QED Vertices and Tree Amplitudes

Gauge redundancy is not yet a theory of charged matter. It tells us that the vector potential $A_\mu$ contains unphysical components, and it tells us that local phase rotations must be compensated by a connection. The next step is to ask what that connection does in perturbation theory. The answer is QED: replace ordinary derivatives by covariant derivatives, expand the action, and read off the vertices.

This page develops the first tree-level rules for scalar QED and spinor QED. The scalar theory is useful because it makes the structure of minimal coupling visible: the covariant derivative produces not only a one-photon vertex, but also a two-photon contact vertex. The spinor theory is closer to ordinary electron–photon physics: its basic interaction is the current vertex $\bar\psi\gamma^\mu\psi A_\mu$, and one-photon exchange already contains the Coulomb force in its nonrelativistic limit.

The conceptual point is sharper than the formulas themselves. Gauge invariance does not merely suggest an interaction. It correlates different diagrams. A photon polarization can be shifted by a multiple of its momentum, and a physical amplitude must not change. At tree level this becomes a concrete algebraic statement: terms proportional to $k_\mu$ cancel after using the equations of motion and after including all diagrams required by minimal coupling.

Required background. Vector fields and gauge redundancy supplies the photon propagator, polarization equivalence, and covariant-derivative convention. Dirac-field quantization and propagators supplies external spinors, spin sums, and the fermion propagator.

Minimal coupling for a charged scalar

A bookkeeping choice independent of charge is whether every momentum is drawn incoming to a vertex or physical incoming/outgoing momenta are followed along a matter line. Compact vertices below use the physical-line convention; the all-incoming translation is given afterward. The invariant check is the Ward identity, not the memorized sign of an isolated vertex.

A complex scalar field has a global phase symmetry

$$\phi\mapsto e^{iq\alpha}\phi, \qquad \phi^*\mapsto e^{-iq\alpha}\phi^*.$$

If $\alpha$ is constant, the free Lagrangian

$$\mathcal L_0=(\partial_\mu\phi)^*(\partial^\mu\phi)-m^2\phi^*\phi$$

is invariant. If $\alpha=\alpha(x)$, the derivative of $\phi$ transforms with an extra term:

$$\partial_\mu\phi\mapsto e^{iq\alpha(x)} \left(\partial_\mu\phi+iq(\partial_\mu\alpha)\phi\right).$$

The cure is to introduce $A_\mu$ and replace $\partial_\mu$ by

$$D_\mu\phi=(\partial_\mu-iqA_\mu)\phi.$$

With $A_\mu\mapsto A_\mu+\partial_\mu\alpha$, one finds

$$D_\mu\phi\mapsto e^{iq\alpha(x)}D_\mu\phi.$$

The gauge-invariant scalar QED Lagrangian is therefore

$$\mathcal L_{\mathrm{scalar\ QED}} =-{1\over4}F_{\mu\nu}F^{\mu\nu} +(D_\mu\phi)^*(D^\mu\phi)-m^2\phi^*\phi, \qquad F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu.$$

Expanding the covariant derivative gives

$$\mathcal L_{\mathrm{scalar\ QED}} =-{1\over4}F_{\mu\nu}F^{\mu\nu} +\partial_\mu\phi^*\partial^\mu\phi-m^2\phi^*\phi +iqA_\mu\left(\phi^*\partial^\mu\phi-(\partial^\mu\phi^*)\phi\right) +q^2A_\mu A^\mu\phi^*\phi.$$

There are two interaction terms. The term linear in $A_\mu$ gives a one-photon scalar vertex. The term quadratic in $A_\mu$ gives a contact vertex with two photons and two scalars. That contact term is often called the seagull vertex because of its shape in diagrams.

For a scalar line with charge flowing from incoming momentum $p$ to outgoing momentum $p'$, and with the photon index $\mu$, the three-point vertex is

$$+iq(p+p')^\mu.$$

For two photons with indices $\mu,\nu$, the four-point scalar–scalar–photon–photon vertex is

$$2iq^2\eta^{\mu\nu}.$$

The factor of $2$ comes from the two identical photon fields in $A_\mu A^\mu$. It is not optional; without it, Ward identities fail.

