7. Quantum Electrodynamics

We saw in The Canonical Formalism how Lagrangian (densities) may be guessed, passed to the interaction picture, and verified against the free fields derived from Lorentz invariance and causality in Quantum Fields and Antiparticles. There were, however, cases we couldn’t handle back in Massless Fields, which include, in particular, massless vector fields. It is this difficulty we will try to handle in this section, which, as it turns out, leads to the most successful quantum field theory to date, namely, the quantum electrodynamics (QED).

The plan to quantize electrodynamics is the following. First, we’ll show that the failed attempt to write down a free massless helicity-\(1\) vector field as Eq.4.7.12 can be remedied by the introduction of a gauge symmetry. The introduction of gauge symmetry turns out to also introduce constraints on the canonical variables, which we’ll try to fix using Dirac’s method discussed in Constraints and Dirac Brackets. Once the constraints have been dealt with, we pass to the interaction picture and write down the Hamiltonian as before. With the propagators calculated using the appropriate commutation relations, the machinery of Feynman diagrams may be applied to calculate S-matrix elements. This is done first theoretically by writing down a recipe, and then applied explicitly to one example – the famous Compton scattering.

7.1. Gauge Invariance and Action

Recall from The failure for vector fields that we’ve tried to write down a quantum vector field \(a_{\mu}(x)\) for massless helicity-\(1\) particle in Eq.4.7.12. The issue, which stems from the obviously non-Lorentz-invariant gauge fixing condition Eq.4.7.14, is that \(a_{\mu}(x)\) fails to be covariant in light of Eq.4.7.18.

From perspective of Lagrangian formalism, however, it’s completely fine that the field itself is not covariant, as long as the action is Lorentz invariant. This consideration motivates the introduction of the co-called mass action \(I_M\), which is required to be invariant under the gauge transformation

(7.1.1)\[A_{\mu}(x) \to A_{\mu}(x) + \p_{\mu} \epsilon(x)\]

Here we switch to the uppercase \(A_{\mu}\) to indicate that it is considered to be a Heisenberg-picture field, i.e., a field that knows everything about the interactions. Note also that an explicit formula for \(I_M\) will only be worked out much later, after we formally apply all sorts of symmetry principles.

Now the variation of \(I_M\) with respect to \(A_{\mu}(x)\) can be written as follows

\[\delta I_M = \int d^4 x \frac{\delta I_M}{\delta A_{\mu}(x)} \p_{\mu} \epsilon(x)\]

It follows from integration-by-parts that in order for \(\delta I_M = 0\), one must have

(7.1.2)\[\p_{\mu} \frac{\delta I_M}{\delta A_{\mu}(x)} = 0\]

Note that this is trivially true if \(I_M\) is a functional of

(7.1.3)\[F_{\mu \nu} = \p_{\mu} A_{\nu} - \p_{\nu} A_{\mu}\]

and its derivatives.

To construct an \(I_M\) that satisfies Eq.7.1.2, we need to couple \(A_{\mu}(x)\) to a conserved \(4\)-current \(J^{\mu}(x)\) in the sense that \(\p_{\mu} J^{\mu} = 0\). This will guarantee

(7.1.4)\[\frac{\delta I_M}{\delta A_{\mu}(x)} = J^{\mu}(x)\]

and hence Eq.7.1.2. Here we’ve secretly absorbed a potential proportional constant into the definition of \(J^{\mu}\), which, in turn, will be absorbed into the definition of charges will be discussed below.

Now recall from Global symmetries that such conserved \(4\)-current can be obtained by assuming symmetries on \(I_M\) in the form of Eq.6.3.3. In QED, we’re specifically interested in variations of the following form

\[\delta \Psi_{\ell}(x) = \ifrak \epsilon(x) q_{\ell} \Psi_{\ell}(x)\]

which is diagonalized into charges \(q_{\ell}\). Following discussions in Lagrangian density preserving symmetry, if we assume

\[I_M = \int d^4 x~\Lscr_M(\Psi_{\ell}(x), \p_{\mu} \Psi_{\ell}(x))\]

such that \(\Lscr_M\) is invariant under \(\Psi_{\ell} \to \Psi_{\ell} + \delta \Psi_{\ell}\) with a constant \(\epsilon(x) = \epsilon\). Then the conserved current, according to Eq.6.3.9, takes the following form

(7.1.5)\[J^{\mu}(x) = -\ifrak \sum_{\ell} \frac{\p \Lscr_M}{\p (\p_{\mu} \Psi_{\ell}(x))} q_{\ell} \Psi_{\ell}(x)\]

Let

(7.1.6)\[Q \coloneqq \int d^3 x~J^0\]

be the conserved charge operator. Then it follows from Eq.6.3.11 (modulo an extremely confusing shuffle of notations)

\[\left[ Q, \Psi_{\ell}(x) \right] = -q_{\ell} \Psi_{\ell}(x)\]

where \(\Psi_{\ell}\) is now officially one of the canonical variables.

To summarize, we have concluded that the matter action \(I_M\) is invariant under the joint (local) transformation

(7.1.7)\[\begin{split}\delta A_{\mu}(x) &= \p_{\mu} \epsilon(x) \\ \delta \Psi_{\ell}(x) &= \ifrak \epsilon(x) q_{\ell} \Psi_{\ell}(x)\end{split}\]

on \(A_{\mu}(x)\) and \(\Psi_{\ell}(x)\). Symmetries like this are known as gauge symmetries.

It turns out that in QED, besides the matter action, we also need a light action which takes the following form

\[I_{\gamma} = -\frac{1}{4} \int d^4 x~F_{\mu \nu} F^{\mu \nu}\]

It’s hard to argue logically for why such a term is needed, but it turns out to work, and it appears also in the Lagrangian density Eq.6.4.6 of massive spin-\(1\) vector fields.

Combining the mass and light actions together, we see that the field equation is given by the variational principle as follows

\[0 = \frac{\delta (I_M + I_{\gamma})}{\delta A_{\nu}} = \p_{\mu} F^{\mu \nu} + J^{\nu}\]

which is recognized as the inhomogeneous Maxwell equations. The homogeneous Maxwell equations

\[\p_{\mu} F_{\nu \kappa} + \p_{\nu} F_{\kappa \mu} + \p_{\kappa} F_{\mu \nu} = 0\]

follows readily from the definition Eq.7.1.3.

To recap our (or rather, Weinberg’s) approach to QED, it starts with a partially successful construction of a massless helicity-\(1\) vector field \(a_{\mu}(x)\) from Eq.4.7.12. Its transformation law Eq.4.7.18, abstracted into Eq.7.1.1, necessitates a conserved coupling field \(J^{\mu}\), which then comes out of a variational principle. In the end, we’ve derived the gauge symmetry Eq.7.1.7 as a consequence of the need of a Lorentz invariant Lagrangian density.

It’s very interesting that the above argument can be totally reverted, as is done in most textbooks. We outline this reversed argument in the dropdown below.

Gauge theoretic approach to the QED Lagrangian

Suppose the particles of interest have some global internal symmetry (e.g. electric charge) of the following form

\[\delta \Psi_{\ell}(x) = \ifrak \epsilon q_{\ell} \Psi_{\ell}(x)\]

with constant \(\epsilon\). The starting point then is the promise to promote these global symmetries to local ones, i.e., gauge symmetries, of the following form

\[\delta \Psi_{\ell}(x) = \ifrak \epsilon(x) q_{\ell} \Psi_{\ell}(x)\]

with function \(\epsilon(x)\).

