8. Path-Integral Methods#

In this chapter we’ll learn about a new way to build quantum field theory that complements the canonical quantization. This is done via a path-integral formalism invented by Feynman. The advantage of the path-integral formalism over the canonical formalism, as we’ll see, is the manifest Lorentz invariance in deriving the propagators, as opposed to the miraculous cancellation of local interactions, such as the Coulomb interaction Eq.7.4.4 discussed in the previous chapter. The disadvantage, on the other hand, is that the verification of the Lorentz invariance of the S-matrix becomes obscure.

Note

The gear is shifted a bit in this chapter in the following two aspects. First, we’ll adopt Dirac’s bra-ket notation, and second, it’s the Schrödinger picture rather than the interaction picture that will be considered alongside the Heisenberg picture.

8.1. Path Integrals for Bosons#

Similar to the canonical formalism Eq.6.1.1, let’s consider Hermitian operators \(Q_a\) and their conjugates \(P_a\) satisfying the following canonical commutation relations

(8.1.1)#\[\begin{split}[Q_a, P_b] &= \ifrak \delta_{ab} \\ [Q_a, Q_b] &= [P_a, P_b] = 0\end{split}\]

We’ll think of \(Q_a\) as (spacetime) coordinates and \(P_a\) as momenta. Moreover, they are considered to be time-independent Schrödinger-picture operators. Note that the commutators are used here since the particles under consideration are bosons. The corresponding theory for fermions will be taken up in the next section.

Since all \(Q_a\)s commute each other, they can be simultaneously diagonalized so that there exist eigenstates \(\ket{q}\) satisfying

(8.1.2)#\[Q_a \ket{q} = q_a \ket{q}\]

Moreover, the eigenstates satisfy the following orthogonality condition

(8.1.3)#\[\braket{q' | q} = \prod_{a} \delta(q'_a - q_a) \eqqcolon \delta(q' - q)\]

and the completeness condition

(8.1.4)#\[1 = \int \prod_a dq_a \ketbra{q}{q}\]

Similar eigenstates exist for \(P_a\)s as well.

It’s important to note that \(\ket{q}\) and \(\ket{p}\) live in the same Hilbert space. Indeed, let’s compute \(\braket{q | p}\) as follows. First, it follows from Eq.8.1.1 that \(P_a\) acts on wave functions in \(q\)-basis as

(8.1.5)#\[P_b = -\ifrak \frac{\p}{\p q_b}\]

Indeed we have

\[\begin{split}\left[ Q_a, P_b \right] \left( f(q) \ket{q} \right) &= Q_a \left( -\ifrak \frac{\p f}{\p q_b} \ket{q} \right) - P_b \left( f(q) q_a \ket{q} \right) \\ &= -\ifrak \frac{\p f}{\p q_b} q_a \ket{q} + \ifrak \frac{\p f}{\p q_b} q_a \ket{q} + \ifrak \delta_{ab} f(q) \ket{q} \\ &= \ifrak \delta_{ab} f(q) \ket{q}\end{split}\]

in agreement with Eq.8.1.1. Next, using the completeness condition, one can write

\[\ket{p} = \int \prod_a dq_a \braket{q | p} \ket{q}\]

It follows then from Eq.8.1.5 and the fact that the \(\ket{q}\)s form a basis that

\[\begin{split}&\int \prod_a dq_a~p_b \braket{q | p} \ket{q} = p_b \ket{p} = P_b \ket{p} = -\ifrak \int \prod_a dq_a \frac{\p \braket{q | p}}{\p q_b} \ket{q} \\ \implies & \frac{\p \braket{q | p}}{\p q_b} = \ifrak p_b \braket{q | p}\end{split}\]

for any \(b\). It follows that

(8.1.6)#\[\braket{q | p} = \prod_a \frac{1}{\sqrt{2\pi}} e^{\ifrak q_a p_a}\]

where the factor \((2\pi)^{-1/2}\) is determined by the normalizing condition \(\braket{p' | p} = \delta(p' - p)\).

8.1.1. The general path integral formula#

To derive the general path integral formula, we need to pass to the Heisenberg picture as follows

(8.1.7)#\[\begin{split}Q_a(t) &= e^{\ifrak Ht} Q_a e^{-\ifrak Ht} \\ P_a(t) &= e^{\ifrak Ht} P_a e^{-\ifrak Ht}\end{split}\]

where the Hamiltonian \(H\) is given as a function of \(P\) and \(Q\). Their eigenstates

(8.1.8)#\[\begin{split}Q_a(t) \ket{t, q} &= q_a \ket{t, q} \\ P_a(t) \ket{t, p} &= p_a \ket{t, p}\end{split}\]

are obviously given by

(8.1.9)#\[\begin{split}\ket{t, q} &= e^{\ifrak Ht} \ket{q} \\ \ket{t, p} &= e^{\ifrak Ht} \ket{p}\end{split}\]

Warning

The eigenstates \(\ket{t, q}\) and \(\ket{t, p}\) given by Eq.8.1.9 are not time-\(t\) evolutions of \(\ket{q}\) and \(\ket{p}\) which, according to Schrödinger’s equation, would be \(e^{-\ifrak Ht} \ket{q}\) and \(e^{-\ifrak Ht} \ket{p}\), respectively.

The time-independent eigenstates satisfy similar orthogonality and completeness conditions as follows

(8.1.10)#\[\begin{split}\braket{t, q' | t, q} &= \delta(q' - q) \\ \braket{t, p' | t, p} &= \delta(p' - p) \\ 1 &= \int \prod_a dq_a \ketbra{t, q}{t, q} \\ 1 &= \int \prod_a dp_a \ketbra{t, p}{t, p}\end{split}\]

Moreover Eq.8.1.6 also carries over

(8.1.11)#\[\braket{t, q | t, p} = \prod_a \frac{1}{\sqrt{2\pi}} e^{\ifrak q_a p_a}\]

Now the key idea in deriving the path integral formula is to evaluate how the eigenstates evolve in infinitesimal time steps \(\tau \to \tau + d\tau\) as follows

(8.1.12)#\[\braket{\tau + d\tau, q' | \tau, q} = \braket{\tau, q' | e^{-\ifrak H d\tau} | \tau, q}\]

In light of Eq.8.1.8, it’ll be convenient to rewrite \(H = H(Q, P)\) in terms of \(Q(t)\) and \(P(t)\) defined by Eq.8.1.7. This is done by the following calculation

(8.1.13)#\[H = H(Q, P) = e^{\ifrak Ht} H(Q, P) e^{-\ifrak Ht} = H(Q(t), P(t))\]

Using the canonical commutation relations Eq.8.1.1, we can make the following assumption without losing any generality.

Assumption

All the \(Q\) operators in \(H\) lie to the left of the \(P\) operators.

Under this assumption, one can expand Eq.8.1.12 for infinitesimal \(d\tau\) using Eq.8.1.8 and Eq.8.1.11 as follows

(8.1.14)#\[\begin{split}\braket{\tau + d\tau, q' | \tau, q} &= \braket{\tau, q' | \exp\left( -\ifrak H(Q(\tau), P(\tau)) d\tau \right) | \tau, q} \\ &= \int \prod_a dp_a \braket{\tau, q' | \exp(-\ifrak H(Q(\tau), P(\tau)) d\tau) | \tau, p} \braket{\tau, p | \tau, q} \\ &= \int \prod_a dp_a \exp(-\ifrak H(q', p) d\tau) \braket{\tau, q' | \tau, p} \braket{\tau, p | \tau, q} \\ &= \int \prod_a \frac{dp_a}{2\pi} \exp\left( -\ifrak H(q', p) d\tau + \ifrak \sum_a (q'_a - q_a) p_a \right)\end{split}\]

Note that the third equality holds only for infinitesimal \(d\tau\), which allows us to pretend that \(e^{-\ifrak H d\tau}\) is linear in \(H\).

Important

The function \(H(q', p)\) in the last expression, or written simply as \(H(q, p)\), is an ordinary function of scalars. In particular, it makes no difference however \(q\) and \(p\) are ordered. It should therefore be remembered that when this process is reversed, i.e., the quantization of a classical Hamiltonian, the quantized Hamiltonian must have all the \(Q\) operators lying to the left of the \(P\) operators.

Now given two time \(t < t'\) with a finite separation, one can divide the time-interval into \(N\) steps

(8.1.15)#\[t < \tau_1 < \tau_2 < \cdots < \tau_N < t'\]

where

\[\tau_i = \frac{t' - t}{N + 1}\]

As \(N \to \infty\), one can apply Eq.8.1.14 to each sub-interval to calculate the transition amplitude

\[\begin{split}&\braket{t', q' | t, q} \\ &= \int \prod_{k=1}^N dq_k \braket{t', q' | t_N, q_N} \braket{t_{N-1}, q_{N-1} | t_{N-2}, q_{N-2}} \cdots \braket{t_1, q_1 | t, q} \\ &= \int \left( \prod_{k=1}^N \prod_a dq_{k, a} \right) \left( \prod_{k=0}^N \prod_a \frac{dp_{k, a}}{2\pi} \right) \exp\left( \ifrak \sum_{k=0}^N \left( -H(q_{k+1}, p_k) d\tau + \sum_a (q_{k+1, a} - q_{k, a}) p_{k, a} \right) \right) \\ &= \int_{\substack{q_a(t) = q_a \\ q_a(t') = q'_a }} \prod_{\tau, a} dq_a(\tau) \prod_{\tau, a} \frac{dp_a(\tau)}{2\pi} \exp\left( \ifrak \int_t^{t'} d\tau \left( -H(q(\tau), p(\tau)) + \sum_a \dot{q}_a(\tau) p_a(\tau) \right) \right)\end{split}\]

with the understanding that \(q_0 = q\) and \(q_{N+1} = q'\). It’s in the last equality where the limit \(N \to \infty\), or equivalently \(d\tau \to 0\), is taken. The integral is taken over all paths from state \(\ket{q}\) at time \(t\) to state \(\ket{q'}\) at time \(t'\), and hence the name – path integral.

It turns out that the same recipe for deriving the general path integral formula above can also be applied to calculate matrix elements of an operator \(\Oscr(P(t), Q(t))\), or more generally a time-ordered product of such operators. Note that in contrast to the Hamiltonian (cf. Eq.8.1.13), we’ve swapped the order of arguments \(Q, P\) in \(\Oscr\). This is, for reasons which will become clear momentarily, due to the following arrangement.

Assumption

All the \(P\) operators in \(\Oscr\) lie to the left of the \(Q\) operators.