A reliable way to check scalar-QED signs is to contract the one-photon vertex with the incoming photon momentum. With $k=p'-p$ and equal-mass scalar legs,

$$k\cdot(p+p')=p'^2-p^2=0$$

on shell. For amplitudes with two external photons, the same cancellation leaves contact terms, and those are precisely canceled by the seagull vertex.

Minimal coupling produces different elementary vertices for scalar and spinor matter. Scalar QED has both the one-photon vertex $+iq(p+p')^\mu$ and the seagull vertex $2iq^2\eta^{\mu\nu}$. Spinor QED has the current vertex $+iq\gamma^\mu$, which becomes $-ie\gamma^\mu$ for an electron of charge $q=-e$.

Spinor QED and the current vertex

For a Dirac field, the gauge-invariant Lagrangian is

$$\mathcal L_{\mathrm{spinor\ QED}} =-{1\over4}F_{\mu\nu}F^{\mu\nu} +\bar\psi(i\gamma^\mu D_\mu-m)\psi.$$

For a field of charge $q$,

$$D_\mu\psi=(\partial_\mu-iqA_\mu)\psi,$$

so

$$\bar\psi(i\gamma^\mu D_\mu-m)\psi =\bar\psi(i\gamma^\mu\partial_\mu-m)\psi +q\bar\psi\gamma^\mu\psi A_\mu.$$

Thus

$$\mathcal L_{\mathrm{int}}=q\bar\psi\gamma^\mu\psi A_\mu.$$

For the electron, $q=-e$, so

$$\mathcal L_{\mathrm{int}}=-e\bar\psi\gamma^\mu\psi A_\mu, \qquad \text{vertex}=-ie\gamma^\mu.$$

The difference between scalar QED and spinor QED is worth pausing over. The scalar kinetic term is quadratic in derivatives, so replacing $\partial$ by $D$ produces both $A\phi\partial\phi$ and $A^2\phi^2$ interactions. The Dirac kinetic term is first order, so minimal coupling produces only one power of $A_\mu$. There is no elementary two-photon–two-fermion seagull vertex in ordinary spinor QED.

The classical current coupled to the photon is

$$j^\mu=q\bar\psi\gamma^\mu\psi.$$

For the electron, $j^\mu=-e\bar\psi\gamma^\mu\psi$. The sign of the current reflects the sign of the charge; the sign of the vertex reflects the convention used in writing $D_\mu$ and the perturbative expansion.

Tree-level QED rulebook

It is useful to fix what a diagram means before assembling amplitudes. We define $\mathcal M$ by

$$\langle f|S|i\rangle =\langle f|i\rangle +i(2\pi)^4\delta^{(4)}(P_f-P_i)\mathcal M_{fi}.$$

Thus the product of propagators, vertices, and external wavefunctions supplied by a connected diagram is $i\mathcal M$. The free internal-line factors are

$$\begin{aligned} \text{charged scalar:}\quad&{i\over p^2-m^2+i\epsilon},\\ \text{charged fermion:}\quad&{i(\gamma\cdot p+m)\over p^2-m^2+i\epsilon},\\ \text{photon in Feynman gauge:}\quad&{-i\eta_{\mu\nu}\over k^2+i\epsilon}. \end{aligned}$$

For physical incoming and outgoing momenta $p$ and $p'$ along a matter line, the vertices are

$$\begin{aligned} A_\mu\phi^*\phi:&\quad +iq(p+p')^\mu,\\ A_\mu A_\nu\phi^*\phi:&\quad 2iq^2\eta^{\mu\nu},\\ A_\mu\bar\psi\psi:&\quad +iq\gamma^\mu. \end{aligned}$$

In an all-incoming convention, if $p_\phi$ labels the $\phi$ leg and $p_{\phi^*}$ labels the $\phi^*$ leg, the scalar three-point rule is $+iq(p_\phi-p_{\phi^*})^\mu$. Since an outgoing scalar of momentum $p'$ is represented by an incoming $\phi^*$ momentum $-p'$, this is the same rule as $+iq(p+p')^\mu$. Incoming photons carry $\varepsilon_\mu$ and outgoing photons carry $\varepsilon_\mu^*$; the complex conjugation is not optional for circular polarization.