Now if the Lagrangian involves any derivatives of the fields \(\Psi_{\ell}(x)\), which is almost certainly the case in reality, then such promotion cannot be done for free because of the following simple rule of differentiation

(7.1.8)\[\delta \p_{\mu} \Psi_{\ell}(x) = \ifrak \epsilon(x) q_{\ell} \p_{\mu} \Psi_{\ell}(x) + \ifrak q_{\ell} \Psi_{\ell}(x) \p_{\mu} \epsilon(x)\]

The cure to this problem is to “invent” a vector field \(A_{\mu}(x)\) with transformation rule

\[\delta A_{\mu}(x) = \p_{\mu} \epsilon(x)\]

and observe that if we introduce the following covariant derivative

(7.1.9)\[D_{\mu} \Psi_{\ell} \coloneqq \p_{\mu} \Psi_{\ell} - \ifrak q_{\ell} A_{\mu} \Psi_{\ell}\]

then Eq.7.1.8 can be replaced by the following

\[\begin{split}\delta D_{\mu} \Psi_{\ell}(x) &= \delta \p_{\mu} \Psi_{\ell}(x) - \ifrak q_{\ell} \left( A_{\mu}(x) \delta \Psi_{\ell}(x) + \Psi_{\ell}(x) \delta A_{\mu}(x) \right) \\ &= \ifrak \epsilon(x) q_{\ell} D_{\mu} \Psi_{\ell}(x)\end{split}\]

It follows that the gauge symmetry can be restored if any dependence of the matter Lagrangian \(\Lscr_M\) on \(\p_{\mu} \Psi_{\ell}\) actually depends on the covariant derivative Eq.7.1.9. In other words, we can write

(7.1.10)\[\Lscr_M = \Lscr_M(\Psi, D_{\mu} \Psi)\]

We can also verify Eq.7.1.4 and Eq.7.1.5 in this setup as follows

\[\begin{split}\frac{\delta I_M}{\delta A_{\mu}(x)} &= \int d^4 y~\frac{\p \Lscr_M(\Psi_{\ell}(y), D_{\mu} \Psi_{\ell}(y))}{\p A_{\mu}(x)} \\ &= \sum_{\ell} \frac{\p \Lscr_M}{\p D_{\mu} \Psi_{\ell}} \left( -\ifrak q_{\ell} \Psi_{\ell}(x) \right) \\ &= -\ifrak \sum_{\ell} \frac{\p \Lscr_M}{\p (\p_{\mu} \Psi_{\ell})} q_{\ell} \Psi_{\ell}(x)\end{split}\]

Comparing with the Lagrangian density of massive spin-\(1\) vector field Eq.6.4.6, we see that the following term

\[-\frac{1}{2} m^2 A_{\mu} A^{\mu}\]

is missing, or \(m=0\). Indeed, the appearance of such quadratic term would obviously violate the postulated gauge symmetry. This is really fantastic since we’ve deduced from just the existence of gauge symmetry that the particle represented by \(A_{\mu}\), which turns out to be photon, is massless. This is in fact one the two greatest predictions of QED that photons are massless – the other one being that they travel at the speed of light, and has been tested against experiment up to great precision.

7.2. Constraints and Gauge Conditions

Recall from the previous section that the QED Lagrangian density splits up into the matter part and the light parts as follows

\[\Lscr = \Lscr_M + \Lscr_{\gamma}\]

where

(7.2.1)\[\Lscr_{\gamma} = -\frac{1}{4} F_{\mu \nu} F^{\mu \nu}\]

and \(\Lscr_M\) involves both \(\Psi\)-field and \(A\)-field. Moreover, in light of Eq.7.1.9 and Eq.7.1.10, there is no involvement of derivatives of \(A\)-field in \(\Lscr_M\). It follows that if we define the field conjugate to \(A_{\mu}\) as follows

(7.2.2)\[\Pi^{\mu} \coloneqq \frac{\p \Lscr}{\p \dot{A}_{\mu}}\]

then just as in the theory of (massive) spin-\(1\) vector fields, we get the following primary constraint

(7.2.3)\[\Pi^0 = \frac{\p \Lscr_{\gamma}}{\p \dot{A}_{\mu}} = \frac{\p \Lscr_{\gamma}}{\p F_{00}} = 0\]

Moreover we use the Euler-Lagrange equation to find the secondary constraint as follows

(7.2.4)\[\p_i \Pi^i = -\p_i \frac{\p \Lscr}{\p F_{i0}} = -\frac{\p \Lscr}{\p A_0} = -J^0\]

where the middle equality is one of the Euler-Lagrange equations Eq.6.2.5 with the time-derivative term dropped out due to the primary constraint, the equality on the left is the definition Eq.7.2.2, and the equality on the right follows from Eq.7.1.4.

Using Eq.7.1.5 and letting \(Q \coloneqq \Psi\) (not to be confused with the charge operator Eq.7.1.6 defined in the previous section) and \(P \coloneqq \delta \Lscr / \delta \dot{\Psi}\) be the conjugate canonical variables, we can write the charge density \(J^0\) as follows

(7.2.5)\[J^0 = -\ifrak \sum_n \frac{\p \Lscr}{\p \dot{\Psi}_n} q_n \Psi_n = -\ifrak \sum_n P^n q_n Q_n\]

Together with Eq.7.2.3 and Eq.7.2.4, we have obtained two constraints

\[\begin{split}\chi_1 &\coloneqq \Pi^0 = 0 \\ \chi_2 &\coloneqq \p_i \Pi^i - \ifrak \sum_n P^n q_n Q_n = 0\end{split}\]

Obviously the Poisson bracket \([\chi_1, \chi_2]_P = 0\). Hence the constraints are of first class as discussed in First class constraints. Moreover, one cannot hope to solve explicitly for \(A_0\), the conjugate of \(\Pi^0\), in terms of the other canonical variables, as we did in the massive case Eq.6.4.9, due to the presence of gauge symmetry Eq.7.1.7.

One way, which is unfortunately not Lorentz invariant, to solve this problem is to “fix the gauge” by imposing an artificial condition on the \(A\)-field so that \(A_0\) may be solved explicitly. [1] It turns out that there are many such conditions that find their use cases under various circumstances. Some of the most common gauge-fixing conditions are listed below

Name	Condition
Lorenz gauge [2]	\(\p_{\mu} A^{\mu} = 0\)
Coulomb gauge	\(\nabla \cdot \Abf = 0\)
Temporal gauge	\(A^0 = 0\)
Axial gauge	\(A^3 = 0\)

Important

We will use the Coulomb gauge unless otherwise specified throughout this chapter. The reason, according to J. Schwinger, is that it is in this gauge that photons come out as helicity-\(1\) particles.

Two things remain to be settled. One is to argue that the \(A\)-field can always be put into the Coulomb gauge, and the other is to solve for \(A^0\) in terms of the other fields.

To address the first point, it suffices, according to Eq.7.1.7, to show that for any \(A = (A_0, \Abf)\), there exists a function \(\lambda(\xbf)\) such that

\[\nabla \cdot (\Abf + \nabla \cdot \lambda) = 0 \iff \nabla^2 \lambda = -\nabla \cdot \Abf\]

The equation on the right is known as Poisson’s equation and can be solved using Green’s function. In fact, the same equation shows up again when solving for \(A^0\), which we now explain.

Combining the secondary constraint Eq.7.2.4 (cf. Eq.7.2.1) with the Coulomb gauge condition, we have

(7.2.6)\[J^0 = \p_i \frac{\delta \Lscr}{\p F_{i0}} = \p_i \frac{\delta \Lscr_{\gamma}}{\p F_{i0}} = -\p_i F^{i0} = -\nabla^2 A^0\]

This equation can be solved explicitly [3] by

(7.2.7)\[A^0(t, \xbf) = \int d^3 y~\frac{J^0(t, \ybf)}{4\pi |\xbf - \ybf|}\]

where the charge density \(J^0\) can be further written in the form of Eq.7.2.5.

Now Eq.7.2.3 and Eq.7.2.7 allow us to get rid of the constrained canonical variables \(A_0\) and \(\Pi^0\), which, as we will show in the next section, removes the first class constraints.

7.3. Quantization in Coulomb Gauge

In the previous section, we imposed a Coulomb gauge condition \(\nabla \cdot \Abf = 0\) and used it to eliminate the constrained variables \(A_0\) and \(\Pi^0\). Now we’re facing a new pair (of families parametrized by spatial coordinates \(\xbf\)) of constraints, listed as follows

(7.3.1)\[\begin{split}\chi_{1 \xbf} &\coloneqq \p_i A^i(\xbf) = 0 \\ \chi_{2 \xbf} &\coloneqq \p_i \Pi^i(\xbf) + J^0(\xbf) = 0\end{split}\]

where we also remember that \(J^0\) can be further expressed in terms of canonical variables as in Eq.7.2.5.

The \(C\) matrix as defined by Eq.6.5.4 is given in this case by

\[\begin{split}C = \begin{bmatrix*} 0 & -\nabla^2 \delta^3(\xbf - \ybf) \\ \nabla^2 \delta^3(\xbf - \ybf) & 0 \end{bmatrix*}\end{split}\]

where, for example, the upper-right entry \(C_{1\xbf, 2\ybf}\) may be calculated as follows

\[\begin{split}C_{1\xbf, 2\ybf} &= [\chi_{1\xbf}, \chi_{2\ybf}]_P \\ &= \int d^3 \zbf \left( \frac{\p (\p_i A^i(\xbf))}{\p A^k(\zbf)} \frac{\p (\p_j \Pi^j(\ybf) + J^0(\ybf))}{\p \Pi_k(\zbf)} - \xbf \leftrightarrow \ybf \right) \\ &= \int d^3 \zbf~\p_i \left( \frac{\p A^i(\xbf)}{\p A^k(\zbf)} \right) \p_j \left( \frac{\p \Pi^j(\ybf)}{\p \Pi_k(\zbf)} \right) \\ &= \int d^3 \zbf~\delta^i_k \delta^{jk} \p_i \delta^3(\xbf - \zbf) \p_j \delta^3(\ybf - \zbf) \\ &= -\nabla^2 \delta^3(\xbf - \ybf)\end{split}\]

Clearly \(C\) is non-singular, and therefore the constraints Eq.7.3.1 are of second class. To apply Dirac’s method, we need the inverse matrix \(C^{-1}\) given by

\[\begin{split}C^{-1} = \begin{bmatrix*} 0 & -\frac{1}{4\pi |\xbf - \ybf|} \\ \frac{1}{4\pi |\xbf - \ybf|} & 0 \end{bmatrix*}\end{split}\]

Indeed, one easily verifies that \(C C^{-1} = 1\) by, for example, the following calculation

\[\int d^3 \zbf~C_{1\xbf, 2\zbf} (C^{-1})_{2\zbf, 1\ybf} = -\int d^3 \zbf~\frac{\nabla^2 \delta^3(\xbf - \zbf)}{4\pi |\zbf - \ybf|} = \delta^3(\xbf - \ybf)\]

where the last equality follows again from the solution to Poisson’s equation.

