass-shell condition \(p^2 + m^2 = 0\).

In light of Eq.4.2.3, we can rewrite Eq.4.3.11 as follows

(4.3.14)#\[\left[ \psi^+_{\mu}(x), \psi^-_{\nu}(y) \right]_{\pm} = \left( \eta_{\mu \nu} - \frac{\p_{\mu} \p_{\nu}}{m^2} \right) \Delta_+(x - y)\]

where \(\Delta_+(x - y)\) is defined by Eq.4.2.4. As in the case of scalar fields, this (anti-)commutator doesn’t vanish even for space-like \(x - y\). Nonetheless, it’s still an even function for space-like separations. The trick, as usual, is to consider a linear combination of \(\psi^+_{\mu}(x)\) and \(\psi^-_{\mu}(x)\) as follows

\[\psi_{\mu}(x) \coloneqq \kappa \psi^+_{\mu}(x) + \lambda \psi^-_{\mu}(x)\]

Now for space-separated \(x\) and \(y\), we can calculate using Eq.4.3.14 and Eq.4.3.10 as follows

\[\begin{split}\left[ \psi_{\mu}(x), \psi_{\nu}(y) \right]_{\pm} &= \kappa\lambda(1 \pm 1) \left( \eta_{\mu \nu} - \frac{\p_{\mu} \p_{\nu}}{m^2} \right) \Delta_+(x-y) \\ \left[ \psi_{\mu}(x), \psi^{\dagger}_{\nu}(y) \right]_{\pm} &= (|\kappa|^2 \pm |\lambda|^2) \left( \eta_{\mu \nu} - \frac{\p_{\mu} \p_{\nu}}{m^2} \right) \Delta_+(x-y)\end{split}\]

For them to vanishes, we see that first of all, we must adopt the top sign, i.e., take the commutator, or in other words, the vector field of spin \(1\) must be bosonic. In addition, we must have \(|\kappa| = |\lambda|\). In fact, by adjusting the phase of the creation/annihilation operators, we can arrange so that \(\kappa = \lambda = 1\). To summarize, we can write a general vector field in the following form

(4.3.15)#\[\psi_{\mu}(x) \coloneqq \psi^+_{\mu}(x) + \psi^-_{\mu}(x) = \psi^+_{\mu}(x) + \psi^{+ \dagger}_{\mu}(x)\]

just like Eq.4.2.6. It’s also obvious that \(\psi_{\mu}(x)\) is Hermitian.

Now if the vector field carries a nonzero (conserved) quantum charge, then one must adjust Eq.4.3.15 as follows

\[\psi_{\mu}(x) \coloneqq \psi^+_{\mu}(x) + \psi^{+ c \dagger}_{\mu}(x)\]

in analogy with Eq.4.2.7 for scalar fields. Finally, we can express the vector field in terms of creation and annihilation operators as follows

(4.3.16)#\[\psi_{\mu}(x) = \sum_{\sigma} \int \frac{d^3 p}{(2\pi)^{3/2} (2p_0)^{1/2}}~\left[ e^{\ifrak p \cdot x} e_{\mu}(\pbf, \sigma) a(\pbf, \sigma) \ + e^{-\ifrak p \cdot x} e_{\mu}^{\ast}(\pbf, \sigma) a^{c \dagger}(\pbf, \sigma) \right]\]

in analogy with Eq.4.2.8 for scalar fields. Finally, let’s calculate the commutator (for general \(x\) and \(y\)) for later use as follows

(4.3.17)#\[\left[ \psi_{\mu}(x), \psi^{\dagger}_{\nu}(y) \right] = \left( \eta_{\mu \nu} - \frac{\p_{\mu} \p_{\nu}}{m^2} \right) \Delta(x-y)\]

where \(\Delta(x-y)\) as defined by Eq.4.2.10.

So far, besides the introduction of the vectors \(e_{\mu}(\pbf, \sigma)\) in Eq.4.3.8 and Eq.4.3.9, the discussion on vector fields looks very much like scalar fields. A key difference, however, stems from the following observation

(4.3.18)#\[e^{\mu}(\pbf, \sigma) p_{\mu} = 0\]

which, in turn, implies that

(4.3.19)#\[\p_{\mu} \psi^{\mu}(x) = 0\]

This condition turns out to be coincide with a so-called “gauge fixing” condition for spin-\(1\) photons in quantum electrodynamics. However, it’s known that photons are massless particles. Therefore we may wonder if a vanishing mass limit \(m \to 0\) may be applied. Now the simplest way to construct a (scalar) interaction density \(\Hscr(x)\) using \(\psi_{\mu}(x)\) is

(4.3.20)#\[\Hscr(x) = J^{\mu}(x) \psi_{\mu}(x)\]

where \(J^{\mu}(x)\) is a \(4\)-vector current. Suppose we fix the in- and out-states in the interaction. Then according to Eq.4.3.16, the rate of (anti-)particle emission is proportional to

\[\sum_{\sigma} \left| \langle J^{\mu} \rangle e_{\mu}^{\ast}(\pbf, \sigma) \right|^2 = \langle J^{\mu} \rangle \langle J^{\nu} \rangle^{\ast} \Pi_{\mu \nu}(\pbf) = \langle J^{\mu} \rangle \langle J^{\nu} \rangle^{\ast} \left( \eta_{\mu \nu} - p_{\mu} p_{\nu} / m^2 \right)\]

where \(\Pi_{\mu \nu}\) is evaluated by Eq.4.3.13, and \(\langle J^{\mu} \rangle\) denotes the matrix element of the current between the fixed in- and out-states. Now this rate blows up at \(m \to 0\) limit unless \(p_{\mu} \langle J^{\mu} \rangle = 0\). This last condition can be translated to spacetime coordinates as follows

(4.3.21)#\[\p_{\mu} J^{\mu}(x) = 0\]

or in other words \(J^{\mu}(x)\) is a conserved current.

4.3.3. The CPT symmetries#

Let’s start with the spatial inversion. First recall from The Lorentz and CPT transformation laws

\[\begin{split}U(\Pcal) a(\pbf, \sigma) U^{-1}(\Pcal) &= \eta^{\ast} a(-\pbf, \sigma) \\ U(\Pcal) a^{c \dagger}(\pbf, \sigma) U^{-1}(\Pcal) &= \eta^c a^{c \dagger}(-\pbf, \sigma)\end{split}\]

It follows that we need to express \(e_{\mu}(-\pbf, \sigma)\) in terms of \(e_{\mu}(\pbf, \sigma)\). To this end, let’s calculate

(4.3.22)#\[\begin{split}e_{\mu}(-\pbf, \sigma) &= {L(-\pbf)_{\mu}}^{\nu} e_{\nu}(0, \sigma) \\ &= {\Pcal_{\mu}}^{\rho} {L(\pbf)_{\rho}}^{\tau} {\Pcal_{\tau}}^{\nu} e_{\nu}(0, \sigma) \\ &= -{\Pcal_{\mu}}^{\rho} {L(\pbf)_{\rho}}^{\tau} e_{\tau}(0, \sigma) \\ &= -{\Pcal_{\mu}}^{\rho} e_{\rho}(\pbf, \sigma)\end{split}\]

It follows that the spatial inversion transformation law is given as follows

(4.3.23)#\[U(\Pcal) \psi_{\mu}(x) U^{-1}(\Pcal) = -\eta^{\ast} {\Pcal_{\mu}}^{\nu} \psi_{\nu}(\Pcal x)\]

under the following assumption

(4.3.24)#\[\eta^c = \eta^{\ast}\]

Omitting further details, the transformation laws for time inversion and charge conjugation are given by

\[\begin{split}U(\Tcal) \psi_{\mu}(x) U^{-1}(\Tcal) &= \zeta^{\ast} {\Pcal_{\mu}}^{\nu} \psi_{\nu}(-\Pcal x) \\ U(\Ccal) \psi_{\mu}(x) U^{-1}(\Ccal) &= \xi^{\ast} \psi_{\mu}^{\dagger}(x)\end{split}\]

under the assumptions

(4.3.25)#\[\zeta^c = \zeta^{\ast}, \quad \xi^c = \xi^{\ast}\]

respectively.

4.4. Dirac Fields#

Here we’ll encounter the first nontrivial representation of the (homogeneous orthochronous) Lorentz group, first discovered by P. Dirac in a completely different (and more physical) context. Our treatment here will be purely mathematical, and will serve as a warm-up for the general representation theory.

4.4.1. Dirac representation and gamma matrices#

Let \(D\) be a representation of the Lorentz group in the sense that \(D(\Lambda_1) D(\Lambda_2) = D(\Lambda_1 \Lambda_2)\). By the discussion in Quantum Lorentz symmetry and ignoring the translation part, we can write \(D(\Lambda)\) up to first order as follows

(4.4.1)#\[D(\Lambda) = 1 + \frac{\ifrak}{2} \omega^{\mu \nu} \Jscr_{\mu \nu} + \cdots\]

where \(\Jscr_{\mu \nu} = -\Jscr_{\nu \mu}\) are (Hermitian) matrices that, according to Eq.1.2.21, satisfy in addition the following Lie-algebraic condition

(4.4.2)#\[\left[ \Jscr_{\mu \nu}, \Jscr_{\rho \kappa} \right] = \ifrak\left( \eta_{\mu \rho} \Jscr_{\nu \kappa} - \eta_{\nu \rho} \Jscr_{\mu \kappa} + \eta_{\kappa \mu} \Jscr_{\rho \nu} - \eta_{\kappa \nu} \Jscr_{\rho \mu} \right)\]

Putting it this way, it may appear hopeless to find any solution to the equation above. Surprisingly, there is in fact a systematic way to find all solutions to Eq.4.4.2, which will be shown in General Fields. The aim of this section, however, is to explain a seemingly unmotivated, but rather ingenious, solution, which also bares a great deal of significance in the quantum theory of electromagnetism.

The trick here is to assume the existence of a set of matrices \(\gamma_{\mu}\) such that

(4.4.3)#\[\left\{ \gamma_{\mu}, \gamma_{\nu} \right\} = 2\eta_{\mu \nu}\]

where the curly bracket denotes the anti-commutator, and is equivalent to the notation \([~,~]_+\) used in previous chapters. Here the right-hand-side, written as a number, should be interpreted as a multiple of the identity matrix. Such matrices \(\gamma_{\mu}\) form a so-called Clifford algebra of the symmetric bilinear form \(\eta_{\mu \nu}\). Then we simply claim that the set of \(\Jscr_{\mu \nu}\) defined by

(4.4.4)#\[\Jscr_{\mu \nu} \coloneqq -\frac{\ifrak}{4} \left[ \gamma_{\mu}, \gamma_{\nu} \right]\]

solves Eq.4.4.2. To see this, let’s first do a preparational calculation as follows

(4.4.5)#\[\begin{split}\left[ \Jscr_{\mu \nu}, \gamma_{\rho} \right] &= -\frac{\ifrak}{4} \left[ \left[ \gamma_{\mu}, \gamma_{\nu} \right], \gamma_{\rho} \right] \\ &= -\frac{\ifrak}{4} \left( \gamma_{\mu}\gamma_{\nu}\gamma_{\rho} - \gamma_{\nu}\gamma_{\mu}\gamma_{\rho} - \gamma_{\rho}\gamma_{\mu}\gamma_{\nu} + \gamma_{\rho}\gamma_{\nu}\gamma_{\mu} \right) \\ &= -\frac{\ifrak}{4} \big( (\gamma_{\mu}\gamma_{\nu}\gamma_{\rho} + \gamma_{\mu}\gamma_{\rho}\gamma_{\nu}) - (\gamma_{\mu}\gamma_{\rho}\gamma_{\nu} + \gamma_{\rho}\gamma_{\mu}\gamma_{\nu}) \phantom{)} \\ &\phantom{(}\qquad\quad - (\gamma_{\nu}\gamma_{\mu}\gamma_{\rho} + \gamma_{\nu}\gamma_{\rho}\gamma_{\mu}) + (\gamma_{\nu}\gamma_{\rho}\gamma_{\mu} + \gamma_{\rho}\gamma_{\nu}\gamma_{\mu}) \big) \\ &= -\frac{\ifrak}{4} \left( 2\gamma_{\mu}\eta_{\nu \rho} - 2\eta_{\mu \rho}\gamma_{\nu} - 2\gamma_{\nu}\eta_{\mu \rho} + 2\eta_{\nu \rho}\gamma_{\mu} \right) \\ &= -\ifrak \eta_{\nu \rho}\gamma_{\mu} + \ifrak \eta_{\mu \rho}\gamma_{\nu}\end{split}\]

Then we can verify Eq.4.4.2, starting from the left-hand-side, as follows

\[\begin{split}\left[ \Jscr_{\mu \nu}, \Jscr_{\rho \kappa} \right] &= -\frac{\ifrak}{4} \left[ \Jscr_{\mu \nu}, \left[ \gamma_{\rho}, \gamma_{\kappa} \right] \right] \\ &= \frac{\ifrak}{4} \left[ \gamma_{\rho}, \left[ \gamma_{\kappa}, \Jscr_{\mu \nu} \right] \right] + \frac{\ifrak}{4} \left[ \gamma_{\kappa}, \left[ \Jscr_{\mu \nu}, \gamma_{\rho} \right] \right] \\ &= \frac{\ifrak}{4} \left[ \gamma_{\rho}, \left( \ifrak \eta_{\nu \kappa} \gamma_{\mu} - \ifrak \eta_{\mu \kappa} \gamma_{\nu} \right) \right] + \frac{\ifrak}{4} \left[ \gamma_{\kappa}, \left( -\ifrak \eta_{\nu \rho} \gamma_{\mu} + \ifrak \eta_{\mu \rho} \gamma_{\nu} \right) \right] \\ &= -\ifrak \eta_{\nu \kappa} \Jscr_{\rho \mu} + \ifrak \eta_{\mu \kappa} \Jscr_{\rho \nu} + \ifrak \eta_{\nu \rho} \Jscr_{\kappa \mu} - \ifrak \eta_{\mu \rho} \Jscr_{\kappa \nu}\end{split}\]

The last expression is easily seen to be equal to the right-hand-side of Eq.4.4.2 using the anti-symmetry of \(\Jscr_{\mu \nu}\).

In fact, the calculation Eq.4.4.5 may be rephrased more compactly as follows

(4.4.6)#\[D(\Lambda) \gamma_{\mu} D^{-1}(\Lambda) = {\Lambda^{\nu}}_{\mu} \gamma_{\nu}\]

or in plain words, \(\gamma_{\mu}\) is a vector.

Using the very definition Eq.4.4.4, one then sees that \(\Jscr_{\mu \nu}\) is an anti-symmetric tensor in the sense that

\[D(\Lambda) \Jscr_{\mu \nu} D^{-1}(\Lambda) = \Lambda^{\rho}_{\mu} \Lambda^{\kappa}_{\nu} \Jscr_{\rho \kappa}\]

Indeed, one can construct all anti-symmetric tensors as follows

\[\begin{split}\Ascr_{\mu \nu \rho} &\coloneqq \gamma_{\mu}\gamma_{\nu}\gamma_{\rho} \pm \text{ signed permutations} \\ \Pscr_{\mu \nu \rho \kappa} &\coloneqq \gamma_{\mu}\gamma_{\nu}\gamma_{\rho}\gamma_{\kappa} \pm \text{ signed permutations}\end{split}\]

There can be no more because we’re constrained by the \(4\)-dimensional spacetime. We note that these anti-symmetric tensors form a complete basis of all matrices that can be constructed out of the \(\gamma\)-matrices. This is because using Eq.4.4.3, any product of \(\gamma\)-matrices can be written as a linear combination of the anti-symmetric tensors with coefficients the metric tensors.

Now we claim that the matrices \(1, \gamma_{\mu}, \Jscr_{\mu \nu}, \Ascr_{\mu \nu \rho}\) and \(\Pscr_{\mu \nu \rho \kappa}\) are all linearly independent.

Counting these linearly independent matrices, we see that there are \(1 + \binom{4}{1} + \binom{4}{2} + \binom{4}{3} + \binom{4}{4} = 16\) of them. It means that the size of the \(\gamma_{\mu}\) matrices is at least \(4 \times 4\).