External wavefunctions and one-photon emission

At tree level, external fields are replaced by on-shell wavefunctions. For a fermion of momentum $p$ and spin $s$,

$$(\gamma\cdot p-m)u_s(p)=0, \qquad \bar u_s(p)(\gamma\cdot p-m)=0.$$

For a photon of momentum $k$ and polarization $\varepsilon_\mu(k)$,

$$k^2=0, \qquad k\cdot\varepsilon(k)=0, \qquad \varepsilon_\mu(k)\sim \varepsilon_\mu(k)+c\,k_\mu.$$

The last relation says that the representative of the polarization vector is gauge-dependent. A physical amplitude must be unchanged by replacing $\varepsilon_\mu$ by $\varepsilon_\mu+c k_\mu$.

The elementary amplitude for a photon to couple to an electron line contains the matrix element

$$\bar u_{s'}(p')\,(-ie\gamma^\mu)\,u_s(p)\,\varepsilon_\mu(k), \qquad k=p'-p,$$

up to the overall convention relating diagram values to $i\mathcal M$. Its Ward check is immediate:

$$k_\mu\bar u_{s'}(p')\gamma^\mu u_s(p) =\bar u_{s'}(p')\gamma\cdot(p'-p)u_s(p).$$

Using the Dirac equation on the left and on the right,

$$\bar u_{s'}(p')\gamma\cdot p' = m\bar u_{s'}(p'), \qquad \gamma\cdot p\,u_s(p)=m u_s(p),$$

so

$$k_\mu\bar u_{s'}(p')\gamma^\mu u_s(p)=0.$$

This is the simplest form of current conservation on an external line. It is also the tree-level seed of the Ward–Takahashi identities.

The external-photon Ward check. Replacing $\varepsilon_\mu(k)$ by $k_\mu$ probes the longitudinal representative of the photon polarization. Current conservation makes this contribution vanish in a physical amplitude.

Photon exchange between charged particles

The simplest nontrivial QED amplitude is one-photon exchange. For two distinguishable charged fermions with electron-like vertex $-ie\gamma^\mu$, the tree contribution is

$$i\mathcal M =\left[\bar u(p_1')(-ie\gamma^\mu)u(p_1)\right] { -i\eta_{\mu\nu}\over q^2+i\epsilon} \left[\bar u(p_2')(-ie\gamma^\nu)u(p_2)\right], \qquad q=p_1'-p_1=p_2-p_2'.$$

Equivalently,

$$\mathcal M =e^2{\left[\bar u(p_1')\gamma^\mu u(p_1)\right] \left[\bar u(p_2')\gamma_\mu u(p_2)\right] \over q^2+i\epsilon},$$

for the stated $S$-matrix and vertex conventions. More generally the numerator carries the product of the two charges, $q_1q_2$. The current–propagator–current structure is

$$J_1^\mu(q)\,D_{\mu\nu}(q)\,J_2^\nu(-q).$$

In a covariant gauge, the extra propagator numerator is proportional to $q_\mu q_\nu$. It drops out because the on-shell currents obey $q_\mu J_1^\mu=q_\nu J_2^\nu=0$. This both checks the momentum assignments and proves that the exchange amplitude is independent of the gauge parameter.

Tree-level photon exchange has the universal structure current–propagator–current. In Feynman gauge the internal photon contributes $-i\eta_{\mu\nu}/(q^2+i\epsilon)$, while each fermion current contributes a spinor bilinear $\bar u\gamma^\mu u$.

The nonrelativistic limit recovers the Coulomb interaction. For a slowly moving fermion,

$$\bar u(p')\gamma^0u(p)\simeq 2m, \qquad \bar u(p')\gamma^iu(p)\ll 2m.$$

The momentum transfer has $q^0\simeq0$, hence

$$q^2=(q^0)^2-\mathbf q^2\simeq-\mathbf q^2.$$

Therefore the dominant exchange is the temporal current coupled through

$${1\over \mathbf q^2}.$$

More explicitly, for charges $q_1$ and $q_2$,

$$\mathcal M_{\mathrm{NR}} \simeq-{4m_1m_2q_1q_2\over\mathbf q^2}.$$

The relativistic Born-amplitude relation $\widetilde V(\mathbf q)=-\mathcal M_{\mathrm{NR}}/(4m_1m_2)$ then gives

$$\widetilde V(\mathbf q)={q_1q_2\over\mathbf q^2}.$$

Fourier transformation yields the Coulomb potential,

$$V(r)={q_1q_2\over4\pi r}.$$

This is a useful sanity check: the same minimal coupling that enforces local gauge invariance also reproduces the familiar long-range electromagnetic force.

