Quarks and leptons: introductory course in modern particle physics

Francis Halzen, Alan D. Martin

Chapter 6 Electrodynamics of Spin $-\frac{1}{2}$ Particles - all with Video Answers

Educators

Chapter Questions

Problem 1

A "spinless" electron can interact with $A^\mu$ only via its charge; the coupling involves $\left(p_f+p_t\right)^\mu$. Show that
$$
\bar{u}_f \gamma^\mu u_i=\frac{1}{2 m} \bar{u}_f\left(\left(p_f+p_i\right)^\mu+i \sigma^{\mu \nu}\left(p_f-p_i\right)_v\right) u_i,
$$
from which it is possible to establish that the physical spin $-\frac{1}{2}$ electron interacts via both its charge and its magnetic moment; see also Exercise 6.2. Equation (6.7) is known as the Gordon decomposition of the current.

Mayukh Banik

Numerade Educator

Problem 2

Show that in the nonrelativistic limit, the Gordon decomposition, (6.7), of the electron current, (6.6), separates the electron interaction with an electromagnetic field $A_\mu$ into a part arising from its charge, $-e$, and a part due to its magnetic moment, $-e / 2 m$. Assume that $A_\mu$ is independent of $t$, so that (6.4) becomes
$$
T_{f i}=-i 2 \pi \delta\left(E_f-E_i\right) \int j_\mu^{f i} A^\mu d^3 x
$$

To identify the magnetic moment interaction $(-\mu \cdot \mathbf{B})$, it suffices to show that
$$
\int\left[-\frac{e}{2 m} \bar{\psi}_f i \sigma_{\mu \nu}\left(p_f-p_i\right)^\nu \psi_i\right] A^\mu d^3 x=\int \psi_A^{f \dagger}\left(\frac{e}{2 m} \boldsymbol{\sigma} \cdot \mathbf{B}\right) \psi_A^j d^3 x,
$$
where $\psi_A$ denotes the upper two (or "large") components of $\psi$; compare with eqs. (5.31) and (5.32).

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Problem 3

Making use of (6.21), prove the trace theorems and the identities (6.24).

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01:22

Problem 4

Assuming a vector-axial vector form of the weak interaction, explain why the electron emitted in the $\mu^{-}$-decay process, $\mu^{-} \rightarrow \mathrm{e}^{-} \bar{\nu}_e \nu_\mu$, must be left-handed. What is the helicity of $\mathrm{e}^{+}$from $\mu^{+}$decay?

Lazar Cvijovic

Numerade Educator

07:24

Problem 5

Use rotation matrix arguments to show that for "spinless" electrons and muons
$$
\mathscr{R}\left(\mathrm{e}^{-} \mathrm{e}^{+} \rightarrow \mu^{-} \mu^{+}\right) \propto \frac{t-u}{s} .
$$

Compare the $s$-channel photon contribution of (4.47).

Abid Hussain

Numerade Educator

Problem 6

Show that the spin-averaged interference term between the two Feynman diagrams for electron-electron scattering is that shown in the table.

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Problem 7

Justify the following relations:
$$
\begin{aligned}
\int \frac{d^3 p^{\prime}}{2 p_0^{\prime}} \delta^{(4)}\left(p+q-p^{\prime}\right) & =\int d^3 p^{\prime} d p_0^{\prime} \delta^{(4)}\left(p+q-p^{\prime}\right) \theta\left(p_0^{\prime}\right) \delta\left(p^{\prime 2}-M^2\right) \\
& =\frac{1}{2 M} \delta\left(\nu+\frac{q^2}{2 M}\right) \\
& =\frac{1}{2 M A} \delta\left(E^{\prime}-E / A\right)
\end{aligned}
$$
where $A=1+(2 E / M) \sin ^2 \frac{\theta}{2}$, and the step function $\theta(x)$ is 1 if $x>0$ and 0 otherwise.
Inserting (6.43) into (6.46) and using (6.47), we obtain
$$
\frac{d \sigma}{d E^{\prime} d \Omega}=\frac{\left(2 \alpha E^{\prime}\right)^2}{q^4}\left\{\cos ^2 \frac{\theta}{2}-\frac{q^2}{2 M^2} \sin ^2 \frac{\theta}{2}\right\} \delta\left(\nu+\frac{q^2}{2 M}\right) .
$$

Using (6.48), we may perform the $d E^{\prime}$ integration and, replacing $q^2$ by (6.44), we finally arrive at the following formula for the differential cross section for $\mathrm{e}^{-} \mu^{-}$ scattering in the laboratory frame:
$$
\left.\frac{d \sigma}{d \Omega}\right|_{\text {lab. }}=\left(\frac{\alpha^2}{4 E^2 \sin ^4 \frac{\theta}{2}}\right) \frac{E^{\prime}}{E}\left\{\cos ^2 \frac{\theta}{2}-\frac{q^2}{2 M^2} \sin ^2 \frac{\theta}{2}\right\}
$$
A powerful technique for exploring the internal structure of a target is to bombard it with a beam of high-energy electrons and to observe the angular distribution and energy of the scattered electrons. Such experiments have repeatedly led to major advances in our understanding of the structure of matter. Starting in Chapter 8, we describe how this method has revealed the structure of the proton. Equation (6.50) plays a central role in the story.