Now we apply Dirac’s recipe Eq.6.5.6 and Eq.6.5.5 to compute the commutators as follows

\[\begin{split}[A_i(\xbf), \Pi_j(\ybf)] &= \ifrak [A_i(\xbf), \Pi_j(\ybf)]_P - \ifrak \int d^3 \zbf \int d^3 \wbf [A_i(\xbf), \chi_{2\zbf}] (C^{-1})_{2\zbf, 1\wbf} [\chi_{1\wbf}, \Pi_j(\ybf)] \\ &= \ifrak \delta_{ij} \delta^3(\xbf - \ybf) + \ifrak \int d^3 \zbf \int d^3 \wbf \left( \p_i \delta^3(\xbf - \zbf) \frac{1}{4\pi |\zbf - \wbf|} \p_j \delta^3(\wbf - \ybf) \right) \\ &= \ifrak \delta_{ij} \delta^3(\xbf - \ybf) + \ifrak \frac{\p^2}{\p x_i \p x_j} \left( \frac{1}{4\pi |\xbf - \ybf|} \right) \\ [A_i(\xbf), A_j(\ybf)] &= [\Pi_i(\xbf), \Pi_j(\ybf)] = 0\end{split}\]

It’s straightforward to check that they are indeed compatible with the constraints Eq.7.3.1.

A formula for \(\Pi^i(\xbf)\)

It’s a real concern that the Coulomb gauge \(\nabla \cdot \Abf = 0\) may have spoiled the very definition Eq.7.2.2 of \(\Pi^i\). To settle this, let’s go back to the root of Lagrangian formalism and redefine \(\Pi^i\) using Eq.6.2.1 as follows

\[\Pi^i \coloneqq \frac{\delta L}{\delta \dot{A}_i}\]

Now if this were well-defined, then for any infinitesimal variation \(\delta \dot{\Abf}\), there should exists a unique \(\bm{\Pscr}\) such that

(7.3.2)\[\delta L = \int d^3 \xbf~\bm{\Pscr} \cdot \delta \dot{\Abf}\]

But this is not true since a substitution \(\bm{\Pscr} \to \bm{\Pscr} + \nabla f\) will also satisfy Eq.7.3.2 for any (scalar) function \(f(\xbf)\) due to the Coulomb gauge condition.

Baring this difficulty in mind, let’s just evaluate the potentially ill-defined \(\Pi^i\) anyway

(7.3.3)\[\Pi^i = \frac{\delta L}{\delta \dot{A}_i} = \frac{\delta \Lscr_{\gamma}}{\delta \dot{A}_i} = \dot{A}^i(\xbf) + \frac{\p A^0(\xbf)}{\p x_i}\]

The miracle is that this \(\Pi^i\) satisfies exactly the second constraint in Eq.7.3.1 (cf. Eq.7.2.6), and hence no ambiguity like \(\nabla f\) can appear. In other words Eq.7.3.3 is the correct formula for \(\Pi^i\).

Next let’s bring in the matter fields introduced in Eq.7.2.5. Due to the appearance of \(J^0\) in the constraints Eq.7.3.1, one can check that while \(\Abf\) commutes with the matter fields, \(\bm{\Pi}\) doesn’t. Indeed, if \(F\) is any function of the matter fields \(Q_n\) and \(P^n\), then we can calculate the Dirac bracket as follows

\[\begin{split}[F, \bm{\Pi}(\zbf)]_D &= -\int d^3 \xbf \int d^3 \ybf~[F, \chi_{2\xbf}]_P \frac{1}{4\pi |\xbf - \ybf|} [\chi_{1\ybf}, \bm{\Pi}(\zbf)]_P \\ &= -\int d^3 \xbf \int d^3 \ybf~[F, J^0(\xbf)]_P \frac{1}{4\pi |\xbf - \ybf|} \nabla \delta^3(\ybf - \zbf) \\ &= -\int d^3 \ybf [F, A^0(\ybf)]_P \nabla \delta^3(\ybf - \zbf) \\ &= [F, \nabla A^0(\zbf)]_P = [F, \nabla A^0(\zbf)]_D\end{split}\]

where we’ve used Eq.7.2.6 in the third equality.

This calculation motivates the following definition a new field

(7.3.4)\[\bm{\Pi}_{\bot} \coloneqq \bm{\Pi} - \nabla A^0 = \dot{\Abf}\]

where the last equality follows from Eq.7.3.3, such that

\[[F, \bm{\Pi}_{\bot}] = 0\]

for any function \(F\) of matter fields. Moreover, one can verify that \(\bm{\Pi}_{\bot}\) satisfies exactly the same commutation relations with \(\Abf\) just as \(\bm{\Pi}\) which we repeat as follows

(7.3.5)\[\begin{split}[A_i(\xbf), (\Pi_{\bot})_j(\ybf)] &= \ifrak \delta_{ij} \delta^3(\xbf - \ybf) + \ifrak \frac{\p^2}{\p x_i \p x_j} \left( \frac{1}{4\pi |\xbf - \ybf|} \right) \\ [A_i(\xbf), A_j(\ybf)] &= [(\Pi_{\bot})_i(\xbf), (\Pi_{\bot})_j(\ybf)] = 0\end{split}\]

Finally, note that the second constraint in Eq.7.3.1 can now be written as follows

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

With all the preparations above, let’s first write down the Hamiltonian in a rather general form

(7.3.7)\[H = \int d^3 x \left( (\Pi_{\bot})_i \dot{A}^i + P_n \dot{Q}^n - \Lscr \right)\]

despite the existence of constraints. For the rest of this chapter, we’ll be considering Lagrangian density of the following form

(7.3.8)\[\Lscr = -\frac{1}{4} F_{\mu\nu} F^{\mu\nu} + J^{\mu} A_{\mu} + \Lscr_{\text{matter}}\]

where \(J^{\mu}\) is the charge density as before, but \(\Lscr_{\text{matter}}\) is different from \(\Lscr_M\) considered in Eq.7.1.10 since it contains only the terms that don’t interact with \(A\), i.e., only the \(Q\) and \(P\) fields.

Note

Both \(J^{\mu}\) and \(\Lscr_{\text{matter}}\) will be further specified later for a theory specific to spin-\(1/2\) fermions, e.g., electrons. In fact, according to [Wei95] (page 349), the QED for spinless particles would require a more complicated Lagrangian density than Eq.7.3.8, but it wasn’t mentioned in the book which physical particle(s) such theory would describe.

Plugging Eq.7.3.8 into Eq.7.3.7, and making use of Eq.7.3.4 and Eq.7.3.3, we can write the Hamiltonian, again with a separation between matter and light, as follows

(7.3.9)\[H = \int d^3 x \left( \bm{\Pi}_{\bot}^2 + \frac{1}{2} (\nabla \times \Abf)^2 - \frac{1}{2} (\bm{\Pi}_{\bot} + \nabla A^0)^2 - \Jbf \cdot \Abf + J^0 A^0 \right) + H_{\text{matter}}\]

where

\[H_{\text{matter}} \coloneqq \int d^3 x \left( P_n \dot{Q}^n - \Lscr_{\text{matter}} \right)\]

Expanding out \((\bm{\Pi}_{\bot} + \nabla A^0)^2\) in Eq.7.3.9 and using integration-by-parts together with Eq.7.3.6 and Eq.7.2.6, we can further rewrite Eq.7.3.9 as follows

(7.3.10)\[H = \int d^3 x \left( \frac{1}{2} \bm{\Pi}_{\bot}^2 + \frac{1}{2} (\nabla \times \Abf)^2 - \Jbf \cdot \Abf + \frac{1}{2} J^0 A^0 \right) + H_{\text{matter}}\]

Note

The term \(\tfrac{1}{2} J^0 A^0\) in Eq.7.3.10 gives nothing but the Coulomb energy as the following calculation (cf. Eq.7.2.7) shows

(7.3.11)\[V_{\text{Coul}} \coloneqq \frac{1}{2} \int d^3 x~J^0 A^0 = \frac{1}{2} \int d^3 x \int d^3 y~\frac{J^0(\xbf) J^0(\ybf)}{4\pi |\xbf - \ybf|}\]

where we’ve suppressed the \(t\)-dependence as usual in the Lagrangian formalism.

