Quantum field theory 1, lecture 23

Quantum electrodynamics

Functional integral for photons

For photons, the field one integrates over in the functional integral is the gauge field \(A_\mu(x)\). The field theory is described by the partition function \[\begin{split} Z_2[J] &= \int DA~\exp\left[iS_2[A]+i\int J^\mu A_\mu \right]\\ &= \int DA ~ \exp\left[i \int d^4x~\left\{-\frac{1}{4} F^{\mu\nu}F_{\mu\nu} + J^\mu A_\mu\right\}\right]. \end{split}\] One can go to momentum space as usual \[A_\mu(x) = \int \frac{d^4 p}{(2\pi)^4}~e^{ipx} A_\mu(p),\] and finds for the term in the exponential \[\begin{split} & \int_x\left\{-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}+J^\mu A_\mu\right\} \\ & =\frac{1}{2}\int \frac{d^4 p}{(2\pi)^4}\left\{-A_\mu(-p) \left(p^2\eta^{\mu\nu}-p^{\mu}p^{\nu} \right) ~A_{\nu}(p)+J^{\mu}(-p) A_{\mu}(p) + A_{\mu}(-p)J^{\mu}(p)\right\}. \end{split}\]

Attempt to invert the inverse propagator and gauge fixing

The next step would now be to perform the Gaussian integral over \(A_\mu\) by completing the square. However, a problem arises here: The “inverse propagator” for the gauge field \[p^2\eta^{\mu\nu}-p^\mu p^\nu = p^2 \mathscr{P}^{\mu\nu}(p),\] is not invertible. We wrote it here in terms of \[\mathscr{P}_{\mu}^{\;\;\nu}(p) = \delta_{\mu}^{\;\;\nu}-\frac{p_\mu p^\nu}{p^2},\] which is in fact a projector to the space orthogonal to \(p_\nu\) \[\mathscr{P}_{\mu}^{\;\;\nu}(p) \mathscr{P}_{\nu}^{\;\;\rho}(p) =\mathscr{P}_{\mu}^{\;\;\rho}(p).\] Here we assume \(p^2\neq 0\), the special case of \(p^2=0\) for freely propagating on-shell photons needs to be discussed separately.

As a projector matrix \(\mathscr{P}_{\mu}^{\;\;\nu}(p)\) has eigenvalues \(0\) and \(1\), only. The momentum \(p_\nu\) is an eigenvector with eigenvalue zero, \[\mathscr{P}_{\mu}^{\;\;\nu}(p) \; p_\nu = 0.\] The field \(A_\nu(p)\) can be decomposed into two parts, \[A_\nu(p) = \frac{i}{e}p_\nu \beta(p) + \hat{A}_\nu(p),\] with \[\hat{A}_{\nu}(p) = \mathscr{P}_{\nu}^{\;\;\rho}(p) A_{\rho}(p),\] such that \(p^\nu \hat{A}_\nu(p) = 0.\) Moreover \[\beta(p)=\frac{e}{ip^2}p^\nu A_\nu(p).\] When acting on \(\hat{A}_\nu(p)\), the projector \(\mathscr{P}_{\mu}^{\;\;\nu}(p)\) is simply the unit matrix.

Recall that gauge transformations shift the field according to \[A_\mu(x) \to A_\mu(x)+\frac{1}{e}\partial_\mu \alpha(x),\] or in momentum space \[A_\mu(p) \to A_\mu(p) + \frac{i}{e}p_\mu \alpha(p).\] One can therefore always perform a gauge transformation such that \(\beta(p) =0\) or \[\partial^\mu A_\mu(x) =0.\] This is known as Lorenz gauge or Landau gauge. We will use this gauge in the following and restrict the functional integral to field configurations that fulfil the gauge condition.