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Problem 8

Show that the cross section for elastic scattering of unpolarized electrons from spinless point-like particles is
$$
\left.\frac{d \sigma}{d \Omega}\right|_{\text {Lab, }}=\left(\frac{\alpha^2}{4 E^2 \sin ^4 \frac{\theta}{2}}\right) \frac{E^{\prime}}{E} \cos ^2 \frac{\theta}{2},
$$
where as before we neglect the mass of the electron. Justify using (6.18) with $L_{\mu \nu}^{\text {muon }}$ replaced by $\left(p+p^{\prime}\right)_\mu\left(p+p^{\prime}\right)_v$. Comparing the cross section with that for $\mathrm{e}^{-} \mu^{-} \rightarrow \mathrm{e}^{-} \mu^{-}$, we see that the $\sin ^2(\theta / 2)$ in $(6.50)$ is due to scattering from the magnetic moment of the muon.

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09:33

Problem 9

Maxwell's equations of classical electrodynamics are, in vacuo,
$$
\begin{array}{ll}
\nabla \cdot \mathbf{E}=\rho, & \nabla \times \mathbf{E}+\frac{\partial \mathbf{B}}{\partial t}=0 \\
\nabla \cdot \mathbf{B}=0, & \nabla \times \mathbf{B}-\frac{\partial \mathbf{E}}{\partial t}=\mathbf{j}
\end{array}
$$
(where we are using Heaviside-Lorentz rationalized units, see Appendix C of Aitchison and Hey). Show that these equations are equivalent to the following covariant equation for $A^\mu$ :
$$
\square^2 A^\mu-\partial^\mu\left(\partial_v A^\nu\right)=j^\mu,
$$
with $j^\mu=(\rho, \mathbf{j})$, and where $A^\mu=(\phi, \mathbf{A})$, the four-vector potential, is related to the electric and magnetic fields by
$$
\mathbf{E}=-\frac{\partial \mathbf{A}}{\partial t}-\nabla \phi, \quad \mathbf{B}=\nabla \times \mathbf{A} .
$$

Further, show that in terms of the antisymmetric field strength tensor
$$
F^{\mu v} \equiv \partial^\mu A^v-\partial^v A^\mu
$$

Maxwell's equations take the compact form
$$
\partial_\mu F^{\mu \nu}=j^p,
$$
and that current conservation, $\partial_\nu j^p=0$, follows as a natural compatibility condition. [Note that $\nabla \times(\nabla \times \mathbf{A})=-\nabla^2 \mathbf{A}+\nabla(\nabla \cdot \mathbf{A})$.]

Carson Merrill

Numerade Educator

Problem 10

Verify that $\mathbf{E}$ and $\mathbf{B}$ in (6.55) are unchanged by the gauge transformation
$$
A_\mu \rightarrow A_\mu^{\prime}=A_\mu+\partial_\mu \chi,
$$
where $\chi$ can be any function of $x$. Use this freedom to write Maxwell's equations in the form
$$
\square^2 A^\mu=j^\mu \quad \text { with } \partial_\mu A^\mu=0 .
$$

The requirement $\partial_\mu A^\mu=0$ is known as the Lorentz condition. However, even after imposing this, there is still some residual freedom in the choice of the potential $A^\mu$. We can still make another gauge transformation,
$$
A_\mu \rightarrow A_\mu^{\prime}=A_\mu+\partial_\mu \Lambda
$$
where $\Lambda$ is any function that satisfies
$$
\square^2 \Lambda=0 .
$$

This last equation ensures that the Lorentz condition is still satisfied.
Turning now from classical to quantum mechanics, we see that the wavefunction $A^\mu$ for a free photon satisfies the equation
$$
\square^2 A^\mu=0,
$$
which has solutions
$$
A^\mu=\varepsilon^\mu(\mathbf{q}) e^{-i q \cdot x} .
$$

The four-vector $\varepsilon^\mu$ is called the polarization vector of the photon. On substituting into the equation, we find that $q$, the four-momentum of the photon, satisfies
$$
q^2=0, \quad \text { that is, } m_\gamma=0 .
$$