As before, let’s first calculate the infinitesimal matrix element as follows

\[\begin{split}\braket{\tau + d\tau, q' | \Oscr(P(\tau), Q(\tau)) | \tau, q} &= \int \prod_a dp_a \braket{\tau, q' | \exp(-\ifrak H d\tau) | \tau, p} \braket{\tau, p | \Oscr | \tau, q} \\ &= \int \prod_a dp_a \exp\left( -\ifrak H(q', p) d\tau \right) \Oscr(p, q) \braket{\tau, q' | \tau, p} \braket{\tau, p | \tau, q} \\ &= \int \prod_a \frac{dp_a}{2\pi} \exp\left( -\ifrak H(q', p) d\tau + \ifrak \sum_a (q'_a - q_a) p_a \right) \Oscr(p, q)\end{split}\]

Consider a time-ordered sequence of operators

\[\Oscr_A(P(t_A), Q(t_A)), \Oscr_B(P(t_B), Q(t_B)), \cdots\]

such that \(t_A > t_B > \cdots\). We can calculate the matrix element of the product of the operators at a finite time difference by dividing the time-interval in the same way as in Eq.8.1.15 and pay attention to the sub-intervals that contains \(t_A, t_B, \cdots\), as follows

\[\begin{split}&\braket{t', q' | \Oscr_A(P(t_A), Q(t_A)) \Oscr_B(P(t_B), Q(t_B)) \cdots | t, q} \\ &\quad = \int_{\substack{q_a(t)=q_a \\ q_a(t')=q'_a}} \prod_{\tau, a} dq_a(\tau) \prod_{\tau, a} \frac{dp_a(\tau)}{2\pi} \Oscr_A(p(t_A), q(t_A)) \Oscr_B(p(t_B), q(t_B)) \cdots \\ &\qquad \times \exp\left( \ifrak \int_t^{t'} d\tau \left( -H(q(\tau), p(\tau)) + \sum_a \dot{q}_a(\tau) p_a(\tau) \right) \right)\end{split}\]

Since the right-hand-side doesn’t rely on the time-ordering, we may replace the product of operators on the left-hand-side with the timed-ordered product as follows

(8.1.16)#\[\begin{split}&\braket{t', q' | T\left\{ \Oscr_A(P(t_A), Q(t_A)) \Oscr_B(P(t_B), Q(t_B)) \cdots \right\} | t, q} \\ &\quad = \int_{\substack{q_a(t)=q_a \\ q_a(t')=q'_a}} \prod_{\tau, a} dq_a(\tau) \prod_{\tau, a} \frac{dp_a(\tau)}{2\pi} \Oscr_A(p(t_A), q(t_A)) \Oscr_B(p(t_B), q(t_B)) \cdots \\ &\qquad \times \exp\left( \ifrak \int_t^{t'} d\tau \left( -H(q(\tau), p(\tau)) + \sum_a \dot{q}_a(\tau) p_a(\tau) \right) \right)\end{split}\]

as long as \(t_A, t_B, \cdots\) are all distinct.

8.1.2. Transition to the S-matrix#

From now on, we will restrict the discussion to quantum field theories where the index \(a\) from the previous section becomes \((\xbf, m)\), where \(\xbf\) is the spatial coordinates and \(m\) denotes other quantum labels such as spin. In this case we rewrite Eq.8.1.16 as follows

\[\begin{split}&\braket{t', q' | T\left\{ \Oscr_A(P(t_A), Q(t_A)), \Oscr_B(P(t_B), Q(t_B)), \cdots \right\} | t, q} \\ &\quad = \int_{\substack{q_m(t, \xbf)=q_m(\xbf) \\ q_m(t', \xbf')=q_m(\xbf')}} \prod_{\tau, \xbf, m} dq_m(\tau, \xbf) \prod_{\tau, \xbf, m} \frac{dp_m(\tau, \xbf)}{2\pi} \Oscr_A(p(t_A), q(t_A)) \Oscr_B(p(t_B), q(t_B)) \cdots \\ &\qquad \times \exp\left( \ifrak \int_t^{t'} d\tau \left( -H(q(\tau), p(\tau)) + \int d^3 x \sum_m \dot{q}_m(\tau, \xbf) p_m(\tau, \xbf) \right) \right)\end{split}\]

Recall that the S-matrix involves matrix elements between in- and out-states, which are states are time \(t = \mp\infty\), respectively. Hence if we write \(\ket{\alpha, \op{in}}\) for the in-state and \(\ket{\beta, \op{out}}\) for the out-state, then the S-matrix element can be written as follows

(8.1.17)#\[\begin{split}&\braket{\beta, \op{out} | T\left\{ \Oscr_A(P(t_A), Q(t_A)), \Oscr_B(P(t_B), Q(t_B)), \cdots \right\} | \alpha, \op{in}} \\ &\quad = \int \prod_{\tau, \xbf, m} dq_m(\tau, \xbf) \prod_{\tau, \xbf, m} \frac{dp_m(\tau, \xbf)}{2\pi} \Oscr_A(p(t_A), q(t_A)) \Oscr_B(p(t_B), q(t_B)) \cdots \\ &\qquad \times \exp\left( \ifrak \int_{-\infty}^{\infty} d\tau \left( -H(q(\tau), p(\tau)) + \int d^3 x \sum_m \dot{q}_m(\tau, \xbf) p_m(\tau, \xbf) \right) \right) \\ &\qquad \times \braket{\beta, \op{out} | q(\infty), \infty} \braket{q(-\infty), -\infty | \alpha, \op{in}}\end{split}\]

where the path integral now has essentially no boundary conditions.

The goal now is to calculate the wave functions \(\braket{\beta, \op{out} | q(\infty), \infty}\) and \(\braket{q(-\infty), -\infty | \alpha, \op{in}}\), if we choose a specific basis for the in- and out-states. It turns out, following discussions in External edges off the mass shell, that it suffices to consider the vacuum state \(\ket{\VAC}\). Moreover we’ll not distinguish between \(\ket{\VAC, \op{in}}\) and \(\ket{\VAC, \op{out}}\) since the calculations will mostly be the same.

The vacuum state, being a state with no particles, can be characterized by

(8.1.18)#\[a(\pbf, \sigma, n) \ket{\VAC} = 0\]

where \(a(\pbf, \sigma, n)\) is the operator that annihilates a particle with momentum \(\pbf\), spin \(z\)-component \(\sigma\), and other quantum numbers \(n\).

For simplicity, we’ll focus on the real scalar field given by Eq.4.2.8 and turned into canonical variables following Eq.6.1.2 as follows

\[\begin{split}\Phi(t, \xbf) &= (2\pi)^{-3/2} (2E)^{-1/2} \int d^3 p \left( e^{\ifrak p \cdot x} a(\pbf) + e^{-\ifrak p \cdot x} a^{\dagger}(\pbf) \right) \\ \Pi(t, \xbf) &= \dot{\Phi}(t, \xbf) = -\ifrak (2\pi)^{-3/2} (E/2)^{1/2} \int d^3 p \left( e^{\ifrak p \cdot x} a(\pbf) - e^{-\ifrak p \cdot x} a^{\dagger}(\pbf) \right)\end{split}\]

where \(E = p_0 = \sqrt{\pbf^2 + m^2}\) on the on-mass-shell energy. From these one can solve for \(a(\pbf)\) as follows

\[\begin{split}a(\pbf) &= (2\pi)^{-3/2} \int d^3 x~e^{-\ifrak p \cdot x} \left( (E/2)^{1/2} \Phi(t, \xbf) + \ifrak (2E)^{-1/2} \Pi(t, \xbf) \right) \\ &= (2\pi)^{-3/2} e^{\ifrak Et} \int d^3 x~e^{\ifrak \pbf \cdot \xbf} \left( (E/2)^{1/2} \Phi(t, \xbf) + \ifrak (2E)^{-1/2} \Pi(t, \xbf) \right)\end{split}\]

where we’ve pulled out the time-dependency since in order to apply it to in- and out-state, we need to take the limits \(t \to \mp\infty\), respectively. More explicitly, one can write

\[a_{\op{in}}(\pbf) = \lim_{t \to -\infty} a(\pbf), \quad a_{\op{out}}(\pbf) = \lim_{t \to \infty} a(\pbf)\]

It turns out that the time limits are not really relevant in calculating the wave functions since \(e^{\ifrak Et}\) is never zero. Hence we’ll continue to just use \(a(\pbf)\) in calculations. In the same vein, define Schrödinger-picture operators

\[\phi(\mp\infty, \xbf) = \lim_{t \to \mp\infty} \Phi(t, \xbf), \quad \pi(\mp\infty, \xbf) = \lim_{t \to \mp\infty} \Pi(t, \xbf)\]

In places where specifying \(t = \mp\infty\) doesn’t matter, we’ll also simply write \(\phi(\xbf)\) and \(\pi(\xbf)\).

Using Eq.8.1.18, one finds a differential equation that the wave functions \(\braket{\mp\infty, \phi(\mp\infty, \xbf) | \VAC}\) must satisfy as follows

(8.1.19)#\[\begin{split}& \braket{\mp\infty, \phi(\mp\infty) | a(\pbf) | \VAC} = 0 \\ \implies & \int d^3 x~e^{\ifrak \pbf \cdot \xbf} \left( \frac{\delta}{\delta \phi(\xbf)} + E(\pbf)\phi(\xbf) \right) \braket{\mp\infty, \phi(\mp\infty, \xbf) | \VAC} = 0\end{split}\]

where we have also used the interpretation of \(\pi(\xbf)\) as variational derivative \(-\ifrak \delta/\delta \phi(\xbf)\) (cf. Eq.8.1.5). Based on the experience of solving an analogous ODE by exponential function, it’s quite natural to postulate a Gaussian solution

(8.1.20)#\[\braket{\mp\infty, \phi(\mp\infty, \xbf) | \VAC} = \Nscr \exp\left( -\frac{1}{2} \int d^3 x~d^3 y~\Escr(\xbf, \ybf) \phi(\xbf) \phi(\ybf) \right)\]

where \(\Nscr\) is a constant. Indeed Eq.8.1.19 becomes equivalent to

\[\begin{split}0 &= \int d^3 x~e^{\ifrak \pbf \cdot \xbf} \left( \int d^3 y~\Escr(\xbf, \ybf) \phi(\ybf) - E(\pbf) \phi(\xbf) \right) \\ &= \int d^3 x~d^3 y~e^{\ifrak \pbf \cdot \xbf} \Escr(\xbf, \ybf) \phi(\ybf) - \int d^3 y~e^{\ifrak \pbf \cdot \ybf} E(\pbf) \phi(\ybf) \\ &= \int d^3 y~\phi(\ybf) \left( \int d^3 x~e^{\ifrak \pbf \cdot \xbf} \Escr(\xbf, \ybf) - e^{\ifrak \pbf \cdot \ybf} E(\pbf) \right)\end{split}\]

For the right-hand-side to vanish for any \(\phi\), the quantity in the parenthesis must vanish. An inverse Fourier transform then gives

(8.1.21)#\[\Escr(\xbf, \ybf) = (2\pi)^{-3} \int d^3 p~e^{\ifrak \pbf \cdot (\xbf - \ybf)} E(\pbf)\]

where we recall once again that \(E(\pbf) = \sqrt{\pbf^2 + m^2}\). This solves Eq.8.1.20 up to an unknown field-independent constant \(\Nscr\), which turns out to be insignificant. Indeed, the same constant \(\Nscr\) also appears in \(\braket{\VAC, \op{out} | \VAC, \op{in}}\) and hence can be eliminated by normalization. More details about this will be discussed in the next section.