It turns out that there exists indeed a solution of Eq.4.4.3 in terms of \(4 \times 4\) matrices, conveniently known as the gamma matrices, which we define as follows

(4.4.7)#\[\begin{split}\gamma_0 \coloneqq -\ifrak \begin{bmatrix*}[r] 0 & 1 \\ 1 & 0 \end{bmatrix*}, \quad \ \bm{\gamma} \coloneqq -\ifrak \begin{bmatrix*}[r] 0 & \bm{\sigma} \\ -\bm{\sigma} & 0 \end{bmatrix*}\end{split}\]

where \(\bm{\sigma} = (\sigma_1, \sigma_2, \sigma_3)\) is made up of the so-called Pauli matrices defined as follows

(4.4.8)#\[\begin{split}\sigma_1 \coloneqq \begin{bmatrix*}[r] 0 & 1 \\ 1 & 0 \end{bmatrix*}, \quad \ \sigma_2 \coloneqq \begin{bmatrix*}[r] 0 & -\ifrak \\ \ifrak & 0 \end{bmatrix*}, \quad \ \sigma_3 \coloneqq \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*}\end{split}\]

Indeed the Pauli matrices make up a solution to not only a \(3\)-dimensional Clifford algebra with respect to the Euclidean inner product, but also an angular momentum representation if multiplied by \(1/2\), or more precisely, a spin-\(1/2\) representation. Note that this representation, in terms of Hermitian matrices, is different from the one given by Eq.1.3.12 with \(\jfrak = 1/2\), in terms of real matrices. One can verify that they differ by a change of basis. Note, however, that as far as the terminology is concerned, one often uses Hermitian and real interchangeably.

Now using the Clifford relations for both gamma and Pauli matrices, we can evaluate Eq.4.4.4 as follows

(4.4.9)#\[\begin{split}\Jscr_{ij} &= -\frac{\ifrak}{4} \left[ \gamma_i, \gamma_j \right] = -\frac{\ifrak}{2} \epsilon_{ij} \gamma_i \gamma_j = -\frac{\ifrak}{2} \epsilon_{ij} \begin{bmatrix} \sigma_i \sigma_j & 0 \\ 0 & \sigma_i \sigma_j \end{bmatrix} = \frac{1}{2} \epsilon_{ijk} \begin{bmatrix} \sigma_k & 0 \\ 0 & \sigma_k \end{bmatrix} \\ \Jscr_{i0} &= -\frac{\ifrak}{4} \left[ \gamma_i, \gamma_0 \right] = -\frac{\ifrak}{2} \gamma_i \gamma_0 = \frac{\ifrak}{2} \begin{bmatrix} \sigma_i & 0 \\ 0 & \sigma_i \end{bmatrix}\end{split}\]

where \(i, j \in \{1,2,3\}\) and \(\epsilon\) is the totally anti-symmetric sign. We see that the representation \(\Jscr_{\mu \nu}\) is in fact reducible. Moreover, we see that that the corresponding representation \(D\) of the Lorentz group given by Eq.4.4.1 is not unitary, since while \(\Jscr_{ij}\) are Hermitian, \(\Jscr_{i0}\) are anti-Hermitian. The fact that \(D\) is not unitary will have consequences when we try to construct the interaction density as in Eq.4.1.3, because products like \(\psi^{\dagger} \psi\) will not be a scalar (see Construction of the Interaction Density for Dirac Fields).

Next let’s consider the parity transformation, i.e., the transformation under spatial inversion, in the context of gamma matrices. In comparison with the transformation laws Eq.1.3.34, we can define

(4.4.10)#\[\begin{split}\beta \coloneqq \ifrak \gamma_0 = \begin{bmatrix*}[r] 0 & 1 \\ 1 & 0 \end{bmatrix*}\end{split}\]

as the parity transformation. Indeed, we clearly have \(\beta^2 = 1\). Moreover, it follows from the Clifford relations Eq.4.4.3 that

(4.4.11)#\[\beta \gamma_i \beta^{-1} = -\gamma_i, \quad \beta \gamma_0 \beta^{-1} = \gamma_0\]

which, in turn, implies that

\[\beta \Jscr_{ij} \beta^{-1} = \Jscr_{ij}, \quad \beta \Jscr_{0i} \beta^{-1} = - \Jscr_{0i}\]

which is consistent with Eq.1.3.34 if we think of \(\Jscr_{ij}\) as the angular momenta and \(\Jscr_{0i}\) as the boosts.

In connection to the non-unitarity of \(D(\Lambda)\), let’s note that since

(4.4.12)#\[\beta \gamma_{\mu}^{\dagger} \beta^{-1} = -\gamma_{\mu}\]

which can be verified by Eq.4.4.11 and Eq.4.4.7, we have \(\beta \Jscr_{\mu \nu}^{\dagger} \beta^{-1} = \Jscr_{\mu \nu}\), and therefore

(4.4.13)#\[\beta D^{\dagger}(\Lambda) \beta^{-1} = D^{-1}(\Lambda)\]

in light of Eq.4.4.1. This identity will be useful when we later construct the interaction density.

At last we’ll introduce yet another special element to the family of gamma matrices, namely \(\gamma_5\), defined as follows

(4.4.14)#\[\begin{split}\gamma_5 \coloneqq -\ifrak \gamma_0 \gamma_1 \gamma_2 \gamma_3 = \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*}\end{split}\]

One nice thing about \(\gamma_5\) is that it anti-commutes with all other \(\gamma\) matrices, and in particular

(4.4.15)#\[\beta \gamma_5 = -\gamma_5 \beta\]

In fact, the collection \(\gamma_0, \gamma_1, \gamma_2, \gamma_3, \gamma_5\) makes exactly a \(5\)-dimensional spacetime Clifford algebra.

4.4.2. Construction of Dirac fields#

As in the case of scalar and vector fields, let’s write the Dirac fields as follows

(4.4.16)#\[\begin{split}\psi^+_{\ell}(x) &= (2\pi)^{-3/2} \sum_{\sigma} \int d^3 p~e^{\ifrak p \cdot x} u_{\ell}(\pbf, \sigma) a(\pbf, \sigma) \\ \psi^{-c}_{\ell}(x) &= (2\pi)^{-3/2} \sum_{\sigma} \int d^3 p~e^{-\ifrak p \cdot x} v_{\ell}(\pbf, \sigma) a^{c \dagger}(\pbf, \sigma)\end{split}\]

Moreover, using \(\Jscr_{ij}\) as given by Eq.4.4.9, we can write the \(\pbf = 0\) conditions Eq.4.1.13 as follows

(4.4.17)#\[\begin{split}\frac{1}{2} \sum_m \bm{\sigma}_{m' m} u_{m \pm}(0, \sigma) &= \sum_{\sigma'} u_{m' \pm}(0, \sigma') \Jbf^{(\jfrak)}_{\sigma' \sigma} \\ -\frac{1}{2} \sum_m \bm{\sigma}_{m' m} v_{m \pm}(0, \sigma) &= \sum_{\sigma'} v_{m' \pm}(0, \sigma') \Jbf^{(\jfrak) \ast}_{\sigma' \sigma}\end{split}\]

Here the signs \(\pm\) correspond to the two identical irreducible representations of \(\Jscr_{ij}\), which is obvious from Eq.4.4.9, while \(m\) and \(m'\) index the Pauli matrices Eq.4.4.8.

Now if we think of \(u_{m \pm}(0, \sigma)\) as matrix elements of a matrix \(U_{\pm}\) and similarly for \(v\), then Eq.4.4.17 can be rewritten compactly in matrix notation as follows

(4.4.18)#\[\begin{split}\tfrac{1}{2} \bm{\sigma} U_{\pm} &= U_{\pm} \Jbf^{(\jfrak)} \\ -\tfrac{1}{2} \bm{\sigma} V_{\pm} &= V_{\pm} \Jbf^{(\jfrak) \ast}\end{split}\]

We recall that both \(\tfrac{1}{2} \bm{\sigma}\) and \(\Jbf^{(\jfrak)}\) (as well as \(-\Jbf^{(\jfrak) \ast}\)) are irreducible representations of the rotation group (or rather, its Lie algebra). We first claim that \(U_{\pm}\) must be isomorphism. Indeed, the kernel of \(U_{\pm}\) is easily seen to be an invariant subspace under the action of \(\Jbf^{(\jfrak)}\), and hence must be null if \(U_{\pm} \neq 0\). On the other hand, the image of \(U_{\pm}\) is an invariant subspace under the action of \(\bm{\sigma}\), and hence must be the whole space if \(U_{\pm} \neq 0\). It follows then that the rank of \(\Jbf^{(\jfrak)}\) and \(U_{\pm}\) must be the same as \(\bm{\sigma}\), which is \(2\). The same argument applies also to \(V_{\pm}\). In particular we must have \(\jfrak = \tfrac{1}{2}\), or in other words, the Dirac fields describe \(\tfrac{1}{2}\)-spin particles.

Note

The mathematical argument above is commonly known as Schur’s lemma.

The matrix form of \(\Jbf^{(1/2)}\) according to Eq.1.3.12 is given By

\[\begin{split}J_1^{(1/2)} = \frac{1}{2} \begin{bmatrix*}[r] 0 & 1 \\ 1 & 0 \end{bmatrix*}, \quad J_2^{(1/2)} = \frac{\ifrak}{2} \begin{bmatrix*}[r] 0 & -1 \\ 1 & 0 \end{bmatrix*}, \quad J_3^{(1/2)} = \frac{1}{2} \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*}\end{split}\]

Comparing with the Pauli matrices Eq.4.4.8, we see that

\[\Jbf^{(1/2)} = \tfrac{1}{2} \bm{\sigma}, \quad -\Jbf^{(1/2) \ast} = \tfrac{1}{2} \sigma_2 \bm{\sigma} \sigma_2\]

Hence the first equation in Eq.4.4.18 may be rewritten as \(\bm{\sigma} U_{\pm} = U_{\pm} \bm{\sigma}\). One can apply Schur’s lemma once again to conclude that \(U_{\pm}\) must be a scalar.

A similar argument can be applied to \(V_{\pm}\) by rewriting the second equation in Eq.4.4.18 as \(\bm{\sigma} (V_{\pm} \sigma_2) = (V_{\pm} \sigma_2) \bm{\sigma}\). Hence \(V_{\pm}\) must be proportional to \(\sigma_2\).

Going back to \(u_{m \pm}(0, \sigma)\) and \(v_{m \pm}(0, \sigma)\) from \(U_{\pm}\) and \(V_{\pm}\), respectively, we have concluded that

\[\begin{split}u_{m \pm}(0, \sigma) &= c_{\pm} \delta_{m \sigma} \\ v_{m \pm}(0, \sigma) &= -\ifrak d_{\pm} (\sigma_2)_{m \sigma}\end{split}\]

where we’ve inserted the extra factor \(-\ifrak\) in front of \(d_{\pm}\) to make the final results look more uniform. We can further unwrap this result in matrix notations as follows

(4.4.19)#\[\begin{split}\begin{alignat*}{2} u(0, 1/2) &= \begin{bmatrix*}[l] c_+ \\ 0 \\ c_- \\ 0 \end{bmatrix*}, \quad u(0, -1/2) &&= \begin{bmatrix*}[l] 0 \\ c_+ \\ 0 \\ c_- \end{bmatrix*} \\ v(0, 1/2) &= \begin{bmatrix*}[l] 0 \\ d_+ \\ 0 \\ d_- \end{bmatrix*}, \quad v(0, -1/2) &&= -\begin{bmatrix*}[l] d_+ \\ 0 \\ d_- \\ 0 \end{bmatrix*} \end{alignat*}\end{split}\]

Now spinors at finite momentum can be determined as usual by Eq.4.1.12 as follows

(4.4.20)#\[\begin{split}u(\pbf, \sigma) &= \sqrt{m/p_0} D(L(p)) u(0, \sigma) \\ v(\pbf, \sigma) &= \sqrt{m/p_0} D(L(p)) v(0, \sigma)\end{split}\]

In general the constants \(c_{\pm}\) and \(d_{\pm}\) may be arbitrary. However, if we further assume the conservation of parity, or in other words, that the fields transform covariantly under the spatial inversion, then the constants, and henceforth the spinors, can be determined uniquely.

To spell out the details, let’s apply the spatial inversion transformation laws from Eq.3.2.10 to Eq.4.4.16 as follows

(4.4.21)#\[\begin{split}& U(\Pcal) \psi^+(x) U^{-1}(\Pcal) \\ &\quad = (2\pi)^{-3/2} \eta^{\ast} \sum_{\sigma} \int d^3 p~e^{\ifrak p \cdot x} u(\pbf, \sigma) a(-\pbf, \sigma) \\ &\quad = (2\pi)^{-3/2} \eta^{\ast} \sum_{\sigma} \int d^3 p~e^{\ifrak p \cdot \Pcal x} u(-\pbf, \sigma) a(\pbf, \sigma) \\ &\quad = (2\pi)^{-3/2} \eta^{\ast} \sum_{\sigma} \int d^3 p~e^{\ifrak p \cdot \Pcal x} \sqrt{m/p_0}~\beta D(L(p)) \beta u(0, \sigma) a(\pbf, \sigma) \\ & U(\Pcal) \psi^{- c}(x) U^{-1}(\Pcal) \\ &\quad = (2\pi)^{-3/2} \eta^c \sum_{\sigma} \int d^3 p~e^{-\ifrak p \cdot \Pcal x} \sqrt{m/p_0}~\beta D(L(p))\beta v(0, \sigma) a^{c \dagger}(\pbf, \sigma)\end{split}\]

Here we’ve evaluated \(u(-\pbf, \sigma)\) and \(v(-\pbf, \sigma)\) in a way similar to Eq.4.3.22. Namely, we’ve used Eq.4.4.20 together with the fact that \(D(\Pcal) = \beta\) defined by Eq.4.4.10.

Now in order for \(U(\Pcal) \psi^+(x) U^{-1}(\Pcal)\) to be proportional to \(\psi^+(x)\), we observe that it would be sufficient (and most likely also necessary) if \(u(0, \sigma)\) is an eigenvector of \(\beta\). Similar argument applies to \(\psi^-(x)\) as well. So let’s suppose

(4.4.22)#\[\begin{split}\beta u(0, \sigma) &= b_+ u(0, \sigma) \\ \beta v(0, \sigma) &= b_- v(0, \sigma)\end{split}\]

with \(b^2_{\pm} = 1\) since \(\beta^2 = 1\). Given this, we can rewrite Eq.4.4.21 as follows

(4.4.23)#\[\begin{split}U(\Pcal) \psi^+(x) U^{-1}(\Pcal) &= \eta^{\ast} b_+ \beta \psi^+(\Pcal x) \\ U(\Pcal) \psi^{- c}(x) U^{-1}(\Pcal) &= \eta^c b_- \beta \psi^{- c}(\Pcal x)\end{split}\]

so that the parity is indeed conserved, or rather, transformed covariantly.

Now using Eq.4.4.22 (and appropriately rescaling), we can rewrite Eq.4.4.19 as follows

\[\begin{split}\begin{alignat*}{2} u(0, 1/2) &= \frac{1}{\sqrt{2}} \begin{bmatrix*}[l] 1 \\ 0 \\ b_+ \\ 0 \end{bmatrix*}, \quad u(0, -1/2) &&= \frac{1}{\sqrt{2}} \begin{bmatrix*}[l] 0 \\ 1 \\ 0 \\ b_+ \end{bmatrix*} \\ v(0, 1/2) &= \frac{1}{\sqrt{2}} \begin{bmatrix*}[l] 0 \\ 1 \\ 0 \\ b_- \end{bmatrix*}, \quad v(0, -1/2) &&= -\frac{1}{\sqrt{2}} \begin{bmatrix*}[l] 1 \\ 0 \\ b_- \\ 0 \end{bmatrix*} \end{alignat*}\end{split}\]

So far by assuming the parity conservation, we’ve managed to reduced the free parameters from \(c_{\pm}, d_{\pm}\) to \(b_{\pm}\). What eventually pins down the spinors is, once again, the causality condition. To see this, let’s try to construct the Dirac field following Eq.4.1.14 as follows

(4.4.24)#\[\psi(x) \coloneqq \kappa \psi^+(x) + \lambda \psi^{- c}(x)\]

As usual, the (anti-)commutator can be calculated using Eq.4.4.16 as follows

(4.4.25)#\[\begin{split}&\left[ \psi_{\ell}(x), \psi_{\ell'}^{\dagger}(y) \right]_{\pm} \\ &\quad = (2\pi)^{-3} \int d^3 p~\left( |\kappa|^2 N_{\ell \ell'}(\pbf) \exp(\ifrak p \cdot (x-y)) \pm |\lambda|^2 M_{\ell \ell'}(\pbf) \exp(-\ifrak p \cdot (x-y)) \right)\end{split}\]

where \(N_{\ell \ell'}\) and \(M_{\ell \ell'}\) are the spin sums defined as follows

(4.4.26)#\[\begin{split}N_{\ell \ell'}(\pbf) &= \sum_{\sigma} u_{\ell}(\pbf, \sigma) u^{\ast}_{\ell'}(\pbf, \sigma) \\ M_{\ell \ell'}(\pbf) &= \sum_{\sigma} v_{\ell}(\pbf, \sigma) v^{\ast}_{\ell'}(\pbf, \sigma)\end{split}\]

To evaluate the spin sums, we first turn back to the zero-momentum case and use Eq.4.4.22 to express the values in terms of \(\beta\) as follows

(4.4.27)#\[\begin{split}\begin{alignat*}{2} N(0) &= \sum_{\sigma} u(0, \sigma) u^{\dagger}(0, \sigma) &&= \frac{1 + b_+ \beta}{2} \\ M(0) &= \sum_{\sigma} v(0, \sigma) v^{\dagger}(0, \sigma) &&= \frac{1 + b_- \beta}{2} \end{alignat*}\end{split}\]

where \(\dagger\) here means transpose conjugation. The easiest way to see this is probably to momentarily forget about the spin \(z\)-component \(\sigma\) so that \(\beta\) as in Eq.4.4.10 behaves like a \(2 \times 2\) matrix with obvious eigenvectors \([1, 1]^T\) for \(b_{\pm} = 1\) and \([1, -1]^T\) for \(b_{\pm} = -1\). Then Eq.4.4.27 can be verified by a direct calculation. Here the superscript \(T\) means taking transpose so we have column vectors.