The polarization vector has four components and yet it describes a spin-1 particle. How does this come about? First, the Lorentz condition, $\partial_\mu A^\mu=0$, gives
$$
q_\mu \varepsilon^\mu=0,
$$
and this reduces the number of independent components of $\varepsilon^\mu$ to three. Moreover, we have to explore the consequences of the additional gauge freedom (6.60). Choose a gauge parameter
$$
\Lambda=i a e^{-i q \cdot x}
$$
with $a$ constant so that (6.61) is satisfied. Substituting this, together with (6.63), into (6.60) shows that the physics is unchanged by the replacement
$$
\varepsilon_\mu \rightarrow \varepsilon_\mu^{\prime}=\varepsilon_\mu+a q_\mu .
$$

In other words, two polarization vectors $\left(\varepsilon_\mu, \varepsilon_\mu^{\prime}\right)$ which differ by a multiple of $q_\mu$ describe the same photon. We may use this freedom to ensure that the time component of $\varepsilon^k$ vanishes, $\varepsilon^0 \equiv 0$; and then the Lorentz condition (6.65) reduces to
$$
\boldsymbol{\varepsilon} \cdot \mathbf{q}=0 .
$$

This (noncovariant) choice of gauge is known as the Coulomb gauge.
From (6.67), we see that there are only two independent polarization vectors and that they are both transverse to the three-momentum of the photon. For example, for a photon traveling along the $z$ axis, we may take
$$
\boldsymbol{\varepsilon}_1=(1,0,0), \quad \varepsilon_2=(0,1,0) .
$$

A free photon is thus described by its momentum $q$ and a polarization vector $\varepsilon_i$. Since $\varepsilon_i$ transforms as a vector, we anticipate that it is associated with a particle of spin 1.

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Problem 11

Determine how the linear combinations
$$
\begin{aligned}
& \boldsymbol{\varepsilon}_R=-\sqrt{\frac{1}{2}}\left(\varepsilon_1+i \varepsilon_2\right) \\
& \boldsymbol{\varepsilon}_L=\sqrt{\frac{1}{2}}\left(\varepsilon_1-i \varepsilon_2\right)
\end{aligned}
$$
transform under a rotation $\theta$ about the $z$ axis. Hence, show that $\varepsilon_R$ and $\varepsilon_L$ describe a photon of helicity +1 and -1 , respectively; $\varepsilon_{R, L}$ are called circular polarization vectors.

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02:27

Problem 12

Show that (in the transverse gauge) the completeness relation is
$$
\sum_{\lambda=R, L}\left(\varepsilon_\lambda\right)_i^*\left(\varepsilon_\lambda\right)_j=\delta_{i j}-\hat{q}_i \hat{q}_j
$$

If $\varepsilon$ were along $q$, it would be associated with a helicity-zero photon. This state is missing because of the transversality condition, $\mathbf{q} \cdot \boldsymbol{\varepsilon}=0$. It can only be absent because the photon is massless. We return to a further discussion of photon polarization vectors in Section 6.13.

Keshav Singh

Numerade Educator

09:55

Problem 14

The condition $\partial_\lambda A^\lambda=0$ does not fully define the propagator. We are at liberty to rewrite wave equation (6.78) as
$$
\left[g^{\nu \lambda} \square^2-\left(1-\frac{1}{\xi}\right) \partial^\nu \partial^\lambda\right] A_\lambda=j^\nu .
$$

In this case, use (6.79) to show that the propagator is
$$
\frac{i}{q^2}\left(-g_{\mu \nu}+(1-\xi) \frac{q_\mu q_r}{q^2}\right) .
$$

The Feynman gauge takes $\xi=1$. But in any case, the extra term in the propagator vanishes in QED calculations in which the virtual photon is coupled to conserved currents which satisfy $q_j j^\mu=q_\nu j^\nu=0$.

Ren Jie Tuieng

Numerade Educator

01:06

Problem 15

For a vector particle of mass $M$, energy $E$, and momentum p along the $z$ axis, show that states of helicity $\lambda$ can be described by polarization vectors
$$
\begin{gathered}
\varepsilon^{(\lambda- \pm 1)}=\mp(0,1, \pm i, 0) / \sqrt{2}, \\
\varepsilon^{(\lambda-0)}=(|\mathbf{p}|, 0,0, E) / M .
\end{gathered}
$$

Narayan Hari

Numerade Educator

Problem 16

Show that the completeness relation is
$$
\sum_\lambda \varepsilon_p^{(\lambda)} * \varepsilon_p^{(\lambda)}=-g_{\mu \nu}+\frac{p_\mu p_v}{M^2}
$$
where the sum is over the three polarization states of the massive vector particle.