Scalar Compton scattering and the seagull term

Consider scalar Compton scattering,

$$\phi(p)+\gamma(k,\varepsilon)\longrightarrow \phi(p')+\gamma(k',\varepsilon'), \qquad p+k=p'+k'.$$

The scalar QED Lagrangian produces three tree diagrams: two exchange diagrams and one contact diagram. The contact diagram is required by gauge invariance.

Scalar Compton scattering contains two scalar-exchange diagrams plus the $A^2\phi^*\phi$ contact interaction. The three diagrams are not separately gauge invariant; their sum is.

The three contributions have the schematic form

$$i\mathcal M_s =\left[+iq(2p+k)^\mu\varepsilon_\mu\right] {i\over(p+k)^2-m^2+i\epsilon} \left[+iq(2p'+k')^\nu\varepsilon_\nu'^*\right],$$ $$i\mathcal M_u =\left[+iq(2p-k')^\nu\varepsilon_\nu'^*\right] {i\over(p-k')^2-m^2+i\epsilon} \left[+iq(2p'-k)^\mu\varepsilon_\mu\right],$$

and

$$i\mathcal M_{\mathrm{contact}} =2iq^2\eta^{\mu\nu}\varepsilon_\mu\varepsilon_\nu'^*.$$

The precise placement of signs depends on whether all momenta are drawn incoming or as physical incoming/outgoing momenta. A reliable way to avoid confusion is to keep the scalar charge flow fixed along the line and then translate to the chosen all-incoming convention only at the end. What does not depend on notation is the Ward identity:

$$\mathcal M(\varepsilon_\mu\to k_\mu)=0, \qquad \mathcal M(\varepsilon_\nu'^*\to k_\nu')=0.$$

The cancellation works because, on shell,

$$k\cdot(2p+k)=(p+k)^2-p^2=(p+k)^2-m^2,$$

and similarly for the other scalar-exchange channel. Thus the numerator of an exchange diagram can cancel its scalar propagator. The leftover local terms are canceled by the contact diagram. This is the practical reason why the $A^2\phi^*\phi$ term cannot be ignored even when one is mainly interested in one-photon vertices.

Spinor Compton scattering

For spinor QED,

$$e(p)+\gamma(k,\varepsilon)\to e(p')+\gamma(k',\varepsilon'), \qquad p+k=p'+k',$$

there are two tree diagrams. There is no seagull vertex because the Dirac Lagrangian is first order in derivatives. With the electron-like vertex $-ie\gamma^\mu$, the amplitude is

$$i\mathcal M_{\mathrm{spinor}} =(-ie)^2\bar u(p') \left[ (\gamma\cdot\varepsilon'^*){i(\gamma\cdot(p+k)+m)\over (p+k)^2-m^2+i\epsilon}(\gamma\cdot\varepsilon) +(\gamma\cdot\varepsilon){i(\gamma\cdot(p-k')+m)\over (p-k')^2-m^2+i\epsilon}(\gamma\cdot\varepsilon'^*) \right]u(p).$$

Here

$$\gamma\cdot\varepsilon=\gamma^\mu\varepsilon_\mu, \qquad \gamma\cdot\varepsilon'^*=\gamma^\mu\varepsilon'^*_{\mu}.$$

The two terms are the two possible orderings of photon absorption and emission along the fermion line. Their matrix order matters because gamma matrices do not commute.

Spinor Compton scattering contains the two possible orderings of photon insertions along the fermion line. Gauge invariance belongs to the sum of the two diagrams, not to either diagram separately.