Quadratic partition function

Now we can easily perform the Gaussian integral, \[\begin{split} Z_2[J] = & \int DA\; \exp\left[\frac{i}{2}\int_p\left\{-\left(A_\mu(-p)-J_\rho(-p)\frac{\mathscr{P}^{\rho}_{\;\;\mu}}{p^2}\right) p^2\mathscr{P}^{\mu\nu} \left(A_\nu(p)-\frac{\mathscr{P}_{\nu}^{\;\;\sigma}}{p^2}J_\sigma(p)\right)\right\}\right] \\ & \times \exp \left[\frac{i}{2}\int_p J^\mu(-p) \frac{\mathscr{P}_{\mu\nu}(p)}{p^2} J^\nu(p)\right]\\ &= \text{const} \times \exp\left[\frac{i}{2}\int_{x,y} J^{\mu}(x)\Delta_{\mu\nu}(x-y) J^{\nu}(y)\right]. \end{split}\] In the last line we used the photon propagator in position space (in Landau gauge) \[\Delta_{\mu\nu}(x-y) = \int \frac{d^4 p}{(2\pi)^4}\;e^{ip(x-y)}\frac{\mathscr{P}_{\mu\nu}(p)}{p^2-i\epsilon}.\] In the last step we have inserted the \(i\epsilon\) term as usual.

Photon propagator in position space

In the free theory one has \[\left\langle A_\mu(x) A_\nu(x) \right\rangle = \frac{1}{i^2}\left(\frac{1}{Z[J]} \frac{\delta^2}{\delta J^\mu(x)\delta J^\nu(y)} Z[J]\right)_{J=0} = \frac{1}{i}\Delta_{\mu\nu}(x-y).\] We use the following graphical notation

or with sources \(iJ^\mu(x)\) at the end points

Free solutions

To describe incoming and outgoing photons we need to discuss free solutions for the gauge field. In momentum space, and for the gauge-fixed field (Landau gauge), the linear equation of motion (Maxwell’s equation) is simply \[p^2 \mathscr{P}_{\mu}^{\;\;\nu}(p) \hat{A}_\nu(p) = p^2 \hat{A}_{\mu}(p) =0.\] Non-trivial solutions satisfy \(p^2 =0\). Without loss of generality we assume now \(p^{\mu} = (E, 0 , 0 , E)\); all other light like momenta can be obtained from this via Lorentz transformations.

Polarizations

Quite generally, a four-vector can be written as \[\hat{A}_{\nu}(p)=\left(b, \, \frac{a_1+a_2}{\sqrt{2}}, \,\frac{-ia_1+ia_2}{\sqrt{2}}, \, c\right).\] From the Landau gauge condition \(p^{\nu}\hat{A}_{\nu} = 0\) it follows that \(b=-c,\) so that one can write \[\hat{A}_\nu(p) = \tilde c \times (-E,0,0,E)+ a_1 \epsilon^{(1)}_{\nu}+a_2 \epsilon^{(2)}_{\nu},\] with \[\epsilon^{(1)}_\nu = \left(0,\frac{1}{\sqrt{2}},\frac{-i}{\sqrt{2}},0\right), \quad\quad\quad \epsilon^{(2)}_\nu = \left(0,\frac{1}{\sqrt{2}},\frac{i}{\sqrt{2}},0\right).\] However, the term \(\sim \tilde c\) is in fact proportional to \(p_\nu = (-E,0,0,E).\) We can do another gauge transformation such that \(\tilde c=0\). This does not violate the Landau gauge condition because of \(p^\nu p_\nu = 0.\) In other words, the photon field has only two independent polarization states, chosen here as positive and negative circular polarizations, or helicities.

Mode expansion

In summary, we can expand free solutions of the photon field like \[A_{\mu}(x) = \sum_{\lambda = 1}^2 \int \frac{d^3 p}{(2\pi)^3} \frac{1}{\sqrt{2E_p}}\left\{a_{\mathbf{p},\lambda}\; \epsilon^{(\lambda)}_{\mu}(p)\;e^{ipx} + a^{\dagger}_{\mathbf{p},\lambda}\; \epsilon_{\mu}^{(\lambda)*}(p) \; e^{-ipx} \right\},\] where \(E_p = |\mathbf{p}|\) is the energy of a photon. The index \(\lambda\) labels the two polarization states.