7.4. Electrodynamics in the Interaction Picture

As before, let’s split the Hamiltonian Eq.7.3.10 into the free part and the interaction part as follows

\[\begin{split}H &= H_0 + V \\ H_0 &= \int d^3 x \left( \frac{1}{2} \bm{\Pi}_{\bot}^2 + \frac{1}{2} (\nabla \times \Abf)^2 \right) + H_{\text{matter}, 0} \\ V &= -\int d^3 x~\Jbf \cdot \Abf + V_{\text{Coul}} + V_{\text{matter}}\end{split}\]

where \(V_{\text{Coul}}\) is defined by Eq.7.3.11 and \(H_{\text{matter}} = H_{\text{matter}, 0} + V_{\text{matter}}\) is the splitting of the matter Hamiltonian into the free and interaction parts.

Before passing to the interaction picture, let’s make one more (potentially confusing) change of notation:

Warning

For the rest of this chapter, we’ll simply write \(\bm{\Pi}\) in place of \(\bm{\Pi}_{\bot}\), which is not the original \(\bm{\Pi}\) as in Eq.7.3.4. Since \(\bm{\Pi}\) and \(\bm{\Pi}_{\bot}\) satisfy the same commutation relations with the other fields such as \(\Abf\) and the matter fields, the only thing we need to keep in mind is that we should use the constraint Eq.7.3.6 rather than the second equation in Eq.7.3.1.

Finally we can introduce the interaction-picture fields \(\abf, \bm{\pi}, q, p\) corresponding to the Heisenberg-picture fields \(\Abf, \bm{\Pi}, Q, P\), respectively. We’ll focus in this section on the \(\abf\) and \(\bm{\pi}\) fields. In fact, the letters \(q, p\) will soon be occupied by something completely different, namely, the momentum-space coordinates. It turns out in the theory for spin-\(1/2\) particles, which will be completely specified in the next section, the matters fields will be named by another letter, so we’ll not run into conflicts of notations.

The interaction-picture free Hamiltonian \(H_0\) takes the following form

(7.4.1)\[H_0 = \int d^3 x \left( \frac{1}{2} \bm{\pi}^2 + \frac{1}{2} (\nabla \times \abf)^2 \right) + H_{\text{matter}, 0}(t)\]

The commutation relations between \(\abf\) and \(\bm{\pi}\) follow directly from Eq.7.3.5 and the definition of interaction-picture fields Eq.2.5.7

(7.4.2)\[\begin{split}[a_i(t, \xbf), \pi_j(t, \ybf)] &= \ifrak \delta_{ij} \delta^3(\xbf - \ybf) + \ifrak \frac{\p^2}{\p x_i \p x_j} \frac{1}{4\pi |\xbf - \ybf|} \\ [a_i(t, \xbf), a_j(t, \ybf)] &= [\pi_i(t, \xbf), \pi_j(t, \ybf)] = 0\end{split}\]

Moreover, they satisfy the following constraints

(7.4.3)\[\begin{split}\nabla \cdot \abf &= 0 \\ \nabla \cdot \bm{\pi} &= 0\end{split}\]

due to the Coulomb gauge condition and Eq.7.3.6, respectively.

For later reference, let’s note that the Coulomb interaction \(V_{\text{Coul}}\) defined by Eq.7.3.11 becomes

(7.4.4)\[V_{\text{Coul}}(t) = \frac{1}{2} \int d^3 x \int d^3 y~\frac{j^0(t, \xbf) j^0(t, \ybf)}{4\pi |\xbf - \ybf|}\]

in the interaction picture, and the interaction part of the full Hamiltonian becomes

(7.4.5)\[V(t) = -\int d^3 x~j^{\mu}(t, \xbf) a_{\mu}(t, \xbf) + V_{\text{Coul}}(t) + V_{\text{matter}}(t)\]

To derive the field equations, we resort to Hamilton’s equations Eq.6.1.7 as follows

\[\begin{split}\dot{a}_i(t,\xbf) &= \ifrak [H_0, a_i(t, \xbf)] \\ &= \ifrak \int d^3 y~\left[ \pi_j(t, \ybf), a_i(t, \xbf) \right] \pi^j(t, \ybf) \\ &= \int d^3 y \left( \delta_{ij} \delta^3(\xbf - \ybf) + \frac{\p^2}{\p x_i \p x_j} \frac{1}{4\pi |\xbf - \ybf|} \right) \pi^j(t, \ybf) \\ &= \pi_i(t, \xbf) - \int d^3 x \left( \frac{\p^2}{\p x_i \p y_j} \frac{1}{4\pi |\xbf - \ybf|} \right) \pi^j(t, \ybf) \\ &= \pi_i(t, \xbf) \\\\ \dot{\pi}_i(t, \xbf) &= \ifrak [H_0, \pi_i(t, \xbf)] \\ &= \ifrak \int d^3 y~[a_j(t, \ybf), \pi_i(t, \xbf)] (\nabla \times \nabla \times \abf(t, \ybf))^j \\ &= -\int d^3 y \left( \delta_{ij} \delta^3(\xbf - \ybf) + \frac{\p^2}{\p x_i \p x_j} \frac{1}{4\pi |\xbf - \ybf|} \right) \left( \nabla (\nabla \cdot \abf(t, \ybf)) - \nabla^2 \abf(t, \ybf) \right)^j \\ &= \nabla^2 a_i(t, \xbf)\end{split}\]

These two equations combined together give the familiar wave equation

(7.4.6)\[\square \abf = 0\]

To upgrade \(\abf\) to a \(4\)-vector field, we must set

(7.4.7)\[a_0 = 0\]

due to the assumption that \(a_0\) shouldn’t depend on the charge density, and it must vanish when charge density vanishes according to Eq.7.2.7.

The general solutions to Eq.7.4.6 and Eq.7.4.7, under the constraint Eq.7.4.3, can be written as follows

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

where \(p_0 = |\pbf|\) and \(e_{\mu}(\pbf, \pm 1)\) are two independent “polarization vectors” satisfying the following familiar conditions (cf. Eq.4.7.14)

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

It follows that \(\ebf(\pbf, \sigma)\) can be normalized as follows

\[\sum_{\sigma = \pm 1} e_i(\pbf, \sigma) e_j^{\ast}(\pbf, \sigma) = \delta_{ij} - \frac{p_i p_j}{|\pbf|^2}\]

Without working out the calculations, we claim, following [Wei95] page 352 –353, that the commutation relations Eq.7.4.2 are satisfied if the operators \(a(\pbf, \sigma)\) and \(a^{\dagger}(\pbf, \sigma)\) satisfy

(7.4.10)\[\begin{split}\left[ a(\pbf, \sigma), a^{\dagger}(\pbf', \sigma') \right] &= \delta^3(\pbf - \pbf') \delta_{\sigma \sigma'} \\ \left[ a(\pbf, \sigma), a(\pbf', \sigma') \right] &= \left[ a^{\dagger}(\pbf, \sigma), a^{\dagger}(\pbf', \sigma') \right] = 0\end{split}\]

Moreover, the free-photon Hamiltonian, i.e., Eq.7.4.1 without the matter term, can be written as follows

\[\begin{split}H_{\gamma, 0} &= \frac{1}{2} \int d^3 x \left( \bm{\pi}^2 + (\nabla \times \abf)^2 \right) \\ &= \frac{1}{2} \int d^3 p~p_0 \sum_{\sigma = \pm 1} \left[ a(\pbf, \sigma), a^{\dagger}(\pbf, \sigma) \right]_+ \\ &= \int d^3 p~p_0 \sum_{\sigma = \pm 1} \left( a^{\dagger}(\pbf, \sigma) a(\pbf, \sigma) + \frac{1}{2} \delta^3(0) \right)\end{split}\]

where the last expression contains one of the infinities in QED.