The Ward identity again comes from differences of inverse propagators. Strip off the common couplings and factors of $i$, and write

$$T^{\nu\mu} =\bar u(p')\left[ \gamma^\nu S_0(p+k)\gamma^\mu +\gamma^\mu S_0(p-k')\gamma^\nu \right]u(p),$$

where

$$S_0(r)={\gamma\cdot r+m\over r^2-m^2}, \qquad S_0^{-1}(r)=\gamma\cdot r-m$$

on the nonsingular internal momenta. Contract the incoming-photon index with $k_\mu$. In the first ordering use

$$\gamma\cdot k=S_0^{-1}(p+k)-S_0^{-1}(p).$$

The second term annihilates $u(p)$, while the first cancels $S_0(p+k)$ and leaves $\bar u(p')\gamma^\nu u(p)$. For the other ordering, momentum conservation gives

$$\gamma\cdot k=S_0^{-1}(p')-S_0^{-1}(p-k').$$

The first term annihilates $\bar u(p')$ and the second cancels the internal propagator with the opposite sign. Consequently,

$$k_\mu T^{\nu\mu} =\bar u(p')\gamma^\nu u(p) -\bar u(p')\gamma^\nu u(p)=0.$$

This telescoping cancellation is the fermionic counterpart of the scalar-QED cancellation among exchange and contact diagrams. Repeating the argument with $k'_\nu$ proves the Ward identity for the outgoing photon.

Spin sums and traces

After writing an amplitude, one often squares it and sums or averages over unobserved spin states. With the standard relativistic normalization,

$$\sum_s u_s(p)\bar u_s(p)=\gamma\cdot p+m, \qquad \sum_s v_s(p)\bar v_s(p)=\gamma\cdot p-m.$$

The basic trace identities are

$$\operatorname{Tr}(\gamma^\mu\gamma^\nu)=4\eta^{\mu\nu},$$

and

$$\operatorname{Tr}(\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma) =4\left( \eta^{\mu\nu}\eta^{\rho\sigma} -\eta^{\mu\rho}\eta^{\nu\sigma} +\eta^{\mu\sigma}\eta^{\nu\rho} \right).$$

For example, a spin-summed fermion current product becomes

$$\sum_{s,s'} \left[\bar u_{s'}(p')\gamma^\mu u_s(p)\right] \left[\bar u_{s'}(p')\gamma^\rho u_s(p)\right]^* = \operatorname{Tr}\left[(\gamma\cdot p'+m)\gamma^\mu(\gamma\cdot p+m)\gamma^\rho\right].$$

Evaluating the trace gives

$$L^{\mu\rho}(p',p) =4\left[p'^\mu p^\rho+p'^\rho p^\mu -\eta^{\mu\rho}(p'\cdot p-m^2)\right].$$

A sum and an average are different operations. For an unpolarized process, sum over every unobserved final spin or helicity and divide by the number of equally populated initial states:

$$\overline{|\mathcal M|^2} ={1\over\prod_{a\in\mathrm{initial}}N_a} \sum_{\substack{\text{initial spins}\\\text{final spins}}} |\mathcal M|^2,$$

where $N_a=2s_a+1$ for a massive particle and $N_a=2$ for a photon. For one-photon exchange between two incoming spin-one-half particles, the average factor is $1/4$. Writing the charges as $q_1,q_2$ and the momentum transfer as $Q$, one obtains the useful trace form

$$\overline{|\mathcal M|^2} ={q_1^2q_2^2\over(Q^2)^2} L_1^{\mu\nu}L_{2,\mu\nu},$$

where each averaged current tensor contains its own factor of $1/2$,

$$L_a^{\mu\nu} ={1\over2}\operatorname{Tr}\!\left[ (\gamma\cdot p_a'+m_a)\gamma^\mu (\gamma\cdot p_a+m_a)\gamma^\nu \right].$$

This separation is a useful defense against a common factor-of-two error: completeness relations perform sums, whereas the prefactor performs the initial-state average.