We can continue the calculation Eq.8.1.17 in the case of vacuum expectation values for real scalar fields as follows

(8.1.22)#\[\begin{split}& \braket{\VAC, \op{out} | \infty, \phi(\infty)} \braket{-\infty, \phi(-\infty) | \VAC, \op{in}} \\ &\quad = |\Nscr|^2 \exp\left( -\frac{1}{2} \int d^3x~d^3y~\Escr(\xbf, \ybf) \left( \phi(\infty, \xbf) \phi(\infty, \ybf) + \phi(-\infty, \xbf) \phi(-\infty, \ybf) \right) \right) \\ &\quad = |\Nscr|^2 \lim_{\epsilon \to 0+} \exp\left( -\frac{\epsilon}{2} \int d^3x~d^3y~\Escr(\xbf, \ybf) \int_{-\infty}^{\infty} d\tau~\phi(\tau, \xbf) \phi(\tau, \ybf) e^{-\epsilon |\tau|} \right)\end{split}\]

and therefore

(8.1.23)#\[\begin{split}& \braket{\VAC, \op{out} | T\left\{ \Oscr_A(\Pi(t_A), \Phi(t_A)), \Oscr(\Pi(t_B), \Phi(t_B)), \cdots \right\} | \VAC, \op{in}} \\ &\quad = |\Nscr|^2 \int \prod_{\tau, \xbf} d\phi(\tau, \xbf) \prod_{\tau, \xbf} \frac{d\pi(\tau, \xbf)}{2\pi} \Oscr_A(\Pi(t_A), \Phi(t_A)) \Oscr_B(\Pi(t_B), \Phi(t_B)) \cdots \\ &\qquad \times \exp\left( \ifrak \int_{-\infty}^{\infty} d\tau \left( -H(\phi(\tau), \pi(\tau)) + \int d^3x~\dot{\phi}(\tau, \xbf) \pi(\tau, \xbf) \right.\right. \\ &\qquad \left.\left. + \frac{\ifrak\epsilon}{2} \int d^3x~d^3y~\Escr(\xbf, \ybf) \phi(\tau, \xbf) \phi(\tau, \ybf) e^{-\epsilon |\tau|} \right) \right)\end{split}\]

Without working out the details, we claim that the only difference in the calculation for general fields is the term after \(\ifrak\epsilon/2\), whose exact form turns out to be insignificant. For later references, the final result is recorded as follows

(8.1.24)#\[\begin{split}& \braket{\VAC, \op{out} | T\left\{ \Oscr_A(P(t_A), Q(t_A)), \Oscr(P(t_B), Q(t_B)), \cdots \right\} | \VAC, \op{in}} \\ &\quad = |\Nscr|^2 \int \prod_{\tau, \xbf, m} dq_m(\tau, \xbf) \prod_{\tau, \xbf} \frac{dp_m(\tau, \xbf)}{2\pi} \Oscr_A(p(t_A), q(t_A)) \Oscr_B(p(t_B), q(t_B)) \cdots \\ &\qquad \times \exp\left( \ifrak \int_{-\infty}^{\infty} d\tau \left( -H(q(\tau), p(\tau)) + \int d^3x \sum_m \dot{q}_m(\tau, \xbf) p_m(\tau, \xbf) + \ifrak\epsilon \text{ terms} \right) \right)\end{split}\]

where the \(\ifrak\epsilon\) terms depend only on \(q\)s.

8.1.3. Lagrangian version of the path integral#

So far the path integral formalism has been developed using the Hamiltonian. Now we’ll develop a version based on the Lagrangian. In fact, the integrand in the exponential power in Eq.8.1.24, leaving alone the \(\ifrak\epsilon\) terms, looks just like the corresponding Lagrangian (cf. Eq.6.1.11). However, there is an important difference, namely, the \(q\) and \(p\) variables in Eq.8.1.24 are independent variables, while in the Lagrangian formalism, they are related by Eq.6.1.14. As we’ll see, it turns out that when the Hamiltonian \(H\) is quadratic in \(p\) and the (timed-ordered) operators \(\Oscr_A, \Oscr_B, \cdots\), are independent of the \(P\)s, one can explicitly evaluate the integral in \(p\) in Eq.8.1.24, which will then produce the Lagrangian version of the path integral.

To spell out the details, let’s write down the (Heisenberg-picture) Hamiltonian in the most general form as follows

(8.1.25)#\[\begin{split}H(Q, P) &= \frac{1}{2} \sum_{n, m} \int d^3x~d^3y~A_{\xbf n, \ybf m}(Q) P_n(\xbf) P_m(\ybf) \\ &\quad + \sum_n \int d^3x~B_{\xbf n}(Q) P_n(\xbf) + C(Q)\end{split}\]

where \(A\) is a real, symmetric, positive matrix. Moreover \(H\) is written in the way that all the \(Q\) operators lie to the left of the \(P\) operators.

Now we can write the power in the exponential in Eq.8.1.24 without the \(\ifrak\epsilon\) terms as follows

(8.1.26)#\[\begin{split}&\int d\tau \left( -H(q(\tau), p(\tau)) + \int d^3x \sum_n \dot{q}_n(\tau, \xbf) p_n(\tau, \xbf) \right) \\ &\quad = -\frac{1}{2} \sum_{n, m} \int d\tau~d\tau'~d^3x~d^3y~A_{\xbf n, \ybf m}(q(\tau)) \delta(\tau - \tau') p_n(\tau, \xbf) p_m(\tau', \ybf) \\ &\qquad - \sum_n \int d\tau~d^3x \left( B_{\xbf n}(q(\tau)) - \dot{q}_n(\tau, \xbf) \right) p_n(\tau, \xbf) - \int d\tau~C(q(\tau))\end{split}\]

where it’s organized so that the first summand on the right-hand-side is quadratic in \(p\), the second is linear, and the third is independent of \(p\). The reason to arrange the power in this form is because of the following (finite-dimensional) Gaussian integral formula.

Gaussian Integral Formula

(8.1.27)#\[\begin{split}&\int_{-\infty}^{\infty} \prod_s d\xi_s \exp\left( -\ifrak \left( \frac{1}{2} \sum_{s, r} \Ascr_{sr} \xi_s \xi_r + \sum_s \Bscr_s \xi_s + \Cscr_s \right) \right) \\ &\quad = \left( \det(\ifrak \Ascr / 2\pi) \right)^{-1/2} \exp\left( -\ifrak \left( \sum_{s, r} \Ascr_{sr} \bar{\xi}_s \bar{\xi}_r + \sum_s \Bscr_s \bar{\xi}_s + \Cscr_s \right) \right)\end{split}\]

where \(\bar{\xi}\) is the (unique) stationary point of the quadratic power given explicitly by

\[\bar{\xi}_s = -\sum_r (\Ascr^{-1})_{sr} \Bscr_r\]

Note

In more general cases where \(H\) is not quadratic in \(P\), approximation techniques such as the stationary phase approximation may be applied.

To figure out the stationary point of the power in Eq.8.1.24 with respect to \(p\), let’s calculate the following variational derivative assuming the \(\ifrak\epsilon\) terms are independent of the \(p\)s

\[\begin{split}&\frac{\delta}{\delta p_n(t, \xbf)} \int_{-\infty}^{\infty} d\tau \left( -H(q(\tau), p(\tau)) + \int d^3y \sum_m \dot{q}_m(\tau, \ybf) p_m(\tau, \ybf) + \ifrak\epsilon \text{ terms} \right) \\ &\quad = - \frac{\delta H}{\delta p_n(t, \xbf)} + \dot{q}_n(t, \xbf)\end{split}\]

It follows that \(\bar{p}\) is stationary if it satisfies Hamilton’s equation

(8.1.28)#\[\dot{q}_n(t, \xbf) = \left. \frac{\delta H}{\delta p_n(t, \xbf)} \right|_{p=\bar{p}}\]

Assuming, in addition, that the (timed-ordered) operators \(\Oscr_A, \Oscr_B, \cdots\), are independent of the \(P\)s, we can evaluate the \(p\)-integral in Eq.8.1.24 using the (infinite-dimensional) Gaussian integral formula Eq.8.1.27 as follows

\[\begin{split}&\int \prod_{\tau, \xbf} \frac{dp_m(\tau, \xbf)}{2\pi} \exp\left( \ifrak \int_{-\infty}^{\infty} d\tau \left( -H(q(\tau), p(\tau)) + \int d^3x \sum_m \dot{q}_m(\tau, \xbf) p_m(\tau, \xbf) + \ifrak\epsilon \text{ terms} \right) \right) \\ &\quad = \left( \det(2\pi\ifrak\Ascr(q)) \right)^{-1/2} \exp\left( \ifrak \int_{-\infty}^{\infty} d\tau \left( L(q(\tau), \dot{q}(\tau)) + \ifrak\epsilon \text{ terms} \right) \right)\end{split}\]

where \(L\) is the Lagrangian defined by

(8.1.29)#\[L(q(\tau), \dot{q}(\tau)) \coloneqq -H(q(\tau), \bar{p}(\tau)) + \int d^3x \sum_m \dot{q}_m(\tau, \xbf) \bar{p}_m(\tau, \xbf)\]

with \(\bar{p}\) satisfying Eq.8.1.28 and

(8.1.30)#\[\Ascr_{\tau \xbf n, \tau' \ybf m}(q) \coloneqq A_{\xbf n, \ybf m}(q(\tau)) \delta(\tau-\tau')\]

is given by Eq.8.1.26.

Finally, we can write down the Lagrangian version of Eq.8.1.24 as follows

(8.1.31)#\[\begin{split}&\braket{\VAC, \op{out} | T\left\{ \Oscr_A(Q(t_A)), \Oscr_B(Q(t_B)), \cdots \right\} | \VAC, \op{in}} \\ &\quad = |\Nscr|^2 \int \prod_{\tau, \xbf, m} dq_m(\tau, \xbf) \Oscr_A(Q(t_A)) \Oscr_B(Q(t_B)) \cdots \\ &\qquad \times \left( \det(2\pi\ifrak\Ascr(q)) \right)^{-1/2} \exp\left( \ifrak \int_{-\infty}^{\infty} d\tau \left( L(q(\tau), \dot{q}(\tau)) + \ifrak\epsilon \text{ terms} \right) \right)\end{split}\]

The rest of this section is devoted to the determination of \(\Ascr(q)\) in various examples.

Scalar fields with non-derivative coupling

Following Eq.6.4.1, consider the following Lagrangian density of a set of (massless) scalar fields \(\Phi_n\) that have only non-derivative interaction \(V\) and are coupled to external currents \(J_n\)

\[\Lscr = -\sum_n \left( \frac{1}{2} \p_{\mu} \Phi_n \p^{\mu} \Phi_n + J_n^{\mu} \p_{\mu} \Phi_n \right) - V(\Phi)\]

The canonical adjoint \(\Pi_n\) is, according to Eq.6.2.1, given by

\[\Pi_n = \frac{\delta \Lscr}{\delta \dot{\Phi}_n} = \dot{\Phi}_n - J_n^0\]

and hence the Hamiltonian is, according to Eq.6.2.6, given by

\[\begin{split}H &= \int d^3x \left( \sum_n \Pi_n \dot{\Phi}_n - \Lscr \right) \\ &= \int d^3x \sum_n \left( \Pi_n (\Pi_n + J_n^0) - \frac{1}{2} (\Pi_n + J_n^0)^2 + \frac{1}{2} (\nabla \Phi_n)^2 + J_n^0 (\Pi_n + J_n^0) + \Jbf_n \cdot \nabla \Phi_n \right) \\ &\quad + \int d^3x~V(\Phi) \\ &= \int d^3x \sum_n \left( \frac{1}{2} (\Pi_n + J^0_n)^2 + \frac{1}{2} (\nabla \Phi_n)^2 + \Jbf_n \cdot \nabla \Phi_n \right) + \int d^3x~V(\Phi)\end{split}\]

Comparing with Eq.8.1.25 and following Eq.8.1.30, we see that

\[\Ascr_{x n, x' n'} = \delta^4(x-x') \delta_{nn'}\]

which is field independent, and therefore can be eliminated in the same way that \(\Nscr\) can be eliminated (cf. Eq.8.1.31).