In the current setup, \(a_{\mathbf{p},\lambda}\) and \(a^{\dagger}_{\mathbf{p},\lambda}\) are simply expansion coefficients, while they become annihilation and creation operators in the operator picture. The non-trivial commutation relation becomes then \[\left[a_{\mathbf{p},\lambda}, a^{\dagger}_{\mathbf{p}^\prime,\lambda^\prime} \right] = (2\pi)^3 \delta^{(3)}(\mathbf{p}-\mathbf{p}^\prime) \delta_{\lambda\lambda^\prime}.\]

LSZ reduction formula for photons

We also need a version of the Lehmann-Symanzik-Zimmermann reduction formula for photons. Recall that for non-relativistic bosons we could replace for the calculation of the interacting part of the S-matrix \[a_{\mathbf{q}}(\infty) \to i\left[-q^0 + \tfrac{\mathbf{q}^2}{2m}+V_0\right] \varphi(q),\] \[a^{\dagger}_{\mathbf{q}}(-\infty) \to i\left[-q^0 + \tfrac{\mathbf{q}^2}{2m}+V_0\right] \varphi^{*}(q).\] For relativistic fields this is in general somewhat more complicated because of renormalization. This will be discussed in more detail in the second part of the course. In the following we will discuss only tree level diagrams where this plays no role. For photons one can replace for outgoing states \[\begin{split} & \sqrt{2E_p}\;a_{\mathbf{p},\lambda}(\infty) \to i\epsilon^{\nu*}_{(\lambda)}(p) \int d^4 x \; e^{-ipx}[-\partial_{\mu}\partial^{\mu}]A_\nu(x) \\ & \sqrt{2E_p}\;a^\dagger_{\mathbf{p},\lambda}(-\infty) \to i\epsilon^{\nu}_{(\lambda)}(p) \int d^4 x \; e^{ipx}[-\partial_{\mu}\partial^{\mu}]A_\nu(x). \end{split}\] These formulas can be used to write S-matrix elements as correlation functions of fields. Note that \([-\partial_{\mu}\partial^{\nu}]\) is essentially the inverse propagator in Landau gauge.

LSZ reduction for Dirac fermions

Finally, let us give the LSZ reduction formulas for Dirac fermions (again neglecting renormalization effects) \[\begin{split} \sqrt{2E_p}b_{\mathbf{p},s}(\infty) & \to i\int d^4 x \, e^{-ipx}\bar{u}_s(\mathbf{p}) (\gamma^\mu \partial_\mu +m) \psi(x), \\ \sqrt{2E_p}d^\dagger_{\mathbf{p},s}(-\infty) & \to -i\int d^4 x \, e^{ipx}\bar{v}_s(\mathbf{p}) (\gamma^\mu \partial_\mu +m) \psi(x), \\ \sqrt{2E_p} d_{\mathbf{p},s}(\infty) & \to -i\int d^4 x \, \bar{\psi}_s(x) (-\gamma^\mu \overleftarrow{\partial}_\mu +m) v_s(\mathbf{p})\,e^{-ipx}, \\ \sqrt{2E_p} b^\dagger_{\mathbf{p},s}(-\infty) & \to i\int d^4 x \, y\bar{\psi}_s(x) (-\gamma^\mu \overleftarrow{\partial}_\mu +m) u_s(\mathbf{p})\,e^{ipx}. \end{split}\] The left-pointing arrows indicate here that these derivatives act to the left (on the field \(\bar{\psi}_s(x)\)). These relations have been obtained as part of the exercises.