Using Eq.4.4.26, Eq.4.4.20, and Eq.4.4.27, we can evaluate the \(N\) and \(M\) matrices in terms of \(\beta\) as follows

(4.4.28)#\[\begin{split}N(\pbf) &= \sum_{\sigma} u(\pbf, \sigma) u^{\dagger}(\pbf, \sigma) \\ &= \frac{m}{p_0} D(L(p)) \left( \sum_{\sigma} u(0, \sigma) u^{\dagger}(0, \sigma) \right) D^{\dagger}(L(p)) \\ &= \frac{m}{2p_0} D(L(p)) (1 + b_+ \beta) D^{\dagger}(L(p)) \\ M(\pbf) &= \frac{m}{2p_0} D(L(p)) (1 + b_- \beta) D^{\dagger}(L(p))\end{split}\]

To go further, we need to invoke the the transformation laws of the gamma matrices and their relatives, e.g., \(\beta\), under \(D(\Lambda)\) from Dirac representation and gamma matrices. More precisely, using the pseudo-unitarity Eq.4.4.13, as well as Eq.4.4.10, Eq.4.4.6, and Eq.1.3.17, we have the following

(4.4.29)#\[\begin{split}D(L(p)) \beta D^{\dagger}(L(p)) &= D(L(p)) D^{-1}(L(p)) \beta = \beta \\ D(L(p)) D^{\dagger}(L(p)) &= D(L(p)) \beta D^{-1}(L(p)) \beta \\ &= \ifrak D(L(p)) \gamma_0 D^{-1}(L(p)) \beta \\ &= \ifrak {\left( L(p) \right)^{\mu}}_0 \gamma_{\mu} \beta \\ &= -\ifrak p^{\mu} \gamma_{\mu} \beta / m = -\ifrak p_{\mu} \gamma^{\mu} \beta / m\end{split}\]

Plugging them into Eq.4.4.28, respectively, we can continue our evaluation as follows

\[\begin{split}N(\pbf) &= \frac{1}{2p_0} \left( -\ifrak p_{\mu} \gamma^{\mu} + b_+ m \right) \beta \\ M(\pbf) &= \frac{1}{2p_0} \left( -\ifrak p_{\mu} \gamma^{\mu} + b_- m \right) \beta\end{split}\]

Now that we’ve finished evaluating the spin sums, we can plug them into Eq.4.4.25 to get the following evaluation of the (anti-)commutator

(4.4.30)#\[\begin{split}\left[ \psi_{\ell}(x), \psi_{\ell'}^{\dagger}(y) \right]_{\pm} &= (2\pi)^{-3} \int \frac{d^3 p}{2p_0}~\big( |\kappa|^2 (-\ifrak p_{\mu} \gamma^{\mu} + b_+ m) e^{\ifrak p \cdot (x-y)} \beta \\ &\quad \pm |\lambda|^2 (-\ifrak p_{\mu} \gamma^{\mu} + b_- m) e^{-\ifrak p \cdot (x-y)} \beta \big)_{\ell \ell'} \\ &= \big( |\kappa|^2 \left( -\ifrak \gamma^{\mu} \p_{x_{\mu}} + b_+ m \right) \Delta_+(x-y) \beta \\ &\quad \pm |\lambda|^2 \left( -\ifrak \gamma^{\mu} \p_{y_{\mu}} + b_- m \right) \Delta_+(y-x) \beta \big)_{\ell \ell'}\end{split}\]

where \(\Delta_+\) is defined by Eq.4.2.4. Recall that \(\Delta_+(x) = \Delta_+(-x)\) for space-like \(x\). Hence for space-like \(x-y\), the following holds

\[\p_{x_{\mu}} \Delta_+(x-y) = \p_{x_{\mu}} \Delta_+(y-x) = -\p_{y_{\mu}} \Delta_+(y-x)\]

It follows that in order for Eq.4.4.30 to vanish for space-separated \(x\) and \(y\), the following must hold

\[\begin{split}|\kappa|^2 \mp |\lambda|^2 &= 0 \\ |\kappa|^2 b_+ \pm |\lambda|^2 b_- &= 0\end{split}\]

We see that first of all, the top sign applies, which means in particular that we must accept the anti-commutator in Eq.4.4.30, or in other words, the Dirac fields must be fermionic. In addition, we must also have \(|\kappa| = |\lambda|\) and \(b_+ + b_- = 0\).

By the usual phase adjustments on the creation and annihilation operators and rescaling, we can arrange so that \(\kappa = \lambda = 1\). Recalling Eq.4.4.15 and replacing \(\psi\) with \(\gamma_5 \psi\) if necessary, we can arrange so that \(b_{\pm} = \pm 1\). Putting these all together, we have evaluated the Dirac field Eq.4.4.24 as follows

(4.4.31)#\[\begin{split}\psi(x) &= \psi^+(x) + \psi^{- c}(x) \\ &= (2\pi)^{-3/2} \sum_{\sigma} \int d^3 p~\left( e^{\ifrak p \cdot x} u(\pbf, \sigma) a(\pbf, \sigma) + e^{-\ifrak p \cdot x} v(\pbf, \sigma) a^{c \dagger}(\pbf, \sigma) \right)\end{split}\]

where the zero-momentum spinors are

(4.4.32)#\[\begin{split}\begin{alignat*}{2} u(0, 1/2) &= \frac{1}{\sqrt{2}} \begin{bmatrix*}[r] 1 \\ 0 \\ 1 \\ 0 \end{bmatrix*}, \quad &&u(0, -1/2) = \frac{1}{\sqrt{2}} \begin{bmatrix*}[r] 0 \\ 1 \\ 0 \\ 1 \end{bmatrix*} \\ v(0, 1/2) &= \frac{1}{\sqrt{2}} \begin{bmatrix*}[r] 0 \\ 1 \\ 0 \\ -1 \end{bmatrix*}, \quad &&v(0, -1/2) = -\frac{1}{\sqrt{2}} \begin{bmatrix*}[r] 1 \\ 0 \\ -1 \\ 0 \end{bmatrix*} \end{alignat*}\end{split}\]

and the spin sums are

(4.4.33)#\[\begin{split}\begin{align*} N(\pbf) &= \frac{1}{2p_0} \left( -\ifrak p^{\mu} \gamma_{\mu} + m \right) \beta \\ M(\pbf) &= \frac{1}{2p_0} \left( -\ifrak p^{\mu} \gamma_{\mu} - m\right) \beta \end{align*}\end{split}\]

and the anti-commutator, calculated by plugging the spin sums into Eq.4.4.25, is

(4.4.34)#\[\left[ \psi_{\ell}(x), \psi_{\ell'}^{\dagger}(y) \right]_+ = \left( \left( -\gamma^{\mu}\p_{\mu} + m \right) \beta \right)_{\ell \ell'} \Delta(x-y)\]

where \(\p_{\mu} = \p_{x_{\mu}}\) and \(\Delta(x-y)\) is defined by Eq.4.2.10.

For \(\psi(x)\) defined by Eq.4.4.31 to transform covariantly under spatial inversion, we recall Eq.4.4.23 to conclude that

(4.4.35)#\[\eta^{\ast} + \eta^c = 0\]

It follows that \(\eta \eta^c = -1\), or in other words, the intrinsic parity of the state consisting of a spin \(1/2\) particle and its antiparticle is odd in the sense of Parity symmetry. The parity transformation law for Dirac fields is as follows

(4.4.36)#\[U(\Pcal) \psi(x) U^{-1}(\Pcal) = \eta^{\ast} \beta \psi(\Pcal x)\]

4.4.3. The CPT symmetries#

The transformation law of Dirac fields under spatial inversion has already been worked out in Eq.4.4.36, so we’re left to work out the transformation laws under time inversion and charge conjugation.

Recall from Space and time inversions that the time-inversion operator \(U(\Tcal)\) is complex anti-linear. Hence to work out the transformation law under time inversion, we’ll need to work out the complex-conjugated spinors \(u^{\ast}(\pbf, \sigma)\) and \(v^{\ast}(\pbf, \sigma)\). Now in light of Eq.4.4.20, and the fact that the spinors are real at zero-momentum, we just need to work out the complex conjugate \(D^{\ast}(L(p))\) in terms of \(D(L(p))\) and the gamma matrices. Now according to Eq.4.4.1, it suffices to work out \(\Jscr^{\ast}_{\mu \nu}\), and finally according to Eq.4.4.4, it suffices to work out \(\gamma^{\ast}_{\mu}\).

Inspecting the explicit forms of the gamma matrices given by Eq.4.4.7 and Eq.4.4.8 we see that \(\gamma_0, \gamma_1, \gamma_3\) are anti-Hermitian while \(\gamma_2\) is Hermitian, or more explicitly

\[\gamma^{\ast}_0 = -\gamma_0, \quad \gamma^{\ast}_1 = -\gamma_1, \quad \gamma^{\ast}_2 = \gamma_2, \quad \gamma^{\ast}_3 = -\gamma_3\]

Using the Clifford algebra relations, this can be written more concisely as follows

(4.4.38)#\[\gamma^{\ast}_{\mu} = \gamma_2 \gamma_{\mu} \gamma_2\]

While this result could’ve be satisfactory in its own right, as we’ll see, it’ll be more convenient to factor out a \(\beta\) matrix. Hence we’re motivated to introduce yet another special matrix

(4.4.39)#\[\begin{split}\Cscr \coloneqq \gamma_2 \beta = -\ifrak \begin{bmatrix*}[r] \sigma_2 & 0 \\ 0 & -\sigma_2 \end{bmatrix*}\end{split}\]

and rewrite Eq.4.4.38 as follows

\[\gamma^{\ast}_{\mu} = \beta \Cscr \gamma_{\mu} \Cscr^{-1} \beta\]

where we also note that \((\Cscr^{-1} \beta)^{-1} = \beta \Cscr\). It follows from Eq.4.4.4 that

\[\Jscr^{\ast}_{\mu \nu} = -\beta\Cscr \Jscr_{\mu \nu} \Cscr^{-1}\beta\]

and hence from Eq.4.4.1 that

(4.4.40)#\[D^{\ast}(L(p)) = 1 - \frac{\ifrak}{2} \omega^{\mu \nu} \Jscr^{\ast}_{\mu \nu} = 1 + \frac{\ifrak}{2} \omega^{\mu \nu} \beta\Cscr \Jscr_{\mu \nu} \Cscr^{-1}\beta = \beta \Cscr D(L(p)) \Cscr^{-1} \beta\]

Using the explicit formula Eq.4.4.39 for \(\Cscr\) as well Eq.4.4.32 for \(u(0, \sigma)\) and \(v(0, \sigma)\), we get the following

(4.4.41)#\[\begin{split}\begin{alignat*}{2} u^{\ast}(\pbf, \sigma) &= \sqrt{m/p_0} D^{\ast}(L(p)) u(0, \sigma) &&= -\beta\Cscr v(\pbf, \sigma) \\ v^{\ast}(\pbf, \sigma) &= \sqrt{m/p_0} D^{\ast}(L(p)) v(0, \sigma) &&= -\beta\Cscr u(\pbf, \sigma) \end{alignat*}\end{split}\]

These relations turns out to be useful for the charge conjugation transformation, but not for the time inversion because the spinors \(u\) and \(v\) are swapped. To remedy this, we notice from Eq.4.4.32 that the \(u\) and \(v\) spinors at zero-momentum are related by \(\gamma_5\) defined by Eq.4.4.14. Moreover, in order to cancel the \(\beta\) matrices at the two ends of right-hand-side of Eq.4.4.40, we can replace \(\pbf\) with \(-\pbf\), which is also desirable as far as the time inversion is concerned in light of Eq.3.2.10. Putting all these considerations together, let’s try the following

\[\begin{split}D^{\ast}(L(-\pbf)) &= D^{\ast}(\Pcal L(p) \Pcal) \\ &= \beta D^{\ast}(L(p)) \beta \\ &= \gamma_5 \beta D^{\ast}(L(p)) \beta \gamma_5 \\ &= \gamma_5 \Cscr D(L(p)) \Cscr^{-1} \gamma_5\end{split}\]

where the third equality holds because of the Clifford relations, namely, \(\gamma_5\) commutes with \(\Jscr_{\mu \nu}\), and hence \(D^{\ast}(L(p))\), and anti-commutes with \(\beta\). Now instead of Eq.4.4.41, we can calculate as follows

(4.4.42)#\[\begin{split}u^{\ast}(-\pbf, \sigma) &= \sqrt{m/p_0} D^{\ast}(L(-\pbf)) u(0, \sigma) \\ &= \sqrt{m/p_0} \gamma_5 \Cscr D^{\ast}(L(p)) \Cscr^{-1} \gamma_5 u(0, \sigma) \\ &= \sqrt{m/p_0} \gamma_5 \Cscr D^{\ast}(L(p)) (-1)^{1/2 + \sigma} u(0, -\sigma) \\ &= (-1)^{1/2 + \sigma} \gamma_5 \Cscr u(\pbf, -\sigma)\end{split}\]

A similar calculation can be done to show that in fact \(v(-\pbf, \sigma)\) satisfies exactly the same conjugation formula.