Nonlinear \(\sigma\)-model

The so-called nonlinear \(\sigma\)-model is described by the following Lagrangian density

\[\Lscr = -\frac{1}{2} \sum_{n, m} \p_{\mu} \Phi_n \p^{\mu} \Phi_m (\delta_{nm} + U_{nm}(\Phi)) - V(\Phi)\]

where the nonlinearity is carried by \(U_{nm}(\Phi)\).

In this case the canonical adjoint \(\Pi_n\) is given by

\[\Pi_n = \frac{\delta \Lscr}{\delta \dot{\Phi}_n} = \sum_m \dot{\Phi}_m (\delta_{nm} + U_{nm}(\Phi))\]

and can be solved in matrix notation as follows

\[\dot{\Phi}_n = \sum_m (1+U(\Phi))^{-1}_{nm} \Pi_m\]

hence the Hamiltonian

\[\begin{split}H &= \int d^3x \left( \sum_n \Pi_n \dot{\Phi}_n - \Lscr \right) \\ &= \int d^3x \sum_{n, m} \left( \frac{1}{2} \Pi_n (1+U(\Phi))^{-1}_{nm} \Pi_m + \frac{1}{2} \nabla \Phi_n \cdot \nabla \Phi_m (1+U(\Phi))^{-1}_{nm} \right) + \int d^3x~V(\Phi)\end{split}\]

In follows that

(8.1.32)#\[\Ascr_{xn, x'n'} = (1+U(\Phi))^{-1}_{nn'} \delta^4(x-x')\]

which obviously depend on \(\Phi\), and therefore cannot be eliminated by the division by the vacuum expectation value. The idea then is to absorb it into the Lagrangian (density) which we now explain.

Looking at Eq.8.1.31, we note the following general identity

(8.1.33)#\[\det\Ascr = \exp \Tr \ln \Ascr\]

for any real symmetric positive \(\Ascr\). To evaluate the logarithm, it’s convenient to discretize the Dirac delta function in Eq.8.1.32 as follows

\[\delta^4(x-x') = \Omega^{-1} \delta_{xx'}\]

where \(\Omega\) denotes an infinitesimal volume in spacetime. It follows that

(8.1.34)#\[(\ln \Ascr)_{xn, x'n'} = \delta_{xx'} \left( -\ln(1+U(\Phi)) - \ln\Omega \right)_{nn'}\]

where \(\ln\Omega\) is understood as a constant multiple as the identity matrix. Next note that the trace of \(\delta_{xx'}\) can be evaluated by

\[\Tr~\delta_{xx'} \cdots = \Omega^{-1} \int d^4x \cdots\]

It follows that

\[\det\Ascr \propto \exp\left( -\Omega^{-1} \int d^4x~\Tr\ln(1+U(\Phi)) \right)\]

where the proportionality constant, coming from the constant \(-\ln\Omega\) in Eq.8.1.34, is field-independent. Plugging into Eq.8.1.31, we see that Lagrangian density receives a correction term

\[\Delta\Lscr = -\frac{\ifrak}{2} \Omega^{-1} \Tr\ln(1+U(\Phi))\]

which unfortunately contains a diverging term \(\Omega^{-1}\). This is known as an ultraviolet divergence since it comes from the infinitesimal spacetime volume. We’ll not address how it may be handled here.

Vector fields

The two examples considered so far admit a Lagrangian without auxiliary fields (cf. Eq.6.2.8). To cover this case, consider the following Lagrangian for a set of non-interacting vector fields (cf. Eq.6.4.6)

(8.1.35)#\[\Lscr = -\sum_n \left( \frac{1}{4} F_{n \mu\nu} F_n^{\mu\nu} + \frac{1}{2} M^2 A_{n \mu} A_n^{\mu} + J_n^{\mu} A_{n\mu} \right)\]

According to Eq.6.4.10, the corresponding Hamiltonian is given by

(8.1.36)#\[\begin{split}H &= \int d^3x \sum_n \left( \frac{1}{2} \bm{\Pi}_n^2 + \frac{1}{2M_n^2} (\nabla \cdot \bm{\Pi}_n)^2 + \frac{1}{M_n^2} J_n^0 \nabla \cdot \bm{\Pi}_n \right. \\ &\qquad \left. + \frac{1}{2} (\nabla \times \Abf_n)^2 + \frac{1}{2} M_n^2 \Abf_n^2 + \frac{1}{2M_n^2} (J_n^0)^2 - \Jbf_n \cdot \Abf_n \right)\end{split}\]

where the terms are ordered in descending power of \(\bm{\Pi}\). Using the following calculation

\[\begin{split}\int d^3x~d^3y~\nabla_i \nabla_j \delta^3(\xbf-\ybf) \bm{\Pi}_n^i(x) \bm{\Pi}_n^j(y) &= -\int d^3x~d^3y~\nabla_j \delta^3(\xbf-\ybf) \p_i \bm{\Pi}_n^i(x) \bm{\Pi}_n^j(y) \\ &= -\int d^3x~d^3y~\delta^3(\xbf-\ybf) \p_i \bm{\Pi}_n^i(x) \p_j \bm{\Pi}_n^j(y) \\ &= -\int d^3x~\left( \nabla \cdot \bm{\Pi}_n \right)^2\end{split}\]

we conclude that

\[\Ascr_{x i n, y j m} = \delta_{nm} \left( \delta_{ij}\delta^4(x-y) - \frac{1}{M_n^2} \nabla_i\nabla_j\delta^4(x-y) \right)\]

which is field-independent. As before, it means that the term \(\det(2\pi\ifrak \Ascr(q))^{-1/2}\) in Eq.8.1.31 plays no role. Nonetheless, the Lagrangian defined by Eq.8.1.29 cannot be the same the original Eq.8.1.35 since the former doesn’t involve the time-component \(A_0\). As a consequence, the Lorentz invariance of Eq.8.1.31 is far from obvious.

To restore the manifest Lorentz invariance, let’s introduce, according to Eq.6.4.9, a correction term to the Hamiltonian \(H \to H + \Delta H\) where

(8.1.37)#\[\Delta H = -\frac{1}{2} \sum_n M_n^2 \int d^3x \left( A_n^0 - M_n^{-2} \nabla \cdot \bm{\Pi}_n - M_n^{-2} J_n^0 \right)^2\]

Moreover, in addition to the integration of \(\Abf_n\) and \(\bm{\Pi}_n\) in Eq.8.1.24, we also integrate over \(A_n^0\). This addition doesn’t really make a difference to the physics since the integrant in \(\Delta H\) being a perfect square means that the integration over \(A_n^0\) will only introduce an insignificant field-independent factor to the matrix element.

Combining Eq.8.1.36 and Eq.8.1.37 together, we have

\[H + \Delta H = \int d^3x \sum_n \left( \frac{1}{2} \bm{\Pi}_n^2 + A_n^0 \nabla \cdot \bm{\Pi}_n + \frac{1}{2} (\nabla \times \Abf_n)^2 + \frac{1}{2} M_n^2 A_n^2 - J_n \cdot A_n \right)\]

We see that the integrand is still quadratic in \(\bm{\Pi}\), whose integration, according to the Gaussian integral formula, can be done by replacing \(\bm{\Pi}_n\) with the solution to Eq.8.1.28, which reads

\[\dot{\Abf}_n = \bm{\Pi}_n - \nabla A_n^0 \iff \bm{\Pi}_n = \dot{\Abf}_n + \nabla A_n^0\]

One can then verify that the Legendre transformed quantity

\[-H - \Delta H + \sum_n \int d^3x~\dot{\Abf}_n \cdot \bm{\Pi}_n\]

indeed recovers the original Lagrangian density Eq.8.1.35.

We see from the above examples that it’s far from obvious to choose the correct Lagrangian in Eq.8.1.31. Moreover, the choice of canonical fields may not be the initial \(q\)s. Examples of this kind include the vector fields discussed above as well as QED which will be discussed later. Under these considerations, let’s rewrite Eq.8.1.31 as follows

(8.1.38)#\[\begin{split}&\braket{\VAC, \op{out} | T\{\Oscr_A(\Psi_A(t_A)), \Oscr_B(\Psi_B(t_B)), \cdots\} | \VAC, \op{in}} \\ &\quad \propto \int \prod_{\tau, \xbf, n} d\psi_n(\tau, \xbf)~\Oscr_A(\psi(t_A)) \Oscr_B(\psi(t_B)) \cdots \\ &\qquad \times \exp\left( \ifrak \int_{-\infty}^{\infty} d\tau \left( L(\psi(\tau), \dot{\psi}(\tau)) + \ifrak\epsilon\text{ terms} \right) \right)\end{split}\]

where the field-independent constants \(|\Nscr|^2\) and the part of \(\det(2\pi\ifrak\Ascr)^{-1/2}\) are suppressed into the proportionality, and the field-dependent part of \(\det(2\pi\ifrak\Ascr)^{-1/2}\) is absorbed into the Lagrangian. In addition, the dependence of the \(\psi\)-fields on the right-hand-side on \(A, B, \cdots\), is suppressed into the index \(n\) in the product measure.

8.1.4. Path integral derivation of Feynman rules#

The vacuum expectation value of a time-ordered product of operators given by Eq.8.1.38 can be evaluated by Feynman diagrams, assuming the propagators have been worked out. However these diagrams may not all be connected. In particular, there is a set of \(2\)-component diagrams: one of them consists of vertices from only the timed-ordered operators, and the other consists of vertices from the interaction density. Such diagrams can be gotten rid of by considering the following normalized vacuum expectation value

(8.1.39)#\[M_{\ell_A, \ell_B, \cdots}(x_A, x_B, \cdots) \coloneqq \frac{ \braket{\VAC, \op{out} | T\{\Psi_{\ell_A}(x_A), \Psi_{\ell_B}(x_B), \cdots\} | \VAC, \op{in}} }{ \braket{\VAC, \op{out} | \VAC, \op{in}} }\]

Here a few notations have changed from the previous sections. Firstly, we’ve used \(\ell_A, \ell_B, \cdots\), instead of \(A, B, \cdots\), to label the time-ordered operators, which will allow us to unify the labels by \(\ell\). Secondly, the argument of the fields has changed from time such as \(t_A\) to spacetime coordinates \(x_A\), which of courses contain \(t_A\) as its time-component.

Now if the Hamiltonian is quadratic in the \(P\)-operators as discussed in Lagrangian version of the path integral, then Eq.8.1.38 implies that Eq.8.1.39 can be rewritten as

(8.1.40)#\[M_{\ell_A, \ell_B, \cdots}(x_A, x_B, \cdots) = \frac{ \int \prod_{x, \ell} d\psi_{\ell}(x)~\psi_{\ell_A}(x_A) \psi_{\ell_B}(t_B) \cdots e^{\ifrak I[\psi]} }{ \int \prod_{x, \ell} d\psi_{\ell}(x)~e^{\ifrak I[\psi]} }\]

where

\[I[\psi] = \int_{-\infty}^{\infty} d\tau \left( L(\psi(\tau), \dot{\psi}(\tau)) + \ifrak\epsilon \text{ terms} \right)\]

is the action.