7.5. The Photon Propagator

As explained in The Feynman Rules, to calculate the S-matrix using Feynman diagrams, one must calculate the propagators defined in Eq.5.1.9. Using Eq.7.4.8 and the commutation relations Eq.7.4.10, the photon propagator can be calculated as follows

\[\begin{split}\underbracket{a_{\mu}(x) a_{\nu}^{\dagger}(y)} &= \theta(x_0 - y_0) (2\pi)^{-3} \int \frac{d^3 p~d^3 p'}{2\sqrt{p_0 p'_0}} \sum_{\sigma, \sigma'} e^{\ifrak (p \cdot x - p' \cdot y)} e_{\mu} e_{\nu}^{\ast} \left[ a, a^{\dagger} \right] + x \leftrightarrow y \\ &= (2\pi)^{-3} \int \frac{d^3 p}{2p_0} \left( \sum_{\sigma} e_{\mu}(\pbf, \sigma) e_{\nu}^{\ast}(\pbf, \sigma) \right) \left( e^{\ifrak p \cdot (x-y)} \theta(x_0 - y_0) + e^{\ifrak p \cdot (y-x) \theta(y_0 - x_0)} \right) \\ &\eqqcolon -\ifrak \Delta_{\mu \nu}(x - y)\end{split}\]

Since the spinor sum

(7.5.1)\[\begin{split}P_{\mu \nu}(\pbf) \coloneqq \sum_{\sigma} e_{\mu}(\pbf, \sigma) e_{\nu}^{\ast}(\pbf, \sigma) = \begin{cases} \delta_{\mu \nu} - p_{\mu} p_{\nu} / |\pbf|^2 & \text{ if } \mu\nu \neq 0 \\ 0 & \text{ otherwise} \end{cases}\end{split}\]

doesn’t depend on \(p_0\), it follows from Eq.5.2.7 and Eq.5.2.13 and the massless condition that

(7.5.2)\[\Delta_{\mu \nu}(x - y) = (2\pi)^{-4} \int d^4 q~\frac{P_{\mu \nu}(\qbf)}{q^2 - \ifrak \epsilon} e^{\ifrak q \cdot (x - y)}\]

where \(q^2 = q_0^2 - |\qbf|^2\) as usual but \(q\) is not constrained to the mass shell.

Using the momentum-space Feynman rules derived in Feynman rules in momentum space, one gets a contribution of

(7.5.3)\[\frac{-\ifrak}{(2\pi)^4} \frac{P_{\mu \nu}(\qbf)}{q^2 - \ifrak \epsilon}\]

for each internal photon line in a Feynman diagram.

Note that the photon line contribution given by Eq.7.5.3 is not Lorentz covariant, which is ultimately a consequence of the \(A\)-field \(a_{\mu}\) (cf. Eq.7.4.8) not being Lorentz covariant. It turns out, rather miraculously, that such Lorentz non-covariance can be countered by yet another Lorentz non-covariant term in the interaction Hamiltonian, namely, the Coulomb interaction Eq.7.4.4

The heuristic for such cancellation starts by rewriting the spinor sum Eq.7.5.1 as follows

\[P_{\mu\nu}(\qbf) = \eta_{\mu\nu} - \frac{q^2 n_{\mu} n_{\nu} - q_0 q_{\mu} n_{\nu} - q_0 q_{\nu} n_{\mu} + q_{\mu} q_{\nu}}{|\qbf|^2}\]

where \(n = (1, 0, 0, 0)\). Here \(q_0\) is completely arbitrary, but its value will be fixed in Feynman diagram evaluations, to be discussed in the next section, by momentum conservation at each vertex.

The point is that the last three terms in the nominator are proportional to \(q_{\mu}\) or \(q_{\nu}\), which means that they will not contribute to the S-matrix due to the coupling \(j^{\mu} a_{\mu}\) in the interaction Eq.7.4.5 and the conservation law \(\p_{\mu} j^{\mu} = 0\). The remaining \(-q^2 n_{\mu} n_{\nu} / |\qbf|^2\), which is non-vanishing only if \(\mu = \nu = 0\), can be plugged into Eq.7.5.2, and then follow the Feynman rules to produce the following contribution to the S-matrix

\[\begin{split}& \frac{\ifrak}{2} \int d^4 x \int d^4 y \left( -\ifrak j^0(x) \right) \left( -\ifrak j^0(y) \right) \frac{-\ifrak}{(2\pi)^4} \int \frac{d^4 q}{|\qbf|^2} e^{\ifrak q \cdot (x-y)} \\ &\quad = -\frac{1}{2} \int d^4 x \int d^4 y~\frac{j^0(x) j^0(y)}{(2\pi)^3} \delta(x_0 - y_0) \int \frac{d^3 q}{|\qbf|^2} e^{\ifrak \qbf \cdot (\xbf - \ybf)} \\ &\quad = -\frac{1}{2} \int d^3 x \int d^3 y~\frac{j^0(t, \xbf) j^0(t, \ybf)}{4\pi |\xbf - \ybf|}\end{split}\]

which is countered by \(V_{\text{Coul}}\) (cf. Eq.7.4.4).

It means that effectively, one can replace \(\Delta_{\mu \nu}(x - y)\) with the following

\[\Delta_{\mu\nu}^{\text{eff}}(x - y) = (2\pi)^{-4} \int d^4 q~\frac{\eta_{\mu \nu}}{q^2 - \ifrak \epsilon} e^{\ifrak q \cdot (x - y)}\]

as long as one forgets about the (instantaneous) Coulomb interaction term \(V_{\text{Coul}}(t)\) in Eq.7.4.5. The corresponding photon line contribution Eq.7.5.3 then takes the following form

(7.5.4)\[\frac{-\ifrak}{(2\pi)^4} \frac{\eta_{\mu\nu}}{q^2 - \ifrak \epsilon}\]

7.6. Feynman Rules for QED

Having settled the photon field in the previous section, we’re now ready to specify the matter field in the Lagrangian density Eq.7.3.8. We’ll be considering the theory describing the interaction between a massive spin-\(1/2\) fermion with charge \(q = -e\), [4] e.g., an electron, and photon. In this case, the Lagrangian density takes the following form

(7.6.1)\[\Lscr = -\frac{1}{4} F^{\mu\nu} F_{\mu\nu} - \bar{\Psi} \left( \gamma^{\mu} (\p_{\mu} + \ifrak e A_{\mu}) + m \right) \Psi\]

It follows from Eq.7.1.4 that the charge density \(J^{\mu}\) takes the following form

\[J^{\mu} = \frac{\p \Lscr}{\p A_{\mu}} = -\ifrak e \bar{\Psi} \gamma^{\mu} \Psi\]

The interaction Eq.7.4.5 now takes the following form

(7.6.2)\[V(t) = \ifrak e \int d^3 x~\bar{\psi}(t, \xbf) \gamma^{\mu} \psi(t, \xbf) a_{\mu}(t, \xbf) + V_{\text{Coul}}(t)\]

where we remember that \(V_{\text{Coul}}\) becomes irrelevant if we take the photon line contribution to the Feynman diagram to be Eq.7.5.4.

Following the general recipe described in Feynman rules in momentum space, let’s spell out the key points, while ignoring routines, in constructing and evaluating Feynman diagrams in QED as follows.

First of all, a Feynman diagram consists of electron lines, photon lines, and \(3\)-valent vertices, each of which has one incoming electron line, one outgoing electron line, and one photon line attached. Electron lines that flow backward in time may also be called positron lines. To each internal line an off-mass-shell \(4\)-momentum is attached. To each external line an on-mass-shell \(4\)-momentum as well as a spin \(z\)-component or helicity, depending on whether it’s an electron line or a photon line, are attached.

Next, to evaluate a Feynman diagram, we associate factors to each vertex, external line, and internal line as follows.

Vertices

To each vertex, we assign a \(\gamma\)-matrix (cf. Eq.4.4.7) index \(\alpha\) to the incoming electron line, and a similar index \(\beta\) to the outgoing electron line, and a spacetime index \(\mu\) to the photon line. Moreover, such a vertex contributes the following factor

\[(2\pi)^4 e (\gamma^{\mu})_{\beta \alpha} \delta^4(k - k' \pm q)\]

where \(k, k'\) are the \(4\)-momentum of the incoming and outgoing electrons, respectively, and \(\pm q\) is the \(4\)-momentum of the incoming/outgoing photon.

External lines

In what follows, we’ll use \(\alpha\) (for incoming) and \(\beta\) (for outgoing) to index Dirac spinors as in Eq.4.4.31, and \(\mu\) to index photon spinors as in Eq.7.4.8.

• For each in-state electron line, include a factor \((2\pi)^{-3/2} u_{\alpha}(\pbf, \sigma)\).

• For each in-state positron line, include a factor \((2\pi)^{-3/2} \bar{v}_{\beta}(\pbf, \sigma)\).

• For each out-state electron line, include a factor \((2\pi)^{-3/2} \bar{u}_{\beta}(\pbf, \sigma)\).

• For each out-state positron line, include a factor \((2\pi)^{-3/2} v_{\alpha}(\pbf, \sigma)\).