With all the preparations above, we can now, using Eq.4.4.31, Eq.3.2.10, and Eq.4.4.42, calculate the spatial inversion transformation laws as follows

\[\begin{split}U(\Tcal) \psi(x) U^{-1}(\Tcal) &= (2\pi)^{-3/2} \sum_{\sigma} \int d^3 p~\big( e^{-\ifrak p \cdot x} u^{\ast}(\pbf, \sigma) U(\Tcal) a(\pbf, \sigma) U^{-1}(\Tcal) \\ &\qquad + e^{\ifrak p \cdot x} v^{\ast}(\pbf, \sigma) U(\Tcal) a^{c \dagger}(\pbf, \sigma) U^{-1}(\Tcal) \big) \\ &= (2\pi)^{-3/2} \sum_{\sigma} (-1)^{1/2 - \sigma} \int d^3 p~\big( e^{-\ifrak p \cdot x} \zeta^{\ast} u^{\ast}(\pbf, \sigma) a(-\pbf, -\sigma) \\ &\qquad + e^{\ifrak p \cdot x} \zeta^c v^{\ast}(\pbf, \sigma) a^{c \dagger}(-\pbf, -\sigma) \big) \\ &= (2\pi)^{-3/2} \sum_{\sigma} (-1)^{1/2 + \sigma} \int d^3 p~\big( e^{-\ifrak p \cdot \Pcal x} \zeta^{\ast} u^{\ast}(-\pbf, -\sigma) a(\pbf, \sigma) \\ &\qquad + e^{\ifrak p \cdot \Pcal x} \zeta^c v^{\ast}(-\pbf, -\sigma) a^{c \dagger}(\pbf, \sigma) \big) \\ &= -(2\pi)^{-3/2} \gamma_5 \Cscr \sum_{\sigma} \int d^3 p~\big( \zeta^{\ast} e^{-\ifrak p \cdot \Pcal x} u(\pbf, \sigma) a(\pbf, \sigma) \\ &\qquad + \zeta^c e^{\ifrak p \cdot \Pcal x} v(\pbf, \sigma) a^{c \dagger}(\pbf, \sigma) \big)\end{split}\]

In order for \(\psi(x)\) to transform nicely under the time inversion, we’re forced to make the following assumption

(4.4.43)#\[\zeta^{\ast} = \zeta^c\]

Under this assumption we’ve finally worked out the time inversion transformation law for Dirac fields

\[U(\Tcal) \psi(x) U^{-1}(\Tcal) = -\zeta^{\ast} \gamma_5 \Cscr \psi(-\Pcal x)\]

Next let’s calculate the charge inversion transformation, using Eq.3.2.10 and Eq.4.4.41, as follows

\[\begin{split}\begin{align*} U(\Ccal) \psi(x) U^{-1}(\Ccal) &= (2\pi)^{-3/2} \sum_{\sigma} \int d^3 p~\left( e^{\ifrak p \cdot x} u(\pbf, \sigma) \xi^{\ast} a^c(\pbf, \sigma) + e^{-\ifrak p \cdot x} v(\pbf, \sigma) \xi^c a^{\dagger}(\pbf, \sigma) \right) \\ &= (2\pi)^{-3/2} \Cscr \beta \sum_{\sigma} \int d^3p~\left( \xi^{\ast} e^{\ifrak p \cdot x} v^{\ast}(\pbf, \sigma) a^c(\pbf, \sigma) + \xi^c e^{-\ifrak p \cdot x} u^{\ast}(\pbf, \sigma) a^{\dagger}(\pbf, \sigma) \right) \end{align*}\end{split}\]

Just as for the time inversion, we are forced to assuming the following condition on the charge conjugation parities

(4.4.44)#\[\xi^{\ast} = \xi^c\]

Under this assumption, we can work out the charge conjugation transformation law as follows

\[U(\Ccal) \psi(x) U^{-1}(\Ccal) = \xi^{\ast} \Cscr \beta \psi^{\ast}(x)\]

Here we’ve used \(\psi^{\ast}(x)\) instead of \(\psi^{\dagger}(x)\) because we don’t want to transpose the spinors, but it should be understood that the \(\ast\) when applied to the creation/annihilation operators are the same as taking the adjoint.

Finally, let’s consider the special case where the spin-\(1/2\) particles are their own antiparticles. These particles are known as Majorana fermions, as already discussed in Parities of elementary particles. According to Eq.4.4.35, Eq.4.4.43 and Eq.4.4.44, we see that the spatial parity of a Majorana fermion must be \(\pm \ifrak\), while the time and charge parity must be \(\pm 1\).

4.4.4. Construction of the interaction density#

As mentioned in Dirac representation and gamma matrices, the fact that the Dirac representation is not unitary means that we cannot construct the interaction density using \(\psi^{\dagger} \psi\) because it won’t be a scalar. Indeed, let’s work out how \(\psi^{\dagger}\) transforms under a (homogeneous orthochronous) Lorentz transformation using Eq.4.1.2 and Eq.4.4.13 as follows

\[\begin{split}U_0(\Lambda) \psi^{\dagger}(x) U_0^{-1}(\Lambda) &= \left( U_0(\Lambda) \psi(x) U_0^{-1}(\Lambda) \right)^{\dagger} \\ &= \left( D^{-1}(\Lambda) \psi(\Lambda x) \right)^{\dagger} \\ &= \psi^{\dagger}(\Lambda x) \left( D^{-1}(\Lambda) \right)^{\dagger} \\ &= \psi^{\dagger}(\Lambda x) \beta D(\Lambda) \beta\end{split}\]

We see that if we define a new adjoint

(4.4.45)#\[\bar{\psi} \coloneqq \psi^{\dagger} \beta\]

then \(\bar{\psi}\) transforms nicely as follows

\[U_0(\Lambda) \bar{\psi}(x) U_0^{-1}(\Lambda) = \bar{\psi}(\Lambda x) D(\Lambda)\]

It follows that we can construct a bilinear form as follows

\[\bar{\psi}(x) M \psi(x)\]

where \(M\) is a \(4 \times 4\) matrix, so that

\[U_0(\Lambda) \bar{\psi}(x) M \psi(x) U_0^{-1}(\Lambda) = \bar{\psi}(\Lambda x) D(\Lambda) M D^{-1}(\Lambda) \psi(\Lambda x)\]

Letting \(M\) to be \(1, \gamma_{\mu}, \Jscr_{\mu \nu}, \gamma_5 \gamma_{\mu}\) or \(\gamma_5\) then produces a scalar, vector, tensor, axial vector or pseudo-scalar, respectively. Here the adjectives “axial” and “pseudo-” refer to the opposite to usual parities under spatial and/or time inversion.

An important example is Fermi’s theory of beta-decay, which involves an interaction density of the following form

\[\bar{\psi}_p \gamma^{\mu} \psi_n \bar{\psi}_e \gamma_{\mu} \psi_{\nu}\]

where \(p, n, e, \nu\) stand for proton, neutron, electron and neutrino, respectively.

4.5. General Fields#

We’ve now seen how scalar, vector, and Dirac fields can be constructed out of specific representations of the (homogeneous orthochronous) Lorentz group. These constructions can be generalized and unified by understanding the general representation theory of the Lorentz group.

4.5.1. General representation theory of the Lorentz group#

The starting point, as in the case of Dirac fields, is the general commutation relation Eq.4.4.2 that the \(\Jscr_{\mu \nu}\) matrices must satisfy. As explained in Quantum Lorentz symmetry, we can rename the \(\Jscr_{\mu \nu}\) matrices as follows

\[\begin{split}\begin{alignat*}{3} \Jscr_1 &\coloneqq \Jscr_{23}, \quad \Jscr_2 &&\coloneqq \Jscr_{31}, \quad \Jscr_3 &&\coloneqq \Jscr_{12} \\ \Kscr_1 &\coloneqq \Jscr_{10}, \quad \Kscr_2 &&\coloneqq \Jscr_{20}, \quad \Kscr_3 &&\coloneqq \Jscr_{30} \end{alignat*}\end{split}\]

and rewrite Eq.4.4.2 as a set of equations as follows

(4.5.1)#\[\begin{split}\left[ \Jscr_i, \Jscr_j \right] &= \ifrak \epsilon_{ijk} \Jscr_k \\ \left[ \Jscr_i, \Kscr_j \right] &= \ifrak \epsilon_{ijk} \Kscr_k \\ \left[ \Kscr_i, \Kscr_j \right] &= -\ifrak \epsilon_{ijk} \Jscr_k\end{split}\]

which correspond to the commutation relations between momentum and boost operators in Eq.1.2.22.

Let’s write \(\bm{\Jscr} \coloneqq \left(\Jscr_1, \Jscr_2, \Jscr_3\right)\) and \(\bm{\Kscr} \coloneqq \left(\Kscr_1, \Kscr_2, \Kscr_3\right)\). It turns out that this Lie algebra generated by \(\bm{\Jscr}\) and \(\bm{\Kscr}\) can be complex linearly transformed into one that splits. The transformation is defined as follows

(4.5.2)#\[\begin{split}\bm{\Ascr} &= \frac{1}{2} \left( \bm{\Jscr} + \ifrak \bm{\Kscr} \right) \\ \bm{\Bscr} &= \frac{1}{2} \left( \bm{\Jscr} - \ifrak \bm{\Kscr} \right)\end{split}\]

so that Eq.4.5.1 is equivalent to the following

(4.5.3)#\[\begin{split}\left[ \bm{\Ascr}_i, \bm{\Ascr}_j \right] &= \ifrak \epsilon_{ijk} \bm{\Ascr}_k \\ \left[ \bm{\Bscr}_i, \bm{\Bscr}_j \right] &= \ifrak \epsilon_{ijk} \bm{\Bscr}_k \\ \left[ \bm{\Ascr}_i, \bm{\Bscr}_j \right] &= 0\end{split}\]

In other words, both \(\bm{\Ascr}\) and \(\bm{\Bscr}\) form a Lie algebra of the \(3\)-dimensional rotation group and they commute each other. It follows then from Representations of angular momentum that representations of the Lie algebra defined by Eq.4.5.3 can be parametrized by two nonnegative (half-)integers \(A\) and \(B\) such that

(4.5.4)#\[\begin{split}\bm{\Ascr}_{a'b', ab} &= \delta_{b'b} \Jbf^{(A)}_{a'a} \\ \bm{\Bscr}_{a'b', ab} &= \delta_{a'a} \Jbf^{(B)}_{b'b}\end{split}\]

where \(a, a' \in \{-A, -A+1, \cdots, A\}\) and \(b, b' \in \{-B, -B+1, \cdots, B\}\), and \(\Jbf^{(A)}\) and \(\Jbf^{(B)}\) are matrices given by Eq.1.3.12. In particular, these representations have dimension \((2A+1)(2B+1)\) and are unitary.

Now each one of these representations gives rise to a representation of the Lorentz group, which will be referred to as the \((A, B)\) representation. As we’ve seen for Dirac fields, these representations are not unitary, because while \(\bm{\Jscr} = \bm{\Ascr} + \bm{\Bscr}\) is Hermitian, \(\bm{\Kscr} = -\ifrak \left( \bm{\Ascr} - \bm{\Bscr} \right)\) is anti-Hermitian. For the corresponding unitary representation of the \(3\)-dimensional rotation group, we recall from Clebsch-Gordan coefficients that it may be split into irreducible components of spin \(\jfrak\), which takes values in the following range

(4.5.5)#\[\jfrak = |A-B|, |A-B| + 1, \cdots, A+B\]

according to Eq.1.3.26. Under this setup, the scalar, vector, and Dirac fields discussed before correspond to the following three scenarios.

Scalar field: \(A = B = 0\). In this case \(\jfrak\) must vanish, and hence no spin is possible.
Vector field: \(A = B = \tfrac{1}{2}\). In this case \(\jfrak\) may be \(0\) or \(1\), which correspond to the time and space components of the vector field, respectively.
Dirac field: \(A = \tfrac{1}{2}, B = 0\) or \(A = 0, B = \tfrac{1}{2}\). In either case \(\jfrak\) must be \(\tfrac{1}{2}\). Indeed, they correspond to the two irreducible components of (the angular momentum part of) the Dirac representation Eq.4.4.9. Therefore the Dirac field may be written in short hand as \(\left( \tfrac{1}{2}, 0 \right) \oplus \left( 0, \tfrac{1}{2} \right)\).

It turns out that any general \((A, B)\) fields can be derived from the above basic ones by taking tensor products and irreducible components. For example \((1, 0)\) and \((0, 1)\) fields can be derived, using again Clebsch-Gordan coefficients, by the following calculation

\[\left( \tfrac{1}{2}, \tfrac{1}{2} \right) \otimes \left( \tfrac{1}{2}, \tfrac{1}{2} \right) = (0, 0) \oplus (0, 1) \oplus (1, 0) \oplus (1, 1)\]

In fact, all \((A, B)\) fields with \(A + B\) being an integer can be obtained in this way by tensoring copies of \(\left( \tfrac{1}{2}, \tfrac{1}{2} \right)\). To get those fields with \(A + B\) being a half-integer, we can consider the following calculation

\[\left( \tfrac{1}{2}, \tfrac{1}{2} \right) \otimes \left( \left(\tfrac{1}{2}, 0\right) \oplus \left(0, \tfrac{1}{2}\right) \right) = \left(0, \tfrac{1}{2}\right) \oplus \left(1, \tfrac{1}{2}\right) \oplus \left(\tfrac{1}{2}, 0\right) \oplus \left(\tfrac{1}{2}, 1\right)\]

4.5.2. Construction of general fields#

We’ve seen that general fields can be indexed by two (half-)integers \(a\) and \(b\), and take the following general form

(4.5.6)#\[\psi_{ab}(x) = (2\pi)^{-3/2} \sum_{\sigma} \int d^3 p~\left( \kappa e^{\ifrak p \cdot x} u_{ab}(\pbf, \sigma) a(\pbf, \sigma) + \lambda e^{-\ifrak p \cdot x} v_{ab}(\pbf, \sigma) a^{c \dagger}(\pbf, \sigma) \right)\]

As usual, let’s translate Eq.4.1.13 in the context of \((A, B)\) representations as follows

\[\begin{split}\sum_{\sigma'} u_{a'b'}(0, \sigma') \Jbf^{(\jfrak)}_{\sigma' \sigma} &= \sum_{a, b} \bm{\Jscr}_{a'b', ab} u_{ab}(0, \sigma) \\ -\sum_{\sigma'} v_{a'b'}(0, \sigma') \Jbf^{(\jfrak) \ast}_{\sigma' \sigma} &= \sum_{a, b} \bm{\Jscr}_{a'b', ab} v_{ab}(0, \sigma)\end{split}\]

Using the fact that \(\bm{\Jscr} = \bm{\Ascr} + \bm{\Bscr}\) and Eq.4.5.4, we can further rewrite these conditions as follows

(4.5.7)#\[\begin{split}\sum_{\sigma'} u_{a'b'}(0, \sigma') \Jbf^{(\jfrak)}_{\sigma' \sigma} &= \sum_a \Jbf^{(A)}_{a'a} u_{ab'}(0, \sigma) + \sum_b \Jbf^{(B)}_{b'b} u_{a'b}(0, \sigma) \\ -\sum_{\sigma'} v_{a'b'}(0, \sigma') \Jbf^{(\jfrak) \ast}_{\sigma' \sigma} &= \sum_a \Jbf^{(A)}_{a'a} v_{ab'}(0, \sigma) + \sum_b \Jbf^{(B)}_{b'b} v_{a'b}(0, \sigma)\end{split}\]

Looking at the equation on the \(u\)-field in Eq.4.5.7, it’s an identity that relates an angular momentum representation of spin \(\jfrak\) on the left to the sum of two independent angular momentum representations on the right. Hence it’s not unreasonable to guess that \(u\) might have something to do with the Clebsch-Gordan coefficients defined by Eq.1.3.20. This turns out to be indeed the case as we now demonstrate. Since \(\bigoplus_{\jfrak} \Jbf^{(\jfrak)} = \Jbf^{(A)} + \Jbf^{(B)}\), where the direct sum is taken over the range Eq.4.5.5 and can be thought of as a block diagonal matrix, we can calculate as follows

(4.5.8)#\[\begin{split}\Jbf^{(\jfrak)}_{\sigma' \sigma} &=\left ( \Psi^{AB~\jfrak}_{\sigma'},~\bigoplus_{\jfrak'} \Jbf^{(\jfrak')} \Psi^{AB~\jfrak}_{\sigma} \right) \\ &= \left( \sum_{a', b'} C^{AB}(\jfrak, \sigma'; a', b') \Psi^{AB}_{a'b'},~\sum_{a, b} C^{AB}(\jfrak, \sigma; a, b) \bigoplus_{\jfrak'} \Jbf^{(\jfrak')} \Psi^{AB}_{ab} \right) \\ &= \sum_{a', b', a, b} C^{AB}(\jfrak, \sigma'; a', b') C^{AB}(\jfrak, \sigma; a, b) \left( \Psi^{AB}_{a'b'},~\left( \Jbf^{(A)} + \Jbf^{(B)} \right) \Psi^{AB}_{ab} \right) \\ &= \sum_{a', b', a, b} C^{AB}(\jfrak, \sigma'; a', b') C^{AB}(\jfrak, \sigma; a, b) \left( \delta_{b'b} \Jbf^{(A)}_{a'a} + \delta_{a'a} \Jbf^{(B)}_{b'b} \right) \\ &= \sum_{a', b'} C^{AB}(\jfrak, \sigma'; a', b') \left( \blue{\sum_a C^{AB}(\jfrak, \sigma; a, b') \Jbf^{(A)}_{a'a} + \sum_b C^{AB}(\jfrak, \sigma; a', b) \Jbf^{(B)}_{b'b}} \right)\end{split}\]

We are now one step away from being able to compare with Eq.4.5.7. Namely we need to bring the coefficients \(C^{AB}(\jfrak, \sigma'; a', b')\) to the left side of the equation. To to this, we recall the following identity of Clebsch-Gordan coefficients

(4.5.9)#\[\sum_{a', b'} C^{AB}(\jfrak', \sigma'; a', b') C^{AB}(\jfrak, \sigma; a', b') = \delta_{\jfrak'\jfrak} \delta_{\sigma'\sigma}\]

which follows from the orthonormality of the states \(\Psi^{AB}_{ab}\) and the reality of the Clebsch-Gordan coefficients as constructed in Clebsch-Gordan coefficients. [3] Now Eq.4.5.8 will be satisfied if we set the blue terms to equal to the following quantity

(4.5.10)#\[\sum_a C^{AB}(\jfrak, \sigma; a, b') \Jbf^{(A)}_{a'a} + \sum_b C^{AB}(\jfrak, \sigma; a', b) \Jbf^{(B)}_{b'b} = \sum_{\sigma''} C^{AB}(\jfrak, \sigma; a', b') \Jbf^{(\jfrak)}_{\sigma'' \sigma}\]

due to Eq.4.5.9.