Suppose, in the same vein as discussed in Perturbation Theory of S-matrix and specifically Eq.2.5.11, the Lagrangian is given by a density \(\Lscr\). Then following the philosophy of perturbation theory, let’s write it as the sum of a free part \(\Lscr_0\) and an interacting part \(\Lscr_1\). In other words

\[L(\psi(\tau), \dot{\psi}(\tau)) = \int d^3x \left( \Lscr_0(\psi(\tau, \xbf), \p_{\mu} \psi(\tau, \xbf)) + \Lscr_1(\psi(\tau, \xbf), \p_{\mu} \psi(\tau, \xbf)) \right)\]

which, in turn, implies that

(8.1.41)#\[\begin{split}I[\psi] &= I_0[\psi] + I_1[\psi] \\ I_0[\psi] &= \int d^4x \left( \Lscr_0(\psi, \p_{\mu} \psi) + \ifrak\epsilon\text{ terms} \right) \\ I_1[\psi] &= \int d^4x~\Lscr_1(\psi, \p_{\mu} \psi)\end{split}\]

Such decomposition then allows us to write the exponential term in Eq.8.1.40 in the following form

(8.1.42)#\[\begin{split}\exp(\ifrak I[\psi]) &= \exp(\ifrak I_0[\psi]) \exp(\ifrak I_1[\psi]) \\ &= \exp(\ifrak I_0[\psi]) \sum_{N=0}^{\infty} \frac{\ifrak^N}{N!} (I_1[\psi])^N\end{split}\]

where we’ve also expanded the second exponential of the interaction action. The reason to do so, or rather, to keep the first exponential of the free action, is that \(I_0[\psi]\) is typically, and will be assumed to be, quadratic. Indeed, an explicit example was worked out for scalar field in Eq.8.1.23, as long as we ignore the term \(e^{-\epsilon |\tau|}\) which spoils the quadraticity only in higher orders of \(\epsilon\).

If we write

(8.1.43)#\[I_0[\psi] = -\frac{1}{2} \int d^4x~d^4x' \sum_{\ell, \ell'} \Dscr_{x \ell, x' \ell'} \psi_{\ell}(x) \psi_{\ell'}(x')\]

then according to Eq.8.1.42, both the denominator and the numerator of Eq.8.1.40 are sums of integrals of the following form

(8.1.44)#\[\Iscr_{\ell_1, \ell_2, \cdots}(x_1, x_2, \cdots) \coloneqq \int \prod_{x, \ell} d\psi_{\ell}(x)~e^{\ifrak I_0[\psi]} \psi_{\ell_1}(x_1) \psi_{\ell_2}(x_2) \cdots\]

where \(I_0[\psi]\) is quadratic. In the case of finite-dimensional integrals, this is a well-known extension of the Gaussian Integral Formula discussed above by integration-by-parts. More formally, this is known as Wick’s theorem which we recall as follows

Wick’s theorem

\[\begin{split}&\int \prod_r d\xi_s~\xi_{s_1} \xi_{s_2} \cdots \xi_{s_{2N}} \exp\left( -\frac{\ifrak}{2} \sum_{s, r} \Dscr_{sr} \xi_s \xi_r \right) \\ &\quad = \left( \det(\ifrak\Dscr / 2\pi) \right)^{-1/2} \sum_{\substack{\text{pairings} \\ \text{of } s_1, \cdots, s_{2N}}} \prod_{\text{pairs}} \left(-\ifrak\Dscr^{-1}\right)_{\text{paired indices}}\end{split}\]

where \(\Dscr\) is a real, symmetric, positive matrix.

Applying Wick’s theorem to Eq.8.1.44 we get

\[\begin{split}\Iscr_{\ell_1, \ell_2, \cdots}(x_1, x_2, \cdots) = (\det(\ifrak\Dscr / 2\pi))^{-1/2} \sum_{\substack{\text{pairings} \\ \text{of fields}}} ~\prod_{\text{pairs}} \left( -\ifrak\Dscr^{-1} \right)_{\text{paired fields}}\end{split}\]

where \(\Dscr\) is given by Eq.8.1.43. Observe that this evaluation, besides the unimportant field-independent factor \((\det(\ifrak\Dscr / 2\pi))^{-1/2}\), can be thought of as a sum over Feynman diagrams where the edges are paired fields that come from either the expansion of \(e^{\ifrak I_1[\psi]}\) or the timed-ordered operators in the denominator of Eq.8.1.40. Moreover, the “propagator” \(-\ifrak\Delta\) can be defined as follows

(8.1.45)#\[\Delta_{\ell_1, \ell_2}(x_1, x_2) \coloneqq \Dscr^{-1}_{x_1 \ell_1, x_2 \ell_2}\]

To invert \(\Dscr\) in spacetime coordinates, let’s rewrite Eq.8.1.45 as an integral equation as follows

\[\int d^4 x_2 \sum_{\ell_2} \Dscr_{x_1 \ell_1, x_2 \ell_2} \Delta_{\ell_2, \ell_3}(x_2, x_3) = \delta^4(x_1-x_3) \delta_{\ell_1 \ell_3}\]

Assuming translation-invariance of the theory, it follows that \(\Dscr\) can be written as a Fourier transform as follows

(8.1.46)#\[\Dscr_{x_1 \ell_1, x_2 \ell_2} \eqqcolon (2\pi)^{-4} \int d^4p~e^{\ifrak p \cdot (x_1-x_2)} \Dscr_{\ell_1 \ell_2}(p)\]

which, in turn, implies

(8.1.47)#\[\Delta_{\ell_1 \ell_2}(x_1, x_2) = (2\pi)^{-4} \int d^4p~e^{\ifrak p \cdot (x_1-x_2)} \Dscr^{-1}_{\ell_1 \ell_2}(p)\]

We conclude the discussion with an example.

Scalar field

Recall from Eq.6.1.12 that the free Lagrangian density takes the following form

\[\Lscr_0 = -\frac{1}{2} \p_{\mu} \phi \p^{\mu} \phi - \frac{1}{2} m^2 \phi^2\]

It follows then from Eq.8.1.23 and Eq.8.1.41 that the free action \(I_0[\phi]\), up to the first order of \(\epsilon\), takes the following form

\[\begin{split}&I_0[\phi] \\ &= -\frac{1}{2} \int d^4x \left( \p_{\mu} \phi \p^{\mu} \phi + m^2 \phi^2 \right) + \frac{1}{2} \ifrak\epsilon \int dt \int d^3x~d^3x'~\Escr(\xbf, \xbf') \phi(t, \xbf) \phi(t, \xbf') \\ &= -\frac{1}{2} \int d^4x~d^4x' \left( \delta^4(x-x') (\p_{\mu} \phi \p^{\mu} \phi + m^2 \phi^2) - \ifrak\epsilon \delta(t-t') \Escr(\xbf, \xbf') \phi(x) \phi(x') \right) \\ &= -\frac{1}{2} \int d^4x~d^4x' \left( \frac{\p^2}{\p x^{\mu} \p x'_{\mu}} \delta^4(x-x') + m^2 \delta^4(x-x') -\ifrak\epsilon \delta(t-t') \Escr(\xbf, \xbf') \right) \phi(x) \phi(x') \\ &= -\frac{1}{2} \int d^4x~d^4x' \left( (2\pi)^{-4} \int d^4p~e^{\ifrak p \cdot (x-x')} \left( p^2 + m^2 - \ifrak\epsilon E(\pbf) \right) \right) \phi(x) \phi(x')\end{split}\]

where in the last equality we’ve also used Eq.8.1.21.

Comparing with Eq.8.1.43 and Eq.8.1.46, we find

\[\Dscr(p) = p^2 + m^2 - \ifrak \epsilon E(\pbf)\]

and therefore the propagator

\[\Delta(x, y) = (2\pi)^{-4} \int d^4p~e^{\ifrak p \cdot (x-y)} \left( p^2 + m^2 - \ifrak\epsilon E(\pbf) \right)^{-1}\]

according to Eq.8.1.47. This recovers the Feynman propagator defined by Eq.5.2.8 and evaluated in Eq.5.2.11.

8.2. Path Integrals for Fermions#

We’ll develop the path integral formalism for fermions in parallel to the theory for bosons. The starting point is the commutation relations between Schrödinger-picture canonical variables

(8.2.1)#\[\begin{split}\{ Q_a, P_b \} &= \ifrak \delta_{ab} \\ \{ Q_a, Q_b \} &= \{ P_a, P_b \} = 0\end{split}\]

where the curly bracket denotes the anti-commutator. This is to be compared with the bosonic commutation relations Eq.8.1.1. As in the bosonic case, the indices \(a, b\) will be replaced by spacetime coordinates as we transit specifically to quantum field theory. A key difference, which will be discussed in more detail in Fermionic zero states, is that, unlike the bosonic canonical variables, the fermionic \(Q\) and \(P\) operators are not Hermitian.

8.2.1. Fermionic zero states#

It follows from Eq.8.2.1 that

\[Q_a^2 = P_a^2 = 0\]

Hence there must exist a ket-state \(\ket{0}\) and a bra-state \(\bra{0}\) such that

(8.2.2)#\[Q_a \ket{0} = \bra{0} P_a = 0\]

Indeed they can be explicitly constructed as follows

\[\begin{split}\ket{0} &\propto \left( \prod_a Q_a \right) \ket{f} \\ \bra{0} &\propto \bra{g} \left( \prod_a P_a \right)\end{split}\]

where \(\ket{f}\) and \(\bra{g}\) can be any states that makes the right-hand-sides nonzero. In particular the zero states are not in general unique. It turns out that in the absence of bosonic degrees of freedom, the zero states are unique up to a scalar, which can be chosen to satisfy the following normalization property

(8.2.3)#\[\braket{0 | 0} = 1\]

Note

The condition Eq.8.2.2 may seem a bit strange given that the fermionic \(Q\) and \(P\) operators are completely interchangeable in light of Eq.8.2.1. However, it cannot be the case that \(Q_a \ket{0} = \bra{0} Q_a = 0\) since it would imply \(\braket{0 | \{Q_a, P_b\} | 0} = 0\) in contradiction with Eq.8.2.3.

Indeed, the relationship between \(Q\) and \(P\) operators may vary. In Dirac’s theory of spin-\(1/2\) particles, we have \(Q_a^{\dagger} = -\ifrak P_a\) in light of Eq.6.4.18 and Eq.4.4.45 (cf. Eq.4.4.7 and Eq.4.4.10). In the theory of ghost field, on the other hand, the \(Q\) and \(P\) operators are not related at all.

8.2.2. Fermionic eigenstates#

In light of Eq.8.2.2, we can think \(P\) as the creation operators and \(Q\) as the annihilation operators. A complete basis of the states can then be obtained from \(\ket{0}\) by applying an arbitrary number of \(P\) operators as follows

(8.2.4)#\[\ket{a_1, a_2, \cdots, a_N} \coloneqq P_{a_1} P_{a_2} \cdots P_{a_N} \ket{0}\]

Note that the basis state is anti-symmetric in the following sense

(8.2.5)#\[\ket{a_1, \cdots, a_i, a_{i+1}, \cdots, a_N} = -\ket{a_1, \cdots, a_{i+1}, a_i, \cdots, a_N}\]

It follows from Eq.8.2.2 that

\[\begin{split}Q_a \ket{a_1, a_2, \cdots, a_N} = \begin{cases} (-1)^{k+1} \ifrak~\ket{a_1, a_2, \cdots, \hat{a}_k, \cdots, a_N} & \text{ if } a = a_k \\ 0 & \text{ if } a \notin \{a_1, a_2, \cdots, a_N\} \end{cases}\end{split}\]

where \(\hat{a}_k\) means that it’s removed from the sequence.