For photons, polarization sums must be handled with care. In a gauge-invariant amplitude, gauge-dependent pieces proportional to $k_\mu$ do not contribute, so one may effectively use

$$\sum_{\lambda\in\mathrm{phys}}\varepsilon_\mu^{(\lambda)}(k) \varepsilon_\nu^{(\lambda)}(k)^* \;\rightsquigarrow\;-\eta_{\mu\nu}$$

inside a complete gauge-invariant squared amplitude. The arrow is deliberate: the equality is not an identity of physical polarization vectors in an arbitrary gauge; it is a shortcut justified by Ward identities. A good practical test is this: before replacing a physical-polarization sum by $-\eta_{\mu\nu}$, verify that the rest of the amplitude vanishes when any external polarization is replaced by its momentum.

Crossing as external-leg bookkeeping

A time-ordered Green function does not know in advance which external legs will be interpreted as incoming particles, outgoing particles, incoming antiparticles, or outgoing antiparticles. These interpretations are imposed by LSZ reduction and by the choice of which pole is approached. Moving a charged external leg from one side of a process to the other turns it into an antiparticle and reverses the sign of the momentum in the analytic amplitude.

Schematically,

$$\mathcal M\bigl(A\to B+\bar f(p)\bigr) \quad\longleftrightarrow\quad \mathcal M\bigl(A+f(-p)\to B\bigr),$$

with the spinor wavefunction changed from a $v$ spinor to a $u$ spinor, or conversely, depending on the direction of the fermion arrow.

The external-wavefunction dictionary is

$$\begin{array}{c|c} \text{external state} & \text{spinor factor}\\ \hline \text{incoming fermion} & u(p)\\ \text{outgoing fermion} & \bar u(p)\\ \text{incoming antifermion} & \bar v(p)\\ \text{outgoing antifermion} & v(p) \end{array}$$

Thus crossing an outgoing antifermion replaces the analytically continued $v(p)$ by an incoming-fermion $u(-p)$ assignment. For processes with several fermions, one must also preserve a fixed ordering of external fermionic operators. Reordering that convention can produce a relative minus sign; it is not captured by the shorthand $p\to-p$ alone.

Crossing is the analytic continuation that moves an external leg across the amplitude. Momentum changes sign, and a particle becomes the corresponding antiparticle. Fermion arrows help keep the spinor factors in the correct order.

This is why Feynman rules are usually formulated with all momenta incoming and with continuous fermion-number arrows. Once the momentum and fermion ordering are fixed, the same analytic function describes several physical channels.

Summary

Minimal coupling converts a global phase symmetry into a local gauge symmetry by replacing ordinary derivatives with covariant derivatives. For a complex scalar, this produces both a one-photon vertex and a two-photon seagull vertex. For a Dirac field, the first-order kinetic term produces the current interaction $\bar\psi\gamma^\mu\psi A_\mu$ and hence the familiar photon–fermion vertex.

Tree amplitudes are assembled from external wavefunctions, propagators, vertices, and momentum conservation. Their most important consistency test is gauge invariance: replacing an external photon polarization by its momentum must give zero. For a single on-shell fermion current this follows directly from the Dirac equation. For scalar Compton scattering it follows only after the exchange diagrams and the seagull diagram are added together.

Spin sums turn squared amplitudes into traces of gamma matrices. Crossing then explains why the same analytic expression can describe several reactions: moving a charged external leg across the amplitude changes a particle into an antiparticle and sends $p\to-p$. These are the practical tools needed before one-loop QED and power counting enter the course.

Common pitfalls

Omitting the scalar seagull. The $A^2\phi^*\phi$ vertex is required by minimal coupling. Exchange graphs alone do not satisfy the two-photon Ward identities.

Testing gauge invariance diagram by diagram. A single exchange graph need not be gauge invariant. Apply the polarization-to-momentum replacement to the complete set of diagrams at the chosen perturbative order.

Using $-\eta_{\mu\nu}$ too early. This replacement is justified only inside a gauge-invariant squared amplitude. Gauge-dependent terms vanish after contraction with conserved currents or after the Ward-related diagrams have been summed.

Confusing sums with averages. Completeness relations sum over spin states; they do not divide by the number of initial states. Insert the initial-state average only once, after the amplitude has been squared.

Mixing charge signs with momentum-flow signs. The convention $D_\mu=\partial_\mu-iqA_\mu$ gives a fermion vertex $+iq\gamma^\mu$; for $q=-e$ this is $-ie\gamma^\mu$. The all-incoming convention changes the scalar momentum labels, not the underlying Ward identity.