Now compare Eq.4.5.10 with Eq.4.5.7, we’ve solved the \(u\)-fields of dimension \((2A+1)(2B+1)\) as follows

(4.5.11)#\[u_{ab}(0, \sigma) = (2m)^{-1/2} C^{AB}(\jfrak, \sigma; a, b)\]

where \((2m)^{-1/2}\) is a conventional coefficient add here to cancel the mass term in Eq.4.1.12 later. Using the fact that

(4.5.12)#\[-\Jbf^{(\jfrak) \ast}_{\sigma'\sigma} = (-1)^{\sigma'-\sigma} \Jbf^{(\jfrak)}_{-\sigma', -\sigma}\]

which can be verified directly using Eq.1.3.12, we can express the \(v\)-fields in terms of the \(u\)-fields as follows

(4.5.13)#\[v_{ab}(0, \sigma) = (-1)^{\jfrak + \sigma} u_{ab}(0, -\sigma)\]

To get the \(u\) and \(v\) fields at finite momentum, we need to invoke the general boost formulae Eq.4.1.12, as well as the \(L\) transformation Eq.1.3.17. Here we’ll think of a boost as a \(1\)-parameter transformation in a given direction \(\hat{\pbf} \coloneqq \pbf / |\pbf|\). It turns out to be neat to use a hyperbolic angle \(\theta\), rather than \(|\pbf|\), defined by

(4.5.14)#\[\cosh\theta = \sqrt{\pbf^2 + m^2} / m, \quad \sinh\theta = |\pbf| / m\]

to parametrize the boost as follows

\[\begin{split}{L(\theta)_0}^0 &= \cosh\theta \\ {L(\theta)_i}^0 &= \hat{\pbf}_i \sinh\theta \\ L(\theta)_{ij} &= \delta_{ij} + \hat{\pbf}_i \hat{\pbf}_j (\cosh\theta - 1)\end{split}\]

The nice thing about this parametrization is that \(L(\theta)\) becomes additive in \(\theta\) in the following sense

(4.5.15)#\[L(\theta') L(\theta) = L(\theta' + \theta)\]

In light of Eq.4.4.1, we can then write

\[D(L(p)) = \exp\left(-\ifrak \theta~\hat{\pbf} \cdot \bm{\Kscr}\right)\]

at least for \(\theta\) infinitesimal. Here the minus sign comes from the fact that we have to bring the upper-index \(\mu=0\) in \(\omega^{\mu\nu}\) down.

For an \((A, B)\) representation, we can further write \(\ifrak \bm{\Kscr} = \bm{\Ascr} - \bm{\Bscr}\), and henceforth

(4.5.16)#\[D(L(p))_{a'b', ab} = \exp\left(-\theta~\hat{\pbf} \cdot \Jbf^{(A)}\right)_{a'a} \exp\left(\theta~\hat{\pbf} \cdot \Jbf^{(B)}\right)_{b'b}\]

since the representation splits into a direct sum of \(\bm{\Ascr}\) and \(\bm{\Bscr}\). Combining Eq.4.5.16 with Eq.4.5.11 and Eq.4.1.12, we obtain the following formula for the \(u\)-field at finite momentum for an \((A, B)\) representation

(4.5.17)#\[u_{ab}(\pbf, \sigma) = \frac{1}{\sqrt{2p_0}} \sum_{a'b'} \left(\exp\left(-\theta~\hat{\pbf} \cdot \Jbf^{(A)}\right)\right)_{a'a} \left(\exp\left(\theta~\hat{\pbf} \cdot \Jbf^{(B)}\right)\right)_{b'b} C^{AB}(\jfrak \sigma; a'b')\]

where we assume implicitly that the spin \(\jfrak\) is within the range Eq.4.5.5, and \(\sigma\) is the corresponding spin \(z\)-component.

Parallel to Eq.4.5.13, we can express the \(v\)-field at finite momentum in terms of the \(u\)-field as follows

(4.5.18)#\[v_{ab}(\pbf, \sigma) = (-1)^{\jfrak + \sigma} u_{ab}(\pbf, -\sigma)\]

The construction of interaction densities for general \((A, B)\) fields relies on Clebsch-Gordan coefficients, and is discussed in some detail in the following dropdown block.

We will now turn to the arguably most interesting causality condition Eq.2.5.14. Indeed, it is this condition that clarifies the correlation between the spin and whether a particle/field is bosonic or fermionic. As before, we need to evaluate the (anti-)commutator between the fields using Eq.4.5.6 as follows

(4.5.21)#\[\left[ \psi_{ab}(x), \psi^{\prime\, \dagger}_{a'b'}(y) \right]_{\pm} = (2\pi)^{-3} \int~\frac{d^3 p}{2p_0}~\pi_{ab,a'b'}(\pbf) \left( \kappa \kappa'^{~\ast} e^{\ifrak p \cdot (x-y}) \pm \lambda \lambda'^{~\ast} e^{-\ifrak p \cdot (x-y)} \right)\]

where \(\pi(\pbf)\) is the (rescaled) spin sum defined by

(4.5.22)#\[(2p_0)^{-1} \pi_{ab,a'b'}(\pbf) \coloneqq \sum_{\sigma} u_{ab}(\pbf, \sigma) u_{a'b'}^{\prime~\ast}(\pbf, \sigma) = \sum_{\sigma} v_{ab}(\pbf, \sigma) v_{a'b'}^{\prime~\ast}(\pbf, \sigma)\]

Here the second equality can be mostly easily seen using Eq.4.5.18. Note also that we are considering the general scenario where \(\psi(x)\) is an \((A, B)\) field, while \(\psi'(x)\) is a possibly different \((A', B')\) field.

Using Eq.4.5.17, we can spell out more details of the spin sum as follows

(4.5.23)#\[\begin{split}\pi_{ab, a'b'}(\pbf) &= \sum_{\bar{a}~\bar{b}} \sum_{\bar{a}'~\bar{b}'} \sum_{\sigma} C^{AB}(\jfrak \sigma; \bar{a} \bar{b}) C^{A'B'}(\jfrak \sigma; \bar{a}' \bar{b}') \\ &\quad \times \left(\exp\left(-\theta~\hat{\pbf} \cdot \Jbf^{(A)}\right)\right)_{\bar{a}a} \left(\exp\left(\theta~\hat{\pbf} \cdot \Jbf^{(B)}\right)\right)_{\bar{b}b} \\ &\quad \times \left(\exp\left(-\theta~\hat{\pbf} \cdot \Jbf^{(A')}\right)\right)_{\bar{a}'a'} \left(\exp\left(\theta~\hat{\pbf} \cdot \Jbf^{(B')}\right)\right)_{\bar{b}'b'}\end{split}\]

This looks horribly complicated, but it has been evaluated by the author in [Wei69]. Without going into the actual calculations, we note the following two facts, which suffice our purposes. The first is that \(\pi_{ab,a'b'}(\pbf)\) is a polynomial \(P\) in \(p\) on the mass shell as follows

(4.5.24)#\[\pi_{ab,a'b'}(\pbf) = P_{ab,a'b'}\left( \sqrt{\pbf^2 + m^2}, \pbf \right)\]

The second is that this polynomial is even or odd depending on the parity of \(2A + 2B'\) as follows

(4.5.25)#\[P_{ab,a'b'}(-p) = (-1)^{2A+2B'} P_{ab,a'b'}(p)\]

Assuming Eq.4.5.24, we note that any \(P_{ab, a'b'}\) can be written in such a way that it’s (at most) linear in the first argument \(\sqrt{\pbf^2 + m^2}\). Changing the content of \(P_{ab, a'b'}\) in Eq.4.5.24, we may then rewrite it as follows

(4.5.27)#\[\pi_{ab, a'b'}(\pbf) = P_{ab, a'b'}(\pbf) + 2 \sqrt{\pbf^2 + m^2} Q_{ab, a'b'}(\pbf)\]

where \(P, Q\) are polynomials in \(\pbf\) that satisfy the following parity conditions

\[\begin{split}P_{ab, a'b'}(-\pbf) &= (-1)^{2A+2B'} P_{ab, a'b'}(\pbf) \\ Q_{ab, a'b'}(-\pbf) &= -(-1)^{2A+2B'} Q_{ab, a'b'}(\pbf)\end{split}\]

Returning to the causality condition Eq.4.5.21, let’s consider space separated \(x\) and \(y\). Up to a Lorentz transformation, we may assume that \(x-y = (0, \xbf-\ybf)\). Under this assumption, we can calculate as follows

\[\begin{split}\left[ \psi_{ab}(x), \psi^{\prime~\dagger}_{a'b'}(y) \right]_{\pm} &= (2\pi)^{-3} \int \frac{d^3 p}{2p_0}~P_{ab, a'b'}(\pbf) \left( \kappa \kappa'^{~\ast} e^{\ifrak p \cdot (x-y)} \pm \lambda \lambda'^{~\ast} e^{-\ifrak p \cdot (x-y)} \right) \\ &\quad + (2\pi)^{-3} \int d^3p~Q_{ab, a'b'}(\pbf) \left( \kappa \kappa'^{~\ast} e^{\ifrak p \cdot (x-y)} \pm \lambda \lambda'^{~\ast} e^{-\ifrak p \cdot (x-y)} \right) \\ &= \kappa\kappa^{\prime~\ast} P_{ab, a'b'}(-\ifrak \nabla) \Delta_+(\xbf - \ybf) \pm \lambda\lambda^{\prime~\ast} P_{ab, a'b'}(\ifrak \nabla) \Delta_+(\ybf - \xbf) \\ &\quad + \kappa\kappa^{\prime~\ast} Q_{ab, a'b'}(-\ifrak \nabla) \delta^3(\xbf - \ybf) \pm \lambda\lambda^{\prime~\ast} Q_{ab, a'b'}(\ifrak \nabla) \delta^3(\ybf - \xbf) \\ &= \left( \kappa\kappa^{\prime~\ast} \pm (-1)^{2A+2B'} \lambda\lambda^{\prime~\ast} \right) P_{ab, a'b'}(-\ifrak \nabla) \Delta_+(\xbf - \ybf) \\ &\quad + \left( \kappa\kappa^{\prime~\ast} \mp (-1)^{2A+2B'} \lambda\lambda^{\prime~\ast} \right) Q_{ab, a'b'}(-\ifrak \nabla) \delta^3(\xbf - \ybf)\end{split}\]

where the derivative \(\nabla\) is always taken with respect to \(x\). Here we’ve also used the fact that \(\Delta_+(x)\) (for space-like \(x\)) and the Dirac delta \(\delta^3(x)\) are even functions. We see that for the (anti-)commutator to vanish for \(\xbf \neq \ybf\), i.e., when \(\delta^3(\xbf - \ybf) = 0\), we must have

(4.5.28)#\[\kappa\kappa^{\prime~\ast} = \mp (-1)^{2A+2B'} \lambda\lambda^{\prime~\ast}\]

Now consider an important special case where \(\psi = \psi'\). It implies in particular that \((A, B) = (A', B')\) and \((\kappa, \lambda) = (\kappa', \lambda')\). In this case we can rewrite Eq.4.5.28 as follows

(4.5.29)#\[|\kappa|^2 = \mp (-1)^{2A+2B} |\lambda|^2 = \mp (-1)^{2\jfrak} |\lambda|^2\]

since \(\jfrak\) differs from \(A+B\) by an integer according to Eq.4.5.5. Hence in addition to the condition \(|\kappa| = |\lambda|\), the field (or rather the particle it describes) must be bosonic, i.e., the bottom sign is taken, if \(\jfrak\) is an integer, and fermionic, i.e., the top sign is taken, if \(\jfrak\) is a half-integer. This is consistent with the corresponding conclusions for scalar , vector, and Dirac fields found in previous sections, and is indeed a great clarification of the relationship between spin and statistics, e.g., Pauli’s exclusion principle.

Back to the general case. We know from Eq.4.5.29 that \(|\kappa'| = |\lambda'|\) and \((-1)^{2A+2B} = (-1)^{2\jfrak} = \mp\), which is the same sign as in Eq.4.5.28. Hence we can rewrite Eq.4.5.28 by dividing both sides by \(|\kappa'|^2 = |\lambda'|^2\) as follows

\[\frac{\kappa}{\kappa'} = (-1)^{2B+2B'} \frac{\lambda}{\lambda'} \implies (-1)^{2B} \frac{\kappa}{\lambda} = (-1)^{2B'} \frac{\kappa'}{\lambda'}\]

Hence we conclude the following relationship between the coefficients \(\kappa\) and \(\lambda\)

\[\lambda = (-1)^{2B} c \kappa\]

where \(c\) is a constant that depends only on the field, or rather, the particle it describes, and not on the specific representation that gives rise to the field. Moreover we note that \(c\) is a phase since \(|\kappa| = |\lambda|\). Hence by adjusting the phase of the creation operator (and correspondingly the annihilation operator), we can arrange so that \(c = 1\).

This marks the end of the discussion about the causality condition on general \((A, B)\). As a result, we’ve obtained the following grand formula for a general (causal) field.

(4.5.30)#\[\psi_{ab}(x) = (2\pi)^{-3/2} \sum_{\sigma} \int d^3 p \left( e^{\ifrak p \cdot x} u_{ab}(\pbf, \sigma) a(\pbf, \sigma) + (-1)^{2B} e^{-\ifrak p \cdot x} v_{ab}(\pbf, \sigma) a^{c \dagger}(\pbf, \sigma) \right)\]

where the spinors \(u_{ab}\) and \(v_{ab}\) are given by Eq.4.5.17 and Eq.4.5.18, respectively.

Same fields given by different representations are physically indifferent

As we’ve seen in Scalar Fields and Spin-0 vector fields, a spin-\(0\) field can arise either as a scalar field, i.e., a \((0, 0)\) field, or from a vector field, i.e., a \(\left( \tfrac{1}{2}, \tfrac{1}{2} \right)\) field. Moreover, the later turns out to be the (first) derivatives of the former, and hence doesn’t produce anything really new. We’ll see here that this is a rather general phenomenon.

Indeed, according to Eq.4.5.5, a spin-\(0\) field can only arise from an \((A, A)\) field. Given a scalar field \(\psi(x)\), one can construct a spin-\(0\) \((A, A)\) field as follows

(4.5.31)#\[\left\{ \p_{\mu_1} \p_{\mu_2} \cdots \p_{\mu_{2A}} \right\} \psi(x)\]

where \(\left\{ \p_{\mu_1} \p_{\mu_2} \cdots \p_{\mu_{2A}} \right\}\) is a traceless (symmetric) product of the partial derivatives. Here the trace of a symmetric tensor \(S_{\mu_1 \mu_2 \cdots \mu_n}\) is defined to be

(4.5.32)#\[\op{tr}(S_{\mu_1 \mu_2 \cdots \mu_n}) \coloneqq \sum_{\mu_1, \mu_2 = 0}^3 \delta_{\mu_1 \mu_2} S_{\mu_1 \mu_2 \cdots \mu_n}\]

We’ll not verify that Eq.4.5.31 indeed transforms as a \((A, A)\) field, but we’ll verify that it at least has the expected dimension \((2A+1)^2\). To this end, we first note that a rank \(2A\) symmetric tensor (in \(4\) dimensions) has dimension

\[\frac{(2A+1)(2A+2)(2A+3)}{3!}\]

which combinatorially is just the number of ways of having four nonnegative integers sum up to \(2A\). Next we must subtract from it the number of traces one can take on these tensors. According to Eq.4.5.32, each trace fixes two indexes, while the rest is still symmetric. Hence the number of traces is the same combinatorial number with \(2A\) replaced by \(2A-2\) as follows

\[\frac{(2A-1)(2A)(2A+1)}{3!}\]

Now the difference of the two number above, which is the dimension of traceless symmetric tensors, is exactly \((2A+1)^2\) as expected.