• For each in-state photon line, include a factor \((2\pi)^{-3/2} (2p_0)^{-1/2} e_{\mu}(\pbf, \sigma)\).

• For each out-state photon line, include a factor \((2\pi)^{-3/2} (2p_0)^{-1/2} e_{\mu}^{\ast}(\pbf, \sigma)\).

Internal lines

In what follows, the same indexing convention as for external lines is used. In fact, to index multiple photon spinors, we’ll use \(\mu\) and \(\nu\).

• For each internal electron line from \(\beta\) to \(\alpha\), include a factor (cf. Eq.5.2.5 and Eq.5.2.15)

  \[\frac{-\ifrak}{(2\pi)^4} \frac{\left( -\ifrak \kslash + m \right)_{\alpha \beta}}{k^2 + m^2 - \ifrak \epsilon}\]

  where the \(\beta\) matrix in Eq.5.2.5 is dropped because of the use of \(\bar{\psi}\), rather than \(\psi^{\dagger}\), in Eq.7.6.2. Moreover, the Feynman slash notation

  \[\kslash \coloneqq \gamma^{\mu} k_{\mu}\]

  is adopted here.

• For each internal photon line between \(\mu\) and \(\nu\), include a factor (cf. Eq.7.5.4)

  \[\frac{-\ifrak}{(2\pi)^4} \frac{\eta_{\mu \nu}}{q^2 - \ifrak \epsilon}\]

The rest of the calculation is the same as any general evaluation of scattering amplitudes using Feynman diagrams, and was discussed in Spacetime Feynman Rules.

Although the full S-matrix calculation can be done in principle by summing up all Feynman diagrams, it quickly becomes computationally infeasible as the number of vertices in the diagram increases. Therefore the calculation is only practically possible if the contribution of a Feynman diagram also diminishes as the complexity grows. Fortunately, this is indeed the case as we now demonstrate.

Consider a connected Feynman diagram [5] with \(V\) vertices, \(I\) internal lines, \(E\) external lines, and \(L\) (independent) loops. Then the following relations hold

(7.6.3)\[\begin{split}L &= I - V + 1 \\ 2I + E &= 3V\end{split}\]

Now the goal is to calculate the constant coefficient involving \(e\) (electric charge) and \(\pi\) (mathematical constant) for any given diagram. According to the explicit Feynman rules described above

• Each vertex contributes a factor \((2\pi)^4 e\).

• Each internal line contributes a factor \((2\pi)^{-4}\).

• Each loop contributes a momentum-space integral (cf. Eq.5.2.16), which, in turn, contributes a factor \(\pi^2\) (cf. Volume of an n-ball).

Note that we don’t include contributions from external lines since they are fixed by the in- and out-states. Multiplying the contributions above all together (and eliminating \(I\) and \(V\) using Eq.7.6.3), the constant coefficient of a Feynman diagram is given as follows

\[\begin{split}(2\pi)^{4V} e^V (2\pi)^{-4I} \pi^{2L} &= (2\pi)^{-4(L-1)} e^{2L+E-2} \pi^{2L} \\ &= (2\pi)^4 e^{E-2} \left( \frac{e^2}{16\pi^2} \right)^L\end{split}\]

It follows that the Feynman diagrams may be organized by an increasing number of (independent) loops so that the higher order terms are suppressed by a power of

\[\frac{e^2}{16\pi^2} \approx 5.81 \times 10^{-4}\]

Finally, note that the Feynman rules developed in this section apply readily to the case where the in and out photon states don’t have definite helicity. More precisely, for photon states with definite helicity, the polarization vectors may be taken to be of the following form (cf. Eq.4.7.11)

\[\begin{split}e(\pbf, \pm 1) = R(\hat{\pbf}) \begin{bmatrix*} 0 \\ 1 / \sqrt{2} \\ \pm \ifrak / \sqrt{2} \\ 0 \end{bmatrix*}\end{split}\]

where \(R(\hat{\pbf})\) rotates \((0, 0, 1)\) to \(\hat{\pbf}\). Now a general polarization vector may be written as

(7.6.4)\[e_{\mu}(\pbf) = \alpha_+ e_{\mu}(\pbf, +1) + \alpha_- e_{\mu}(\pbf, -1)\]

where \(|\alpha_+|^2 + |\alpha_-|^2 = 1\). The polarization is called circular if \(\alpha_+ = 0\) or \(\alpha_- = 0\), and linear if \(|\alpha_+| = |\alpha_=| = 1/\sqrt{2}\), just as in Polarization of photons. Note, in particular, that

(7.6.5)\[e_{\mu} {e^{\mu}}^{\ast} = 1\]

7.7. Compton Scattering

In this section, we’ll go through some really exciting calculations, utilizing many tools we’ve learned so far, on the scattering amplitudes between a photon and an electron, culminating in the famous estimation of the electron radius.

The in-state consists of a on-mass-shell electron labeled by \(4\)-momentum \(p_{\mu}\) and spin \(z\)-component \(\sigma\) and a on-mass-shell photon labeled by \(4\)-momentum \(k_{\mu}\) and polarization vector \(e_{\mu}\). The out-state consists similarly of a on-mass-shell electron labeled by \(p'_{\mu}\) and \(\sigma'\) and a on-mass-shell photon labeled by \(k'_{\mu}\) and \(e'_{\mu}\). In the lowest order of the electric charge \(e\), there are just two Feynman diagrams, shown as follows

Fig. 7.7.1 The two lowest-order Feynman diagrams for the electron-photon scattering. The solid lines are electron lines, and the wavy lines are photon lines.

The corresponding S-matrix element is given as follows

\[\begin{split}S(\pbf', \sigma'; \kbf', e',~\pbf, \sigma; \kbf, e) &= \sum_{\alpha, \beta, \alpha', \beta'} \frac{u_{\alpha}(\pbf, \sigma)}{(2\pi)^{3/2}} \frac{e_{\mu}(\kbf)}{(2\pi)^{3/2} \sqrt{2k_0}} \frac{\bar{u}_{\beta'}(\pbf', \sigma')}{(2\pi)^{3/2}} \frac{{e'_{\nu}}^{\ast}(\kbf')}{(2\pi)^{3/2} \sqrt{2k'_0}} \\ &\quad \times \int d^4 q~\frac{-\ifrak}{(2\pi)^4} \frac{\left(-\ifrak \qslash + m\right)_{\alpha' \beta}}{q^2 + m^2 - \ifrak \epsilon} \\ &\qquad \times \Big( (2\pi)^4 e (\gamma^{\mu})_{\beta \alpha} \delta^4(q-p-k) (2\pi)^4 e (\gamma^{\nu})_{\beta' \alpha'} \delta^4(p'+k'-q) \\ &\qquad + (2\pi)^4 e (\gamma^{\mu})_{\beta' \alpha'} \delta^4(p'-q-k) (2\pi)^4 e (\gamma^{\nu})_{\beta \alpha} \delta^4(q-p-k') \Big)\end{split}\]

Evaluating the (trivial) integral, using obvious shorthand notations, and with matrix multiplication rules understood, the amplitude \(S = S(\pbf', \sigma'; \kbf', e',~\pbf, \sigma; \kbf, e)\) may be written as follows

\[\begin{split}S &= \frac{-\ifrak e^2 \delta^4(p+k-p'-k')}{(2\pi)^2 2\sqrt{k_0 k'_0}} \\ &\quad \times \bar{u}(\pbf', \sigma') \left( \cancel{{e'}^{\ast}} \frac{-\ifrak \left(\pslash + \kslash \right) + m}{(p+k)^2 + m^2} \eslash + \eslash \frac{-\ifrak \left( \pslash - \cancel{k'} \right) + m}{(p - k')^2 + m^2} \cancel{{e'}^{\ast}} \right) u(\pbf, \sigma) \\\end{split}\]

Using the mass-shell conditions \(p^2 = m^2\) and \(k^2 = 0\), the denominators in the big parenthesis may be simplified as follows

\[\begin{split}(p + k)^2 + m^2 &= 2p \cdot k \\ (p - k')^2 + m^2 &= -2p \cdot k'\end{split}\]

Under the assumption that the out-state is different from the in-state, we can split the amplitude \(S\) according to Eq.2.4.4 as follows

\[S = -2\pi\ifrak \delta^4(p'+k'-p-k) M\]

where

(7.7.1)\[\begin{split}M &= \frac{e^2}{(2\pi)^3 4\sqrt{k_0 k'_0}} \bar{u}(\pbf', \sigma') \left( \cancel{{e'}^{\ast}} \frac{-\ifrak \left( \pslash + \kslash \right) + m}{p \cdot k} \eslash \right. \\ &\qquad \left. + \eslash \frac{\ifrak \left( \pslash - \cancel{k'} \right) - m}{p \cdot k'} \cancel{{e'}^{\ast}} \right) u(\pbf, \sigma)\end{split}\]