Similarly, the dual basis can be obtained from \(\bra{0}\) as follows

(8.2.6)#\[\bra{a_1, a_2, \cdots, a_N} \coloneqq \bra{0} (-\ifrak Q_{a_N}) \cdots (-\ifrak Q_{a_2}) (-\ifrak Q_{a_1})\]

The reason to define the dual vector this way is to realize the following normalization condition

(8.2.7)#\[\begin{split}\braket{b_1, b_2, \cdots, b_M | a_1, a_2, \cdots, a_N} &= \braket{0 | (-\ifrak Q_{b_M}) \cdots (-\ifrak Q_{b_1}) P_{a_1} \cdots P_{a_N} | 0} \\ &= \begin{cases} 0 & \text{ if } \{ b_1, b_2, \cdots, b_M \} \neq \{ a_1, a_2, \cdots, a_N \} \text{ as sets} \\ 1 & \text{ if } M=N \text{ and } b_1 = a_1, b_2 = a_2, \cdots, b_M = a_N \end{cases}\end{split}\]

The cases when \(\{ b_1, b_2, \cdots, b_M \}\) is a permutation of \(\{ a_1, a_2, \cdots, a_N \}\) can be covered using Eq.8.2.5.

The issue with the ket and bra-states defined by Eq.8.2.4 and Eq.8.2.6, respectively, is that they are not eigenstates of \(Q\) or \(P\). In fact, in sharp contrast to the bosonic case (cf. Eq.8.1.2), there cannot be any eigenstate of, say, all \(Q\) operators with nonzero (numeric) eigenvalues in the following sense

(8.2.8)#\[\begin{split}Q_a \ket{q} &= q_a \ket{q} \\ \bra{q} Q_a &= \bra{q} q_a\end{split}\]

Indeed, the fermionic commutation relation Eq.8.2.1 would demand

\[q_a q_b + q_b q_a = 0\]

which cannot be satisfied if \(q_a, q_b\) are nonzero complex numbers. It turns out that the solution to this difficulty, which may seem to be artificial, is to introduce a new set of “numbers” \(q_a\), known as Grassmann numbers which satisfy the following anti-commutation relations

\[\{ q_a, q_b \} = \{ q_a, Q_b \} = \{ q_a, P_b \} = 0\]

Now the fermionic eigenstate equation Eq.8.2.8 as well as its dual can be solved by the following

(8.2.9)#\[\begin{split}\ket{q} &\coloneqq \exp\left( -\ifrak \sum_a P_a q_a \right) \ket{0} \\ \bra{q} &\coloneqq \bra{0} \left( \prod_a Q_a \right) \exp\left( \ifrak \sum_a P_a q_a \right)\end{split}\]

where the exponential is defined using its Taylor expansion.

Warning

In the definition of \(\bra{q}\) there is a sign ambiguity depending on the ordering of the product of the \(Q\) operators.
The ket-state \(\ket{q}\) (e.g. \(\ket{0}\)) is not necessarily the adjoint of the corresponding bra-state \(\bra{q}\) (e.g. \(\bra{0}\)) since \(Q\) is not Hermitian.

Moreover the scalar product of the ket and bra \(Q\)-eigenstates can be evaluated as follow

(8.2.10)#\[\begin{split}\braket{q' | q} &= \braket{0 | \left( \prod_a Q_a \right) \exp\left( \ifrak \sum_b P_b \left( q'_b - q_b \right) \right) | 0} \\ &= \braket{0 | \left( \prod_a Q_a \right) \prod_b \left( 1 + \ifrak P_b (q'_b - q_b) \right) | 0} \\ &= \prod_a \left( q_a - q'_a \right)\end{split}\]

where in the last step, we’ve used Eq.8.2.1 to move the \(Q\) operators to the right of the \(P\) operators. Though not obvious at the moment, the right-hand-side will work as a delta function in fermionic integrals (cf. the bosonic case Eq.8.1.3).

The eigenstate of the \(P\) operators satisfying

\[\begin{split}P_a \ket{p} &= p_a \ket{p} \\ \bra{p} P_a &= \bra{p} p_a\end{split}\]

can be constructed in a way similar to Eq.8.2.9 as follows

(8.2.11)#\[\begin{split}\ket{p} &= \exp\left( -\ifrak\sum_a Q_a p_a \right)\left( \prod_a P_a \right) \ket{0} \\ \bra{p} &= \bra{0} \exp\left( \ifrak\sum_a Q_a p_a \right)\end{split}\]

where the order of the product \(\prod_a P_a\) is, by convention, the same as the one in Eq.8.2.9.

The scalar product between the \(P\)-eigenstates can be similarly evaluated to the following

(8.2.12)#\[\braket{p' | p} = \prod_a \left( p'_a - p_a \right)\]

In analogy to Eq.8.1.6, let’s calculate the scalar products between \(Q\) and \(P\)-eigenstates as follows

(8.2.13)#\[\begin{split}\braket{q | p} &= \braket{q | \exp\left( -\ifrak \sum_a Q_a p_a \right) \left( \prod_a P_a \right) | 0} \\ &= \exp\left( -\ifrak \sum_a q_a p_a \right) \braket{q | \prod_a P_a | 0} \\ &= \exp\left( -\ifrak \sum_a q_a p_a \right) \braket{0 | \left( \prod_a Q_a \right) \exp\left( \ifrak \sum_a P_a q_a \right) \left( \prod_a P_a \right) | 0} \\ &= \exp\left( -\ifrak \sum_a q_a p_a \right) \braket{0 | \left( \prod_a Q_a \right) \left( \prod_a P_a \right) | 0} \\ &= \ifrak^N (-1)^{N(N+1)/2} \exp\left( -\ifrak \sum_a q_a p_a \right)\end{split}\]

where Eq.8.2.6, Eq.8.2.7, and Eq.8.2.3 are used in the last equality. Here \(N\) is the number of \(Q_a\)s, which is the same as the number of \(P_a\)s. Similarly, but more simply, we have

(8.2.14)#\[\begin{split}\braket{p | q} &= \braket{p | \exp\left( -\ifrak\sum_a P_a q_a \right) | 0} \\ &= \exp\left( -\ifrak\sum_a p_a q_a \right) \braket{p | 0} \\ &= \exp\left( -\ifrak\sum_a p_a q_a \right) \braket{0 | \exp\left( \ifrak\sum_a Q_a p_a \right) | 0} \\ &= \exp\left( -\ifrak\sum_a p_a q_a \right)\end{split}\]

We end this section with the note that the states \(\ket{q}\) are complete in the following sense. If we expand \(\ket{q}\) in Eq.8.2.9 as a power series in products of the \(q_a\)s, then the coefficients span the whole space of states defined by Eq.8.2.4.

8.2.3. Fermionic calculus#

Since the eigenvalues of fermionic eigenstates are Grassmann numbers rather than ordinary (complex) numbers, we need a framework to do calculus, in particular integration, for functions of Grassmann variables. It turns out that the fermionic integration can be formalized as the so-called Berezin integration, which can be determined by just two rules. Writing \(\xi\) for generic Grassmann variables, the first rule consists of the evaluation of the integral on a single monomial

(8.2.15)#\[\int \left( d\xi_N \cdots d\xi_2 d\xi_1 \right) \xi_1 \xi_2 \cdots \xi_N \xi_{N+1} \cdots \xi_M = \xi_{N+1} \cdots \xi_M\]

and the second rule states that the integral is linear in both summation and multiplication by ordinary numbers. Here the ordering of the “differentials” \(d\xi_i\) in Eq.8.2.15 is made so that the integral can be evaluated in steps as follows

\[\int d\xi_N \cdots d\xi_2 d\xi_1~f(\xi) = \int d\xi_N \cdots \int d\xi_2 \int d\xi_1~f(\xi)\]

It turns out to be very convenient to introduce yet another anti-commutativity relation as follows

(8.2.16)#\[\{\xi_i, d\xi_j\} = 0\]

so we can move the integrand to the left of the “volume element” \(\prod_n d\xi_n\) at the cost of a sign. Under this convention, it’s straightforward to show that given an arbitrary function \(g(\xi')\) of Grassmann variables that are not integrated, the following two formulae hold

\[\begin{split}\int \left( \prod_n d\xi_n \right) \left( f(\xi) g(\xi') \right) &= \left( \int \left( \prod_n d\xi_n \right) f(\xi) \right) g(\xi') \\ \int g(\xi') \left( \prod_{n=1}^N d\xi_n \right) f(\xi) &= \int \left( \prod_{n=1}^N d\xi_n \right) \left( g((-1)^N \xi') f(\xi) \right) \\ &= g(\xi') \int \left( \prod_{n=1}^N d\xi_n \right) f(\xi)\end{split}\]

Another important formula in Berezin integration is to describe how the integral transforms under a (linear) change of variables. Consider the following transformation

\[\xi_n \to \xi'_n = \sum_m \Sscr_{nm} \xi_m\]

where \(\Sscr = \left( \Sscr_{nm} \right)\) is a non-singular matrix of ordinary numbers. It follows that

\[\prod_n \xi'_n = \left( \det\Sscr \right) \prod_n \xi_n\]

and henceforth

(8.2.17)#\[\int \left( \prod_n d\xi'_n \right) f = \left( \det\Sscr \right)^{-1} \int \left( \prod_n d\xi_n \right) f\]

As an application of this formalism, we’ll establish a fermionic analog of the completeness condition Eq.8.1.4. First, note that any state \(\ket{f}\) can be written as an integral

(8.2.18)#\[\ket{f} = \int \left( \prod_a dq_a \right) \ket{q} f(q)\]

where \(f(q)\) is a polynomial in the Grassmann variables \(q\).

It follows from Eq.8.2.10 and Eq.8.2.16 that

(8.2.19)#\[\begin{split}\braket{q' | f} &= \int \braket{q' | q} \left( \prod_{n=1}^N dq_n \right) f(q) \\ &= \int \left( \prod_{n=1}^N \left( q_n - q'_n \right) \right) \left( \prod_{n=1}^N dq_n \right) f(q) \\ &= (-1)^N \int \left( \prod_{n=1}^N dq_n \right) \left( \prod_{n=1}^N \left( q_n - q'_n \right) \right) f(q) \\ &= (-1)^N \int \left( \prod_{n=1}^N dq_n \right) \left( \prod_{n=1}^N \left( q_n - q'_n \right) \right) f(q') \\ &= (-1)^N f(q')\end{split}\]

where the easiest way to justify the second-to-last equality is to write \(f(q) = f(q' + (q - q'))\) and expand it in powers of \(q-q'\), so that only the zeroth order term survive due to the product \(\prod_n \left( q_n - q'_n \right)\) to the left.