In general a spin-\(\jfrak\) field can arise from an \((A, B)\) field as long as the triangle inequality

(4.5.33)#\[|A-B| \leq \jfrak \leq A+B\]

is satisfied. We claim that the \((A, B)\) representation is the same as the tensor product of the \((\jfrak, 0)\) representation and the \((B, B)\) representation. Indeed, following Clebsch-Gordan coefficients, we see that the later is a direct sum of \((A, B)\) representations as long as

\[|B-\jfrak| \leq A \leq B+\jfrak\]

but this triangle inequality is exactly the same as Eq.4.5.33. Hence the claim is proved.

By the same argument as in the spin-\(0\) case, we conclude that any spin-\(\jfrak\) field can be written as

\[\left\{ \p_{\mu_1} \p_{\mu_2} \cdots \p_{\mu_{2B}} \right\} \psi_{\sigma}(x)\]

where \(\psi_{\sigma}(x)\) is the \((\jfrak, 0)\) field. Swapping the role of \(A\) and \(B\), one can also write it as rank \(2A\) traceless derivatives of the \((0, \jfrak)\) field.

4.5.3. The CPT symmetries#

The calculations of space, time, and charge conjugation transformations in the general case is essentially the same as for the Dirac field. In particular, instead of reverting the \(3\)-momentum in Dirac spinors as in Eq.4.4.21, we need to do it for general \((A, B)\) spinors Eq.4.5.17, which involves the Clebsch-Gordan coefficients.

Without going to the details, we list the relevant symmetry properties of Clebsch-Gordan coefficients as follows

(4.5.34)#\[\begin{split}C^{AB}(\jfrak, \sigma; a, b) &= (-1)^{A+B-\jfrak} C^{BA}(\jfrak, \sigma; b, a) \\ C^{AB}(\jfrak, \sigma; a, b) &= (-1)^{A+B-\jfrak} C^{AB}(\jfrak, -\sigma; -a, -b)\end{split}\]

The first relation is proved in [Wei00] page 124, and the second relation can be deduced from the time reversal transformation law Eq.1.3.37.

Consider first the spatial inversion. Combining Eq.4.5.34 with Eq.4.5.17, one obtains the following relations on the spinors

(4.5.35)#\[\begin{split}u^{AB}_{ab}(-\pbf, \sigma) &= (-1)^{A+B-\jfrak} u^{BA}_{ba}(\pbf, \sigma) \\ v^{AB}_{ab}(-\pbf, \sigma) &= (-1)^{A+B-\jfrak} v^{BA}_{ba}(\pbf, \sigma)\end{split}\]

We can then calculate the spatial conjugation as follows

\[\begin{split}U(\Pcal) \psi^{AB}_{ab}(x) U^{-1}(\Pcal) &= (2\pi)^{-3/2} \sum_{\sigma} \int d^3 p \Big( \eta^{\ast} e^{\ifrak p \cdot x} u^{AB}_{ab}(\pbf, \sigma) a(-\pbf, \sigma) \\ &\qquad + (-1)^{2B} \eta^c e^{-\ifrak p \cdot x} v^{AB}_{ab}(\pbf, \sigma) a^{c \dagger}(-\pbf, \sigma) \Big) \\ &= (2\pi)^{-3/2} \sum_{\sigma} \int d^3 p~(-1)^{A+B-\jfrak} \Big( \blue{\eta^{\ast} e^{\ifrak p \cdot \Pcal x} u_{ba}(\pbf, \sigma) a(\pbf, \sigma)} \\ &\qquad \blue{ + (-1)^{2B} \eta^c e^{-\ifrak p \cdot \Pcal x} v_{ba}(\pbf, \sigma) a^{c \dagger}(\pbf, \sigma)} \Big)\end{split}\]

In order for the blue terms to be proportional to the corresponding terms in, in this case, a \((B, A)\) field, we must have

(4.5.36)#\[(-1)^{2B} \eta^c = (-1)^{2A} \eta^{\ast} \iff \eta^c = (-1)^{2\jfrak} \eta^{\ast}\]

Under this assumption, we can complete the transformation law for spatial inversion as follows

(4.5.37)#\[U(\Pcal) \psi^{AB}_{ab}(x) U^{-1}(\Pcal) = \eta^{\ast} (-1)^{A+B-\jfrak} \psi^{BA}_{ba}(\Pcal x)\]

which recovers the cases of scalar field Eq.4.2.11 with \(A=B=\jfrak=0\), vector field Eq.4.3.23 with \(A=B=1/2\) and \(\jfrak=1\), and Dirac field Eq.4.4.36 with \((A, B)=(0, 1/2)\) or \((1/2, 0)\) and \(\jfrak=1/2\), where \(\beta\), as defined by Eq.4.4.10, serves the function of swapping \(A\) and \(B\).

Next consider the time inversion. As for the spatial inversion, we’ll need, using Eq.4.5.17 and Eq.4.5.34, the following identities

(4.5.38)#\[\begin{split}& u^{AB \ast}_{ab}(-\pbf, -\sigma) \\ &\quad= \frac{1}{\sqrt{2p_0}} \sum_{a'b'} \left(\exp\left(\theta~\hat{\pbf} \cdot \Jbf^{(A) \ast}\right)\right)_{-a', a} \left(\exp\left(-\theta~\hat{\pbf} \cdot \Jbf^{(B) \ast}\right)\right)_{-b', b} \\ &\quad\qquad \times C^{AB}(\jfrak, -\sigma; -a', -b') \\ &\quad= \frac{(-1)^{A+B-\jfrak}}{\sqrt{2p_0}} \sum_{a'b'} \left(\exp\left(\theta~\hat{\pbf} \cdot \Jbf^{(A) \ast}\right)\right)_{-a', a} \left(\exp\left(-\theta~\hat{\pbf} \cdot \Jbf^{(B) \ast}\right)\right)_{-b', b} \\ &\quad\qquad \times C^{AB}(\jfrak, \sigma; a', b') \\ &\quad= \frac{(-1)^{A+B-\jfrak}}{\sqrt{2p_0}} \sum_{a'b'} (-1)^{a-a'+b-b'} \left( \exp(-\theta~\hat{\pbf} \cdot \Jbf^{(A)}) \right)_{a', -a} \left( \exp(\theta~\hat{\pbf} \cdot \Jbf^{(B)}) \right)_{b', -b} \\ &\quad\qquad \times C^{AB}(\jfrak, \sigma; a', b') \\ &\quad= (-1)^{A+B+a+b-\sigma - \jfrak} u^{AB}_{-a, -b}(\pbf, \sigma)\end{split}\]

where it’s convenient for the third equality to reformulate Eq.4.5.12 as follows

\[\Jbf^{(\jfrak) \ast} = -C \Jbf^{(\jfrak)} C^{-1}, \quad \text{where}~~C_{\sigma' \sigma} = (-1)^{\jfrak - \sigma} \delta_{\sigma', -\sigma}\]

Since \(v^{AB}_{ab}(\pbf, \sigma)\) is related to \(u^{AB}_{ab}(\pbf, \sigma)\) by Eq.4.5.18, we can derive the \(v\)-counterpart of Eq.4.5.38, using Eq.4.5.18 and Eq.4.5.38, as follows

(4.5.39)#\[\begin{split}v^{AB \ast}_{ab}(-\pbf, -\sigma) &= (-1)^{\jfrak - \sigma} u^{AB \ast}_{ab}(-\pbf, \sigma) \\ &= (-1)^{A+B+a+b} u^{AB}_{-a, -b}(\pbf, -\sigma) \\ &= (-1)^{A+B+a+b-\sigma - \jfrak} v^{AB}_{-a, -b}(\pbf, \sigma)\end{split}\]

Remembering that \(U(\Tcal)\) is anti-unitary, we can calculate the time inversion transformation, using Eq.4.5.30, Eq.3.2.10, Eq.4.5.38, and Eq.4.5.39, as follows

(4.5.40)#\[\begin{split}U(\Tcal) \psi^{AB}_{ab}(x) U^{-1}(\Tcal) &= (2\pi)^{-3/2} \sum_{\sigma} \int d^3 p~(-1)^{\jfrak - \sigma} \Big( \zeta^{\ast} e^{-\ifrak p \cdot x} u^{AB \ast}_{ab}(\pbf, \sigma) a(-\pbf, -\sigma) \\ &\qquad + (-1)^{2B} \zeta^c e^{\ifrak p \cdot x} v^{AB \ast}_{ab}(\pbf, \sigma) a^{c \dagger}(-\pbf, -\sigma) \Big) \\ &= (2\pi)^{-3/2} \sum_{\sigma} \int d^3 p~(-1)^{\jfrak - \sigma} \Big( \zeta^{\ast} e^{-\ifrak p \cdot \Pcal x} u^{AB \ast}_{ab}(-\pbf, -\sigma) a(\pbf, \sigma) \\ &\qquad + (-1)^{2B} \zeta^c e^{\ifrak p \cdot \Pcal x} v^{AB \ast}_{ab}(-\pbf, -\sigma) a^{c \dagger}(\pbf, \sigma) \Big) \\ &= (2\pi)^{-3/2} \sum_{\sigma} \int d^3 p~(-1)^{A+B+a+b-2\sigma} \Big( \zeta^{\ast} e^{-\ifrak p \cdot \Pcal x} u^{AB}_{-a, -b}(\pbf, \sigma) \\ &\qquad + (-1)^{2B} \zeta^c e^{\ifrak p \cdot \Pcal x} v^{AB}_{-a, -b}(\pbf, \sigma) a^{c \dagger}(\pbf, \sigma) \Big) \\ &= (-1)^{A+B+a+b-2\jfrak} \zeta^{\ast} \psi^{AB}_{-a, -b}(-\Pcal x)\end{split}\]

where the last equality assumes the following symmetry on the time-reversal parity

(4.5.41)#\[\zeta^{\ast} = \zeta^c\]

At this point, we’re pretty proficient at (and tired of) this kind of calculation. Hence we’ll not spell out the (rather similar) details for the charge conjugation symmetry, but rather list the result as follows

(4.5.42)#\[U(\Ccal) \psi^{AB}_{ab}(x) U^{-1}(\Ccal) = (-1)^{-2A-a-b-\jfrak} \xi^{\ast} \psi^{BA \dagger}_{-b, -a}(x)\]

under the following assumption on the charge-reversal parity

(4.5.43)#\[\xi^{\ast} = \xi^c\]

4.6. The CPT Theorem#

With all the hard work we’ve done in General Fields, we can now reward ourselves a bit with the celebrated CPT theorem which is stated as follows

For an appropriate choice of the inversion phases \(\eta\) (space), \(\zeta\) (time), and \(\xi\) (charge), the product \(U(CPT)\) is conserved.

We’ll skip over the special case of scalar, vector, and Dirac fields, and jump directly into the general, and in fact simpler, case of \((A, B)\) fields. Using Eq.4.5.40, Eq.4.5.37, and Eq.4.5.42, we have the following

(4.6.1)#\[\begin{split}U(CPT) \psi^{AB}_{ab}(x) U^{-1}(CPT) &= (-1)^{A+B+a+b-2\jfrak} \zeta^{\ast} U(CP) \psi^{AB}_{-a, -b}(-\Pcal x) U^{-1}(CP) \\ &= (-1)^{a+b-\jfrak} \zeta^{\ast} \eta^{\ast} U(C) \psi^{BA}_{-b, -a}(-x) U^{-1}(C) \\ &= (-1)^{-2B} \zeta^{\ast} \eta^{\ast} \xi^{\ast} \psi^{AB \dagger}_{ab}(-x)\end{split}\]

Hence if we assume the following condition on the inversion parities

Assumption on the inversion parities

(4.6.2)#\[\zeta~\eta~\xi = 1\]

then we can rewrite Eq.4.6.1 as follows

\[U(CPT) \psi^{AB}_{ab}(x) U^{-1}(CPT) = (-1)^{2B} \psi^{AB \dagger}_{ab}(-x)\]

A few words are needed, however, to justify the seemingly strange assumption on the product of inversion parities. Indeed, it is physically meaningless to specify any inversion parity for a single species of particles because it’s just a phase. The only conditions that we’ve seen on the parities come from pairs of particles and their antiparticles, notably Eq.4.5.36, Eq.4.5.41, and Eq.4.5.43. We saw that the time and charge inversion parities are the same between the particle and its antiparticle, respectively. However, a sign \((-1)^{2\jfrak}\) is involved in the space inversion parity. So if we impose Eq.4.6.2 on one particle species, then it will fail on its antiparticle species if the particle in question is a fermion! We’re eventually saved by the fact that the interaction density must involve an even number of fermions (cf. discussions in Causality and antiparticles). In any case Eq.4.6.2 is a rather sloppy assumption, which cannot hold in general, but it also doesn’t make a difference to the CPT theorem.

Now suppose the interaction density \(\Hscr(x)\) is defined by Eq.4.5.19 as a linear combination of monomials like

\[\psi^{A_1 B_1}_{a_1 b_1}(x) \psi^{A_2 B_2}_{a_2 b_2}(x) \cdots \psi^{A_n B_n}_{a_n b_n}(x)\]

Hence in light of Eq.4.5.20, we know that both \(A_1 + A_2 + \cdots + A_n\) and \(B_1 + B_2 + \cdots + B_n\) must be integers, for otherwise they cannot be coupled to a spinless state. It follows then the following CPT transformation law on the interaction density

\[U(CPT) \Hscr(x) U^{-1}(CPT) = -\Hscr(-x)\]

Recall from Eq.2.5.6 and Eq.2.5.11 that the interaction term \(V = \int d^3 x~\Hscr(0, \xbf)\) satisfies the following

\[U(CPT) V U^{-1}(CPT) = -\int d^3 x~\Hscr(-x) = V\]

Since the CPT symmetry is clearly conserved for free particles, it is also conserved in interactions according to Eq.2.2.2.

4.7. Massless Fields#

So far the story about quantum fields has been a 100% success. We’ve namely found the general formula Eq.4.5.30 for any field that represents a massive particle. However, such success will come to an end when we consider instead massless particles as we’ll see in this section. This should not come as a surprise though since we’ve see in Eq.4.3.13 for example, that the spin sum blows up in the massless limit \(m \to 0\).

Let’s nonetheless kickstart the routine of constructing fields as follows, and see where the problem should arise.

(4.7.1)#\[\psi_{\ell}(x) = (2\pi)^{-3/2} \sum_{\sigma} \int d^3 p \left( \kappa e^{\ifrak p \cdot x} u_{\ell}(\pbf, \sigma) a(\pbf, \sigma) + \lambda e^{-\ifrak p \cdot x} v_{\ell}(\pbf, \sigma) a^{c \dagger}(\pbf, \sigma) \right)\]

This is reasonable because the translation symmetry is the same for massive and massless particles, and hence Eq.4.1.8 and Eq.4.1.9 apply.