Now the differential cross-section \(d\sigma\) (not to be confused with spin \(z\)-component \(\sigma\)) according to Eq.2.4.10 is

(7.7.2)\[d\sigma = (2\pi)^4 u^{-1} \delta^4(p'+k'-p-k) |M|^2 d^3 p' d^3 k'\]

where \(u\) denotes the relative velocity between the in-state electron and photon which, according to Eq.2.4.11, is given by

(7.7.3)\[u = \frac{|p \cdot k|}{p_0 k_0}\]

Here remember that \(u\) is not physical, albeit being called velocity, as it may exceed \(1\). It turns out to be convenient to work in the so-called “laboratory frame” where the initial electron is at rest, i.e.,

(7.7.4)\[p_0 = m, \quad \pbf = 0\]

In this frame Eq.7.7.3 becomes simply

\[u = 1\]

Next, to handle \(\delta^4(p'+k'-p-k)\), it’ll be convenient to introduce the following variables that represent photon energies

(7.7.5)\[\begin{split}\omega &\coloneqq k_0 = |\kbf| = -p \cdot k / m \\ \omega' &\coloneqq k'_0 = |\kbf'| = -p \cdot k' / m\end{split}\]

Splitting the energy-momentum delta function into the energy part and the momentum part

\[\delta^4(p'+k'-p-k) = \delta(p'_0 + k'_0 - p_0 - k_0) \delta^3(\pbf' + \kbf' - \pbf - \kbf)\]

and using the laboratory frame Eq.7.7.4, we see that the momentum delta function

\[\delta^3(\pbf' + \kbf' - \pbf - \kbf) d^3 p' = \delta^3(\pbf' + \kbf' - \kbf) d^3 p'\]

serves the purpose of equating \(\pbf' = \kbf - \kbf'\) and eliminating \(d^3 p'\) from \(d\sigma\) in Eq.7.7.2. It follows that we can write

\[\begin{split}p'_0 &= \sqrt{{\pbf'}^2 + m^2} \\ &= \sqrt{(\kbf - \kbf')^2 + m^2} \\ &= \sqrt{\omega^2 + {\omega'}^2 - 2\omega\omega' \cos\theta + m^2}\end{split}\]

where \(\theta\) denotes the angle between \(\kbf\) and \(\kbf'\), and the energy delta function

\[\begin{split}\delta(p'_0 + k'_0 - p_0 - k_0) d^3 k' &= \delta\left( \sqrt{\omega^2 + {\omega'}^2 - 2\omega\omega' \cos\theta + m^2} + \omega' - m - \omega \right) {\omega'}^2 d\omega' d\Omega \\ &= \frac{\delta(\omega' - \omega_c(\theta))}{\p_{\omega'} \left( \sqrt{\omega^2 + {\omega'}^2 - 2\omega\omega' \cos\theta + m^2} + \omega' \right)} {\omega'}^2 d\omega' d\Omega \\ &= \frac{\delta(\omega' - \omega_c(\theta))}{1 + (\omega' - \omega \cos\theta) / p'_0} {\omega'}^2 d\omega' d\Omega \\ &= \delta(\omega' - \omega_c(\theta)) \frac{p'_0 {\omega'}^3}{m\omega} d\omega' d\Omega\end{split}\]

where \(\Omega\) is the solid angle into which the out-state photon is scattered and the solution to

\[\sqrt{\omega^2 + {\omega'}^2 - 2\omega\omega' \cos\theta + m^2} + \omega' - m - \omega = 0\]

is given by

(7.7.6)\[\omega' = \frac{m\omega}{m + \omega(1 - \cos\theta)} \eqqcolon \omega_c(\theta)\]

Plugging our evaluation of \(\delta^4(p'+k'-p-k)\) into Eq.7.7.2, we get

(7.7.7)\[d\sigma = (2\pi)^4 |M|^2 \frac{p'_0 {\omega'}^3}{m \omega} d\Omega\]

where \(p'_0 = m + \omega - \omega'\) and \(\omega'\) given by the right-hand-side of Eq.7.7.6.

Now let’s evaluate \(|M|^2\) where \(M\) is given by Eq.7.7.1. In fact, we’ll sum over the spin \(z\)-component as in the spin sum evaluations. Besides the fact that such summation makes the calculation easier, it can be justified by noting that spin \(z\)-component is usually not measured in experiments. Instead of working with the specific \(M\) in Eq.7.7.1, let’s first compute more generally with an arbitrary matrix \(A\), using the Dirac field spin sum formula Eq.4.4.33, the following quantity

\[\begin{split}\sum_{\sigma, \sigma'} \left| \bar{u}(\pbf', \sigma') A u(\pbf, \sigma) \right|^2 &= \sum_{\sigma, \sigma'} \left( \bar{u}(\pbf', \sigma') A u(\pbf, \sigma) \right) \left( u^{\dagger}(\pbf, \sigma) A^{\dagger} \bar{u}^{\dagger}(\pbf', \sigma') \right) \\ &= \sum_{\sigma, \sigma'} \left( \bar{u}(\pbf', \sigma') A u(\pbf, \sigma) \right) \left( \bar{u}(\pbf, \sigma) \beta A^{\dagger} \beta u(\pbf', \sigma') \right) \\ &= \sum_{\sigma, \sigma', \alpha, \beta, \gamma, \delta} A_{\beta \alpha} u_{\alpha}(\pbf, \sigma) \bar{u}_{\gamma}(\pbf, \sigma) \left( \beta A^{\dagger} \beta \right)_{\gamma \delta} u_{\delta}(\pbf', \sigma') \bar{u}_{\alpha}(\pbf', \sigma') \\ &= \Tr\left( A \frac{-\ifrak \pslash + m}{2p_0} \beta A^{\dagger} \beta \frac{-\ifrak \cancel{p'} + m}{2p'_0} \right)\end{split}\]

Applying this calculation to Eq.7.7.1, and using Eq.7.7.5 and Eq.4.4.12, we get

\[\begin{split}\sum_{\sigma, \sigma'} |M|^2 &= \frac{e^4}{64 (2\pi)^6 \omega \omega' p_0 p'_0} \\ &\quad \times \Tr\left( \left( \cancel{{e'}^{\ast}} \frac{-\ifrak \left( \pslash + \kslash \right) + m}{p \cdot k} \eslash + \eslash \frac{\ifrak \left( \cancel{p'} - \cancel{k'} \right) - m}{p \cdot k'} \cancel{{e'}^{\ast}} \right) (-\ifrak \pslash + m) \right. \\ &\qquad \left. \times \left( \cancel{e^{\ast}} \frac{-\ifrak \left( \pslash + \kslash \right) + m}{p \cdot k} \cancel{e'} + \cancel{e'} \frac{\ifrak \left( \cancel{p'} - \cancel{k'} \right)- m}{p \cdot k'} \cancel{e^{\ast}} \right) (-\ifrak \cancel{p'} + m) \right)\end{split}\]

A few tricks can be applied to simplify such a complicated expression. First, recall that the Coulomb gauge condition Eq.7.4.9 implies \(e_0 = 0\), and the laboratory frame condition Eq.7.7.4 implies \(\pbf = 0\). It follows that

(7.7.8)\[e \cdot p = e^{\ast} \cdot p = e' \cdot p = {e'}^{\ast} \cdot p = 0\]

This, together with the Clifford algebra relation Eq.4.4.3, in turn, implies the following

\[\begin{split}\left( -\ifrak \pslash + m \right) \eslash \left( -\ifrak \pslash + m \right) &= \eslash \left( \ifrak \pslash + m \right) \left( -\ifrak \pslash + m \right) \\ &= \eslash \left( \pslash^2 + m^2 \right) \\ &= \eslash \left( p^2 + m^2 \right) = 0\end{split}\]

and likewise for \(\cancel{e^{\ast}}, \cancel{e'},\) and \(\cancel{{e'}^{\ast}}\). This allows the following simplification