Plugging Eq.8.2.19 into Eq.8.2.18 we have

(8.2.20)#\[\ket{f} = (-1)^N \int \left( \prod_{n=1}^N dq_n \right) \ket{q} \braket{q | f} \implies 1 = \int \left( \prod_a -dq_a \right) \ketbra{q}{q}\]

which is the fermionic version of Eq.8.1.4. The same calculation can be done to the \(P\)-eigenstates to get the following

(8.2.21)#\[1 = \int \left( \prod_a dp_a \right) \ketbra{p}{p}\]

8.2.4. The general path integral formula and transition to S-matrix#

Let \(H\) be the (full) Hamiltonian. Then just as in the bosonic case (cf. Eq.8.1.7), we define Heisenberg-picture operators

\[\begin{split}Q_a(t) &= e^{\ifrak Ht} Q_a e^{-\ifrak Ht} \\ P_a(t) &= e^{\ifrak Ht} P_a e^{-\ifrak Ht}\end{split}\]

with right and left-eigenstates defined as follows

\[\begin{split}\begin{alignat*}{3} \ket{t, q} &\coloneqq e^{\ifrak Ht} \ket{q}, \qquad &&\ket{t, p} &&\coloneqq e^{\ifrak Ht} \ket{p} \\ \bra{t, q} &\coloneqq \bra{q} e^{-\ifrak Ht}, \qquad &&\bra{t, p} &&\coloneqq \bra{p} e^{-\ifrak Ht} \end{alignat*}\end{split}\]

Here we recall that \(H\) must contain an even number of fermionic fields and therefore commute with any Grassmann numbers. Just as in the bosonic case Eq.8.1.10, the time-dependent eigenstates satisfy the obviously analogous time-independent scalar product formulae Eq.8.2.10, Eq.8.2.12, Eq.8.2.13, Eq.8.2.14 and orthogonality conditions Eq.8.2.20, Eq.8.2.21.

Assuming \(H = H(P, Q)\) is arranged so that all the \(P\) operators lie to the left of the \(Q\) operators, we can calculate the infinitesimal transition amplitude in parallel to the bosonic case Eq.8.1.14 (except for the ordering \(Q\) and \(P\) which is merely a matter of convenience) as follows

\[\begin{split}\braket{\tau+d\tau, q' | \tau, q} &= \braket{\tau, q' | \exp(-\ifrak H(P, Q)) d\tau | \tau, q} \\ &= \int \prod_a dp_a \braket{\tau, q' | \tau, p} \braket{\tau, p | \exp(-\ifrak H(P, Q)) d\tau | \tau, q} \\ &= \int \prod_a dp_a \braket{\tau, q' | \tau, p} \braket{\tau, p | \tau, q} \exp(-\ifrak H(p, q) d\tau) \\ &\propto \int \prod_a dp_a \exp\left( \ifrak\sum_a p_a(q'_a - q_a) - \ifrak H(p, q) d\tau \right)\end{split}\]

where in the last quantity we’ve thrown away an insignificant field-independent phase factor (coming from Eq.8.2.13), and hence the proportionality is used instead of equality.

Now given a sequence of time-ordered operators, the matrix element analogous to the bosonic Eq.8.1.16 is given by

\[\begin{split}&\braket{t', q' | T\left\{ \Oscr_A(P(t_A), Q(t_A)), \Oscr_B(P(t_B), Q(t_B)), \cdots \right\} | t, q} \\ &\quad \propto \int_{\substack{q_a(t)=q_a} \\ q_a(t')=q'_a} \prod_{\tau, a} dq_a(\tau) dp_a(\tau)~\Oscr_A(p(t_A), q(t_A)) \Oscr_B(p(t_B), q(t_B)) \cdots \\ &\qquad \times \exp\left( \ifrak\int_t^{t'} d\tau \left( -H(p(\tau), q(\tau)) + \sum_a p_a(\tau) \dot{q}_a(\tau) \right) \right)\end{split}\]

Note that, unlike the bosonic case, an extra sign is added to each permutation of the fermionic operators demanded by the time-ordering operator \(T\).

Transitioning to quantum field theory, we get the fermionic analog of Eq.8.1.24 as follows

(8.2.22)#\[\begin{split}&\braket{\VAC, \op{out} | T\left\{ \Oscr_A(P(t_A), Q(t_B)), \Oscr_B(P(t_B), Q(t_B)), \cdots \right\} | \VAC, \op{in}} \\ &\quad \propto \int \prod_{\tau, \xbf, m} dq_m(\tau, \xbf) \prod_{\tau, \xbf, m} dp_m(\tau, \xbf)~\Oscr_A(p(t_A), q(t_B)) \Oscr_B(p(t_B), q(t_B)) \cdots \\ &\qquad \times \exp\left( \ifrak \int_{-\infty}^{\infty} d\tau \left( -H(p(\tau), q(\tau)) + \int d^3x \sum_m p_m(\tau, \xbf) \dot{q}_m(\tau, \xbf) + \ifrak \epsilon \text{ terms} \right) \right)\end{split}\]

where we’ve, once again, omitted the details of the \(\ifrak\epsilon\) terms. As in the bosonic case, they come from the vacuum wave functions (cf. Eq.8.1.23).

The next step in the bosonic case, as discussed in Lagrangian version of the path integral, is to assume that the Hamiltonian \(H(p, q)\) is quadratic in \(p\) and hence the integral in \(p\)-fields can be evaluated using the Gaussian integral formula (cf. Eq.8.1.31). This is not the case for fermions. Unlike the bosonic case (cf. Eq.8.1.28), the canonical conjugate \(p\) is unrelated to \(\dot{q}\). In fact, for each fermion that carries a nonzero quantum number, there are equal number of \(p\)s and \(q\)s. In particular, the free Hamiltonian \(H_0\) is bilinear in \(p\) and \(q\) so that

(8.2.23)#\[\begin{split}&\int_{-\infty}^{\infty} d\tau \left( -H_0(p(\tau), q(\tau)) + \int d^3x \sum_m p_m(\tau, \xbf) \dot{q}_m(\tau, \xbf) + \ifrak\epsilon\text{ terms} \right) \\ &\quad = -\sum_{m, n} \int d^4x~d^4y~\Dscr_{x m, y n} p_m(x) q_n(y)\end{split}\]

where \(\Dscr\) is a matrix in ordinary numbers.

Expanding both the time-ordered operators and the interaction Hamiltonian \(V = H - H_0\) in a power series in \(p\)s and \(q\)s, the right-hand-side of Eq.8.2.22 becomes a sum of Gaussian-like integrals and can be evaluated as follows

(8.2.24)#\[\begin{split}&\Iscr_{m_1 n_1 m_2 n_2 \cdots m_N n_N}(x_1, y_1, x_2, y_2, \cdots, x_N, y_N) \\ &\quad = \int \prod_{\tau, \xbf, m} dq_m(\tau, \xbf) \prod_{\tau, \xbf, m} dp_m(\tau, \xbf) p_{m_1}(x_1) q_{n_1}(y_1) \cdots p_{m_N}(x_N) q_{n_N}(y_N) \\ &\qquad \times \exp\left( -\ifrak \sum_{m, n} \int d^4x~d^4y~\Dscr_{xm, yn} p_m(x) q_n(y) \right) \\ &\quad \propto \sum_{\sigma \in \op{Perm}_N} (-1)^{\op{sign}(\sigma)} \prod_{k=1}^N \left( -\ifrak \Dscr^{-1} \right)_{x_k m_k,~y_{\sigma(k)} m_{\sigma(k)}}\end{split}\]

where \(\op{Perm}_N\) denotes the permutation group of \(N\) symbols. This reproduces the Feynman rules if we regard \(\left(\Dscr^{-1}\right)_{xm, yn}\) as the propagator between \(p_m(x)\) and \(q_n(y)\).

Derivation of the integral evaluation in Eq.8.2.24

Consider a generating function defined as follows

(8.2.25)#\[\begin{split}\Iscr(f, g) &\coloneqq \int \prod_{\tau, \xbf, m} dq_m(\tau, \xbf) \prod_{\tau, \xbf, m} dp_m(\tau, \xbf) \\ &\quad \times \exp\left( -\ifrak\sum_{m, n} \int d^4x~d^4y~\Dscr_{xm, yn} p_m(x) q_n(y) \right. \\ &\qquad \left. -\ifrak\sum_m \int d^4x~p_m(x) f_m(x) - \ifrak\sum_n \int d^4y~g_n(y) q_n(y) \right)\end{split}\]

where \(f, g\) are generic Grassmann numbers. Next, consider the following (translational) change of variables

\[\begin{split}p'_m(x) &= p_m(x) - \sum_n \int d^4y~g_n(y) \left( \Dscr^{-1} \right)_{yn, xm} \\ q'_n(y) &= q_n(y) - \sum_m \int d^4x \left( \Dscr^{-1} \right)_{yn, xm} f_m(x)\end{split}\]

Obviously the integral doesn’t change by this change of variables. Let’s calculate the change of exponential power in Eq.8.2.25 (without the common factor \(-\ifrak\)) as follows

\[\begin{split}&\sum_{m,n} \int d^4x~d^4y~\Dscr_{xm,yn} p'_m(x) q'_n(y) + \sum_m \int d^4x~p'_m(x) f_m(x) + \sum_n \int d^4y~g_n(y) q'_n(y) \\ &= \sum_{m,n} \int d^4x~d^4y~\Dscr_{xm,yn} p_m(x) q_n(y) - \sum_{m,n,k} \int d^4x~d^4y~d^4w~\Dscr_{xm,yn} p_m(x) \left(\Dscr^{-1}\right)_{yn,wk} f_k(w) \\ &\quad - \sum_{m,n,k} \int d^4x~d^4y~d^4w~\Dscr_{xm,yn} g_k(w) \left(\Dscr^{-1}\right)_{wk,xm} q_n(y) \\ &\quad + \sum_{m,n,k,r} \int d^4x~d^4y~d^4w~d^4v~\Dscr_{xm,yn} g_k(w) \left(\Dscr^{-1}\right)_{wk,xm} \left(\Dscr^{-1}\right)_{yn,vr} f_r(v) \\ &\quad + \sum_m \int d^4x~p_m(x) f_m(x) - \sum_{m,n} \int d^4x~d^4y~g_n(y) \left(\Dscr^{-1}\right)_{yn,xm} f_m(x) \\ &\quad + \sum_n \int d^4y~g_n(y) q_n(y) - \sum_{m,n} \int d^4x~d^4y~g_n(y) \left(\Dscr^{-1}\right)_{yn,xm} f_m(x) \\ &= \sum_{m,n} \int d^4x~d^4y~\Dscr_{xm,yn} p_m(x) q_n(y) - \sum_{m,n} \int d^4x~d^4y~\left(\Dscr^{-1}\right)_{yn,xm} g_n(y) f_m(x)\end{split}\]

It follows that Eq.8.2.25 can be rewritten as follows

(8.2.26)#\[\begin{split}\Iscr(f, g) &= \exp\left( \ifrak \sum_{m,n} \int d^4x~d^4y \left(\Dscr^{-1}\right)_{yn,xm} g_n(y) f_m(x) \right) \\ &\times \int \prod_{\tau, \xbf, m} dq_m(\tau, \xbf) \prod_{\tau, \xbf, m} dp_m(\tau, \xbf) \\ &\times \exp\left( -\ifrak \sum_{m,n} \int d^4x~d^4y~\Dscr_{xm,yn} p_m(x) q_n(y) \right)\end{split}\]

Now the integral Eq.8.2.25 is called a generating function since if the exponential is expanded as a power series in \(f\) and \(g\), then the coefficient of

(8.2.27)#\[f_{m_1} g_{n_1} f_{m_2} g_{n_2} \cdots f_{m_N} g_{n_N}\]

is, up to a constant factor independent of \(x,y,m,n\), precisely the integral Eq.8.2.24. Expanding Eq.8.2.26 also in a power series of \(f\) and \(g\), the coefficient of Eq.8.2.27 then gives the final result in Eq.8.2.24.