Next, using the general transformation laws Eq.4.1.5 for creation and annihilation operators, as well as the \(D\) matrix Eq.1.3.30 for massless particles, we can infer the homogeneous Lorentz transformation laws as follows

(4.7.2)#\[\begin{split}U(\Lambda) a^{\dagger}(\pbf, \sigma) U^{-1}(\Lambda) &= \sqrt{\frac{(\Lambda p)_0}{p_0}} \exp(\ifrak \sigma \theta(\Lambda, p)) a^{\dagger}(\pbf_{\Lambda}, \sigma) \\ U(\Lambda) a(\pbf, \sigma) U^{-1}(\Lambda) &= \sqrt{\frac{(\Lambda p)_0}{p_0}} \exp(-\ifrak \sigma \theta(\Lambda, p)) a(\pbf_{\Lambda}, \sigma)\end{split}\]

Just as in the massive case, we’d like \(\psi_{\ell}(x)\) to satisfy the following transformation law

(4.7.3)#\[U(\Lambda) \psi_{\ell}(x) U^{-1}(\Lambda) = \sum_{\ell'} D_{\ell \ell'}(\Lambda^{-1}) \psi_{\ell'}(\Lambda x)\]

To see what conditions the spinors must satisfy, let’s first expand the left-hand-side as follows

\[\begin{split}U(\Lambda) \psi_{\ell}(x) U^{-1}(\Lambda) &= (2\pi)^{-3/2} \sum_{\sigma} \int d^3 p \Big( \kappa e^{\ifrak p \cdot x} \blue{u_{\ell}(\pbf, \sigma) \sqrt{\frac{(\Lambda p)_0}{p_0}} \exp(-\ifrak \sigma \theta(\Lambda, p))} a(\pbf_{\Lambda}, \sigma) \\ &\qquad + \lambda e^{-\ifrak p \cdot x} \blue{v_{\ell}(\pbf, \sigma) \sqrt{\frac{(\Lambda p)_0}{p_0}} \exp(\ifrak \sigma \theta(\Lambda, p))} a^{c \dagger}(\pbf_{\Lambda}, \sigma) \Big)\end{split}\]

Then the right-hand-side as follows

\[\begin{split}\sum_{\ell'} D_{\ell \ell'}(\Lambda^{-1}) \psi_{\ell'}(\Lambda x) &= (2\pi)^{-3/2} \sum_{\sigma} \int \frac{d^3 p}{p_0}~p_0 \sum_{\ell'} \Big( \kappa e^{\ifrak p \cdot \Lambda x} D_{\ell \ell'}(\Lambda^{-1}) u_{\ell'}(\pbf, \sigma) a(\pbf, \sigma) \\ &\qquad + \lambda e^{-\ifrak p \cdot \Lambda x} D_{\ell \ell'}(\Lambda^{-1}) v_{\ell'}(\pbf, \sigma) a^{c \dagger}(\pbf, \sigma) \Big) \\ &= (2\pi)^{-3/2} \sum_{\sigma} \int d^3 p~\blue{\frac{(\Lambda p)_0}{p_0} \sum_{\ell'}} \Big( \kappa e^{\ifrak p \cdot x} \blue{D_{\ell \ell'}(\Lambda^{-1}) u_{\ell'}(\pbf_{\Lambda}, \sigma)} a(\pbf_{\Lambda}, \sigma) \\ &\qquad + \lambda e^{-\ifrak p \cdot x} \blue{D_{\ell \ell'}(\Lambda^{-1}) v_{\ell'}(\pbf_{\Lambda}, \sigma)} a^{c \dagger}(\pbf_{\Lambda}, \sigma) \Big)\end{split}\]

Equating the coefficients of \(a(\pbf_{\Lambda, \sigma})\) and \(a^{c \dagger}(\pbf_{\Lambda}, \sigma)\) (i.e., the blue terms), and inverting \(D_{\ell \ell'}(\Lambda^{-1})\) as in Eq.4.1.6, we get the following conditions on the spinors

(4.7.4)#\[\begin{split}\exp(\ifrak \sigma \theta(\Lambda, p)) u_{\ell'}(\pbf_{\Lambda}, \sigma) &= \sqrt{\frac{p_0}{(\Lambda p)_0}} \sum_{\ell} D_{\ell' \ell}(\Lambda) u_{\ell}(\pbf, \sigma) \\ \exp(-\ifrak \sigma \theta(\Lambda, p)) v_{\ell'}(\pbf_{\Lambda}, \sigma) &= \sqrt{\frac{p_0}{(\Lambda p)_0}} \sum_{\ell} D_{\ell' \ell}(\Lambda) v_{\ell}(\pbf, \sigma)\end{split}\]

The next step is to take the massless analogy to the boost operator in the massive case. Namely, if we let \(\Lambda = L(p)\) be the chosen Lorentz transformation that takes the standard \(k = (1, 0, 0, 1)\) to \(p\), then \(\theta(\Lambda, p) = 0\). Taking \(p = k\) in Eq.4.7.4, we obtain the following

(4.7.5)#\[\begin{split}u_{\ell'}(\pbf, \sigma) &= \frac{1}{\sqrt{p_0}} \sum_{\ell} D_{\ell' \ell}(L(p)) u_{\ell}(\kbf, \sigma) \\ v_{\ell'}(\pbf, \sigma) &= \frac{1}{\sqrt{p_0}} \sum_{\ell} D_{\ell' \ell}(L(p)) v_{\ell}(\kbf, \sigma)\end{split}\]

Next, in analogy to the rotation transformation, let’s consider a little group element \(W\) that fixes \(k\). In this case Eq.4.7.4 takes the following form

(4.7.6)#\[\begin{split}\exp(\ifrak \sigma \theta(W, k)) u_{\ell'}(\kbf, \sigma) &= \sum_{\ell} D_{\ell' \ell}(W) u_{\ell}(\kbf, \sigma) \\ \exp(-\ifrak \sigma \theta(W, k)) v_{\ell'}(\kbf, \sigma) &= \sum_{\ell} D_{\ell' \ell}(W) v_{\ell}(\kbf, \sigma)\end{split}\]

Recall from Massless particle states that any \(W\) can be written in the following form

(4.7.7)#\[W(a, b, \theta) = S(a, b) R(\theta)\]

where \(S(a, b)\) is defined by Eq.1.3.27, and \(R(\theta)\) is defined by Eq.1.3.28. Considering separately the two cases \(W(0, 0, \theta) = R(\theta)\) and \(W(a, b, 0) = S(a, b)\), we get the following two consequences of Eq.4.7.4

(4.7.8)#\[\begin{split}e^{\ifrak \sigma \theta} u_{\ell'}(\kbf, \sigma) &= \sum_{\ell} D_{\ell' \ell}(R(\theta)) u_{\ell}(\kbf, \sigma) \\ u_{\ell'}(\kbf, \sigma) &= \sum_{\ell} D_{\ell' \ell}(S(a, b)) u_{\ell}(\kbf, \sigma)\end{split}\]

Similar constraints hold for \(v\) as well, but we’ll not bother to write them down.

It turns out, however, that these conditions can never be satisfied! To illustrate the difficulties, we’ll first consider the case of vector fields, both as a warm-up and for later references when we’ll analyze the electromagnetic theory. Then we’ll show that the difficulties persist to the general case of arbitrary \((A, B)\) fields.

4.7.1. The failure for vector fields#

For vector field \({D_{\mu}}^{\nu}(\Lambda) = {\Lambda_{\mu}}^{\nu}\) as in the massive case. As a convention, let’s write

\[u_{\mu}(\pbf, \sigma) \eqqcolon \frac{1}{\sqrt{2p_0}} e_{\mu}(\pbf, \sigma)\]

Since \({D_{\mu}}^{\nu}(\Lambda)\) is real, it follows from Eq.4.7.4 that \(v\) satisfies equations that are complex conjugate to those that \(u\) satisfies. Hence \(v_{\mu}(\pbf, \sigma) = u^{\ast}_{\mu}(\pbf, \sigma)\).

Now we can translate Eq.4.7.5 from a boosting formula for \(u\) to one for \(e\) as follows

(4.7.9)#\[e_{\mu}(\pbf, \sigma) = {L(p)_{\mu}}^{\nu} e_{\nu}(\kbf, \sigma)\]

Moreover Eq.4.7.8 can be translated to conditions on \(e\) as well as follows

(4.7.10)#\[\begin{split}e^{\ifrak \sigma \theta} e_{\mu}(\kbf, \sigma) &= {R(\theta)_{\mu}}^{\nu} e_{\nu}(\kbf, \sigma) \\ e_{\mu}(\kbf, \sigma) &= {S(a, b)_{\mu}}^{\nu} e_{\nu}(\kbf, \sigma)\end{split}\]

Using the explicit formula Eq.1.3.28 for \(R(\theta)\), we can derive from the \(R(\theta)\) part of Eq.4.7.10 that the helicity \(\sigma = \pm 1\), and moreover,

(4.7.11)#\[\begin{split}e_{\mu}(\kbf, \pm 1) = \frac{1}{\sqrt{2}} \begin{bmatrix*}[r] 0 \\ 1 \\ \pm \ifrak \\ 0 \end{bmatrix*}\end{split}\]

up to normalization. However, by the explicit formula Eq.1.3.27 for \(S(a, b)\), we see that the \(S(a, b)\) part of Eq.4.7.10 then requires

\[a \pm \ifrak b = 0\]

which is impossible for any real \(a, b\) that are not both zero.

For reasons that will be justified later, it’s nonetheless legitimate to adopt the vectors \(e_{\mu}\) as defined by Eq.4.7.11 as the spinors, as well as the condition \(\kappa = \lambda = 1\) as in the case of massive vector fields. With these assumptions, we can rename \(\psi\) by \(a\) (as it’ll correspond to the electromagnetic potential which is conventionally named by \(a\)), and rewrite Eq.4.7.1 as follows

(4.7.12)#\[a_{\mu}(x) = (2\pi)^{-3/2} \int \frac{d^3 p}{\sqrt{2p_0}} \sum_{\sigma = \pm 1} \left( e^{\ifrak p \cdot x} e_{\mu}(\pbf, \sigma) a(\pbf, \sigma) + e^{-\ifrak p \cdot x} e^{\ast}_{\mu}(\pbf, \sigma) a^{c \dagger}(\pbf, \sigma) \right)\]

Warning

The notations are getting slightly out of hands here. Namely, we’ve used \(a\) for at least three different things in one place: the vector field \(a_{\mu}\), the parameter in \(S(a, b)\), and the creation operator \(a(\pbf, \sigma)\). There will actually be a fourth place where \(a\) is used as the spin \(z\)-component in an \((A, B)\) field. We can only hope that the context will make it clear what \(a\) (or \(b\)) really represents.

As for massive vector fields, we’ll search for field equations that \(a_{\mu}(x)\) must satisfy. First of all, it satisfies obviously the (massless) Klein-Gordon equation

(4.7.13)#\[\square a_{\mu}(x) = 0\]

which is nothing but an incarnation of the mass-shell condition \(p_0^2 = \pbf^2\). Then let’s consider the massless analog to the gauge-fixing condition Eq.4.3.19. To this end, we claim that \(e_0(\kbf, \pm 1) = 0\) and \(\kbf \cdot \ebf(\kbf, \pm 1) = 0\) imply the following

(4.7.14)#\[\begin{split}e_0(\pbf, \pm 1) &= 0 \\ \pbf \cdot \ebf(\pbf, \pm 1) &= 0\end{split}\]

in analogy to Eq.4.3.18 by the following argument. First, note that \(e_{\mu}(\pbf, \pm 1)\) can be obtained from \(e_{\mu}(\kbf, \pm 1)\) by applying \(L(p)\) as in Eq.4.7.9. Second, \(L(p)\) can be decomposed into a boost along the \(z\)-axis as in Eq.1.3.32 followed by a \(3\)-rotation. Finally, we conclude Eq.4.7.14 by noting that \(e_{\mu}(\kbf, \pm 1)\) is unaffected by boosts along the \(z\)-axis, and the dot product is preserved by any \(3\)-rotation. In contrast to the massless case Eq.4.3.18, we have a stronger constraint here because the helicity \(0\) spinor is missing.

Using Eq.4.7.12, the corresponding constraints on \(a_{\mu}(x)\) is the following

(4.7.15)#\[\begin{split}a_0(x) &= 0 \\ \nabla \cdot \abf(x) &= 0\end{split}\]

which is clearly not Lorentz invariant.

But let’s calculate \(U(\Lambda) a_{\mu}(x) U^{-1}(\Lambda)\) anyway and see to some extent Eq.4.7.3 fails. To this end, we’ll need to calculate the action of the \(D\) matrix on the spinors, and we’ll first do this for \(e_{\mu}(k, \pm 1)\), using Eq.4.7.7 and Eq.1.3.27, as follows

(4.7.16)#\[\begin{split}{D_{\mu}}^{\nu}(W(a, b, \theta)) e_{\nu}(\kbf, \pm 1) &= {S(a, b)_{\mu}}^{\lambda} {R(\theta)_{\lambda}}^{\nu} e_{\nu}(\kbf, \pm 1) \\ &= e^{\pm \ifrak \theta} {S(a, b)_{\mu}}^{\lambda} e_{\lambda}(\kbf, \pm 1) \\ &= e^{\pm \ifrak \theta} \left( e_{\mu}(\kbf, \pm 1) + \frac{a \pm \ifrak b}{\sqrt{2}} k_{\mu} \right)\end{split}\]

Next we recall the little group element \(W(\Lambda, p) = L^{-1}(\Lambda p) \Lambda L(p)\) by definition. Plugging into Eq.4.7.16, we obtain the following

(4.7.17)#\[{\Lambda_{\mu}}^{\nu} e_{\nu}(\pbf, \pm 1) = e^{\pm \ifrak \theta(\Lambda, p)} \left(e_{\mu}(\pbf_{\Lambda}, \pm 1) + (\Lambda p)_{\mu} \Omega_{\pm}(\Lambda, p)\right)\]

where

\[\Omega_{\pm}(\Lambda, p) \coloneqq \frac{a(\Lambda, p) \pm \ifrak b(\Lambda, p)}{\sqrt{2}}\]

is the extra term that makes it different from Eq.4.7.4 which would have been satisfied if Eq.4.7.3 holds.

To most conveniently utilize Eq.4.7.17, let’s calculate \({D_{\mu}}^{\nu} \left( U(\Lambda) a_{\nu}(x) U^{-1}(\Lambda) \right)\), using Eq.4.7.2, Eq.4.7.12, and Eq.4.7.17, as follows

\[\begin{split}&{D_{\mu}}^{\nu}(\Lambda) \left(U(\Lambda) a_{\nu}(x) U^{-1}(\Lambda)\right) \\ &\quad = (2\pi)^{-3/2} \int \frac{d^3 p}{\sqrt{2p_0}} \sum_{\sigma = \pm 1} \Big( e^{\ifrak p \cdot x} {\Lambda_{\mu}}^{\nu} e_{\nu}(\pbf, \sigma) \sqrt{\frac{(\Lambda p)_0}{p_0}} e^{-\ifrak \sigma \theta(\Lambda, p)} a(\pbf_{\Lambda}, \sigma) \\ &\qquad + e^{-\ifrak p \cdot x} {\Lambda_{\mu}}^{\nu} e^{\ast}_{\nu}(\pbf, \sigma) \sqrt{\frac{(\Lambda p)_0}{p_0}} e^{\ifrak \sigma \theta} a^{c \dagger}(\pbf_{\Lambda}, \sigma) \Big) \\ &\quad = (2\pi)^{-3/2} \int \frac{d^3 p}{p_0} \sqrt{\frac{(\Lambda p)_0}{2}} \sum_{\sigma = \pm 1} \Big( e^{\ifrak p \cdot x} e_{\mu}(\pbf_{\Lambda}, \sigma) a(\pbf_{\Lambda}, \sigma) + e^{-\ifrak p \cdot x} e^{\ast}_{\mu}(\pbf_{\Lambda}, \sigma) a^{c \dagger}(\pbf_{\Lambda}, \sigma) \\ &\qquad + (\Lambda p)_{\mu} \left( e^{\ifrak p \cdot x} \Omega_{\sigma}(\Lambda, p) a(\pbf_{\Lambda}, \sigma) + e^{-\ifrak p \cdot x} \Omega^{\ast}_{\sigma}(\Lambda, p) a^{c \dagger}(\pbf_{\Lambda}, \sigma) \right) \Big) \\ &\quad = (2\pi)^{-3/2} \int \frac{d^3 p}{\sqrt{2p_0}} \sum_{\sigma = \pm 1} \Big( e^{\ifrak p \cdot \Lambda x} e_{\mu}(\pbf, \pm 1) a(\pbf, \sigma) + e^{-\ifrak p \cdot \Lambda x} e^{\ast}_{\mu}(\pbf, \pm 1) a^{c \dagger}(\pbf, \sigma) \\ &\qquad + p_{\mu} \left( e^{\ifrak p \cdot \Lambda x} \Omega_{\sigma}(\Lambda, \Lambda^{-1} p) + e^{-\ifrak p \cdot \Lambda x} \Omega^{\ast}_{\sigma}(\Lambda, \Lambda^{-1} p) a^{c \dagger}(\pbf, \sigma) \right) \Big) \\ &\quad= a_{\mu}(\Lambda x) + \frac{\p}{\p \left( (\Lambda x)_{\mu} \right)} \Omega(\Lambda, x)\end{split}\]

where \(\Omega(\Lambda, x)\) is a linear combination of creation and annihilation operators, whose precise form is not important here. Finally, moving \(D(\Lambda)\) to the right-hand-side, we obtain the following variation of Eq.4.7.3

(4.7.18)#\[U(\Lambda) a_{\mu}(x) U^{-1}(\Lambda) = {D(\Lambda^{-1})_{\mu}}^{\nu} a_{\nu}(\Lambda x) + \p_{\mu} \Omega(\Lambda, x)\]

which the massless vector field Eq.4.7.12 actually satisfies.