Treating crossing as only $p\to-p$. A crossed fermion also changes between $u$ and $v$ wavefunctions, and a fixed ordering of external fermions must be maintained. Any relative Fermi sign comes from that ordering, not from an arbitrary sign attached to crossing.

Exercises

Exercise 1

Derive the scalar QED three-point and four-point vertices from

$$\mathcal L=(D_\mu\phi)^*(D^\mu\phi)-m^2\phi^*\phi, \qquad D_\mu=\partial_\mu-iqA_\mu.$$

Use the convention that the scalar enters with momentum $p$ and leaves with momentum $p'$.

Solution

Expanding the covariant derivative gives

$$(D_\mu\phi)^*(D^\mu\phi) =(\partial_\mu\phi^*+iqA_\mu\phi^*) (\partial^\mu\phi-iqA^\mu\phi).$$

Thus

$$(D_\mu\phi)^*(D^\mu\phi) =\partial_\mu\phi^*\partial^\mu\phi +iqA_\mu\left(\phi^*\partial^\mu\phi-(\partial^\mu\phi^*)\phi\right) +q^2A_\mu A^\mu\phi^*\phi.$$

The term linear in $A_\mu$ gives the one-photon vertex. With the standard scalar-line momentum assignment, the two derivative terms combine into $(p+p')^\mu$. Including the factor of $i$ from $iS_{\mathrm{int}}$, the vertex is

$$+iq(p+p')^\mu.$$

The quadratic term gives a vertex with two photon fields. Differentiating twice with respect to $A_\mu$ and $A_\nu$ gives a factor $2\eta^{\mu\nu}$, and the factor of $i$ from $iS_{\mathrm{int}}$ gives

$$2iq^2\eta^{\mu\nu}.$$

Exercise 2

Show that the on-shell fermion current

$$J^\mu=\bar u(p')\gamma^\mu u(p)$$

obeys

$$(p'-p)_\mu J^\mu=0.$$

Solution

Let $k=p'-p$. Then

$$k_\mu J^\mu =\bar u(p')\gamma\cdot(p'-p)u(p) =\bar u(p')\gamma\cdot p'\,u(p)-\bar u(p')\gamma\cdot p\,u(p).$$

The Dirac equations are

$$(\gamma\cdot p-m)u(p)=0, \qquad \bar u(p')(\gamma\cdot p'-m)=0.$$

Therefore

$$\bar u(p')\gamma\cdot p'=m\bar u(p'), \qquad \gamma\cdot p\,u(p)=m u(p).$$

Substituting,

$$k_\mu J^\mu =m\bar u(p')u(p)-m\bar u(p')u(p)=0.$$

This is the tree-level Ward identity for a photon attached to an external on-shell fermion line.

Exercise 3

Using the trace identities in the text, prove that

$$\operatorname{Tr}\left[(\gamma\cdot p'+m)\gamma^\mu(\gamma\cdot p+m)\gamma^\rho\right] =4\left[p'^\mu p^\rho+p'^\rho p^\mu -\eta^{\mu\rho}(p'\cdot p-m^2)\right].$$

Solution

Expand the trace:

$$\operatorname{Tr}\left[(\gamma\cdot p'+m)\gamma^\mu(\gamma\cdot p+m)\gamma^\rho\right] = p'_\alpha p_\beta \operatorname{Tr}(\gamma^\alpha\gamma^\mu\gamma^\beta\gamma^\rho) +m^2\operatorname{Tr}(\gamma^\mu\gamma^\rho),$$

because the terms with three gamma matrices have zero trace. Now use

$$\operatorname{Tr}(\gamma^\alpha\gamma^\mu\gamma^\beta\gamma^\rho) =4\left(\eta^{\alpha\mu}\eta^{\beta\rho} -\eta^{\alpha\beta}\eta^{\mu\rho} +\eta^{\alpha\rho}\eta^{\mu\beta}\right)$$