\[\begin{split}\sum_{\sigma, \sigma'} |M|^2 &= \frac{-e^4}{64 (2\pi)^6 \omega \omega' p_0 p'_0} \Tr\left( \left( \frac{\cancel{{e'}^{\ast}} \kslash \eslash}{p \cdot k} + \frac{\eslash \cancel{k'} \cancel{{e'}^{\ast}}}{p \cdot k'} \right) \left( -\ifrak \pslash + m \right) \right. \\ &\quad \left. \times \left( \frac{\cancel{e^{\ast}} \kslash \cancel{e'}}{p \cdot k} + \frac{\cancel{e'} \cancel{k'} \cancel{e^{\ast}}}{p \cdot k'} \right) \left( -\ifrak \cancel{p'} + m \right) \right) \\ &= \frac{e^4}{64 (2\pi)^6 \omega \omega' p_0 p'_0} \left( \frac{\Tr\left( \cancel{{e'}^{\ast}} \kslash \eslash \pslash \cancel{e^{\ast}} \kslash \cancel{e'} \cancel{p'} \right)}{(p \cdot k)^2} \right. \\ &\quad + \frac{\Tr\left( \cancel{{e'}^{\ast}} \kslash \eslash \pslash \cancel{e'} \cancel{k'} \cancel{e^{\ast}} \cancel{p'} \right)}{(p \cdot k)(p \cdot k')} + \frac{\Tr\left( \eslash \cancel{k'} \cancel{{e'}^{\ast}} \pslash \cancel{e^{\ast}} \kslash \cancel{e'} \cancel{p'} \right)}{(p \cdot k')(p \cdot k)} \\ &\quad + \frac{\Tr\left( \eslash \cancel{k'} \cancel{{e'}^{\ast}} \pslash \cancel{e'} \cancel{k'} \cancel{e^{\ast}} \cancel{p'} \right)}{(p \cdot k')^2} - \frac{m^2 \Tr\left( \cancel{{e'}^{\ast}} \kslash \eslash \cancel{e^{\ast}} \kslash \cancel{e'} \right)}{(p \cdot k)^2} \\ &\quad - \frac{m^2 \Tr\left( \cancel{{e'}^{\ast}} \kslash \eslash \cancel{e'} \cancel{k'} \cancel{e^{\ast}} \right)}{(p \cdot k)(p \cdot k')} - \frac{m^2 \Tr\left( \eslash \cancel{k'} \cancel{{e'}^{\ast}} \cancel{e^{\ast}} \kslash \cancel{e'} \right)}{(p \cdot k')(p \cdot k)} \\ &\quad - \left. \frac{m^2 \Tr\left( \eslash \cancel{k'} \cancel{{e'}^{\ast}} \cancel{e'} \cancel{k'} \cancel{e^{\ast}} \right)}{(p \cdot k')^2} \right)\end{split}\]

where the last equality uses the fact that the trace of a product of odd number of gamma matrices vanishes.

Further simplifications can be achieved by assuming, on top of previous assumptions including Eq.7.7.8 and Eq.7.6.5, that incoming and outgoing photons are linearly polarized, i.e.,

\[e_{\mu} = e_{\mu}^{\ast} \quad \text{and} \quad e'_{\mu} = {e'_{\mu}}^{\ast}\]

in which case one can, modulo several rather tedious calculations, arrive at the following simple form

(7.7.9)\[\sum_{\sigma, \sigma'} |M|^2 = \frac{e^4}{64 (2\pi)^6 \omega \omega' p_0 p'_0} \left( \frac{8(k \cdot k')^2}{(k \cdot p)(k' \cdot p)} + 32(e \cdot e')^2 \right)\]

Using Eq.7.7.5, Eq.7.7.6, and the fact that \(\theta\) is the angle between \(\kbf\) and \(\kbf'\), one have the following identities

\[\begin{split}k \cdot p &= -m \omega \\ k' \cdot p &= -m \omega' \\ k \cdot k' &= \omega \omega' (\cos\theta - 1) = m(\omega' - \omega)\end{split}\]

which can then be plugged into Eq.7.7.9 and be combined with Eq.7.7.7 to obtain the following averaged differential cross-section for linearly polarized photons

(7.7.10)\[\begin{split}\frac{1}{2} \sum_{\sigma, \sigma'} d\sigma(\pbf, \sigma; \kbf, e \to \pbf', \sigma'; \kbf', e') &= \frac{e^4 {\omega'}^2}{128 (2\pi)^2 m^2 \omega^2} \left( \frac{8(\omega - \omega')^2}{\omega \omega'} + 32(e \cdot e')^2 \right) d\Omega \\ &= \frac{e^4 {\omega'}^2}{64 \pi^2 m^2 \omega^2} \left( \frac{\omega}{\omega'} + \frac{\omega'}{\omega} - 2 + 4(e \cdot e')^2 \right) d\Omega\end{split}\]

This formula is known as the Klein-Nishina formula for linearly polarized photons.

Now if the incoming photon is prepared without any particular polarization (which usually is the case in experiments/observations), then one should further average over the two independent polarization vectors to get the following

(7.7.11)\[\frac{1}{4} \sum_{\sigma, \sigma', e} d\sigma(\pbf, \sigma; \kbf, e \to \pbf', \sigma'; \kbf', e') = \frac{e^4 {\omega'}^2}{64 \pi^2 m^2 \omega^2} \left( \frac{\omega}{\omega'} + \frac{\omega'}{\omega} - 2\left(\hat{\kbf} \cdot e'\right)^2 \right) d\Omega\]

Indeed, if we write \(e_1, e_2\) for the two orthogonal, linearly polarized, polarization vectors of the incoming photon, then using Eq.7.5.1 we have

\[\begin{split}\sum_e (e \cdot e')^2 &= \left( \sum_{i=1}^3 {e_1}_i e'_i \right)^2 + \left( \sum_{i=1}^3 {e_2}_i e'_i \right)^2 \\ &= \sum_{i=1}^3 \left( {e^2_1}_i + {e^2_2}_i \right) {e'_i}^2 + 2 \sum_{i \neq j} \left( {e_1}_i {e_1}_j + {e_2}_i {e_2}_j \right) e'_i e'_j \\ &= \sum_{i=1}^3 \left( 1 - \hat{\kbf}^2_i \right) {e'_i}^2 - 2 \sum_{i \neq j} \hat{\kbf}_i \hat{\kbf}_j e'_i e'_j \\ &= 1 - \left( \hat{\kbf} \cdot e' \right)^2\end{split}\]

Note

It follows from Eq.7.7.11 that for randomly polarized incoming photon, the scattered photon is preferably polarized in the direction perpendicular to \(\hat{\kbf}\) – the direction of the incoming photon. Since the polarization of the scattered photon is also perpendicular to the direction of the outgoing photon, it must be in the direction perpendicular to the plane of scattering. According to [Wei95] page 368, such phenomenon is indeed observed from eclipsing binary stars. See Chandrasekhar polarization.

Finally let’s average also over the polarizations of the outgoing photon using the following identity

\[\sum_{e, e'} (e \cdot e')^2 = 1 + \cos^2 \theta\]

where we recall that \(\theta\) is the angle between \(\kbf\) and \(\kbf'\). Plugging this into Eq.7.7.10, we get

\[\frac{1}{8} \sum_{\sigma, \sigma', e, e'} d\sigma(\pbf, \sigma; \kbf, e \to \pbf', \sigma'; \kbf', e') = \frac{e^4 {\omega'}^2}{64 \pi^2 m^2 \omega^2} \left( \frac{\omega}{\omega'} + \frac{\omega'}{\omega} - 1 + \cos^2 \theta \right) d\Omega\]

Note that it’s customary to multiply both sides of the above equality by \(2\) since the polarizations \(e\) and \(e'\) are correlated, as observed above.

In the low-energy limit \(\omega \ll m\), we have \(\omega \approx \omega'\) by Eq.7.7.6. It follows that

\[\frac{1}{4} \sum_{\sigma, \sigma', e, e'} d\sigma(\pbf, \sigma; \kbf, e \to \pbf', \sigma'; \kbf', e') = \frac{e^4}{32 \pi^2 m^2} \left( 1 + \cos^2\theta \right) d\Omega\]

Integrating over the solid angle \(d\Omega\), we get the so-called Thomson cross-section

\[\begin{split}\sigma_T &\coloneqq \frac{1}{4} \int \sum_{\sigma, \sigma', e, e'} d\sigma \\ &= \frac{e^4}{32 \pi^2 m^2} \int \left( 1 + \cos^2\theta \right) d\Omega \\ &= \frac{e^4}{32 \pi^2 m^2} \int_0^{2\pi} d\phi \int_0^{\pi} d\theta \left( 1 + \cos^2\theta \right) \sin\theta \\ &= \frac{e^4}{6\pi m^2}\end{split}\]

Footnotes

[1]

The more modern way to quantize a field theory with gauge symmetry is via the so-called BRST quantization, which is Lorentz invariant. We will come back to it much later when we discuss non-Abelian gauge symmetries.

[2]

It’s really unfortunate for L. Lorenz to work in the same field as H. Lorentz and be completely overshadowed. Apparently Weinberg thought this gauge condition was named after the more famous Nobel laureate.

[3]

My favorite solution to Poisson’s equation is given by Feynman in his lecture on electric field.

[4]

If using the same letter \(e\) for the spinor and the mathematical constant was not confusing enough, it should be now by using it also for the electric charge.

[5]

Recall from Cluster Decomposable Hamiltonians that only connected diagrams contribute to the S-matrix due to the cluster decomposition principle.