It follows, using integration by parts, that one can construct interaction densities by coupling \(j^{\mu}(x) a_{\mu}(x)\) as long as \(\p_{\mu} j^{\mu}(x) = 0\). This is completely parallel to Eq.4.3.20 and Eq.4.3.21 for massive vector fields, and hence partially justifies the choice of the spinors in Eq.4.7.11 although it doesn’t satisfy Eq.4.7.10.

Another byproduct of Eq.4.7.18 is the observation that although \(a_{\mu}\) fails to be a vector, one can easily construct a \(2\)-tensor as follows

(4.7.19)#\[f_{\mu \nu} = \p_{\mu} a_{\nu} - \p_{\nu} a_{\mu}\]

which satisfies

\[U(\Lambda) f_{\mu \nu} U^{-1}(\Lambda) = {D_{\mu}}^{\lambda}(\Lambda^{-1}) {D_{\nu}}^{\sigma}(\Lambda^{-1}) f_{\lambda \sigma}\]

by Eq.4.7.18. In fact, using Eq.4.7.13 and Eq.4.7.15, one can show that \(f_{\mu \nu}\) satisfies the vacuum Maxwell equations

\[\begin{split}\p_{\mu} f^{\mu \nu} &= 0 \\ \epsilon^{\rho \tau \mu \nu} \p_{\tau} f_{\mu \nu} &= 0\end{split}\]

where \(\epsilon^{\rho \tau \mu \nu}\) is the totally anti-symmetric sign. Indeed \(f_{\mu \nu}\) is the quantization of the electromagnetic field, while \(a_{\mu}\) is the quantization of the electromagnetic potential.

Build quantum fields with \(f_{\mu \nu}\)

Since \(f_{\mu \nu}\) is a tensor, one might try to build a (causal) quantum field with \(f_{\mu \nu}\), instead of \(a_{\mu}\). The causality condition Eq.2.5.14 demands that the (anti-)commutator \(\left[ f_{\mu \nu}(x), f^{\dagger}_{\rho \tau}(y) \right]_{\pm}\) must vanish for space-separated \(x\) and \(y\). As in the other cases, some preliminary calculations are in order. First we calculate the spin sum, using Eq.4.7.9, as follows

(4.7.20)#\[\sum_{\sigma = \pm 1} \ebf_i(\kbf, \sigma) \ebf_j^{\ast}(\kbf, \sigma) = \delta_{ij} - \kbf_i \kbf_j \implies \sum_{\sigma = \pm 1} \ebf_i(\pbf, \sigma) \ebf_j^{\ast}(\pbf, \sigma) = \delta_{ij} - \frac{\pbf_i \pbf_j}{\pbf^2}\]

where the first equality can be directly verified using Eq.4.7.11. In particular, we notice that the spin sum is real – a neat fact that will be used in the next calculation. Moreover, the spin sum vanishes if any index is \(0\) because of Eq.4.7.14.

Next we calculate, using Eq.4.7.12, Eq.3.2.5, and Eq.4.7.20, the (anti-)commutator between the derivatives of the components of an \(a\)-field, with coefficients \(\kappa\) and \(\lambda\) as in Eq.4.7.1 restored, as follows

\[\begin{split}\left[ \p_{\mu} a_{\nu}(x), \p_{\rho} a^{\dagger}_{\tau}(y) \right]_{\pm} &= (2\pi)^{-3} \int \frac{d^3 p~d^3 p'}{2\sqrt{p_0 p'_0}} \\ &\quad \times \sum_{\sigma, \sigma' = \pm 1} \Big( |\kappa|^2 e^{\ifrak p \cdot x - \ifrak p' \cdot y} p_{\mu} p'_{\rho} e_{\nu}(\pbf, \sigma) e^{\ast}_{\tau}(\pbf', \sigma') \left[ a(\pbf, \sigma), a^{\dagger}(\pbf', \sigma') \right]_{\pm} \\ &\qquad + |\lambda|^2 e^{-\ifrak p \cdot x + \ifrak p' \cdot y} p_{\mu} p'_{\rho} e^{\ast}_{\nu}(\pbf, \sigma) e_{\tau}(\pbf', \sigma') \left[ a^{c \dagger}(\pbf, \sigma), a^c(\pbf', \sigma') \right]_{\pm} \Big) \\ &= (2\pi)^{-3} \int \frac{d^3 p}{2p_0} \sum_{\sigma = \pm 1} p_{\mu} p_{\rho} \Big( |\kappa|^2 e^{\ifrak p \cdot (x-y)} e_{\nu}(\pbf, \sigma) e^{\ast}_{\tau}(\pbf, \sigma) \\ &\quad \pm |\lambda|^2 e^{-\ifrak p \cdot (x-y)} e^{\ast}_{\nu}(\pbf, \sigma) e_{\tau}(\pbf, \sigma) \Big) \\ &= (2\pi)^{-3} \int \frac{d^3 p}{2p_0} \blue{p_{\mu} p_{\rho} (1 - \delta_{0\nu})(1 - \delta_{0\tau}) \left( \delta_{\nu \tau} - \frac{p_{\nu} p_{\tau}}{\pbf^2} \right)} \\ &\quad \times \left( |\kappa|^2 e^{\ifrak p \cdot (x-y)} \pm |\lambda|^2 e^{-\ifrak p \cdot (x-y)} \right)\end{split}\]

Notice that the blue terms are the only terms that involve the indexes \(\mu, \nu, \rho\), and \(\tau\). For the convenience of notations, let’s call it \(P_{\mu \nu \rho \tau}\). Now we can calculate the (anti-)commutator as follows

(4.7.21)#\[\begin{split}\left[ f_{\mu \nu}(x), f^{\dagger}_{\rho \tau}(y) \right]_{\pm} &= \left[ \p_{\mu} a_{\nu}(x) - \p_{\nu} a_{\mu}(x), \p_{\rho} a^{\dagger}_{\tau}(y) - \p_{\tau} a^{\dagger}_{\rho}(y) \right] \\ &= \left[ \p_{\mu} a_{\nu}(x), \p_{\rho} a^{\dagger}_{\tau}(y) \right] - \left[ \p_{\mu} a_{\nu}(x), \p_{\tau} a^{\dagger}_{\rho}(y) \right] \\ &\quad - \left[ \p_{\nu} a_{\mu}(x), \p_{\rho} a^{\dagger}_{\tau}(y) \right] + \left[ \p_{\nu} a_{\mu}(x), \p_{\tau} a^{\dagger}_{\rho}(y) \right] \\ &= (2\pi)^{-3} \int \frac{d^3 p}{2p_0} \left( P_{\mu \nu \rho \tau} - P_{\mu \nu \tau \rho} - P_{\nu \mu \rho \tau} + P_{\nu \mu \tau \rho} \right) \\ &\quad \times \left( |\kappa|^2 e^{\ifrak p \cdot (x-y)} \pm |\lambda|^2 e^{-\ifrak p \cdot (x-y)} \right)\end{split}\]

It remains to calculate the following

\[\begin{split}P_{\mu \nu \rho \tau} - P_{\mu \nu \tau \rho} - P_{\nu \mu \rho \tau} + P_{\nu \mu \tau \rho} &= p_{\mu} p_{\rho} \left( \delta_{\nu \tau} - \delta_{0 \nu}\delta_{0 \tau} - (1 - \delta_{0 \nu})(1 - \delta_{0 \tau}) \frac{p_{\nu} p_{\tau}}{\pbf^2} \right) \\ &\phantom{=} - p_{\mu} p_{\tau} \left( \delta_{\nu \rho} - \delta_{0 \nu}\delta_{0 \rho} - (1 - \delta_{0 \nu})(1 - \delta_{0 \rho}) \frac{p_{\nu} p_{\rho}}{\pbf^2} \right) \\ &\phantom{=} - p_{\nu} p_{\rho} \left( \delta_{\mu \tau} - \delta_{0 \mu}\delta_{0 \tau} - (1 - \delta_{0 \mu})(1 - \delta_{0 \tau}) \frac{p_{\mu} p_{\tau}}{\pbf^2} \right) \\ &\phantom{=} + p_{\nu} p_{\tau} \left( \delta_{\mu \rho} - \delta_{0 \mu}\delta_{0 \rho} - (1 - \delta_{0 \mu})(1 - \delta_{0 \rho}) \frac{p_{\mu} p_{\rho}}{\pbf^2} \right) \\ &= p_{\mu} p_{\rho} (\delta_{\nu \tau} - 2\delta_{0 \nu}\delta_{0 \tau}) - p_{\mu} p_{\tau} (\delta_{\nu \rho} - 2\delta_{0 \nu}\delta_{0 \rho}) \\ &\phantom{=} - p_{\nu} p_{\rho} (\delta_{\mu \tau} - 2\delta_{0 \mu}\delta_{0 \tau}) + p_{\nu} p_{\tau} (\delta_{\mu \rho} - 2\delta_{0 \mu}\delta_{0 \rho}) \\ &= \eta_{\nu \tau} p_{\mu} p_{\rho} - \eta_{\nu \rho} p_{\mu} p_{\tau} - \eta_{\mu \tau} p_{\nu} p_{\rho} + \eta_{\mu \rho} p_{\nu} p_{\tau}\end{split}\]

Finally we can finish Eq.4.7.21 as follows

\[\begin{split}\left[ f_{\mu \nu}(x), f^{\dagger}_{\rho \tau}(y) \right]_{\pm} &= -(2\pi)^{-3} \left( \eta_{\nu \tau} \p_{\mu} \p_{\rho} - \eta_{\nu \rho} \p_{\mu} \p_{\tau} - \eta_{\mu \tau} \p_{\nu} \p_{\rho} + \eta_{\mu \rho} \p_{\nu} \p_{\tau} \right) \\ &\quad \times \int \frac{d^3 p}{2p_0} \left( |\kappa|^2 e^{\ifrak p \cdot (x-y)} \pm |\lambda|^2 e^{-\ifrak p \cdot (x-y)} \right)\end{split}\]

For this expression to vanish, we must take the bottom sign, i.e., it’s a commutator. Moreover, we must have \(|\kappa| = |\lambda|\) since \(\Delta_+(x-y) = \Delta_+(y-x)\) if \(x-y\) is space-like. By further rescaling the creation operator if necessary, we may assume \(\kappa = \lambda = 1\), which partially justifies the same choice taken for the \(a\)-field Eq.4.7.12. In particular, the massless helicity \(\pm 1\) (vector) field must be bosonic, as expected.

4.7.2. The failure for general fields#

The problem that massless fields cannot be made to satisfy the transformation law Eq.4.7.3 is not specific to vector fields. Indeed, we’ll show in this section that the same problem persists to all \((A, B)\) representations.

Recall from Eq.4.5.2, and Eq.4.5.4, that we can explicitly write the \(\Jscr_{\mu \nu}\) matrix as follows

(4.7.22)#\[\begin{split}\left( \Jscr_{ij} \right)_{a'b', ab} &= \epsilon_{ijk} \left( \delta_{b'b} \left( \Jbf^{(A)}_k \right)_{a'a} + \delta_{a'a} \left( \Jbf^{(B)}_k \right)_{b'b} \right) \\ \left( \Jscr_{k0} \right)_{a'b', ab} &= -\ifrak \left( \delta_{b'b} \left( \Jbf^{(A)}_k \right)_{a'a} - \delta_{a'a} \left( \Jbf^{(B)}_k \right)_{b'b} \right)\end{split}\]

It follows from Eq.4.4.1, and the fact according to Eq.1.3.28 that \(R(\theta)\) is the rotation in the \(xy\)-plane, that

\[D(R(\theta)) = 1 + \ifrak \theta \Jscr_{12}\]

for \(\theta\) infinitesimal. The linearized version of Eq.4.7.8, as well as its \(v\)-counterpart, in this case are given by the following

\[\begin{split}\sigma u_{a'b'}(\kbf, \sigma) &= \sum_{a'b'} \left( \Jscr_{12} \right)_{a'b', ab} u_{ab}(\kbf, \sigma) \\ &= \sum_{ab} (a' + b') \delta_{a'a} \delta_{b'b} u_{ab}(\kbf, \sigma) \\ &= (a' + b') u_{a'b'}(\kbf, \sigma) \\ -\sigma v_{a'b'}(\kbf, \sigma) &= (a' + b') v_{a'b'}(\kbf, \sigma)\end{split}\]

It follows that

(4.7.23)#\[\begin{split}\begin{alignat*}{2} a' + b' &= \sigma \quad &&\text{if}~~ u_{a'b'}(\kbf, \sigma) \neq 0 \\ a' + b' &= -\sigma \quad &&\text{if}~~ v_{a'b'}(\kbf, \sigma) \neq 0 \end{alignat*}\end{split}\]

Next let’s linearize the \(S(a, b)\)-part of Eq.4.7.8 using the explicit formula Eq.1.3.27 for \(S(a, b)\) as follows

\[\begin{split}0 &= \sum_{a, b} \left( \Jscr_{01} + \Jscr_{31} \right)_{a'b', ab} u_{ab}(\kbf, \sigma) \\ &= \sum_{a, b} \left( \ifrak \delta_{b'b} \left( \Jbf^{(A)}_1 \right)_{a'a} - \ifrak \delta_{a'a} \left( \Jbf^{(B)}_1 \right)_{b'b} + \delta_{b'b} \left( \Jbf^{(A)}_2 \right)_{a'a} + \delta_{a'a} \left( \Jbf^{(B)}_2 \right)_{b'b} \right) u_{ab}(\kbf, \sigma) \\ &= \sum_{a} \left( \ifrak \Jbf^{(A)}_1 + \Jbf^{(A)}_2 \right)_{a'a} u_{ab'}(\kbf, \sigma) + \sum_{b} \left( -\ifrak \Jbf^{(B)}_1 + \Jbf^{(B)}_2 \right)_{b'b} u_{a'b}(\kbf, \sigma) \\ 0 &= \sum_{a, b} \left( \Jscr_{02} + \Jscr_{32} \right)_{a'b', ab} u_{ab}(\kbf, \sigma) \\ &= \sum_{a, b} \left( \ifrak \delta_{b'b} \left( \Jbf^{(A)}_2 \right)_{a'a} - \ifrak \delta_{a'a} \left( \Jbf^{(B)}_2 \right)_{b'b} - \delta_{b'b} \left( \Jbf^{(A)}_1 \right)_{a'a} - \delta_{a'a} \left( \Jbf^{(B)}_1 \right)_{b'b} \right) u_{ab}(\kbf, \sigma) \\ &= \sum_{a} \left( -\Jbf^{(A)}_1 + \ifrak \Jbf^{(A)}_2 \right)_{a'a} u_{ab'}(\kbf, \sigma) - \sum_{b} \left( \Jbf^{(B)}_1 + \ifrak \Jbf^{(B)}_2 \right)_{b'b} u_{a'b}(\kbf, \sigma)\end{split}\]

which correspond to taking \(a\) infinitesimal and \(b = 0\), and \(b\) infinitesimal and \(a = 0\), respectively. Since \(u_{ab}\) are linearly independent, the above constraints are equivalent to the following

\[\begin{split}\sum_{a} \left( \Jbf^{(A)}_1 - \ifrak \Jbf^{(A)}_2 \right)_{a'a} u_{ab'}(\kbf, \sigma) &= 0 \\ \sum_{b} \left( \Jbf^{(B)}_1 + \ifrak \Jbf^{(B)}_2 \right)_{b'b} u_{a'b}(\kbf, \sigma) &= 0\end{split}\]

In order for \(u_{ab'}\) to be annihilated by the lowering operator, and for \(u_{a'b}\) to be annihilated by the raising operator, we must have

(4.7.24)#\[a' = -A, \quad b' = B\]

if \(u_{a'b'}(\kbf, \sigma) \neq 0\). Combining Eq.4.7.23 and Eq.4.7.24, we conclude that for the \(u\)-spinor to not vanish, we must have

(4.7.25)#\[\sigma = B - A\]

It follows that a general massless \((A, B)\) field, according to Eq.4.5.6, can only destroy particles of helicity \(B-A\). Similar argument can be applied to the \(v\)-spinor to conclude that the field can only create antiparticles of helicity \(A-B\).

As a special case, we see once again that a massless helicity \(\pm 1\) field cannot be constructed as a vector field, i.e., a \(\left( \tfrac{1}{2}, \tfrac{1}{2} \right)\) field, because such vector field must be scalar by Eq.4.7.25. Indeed, the simplest massless helicity \(\pm 1\) field must be a \((1, 0) \oplus (0, 1)\) field, which is nothing but the anti-symmetric \(2\)-tensor \(f_{\mu \nu}\) defined by Eq.4.7.19.

Footnotes