and

$$\operatorname{Tr}(\gamma^\mu\gamma^\rho)=4\eta^{\mu\rho}.$$

Therefore

$$p'_\alpha p_\beta \operatorname{Tr}(\gamma^\alpha\gamma^\mu\gamma^\beta\gamma^\rho) =4\left(p'^\mu p^\rho-(p'\cdot p)\eta^{\mu\rho}+p'^\rho p^\mu\right),$$

and the mass term adds $4m^2\eta^{\mu\rho}$. Combining terms gives

$$4\left[p'^\mu p^\rho+p'^\rho p^\mu -\eta^{\mu\rho}(p'\cdot p-m^2)\right].$$

Exercise 4

In scalar QED, explain why the seagull graph is required for gauge invariance of scalar Compton scattering. You do not need to compute the full amplitude; show how replacing one polarization vector by its momentum cancels a scalar propagator.

Solution

Consider the part of the $s$-channel diagram where the incoming photon of momentum $k$ attaches to an incoming on-shell scalar of momentum $p$. The vertex contains

$$+iq(2p+k)^\mu\varepsilon_\mu.$$

Replace $\varepsilon_\mu$ by $k_\mu$. Then

$$k\cdot(2p+k)=2p\cdot k+k^2.$$

Since the external photon is on shell, $k^2=0$, and since the scalar is on shell, $p^2=m^2$. Hence

$$2p\cdot k+k^2=(p+k)^2-p^2=(p+k)^2-m^2.$$

This is the inverse of the scalar propagator in that channel. Thus the Ward variation of the exchange diagram collapses the internal scalar propagator and leaves a local term. The corresponding variation of the other exchange diagram leaves another local term. These local remnants are canceled by the variation of the contact diagram from

$$q^2A_\mu A^\mu\phi^*\phi.$$

Therefore the exchange diagrams alone are not gauge invariant; the complete minimal-coupling set is.

Exercise 5

Two distinguishable spin-one-half particles of charges $q_1,q_2$ scatter by one-photon exchange. Starting from

$$\mathcal M={q_1q_2\over Q^2} \left[\bar u(p_1')\gamma^\mu u(p_1)\right] \left[\bar u(p_2')\gamma_\mu u(p_2)\right],$$

write the unpolarized squared amplitude as a product of two traces. Identify the initial-state average.

Solution

There are two spin states for each incoming particle, so the initial-state average is $1/(2\cdot2)=1/4$. Summing final spins and using the completeness relations gives

$$\begin{aligned} \overline{|\mathcal M|^2} ={q_1^2q_2^2\over(Q^2)^2}{1\over4} &\operatorname{Tr}\!\left[ (\gamma\cdot p_1'+m_1)\gamma^\mu (\gamma\cdot p_1+m_1)\gamma^\nu \right]\\ \times{}&\operatorname{Tr}\!\left[ (\gamma\cdot p_2'+m_2)\gamma_\mu (\gamma\cdot p_2+m_2)\gamma_\nu \right]. \end{aligned}$$

Equivalently, absorb one factor of $1/2$ into each current tensor. No average is taken over final spins unless the measurement explicitly prepares an incoherent ensemble of final states; unobserved final spins are summed.

Further reading

• Coleman, Sidney. Lectures of Sidney Coleman on Quantum Field Theory. Edited by Bryan Gin-ge Chen, David Derbes, David Griffiths, Brian Hill, Richard Sohn, and Yuan-Sen Ting. World Scientific, 2019. The electrodynamics chapters develop minimal coupling and low-order spinor-QED calculations.
• Schwartz, Matthew D. Quantum Field Theory and the Standard Model. Cambridge University Press, 2014. The chapters on QED scattering and gauge invariance give detailed tree-amplitude and Ward-identity derivations.
• Srednicki, Mark. Quantum Field Theory. Cambridge University Press, 2007. The scalar- and spinor-electrodynamics chapters derive the vertices, spin sums, and Ward identities used here.
• Weinberg, Steven. The Quantum Theory of Fields, Volume I: Foundations. Cambridge University Press, 1995. Chapters 6 and 8 develop scattering rules, electrodynamics, and gauge invariance.
• Zee, A. Quantum Field Theory in a Nutshell. 2nd ed., Princeton University Press, 2010. The discussions of electron scattering and diagrammatic gauge invariance emphasize the physical Ward checks.