The factors of \(i\) will typically cancel out in claculations of \(S\)-matrix elements with similar factors that come with propagators. For anti-fermion lines there is an additional factor \((-1)\) in the propagator because the momentum is there counted opposite to the fermion flow direction.

Action and partition function

We are now ready to formulate the Feynman rules for a perturbative treatment of quantum electrodynamics. The microscopic action is \[\begin{split} S &= \int d^4 x \left\{-\frac{1}{4} F^{\mu\nu}F_{\mu\nu} - \bar{\psi}\gamma^\mu (\partial_\mu - ieA_\mu)\psi - m\bar{\psi}\psi \right\}\\ &= S_2[\bar{\psi},\psi, A]+ i e \int d^4 x \; \bar{\psi}\gamma^\mu A_\mu \psi. \end{split}\] The last term is cubic in the fields \(\bar{\psi},\psi\) and \(A_\mu\), while all others terms are quadratic. We will perform a perturbative expansion in the electric charge \(e\).

Let us write the partition function as \[Z[\bar{\eta},\eta,J] = \int D\bar{\psi} D\psi DA \; \exp\left[iS[\bar{\psi},\psi,A]+ i\int\left\{\bar{\eta}\psi +\bar{\psi}\eta +J^\mu A_\mu\right\} \right]\] with \(\bar{\eta}\psi = \bar{\eta}_\alpha \psi_\alpha\) where \(\alpha = 1, \ldots, 4\) sums over spinor components. Formally, one can write \[Z[\bar{\eta},\eta,J] = \text{exp} \left[-e \int d^4 x \left(\frac{1}{i}\frac{\delta}{\delta J^\mu (x)}\right)\left(i\frac{\delta}{\delta \eta_\alpha(x)}\right)(\gamma^\mu)_{\alpha\beta}\left(\frac{1}{i}\frac{\delta}{\delta \bar{\eta}_\beta(x)} \right)\right] Z_2[\bar{\eta},\eta,J],\] with quadratic partition function \[\begin{split} Z_2 &= \int D\bar{\psi} D\psi DA \; \exp\left[iS_2[\bar{\psi},\psi,A] + i\int\left\{\bar{\eta}\psi +\bar{\psi}\eta +J^\mu A_\mu\right\} \right]\\ &= \exp \left[i \int d^4 x d^4 y \; \bar{\eta}(x) S(x-y) \eta(y) \right] \times \exp \left[{\frac{i}{2}\int d^4 x d^4 y \; J^\mu(x) \Delta_{\mu\nu}(x-y) J^\nu(y)}\right]. \end{split}\]

Propagator for Dirac fermions

We have used here also the propagator for Dirac fermions \(S_{\alpha\beta}(x-y)\) introduced previously. We can now calculate S-matrix elements by first expressing them as correlation functions which get then evaluated in a perturbative expansion of the functional integral. These perturbative expressions have an intuitive graphical representation as we have briefly discussed before. We concentrate here on tree diagrams for which renormalization is not needed yet.

The correlation function of two Dirac fields can also be expressed in terms of the Dirac propagator, \[\langle \psi_\alpha(x) \bar{\psi}_\beta(y) \rangle = \frac{1}{Z_2}\left( \frac{1}{i} \frac{\delta}{\delta \bar{\eta}_\alpha(x)}\right) \left(i \frac{\delta}{\delta \eta_\beta (y)}\right) Z_2 {\Big |}_{\bar{\eta}=\eta=J=0} = \frac{1}{i} S_{\alpha \beta}(x-y).\] We introduce a graphical representation for this, as well,

With sources \(i\bar{\eta}_\alpha(x)\) and \(i\eta_\beta(y)\) at the end this would be

The conventions are such that the arrow points away from the source \(\eta\) and to the source \(\bar{\eta}\). It can also be seen as denoting the direction of fermions while anti-fermions move against the arrow direction. The Dirac indices \(\alpha,\beta\) are sometimes left implicit when there is no doubt about them.