As an example, consider the free-particle action of spin-\(1/2\) particles as follows (cf. Eq.6.4.17 and Eq.6.4.19)

\[\begin{split}&\int_{-\infty}^{\infty} d\tau \left( -H_0(p(\tau), q(\tau)) + \int d^3x \sum_m p_m(\tau, \xbf) \dot{q}_m(\tau, \xbf) \right) \\ &\quad = -\int d^4x~\bar{\psi}(x) \left( \gamma^{\mu} \p_{\mu} + M \right) \psi(x)\end{split}\]

The canonical variables are given by (cf. Eq.6.4.18)

\[\begin{split}q_m(x) &= \psi_m(x) \\ p_m(x) &= -\left( \bar{\psi}(x) \gamma^0 \right)_m\end{split}\]

Comparing with Eq.8.2.23 we have

\[\begin{split}\Dscr_{xm, yn} &= \left( \gamma^0 \left( \gamma^{\mu} \p_{x_{\mu}} + M - \ifrak\epsilon \right) \right)_{mn} \delta^4(x-y) \\ &= \int \frac{d^4 k}{(2\pi)^4} \left( \gamma^0 \left( \ifrak \gamma^{\mu} k_{\mu} + M -\ifrak\epsilon \right) \right)_{mn} e^{\ifrak k \cdot (x-y)}\end{split}\]

where we’ve omitted the details about the \(\ifrak\epsilon\) term since it’s not important here and can be worked out as the vacuum wave function in the same way as for (bosonic) scalar field (cf. Eq.8.1.22). It follows that the propagator can be written as follows

\[\begin{split}\left(\Dscr^{-1}\right)_{xm, yn} &= \int \frac{d^4k}{(2\pi)^4} \left( \left( \ifrak\gamma^{\mu} k_{\mu} + M -\ifrak\epsilon \right)^{-1} (-\gamma^0) \right)_{mn} e^{\ifrak k \cdot (x-y)} \\ &= \int \frac{d^4k}{(2\pi)^4} \frac{\left( \left( -\gamma^{\mu} k_{\mu} + M \right) (-\gamma^0) \right)_{mn}}{k^2 + M^2 - \ifrak\epsilon} e^{\ifrak k \cdot (x-y)}\end{split}\]

which recovers Eq.5.2.13 and Eq.5.2.5, except for that we’ve calculated here the propagator between \(\psi\) and \(-\bar{\psi} \gamma^0\), rather than \(\psi^{\dagger}\).

To illustrate how path-integral method works in this case, let’s consider the following interacting Lagrangian density

\[\Lscr = -\bar{\psi} \left( \gamma^{\mu} \p_{\mu} + M + \Gamma \right) \psi\]

where \(\Gamma(x)\) represents the interaction of the fermion with an external field. Applying Eq.8.2.22, we can calculate the vacuum persistence amplitude (in matrix notation) as follows

(8.2.28)#\[\begin{split}\braket{\VAC, \op{out} | \VAC, \op{in}}_{\Gamma} &\propto \int \prod_{\tau, \xbf, m} dq_m(\tau, \xbf) \prod_{\tau, \xbf, m} dp_m(\tau, \xbf, m) \\ &\quad \times \exp\left( -\ifrak \int d^4x~p^T \gamma^0 (\gamma^{\mu} \p_{\mu} + M + \Gamma - \ifrak\epsilon) q \right) \\ &= \int \prod_{\tau, \xbf, m} dq_m(\tau, \xbf) \prod_{\tau, \xbf, m} dp_m(\tau, \xbf, m) \\ &\quad \times \exp\left( -\ifrak \sum_{m, n} \int d^4x~d^4y~p_m(x) q_n(y) \Kscr(\Gamma)_{xm, yn} \right)\end{split}\]

where

(8.2.29)#\[\Kscr(\Gamma)_{xm, yn} \coloneqq \left( \gamma^0 \left( \gamma^{\mu} \p_{x_{\mu}} + M + \Gamma(x) - \ifrak\epsilon \right) \right)_{mn} \delta^4(x-y)\]

Observe that the following change of variables

\[q'_m(x) \coloneqq \sum_n \int d^4y~\Kscr(\Gamma)_{xm, yn} q_n(y)\]

will make the integral in Eq.8.2.28 independent of \(\Gamma\). Hence it follows from Eq.8.2.17 that

(8.2.30)#\[\braket{\VAC, \op{out} | \VAC, \op{in}}_{\Gamma} \propto \det\Kscr(\Gamma)\]

To see that this calculation is consistent with the Feynman rules derived in Momentum Space Feynman Rules, rewrite Eq.8.2.29 as follows

\[\Kscr(\Gamma) = \Dscr + \Gscr(\Gamma)\]

where

\[\Gscr(\Gamma)_{xm, yn} \coloneqq \left( \gamma^0 \Gamma(x) \right)_{mn} \delta^4(x-y)\]

It follows then (cf. Eq.8.1.33)

\[\begin{split}\braket{\VAC, \op{out} | \VAC, \op{in}}_{\Gamma} &\propto \det\left( \Dscr \left( 1 + \Dscr^{-1} \Gscr(\Gamma) \right) \right) \\ &= \left(\det\Dscr\right) \exp\left( \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \Tr\left( \Dscr^{-1} \Gscr(\Gamma) \right) \right)\end{split}\]

which is exactly what one would expect from a Feynman diagram calculation. Indeed, the Feynman rules demand that each vertex contributes a factor \(-\ifrak\Gscr(\Gamma)\) and each (internal) edge contributes a factor \(-\ifrak\Dscr\). In this example all connected Feynman diagram are simply loops. A loop with \(n\) vertices then contributes a factor \(1/n\) accounting for the cyclic symmetry. Finally the extra sign in \((-1)^{n+1}\) accounts for the fermionic loop.

It turns out that path integral calculations like Eq.8.2.30 do much more than just recovering perturbative calculations. But we’ll only come back to this much later.

8.2.5. Path-integral formulation of QED#

In Quantum Electrodynamics, we derived the photon propagator Eq.7.5.4 in a rather cumbersome way by working with non-Lorentz-invariant \(A\)-fields (cf. Eq.7.4.7 and Eq.7.4.8). In this section, we will re-derive Eq.7.5.4 in a cleaner manner using path integrals.

Recall from Eq.7.3.10 that the QED Hamiltonian can be written as follows

\[H = \int d^3x \left( \frac{1}{2} \bm{\Pi}^2_{\bot} + \frac{1}{2} (\nabla \times \Abf)^2 - \Jbf \cdot \Abf \right) + V_{\op{Coul}} + H_{\op{matter}}\]

where the Coulomb potential \(V_{\op{Coul}}\) is given by Eq.7.3.11, under the Coulomb gauge condition

(8.2.31)#\[\nabla \cdot \Abf = 0\]

and the constraint (cf. Eq.7.3.6)

(8.2.32)#\[\nabla \cdot \bm{\Pi}_{\bot} = 0\]

The general path integral calculation of the vacuum expectation value of a timed-ordered product of operators, whether the bosonic Eq.8.1.24 or the fermionic Eq.8.2.22 can be applied to the QED Hamiltonian to give the following

(8.2.33)#\[\begin{split}&\braket{T\{\Oscr_A \Oscr_B \cdots\}}_{\VAC} = \int \prod_{x, i} da_i(x) \prod_{x, i} d\pi_i(x) \prod_{x, \ell} d\psi_{\ell}(x)~\Oscr_A \Oscr_B \cdots \\ &\qquad \times \exp\left( \ifrak \int d^4x \left( \bm{\pi} \cdot \dot{\abf} - \frac{1}{2} \bm{\pi}^2 - \frac{1}{2}(\nabla \times \abf)^2 + \jbf \cdot \abf + \Lscr_M \right) - \ifrak \int dt~V_{\op{Coul}}(t) \right) \\ &\qquad \times \left( \prod_x \delta(\nabla \cdot \abf(x)) \right) \left( \prod_x \delta(\nabla \cdot \bm{\pi}(x)) \right)\end{split}\]

where the lowercase fields represent simply variables of integration, rather than interaction-picture fields as in Quantum Electrodynamics, the index \(i\) runs through the spatial \(1,2,3\) as usual, the matter field variables are represented by \(\psi_{\ell}\), and the last two delta functions are introduced to enforce the constraints Eq.8.2.31 and Eq.8.2.32.

First thing to observe is that since \(\Lscr_M\) doesn’t involve \(\bm{\pi}\) (cf. Eq.7.6.1 and Eq.6.4.17) and the power of the exponential is quadratic in \(\bm{\pi}\), it can be integrated out by the Gaussian integral formula Eq.8.1.27. More explicitly, it amounts to substitute \(\bm{\pi}\) with the solution to the stationary point of the quadratic power, which is \(\bm{\pi} = \dot{\abf}\), so we can rewrite Eq.8.2.33 as follows

(8.2.34)#\[\begin{split}&\braket{T\{\Oscr_A \Oscr_B \cdots\}}_{\VAC} = \int \prod_{x, i} da_i(x) \prod_{x, \ell} d\psi_{\ell}(x) \Oscr_A \Oscr_B \cdots \\ &\qquad \times \exp\left( \ifrak \int d^4x \left( \frac{1}{2} \dot{\abf}^2 - \frac{1}{2} \bm{\pi}^2 - \frac{1}{2} (\nabla \times \abf)^2 + \jbf \cdot \abf + \Lscr_M \right) - \ifrak \int dt~V_{\op{Coul}}(t) \right) \\ &\qquad \times \left( \prod_x \delta(\nabla \cdot \abf(x)) \right)\end{split}\]

Next we’d like to restore manifest Lorentz invariance by integrating also over \(a_0\). The trick is to consider the following quantity

(8.2.35)#\[\int d^4x \left( -a_0(x) j_0(x) + \frac{1}{2} \left(\nabla a_0(x)\right)^2 \right)\]

whose exponential’s (path) integral over \(a_0(x)\) can be done by setting \(a_0(x)\) to the stationary point. In other words \(a_0(x)\) should solve the following differential equation

\[j_0(x) + \nabla^2 a_0(x) = 0\]

This is a rather familiar equation (cf. Eq.7.2.6), whose solution, given by Eq.7.2.7, is reproduced here

\[a_0(t, \xbf) = \int d^3y \frac{j_0(t, \ybf)}{4\pi|\xbf-\ybf|}\]

Plugging it back to Eq.8.2.35, we get nothing but the Coulomb action. It follows that the integral in the exponential in Eq.8.2.34 can be rewritten as follows (cf. Eq.7.3.3)

\[\int d^4x \left( \frac{1}{2} \dot{\abf}^2 - \frac{1}{2} \bm{\pi}^2 - \frac{1}{2} (\nabla \times \abf)^2 + \jbf \cdot \abf + \Lscr_M - a_0 j_0 + \frac{1}{2} \left(\nabla a_0\right)^2 \right)\]