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Problem 17

Verify (6.100) by making use of the Fourier transform
$$
\frac{1}{|\mathbf{q}|^2}=\int d^3 x e^{i \mathbf{q} \cdot \mathbf{x}} \frac{1}{4 \pi|\mathbf{x}|} .
$$

Finally, by inspection of (6.95), we see that the division of $-g^{\mu \nu} / q^2$ into a transverse propagating contribution and a longitudinal/scalar static contribution is not a Lorentz covariant separation. Only the sum forms a covariant photon propagator.

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03:33

Problem 18

EXERCISE 6.18 Show that, individually, the amplitudes $\Re_1$ and $\Re_2$ are not gauge invariant but that their sum indeed satisfies (6.108).

It is instructive to work through the calculation of the Compton scattering amplitude in detail. For simplicity, we neglect the mass of the electron, and so the invariant variables for $\gamma(k)+\mathrm{e}(p) \rightarrow \gamma\left(k^{\prime}\right)+\mathrm{e}\left(p^{\prime}\right)$ are
$$
\begin{aligned}
& s=(k+p)^2=2 k \cdot p=2 k^{\prime} \cdot p^{\prime} \\
& t=\left(k-k^{\prime}\right)^2=-2 k \cdot k^{\prime}=-2 p \cdot p^{\prime} \\
& u=\left(k-p^{\prime}\right)^2=-2 k \cdot p^{\prime}=-2 p \cdot k^{\prime} .
\end{aligned}
$$

The two invariant amplitudes, (6.103) and (6.104), are
$$
\begin{aligned}
& \mathscr{R}_1=\varepsilon_v^{\prime *} \varepsilon_\mu e^2 \bar{u}\left(p^{\prime}\right) \gamma^\nu(\not p+k) \gamma^\mu u(p) / s, \\
& \Re_2=\varepsilon_v^{\prime *} \varepsilon_\mu e^2 \bar{u}\left(p^{\prime}\right) \gamma^\mu\left(\not p-k^{\prime}\right) \gamma^\nu u(p) / u .
\end{aligned}
$$

To obtain the unpolarized cross section, we must average/sum $\left|\Re_1+\mathscr{R}_2\right|^2$ over the initial/final electron and photon spins. Fortunately, this is not as difficult as it first appears. For physical photons, (6.98) applies, and we can make the replacement
$$
\sum_T \varepsilon_\mu^{T *} \varepsilon_{\mu^{\prime}}^T \rightarrow-g_{\mu \mathrm{F}^{\prime}}
$$
where $T$ denotes transverse. We have a similar completeness relation for the outgoing photon states, $\varepsilon^{\prime}$. Thus, for example,
$$
\overline{\left|\mathscr{R}_1\right|^2}=\frac{e^4}{4 s^2} \sum_{s, s^{\prime}}\left(\bar{u}^{\left(s^{\prime}\right)} \gamma^\nu(\not p+k) \gamma^\mu u^{(s)}\right)\left(\bar{u}^{(s)} \gamma_\mu(p+k) \gamma_v u^{\left(s^{\prime}\right)}\right) \text {. }
$$

The factor $\frac{1}{4}$ is due to averaging over the initial electron and photon spins. The spinor completeness relation, $(5.47)$, allows the sum over $u \bar{u}$ states to be performed (just as we described for $\mathrm{e}^{-} \mu^{-}$scattering in Section 6.3), and we find
$$
\begin{aligned}
& \overline{\left|\Re_1\right|^2}=\frac{e^4}{4 s^2} \operatorname{Tr}(\underbrace{\check{p}^{\prime} \gamma^\nu}_{-2 p^{\prime}}(\not p+k) \underbrace{\gamma^\mu p \gamma_\mu}_{-2 \not p}(\not p+k) \gamma_\nu^{\prime}) \\
& =\frac{e^4}{s^2} \operatorname{Tr}\left(p^{\prime} k p k\right) \\
& =\frac{4 e^4}{s^2} 2\left(p^{\prime} \cdot k\right)(p \cdot k) \\
& =2 e^4\left(-\frac{u}{s}\right) \text {. } \\
&
\end{aligned}
$$
where we have made use of (6.24) and (6.22). Similarly, we obtain
$$
\begin{aligned}
\overline{\left|\Re_2\right|^2} & =2 e^4\left(-\frac{s}{u}\right), \\
\overline{\pi_1 \Re_2^*} & =0 .
\end{aligned}
$$
Thus, the spin-averaged Compton amplitude is
$$
\overline{\mid \Im U^2}=\overline{\left|9 R_1+9 \pi_2\right|^2}=2 e^4\left(-\frac{u}{s}-\frac{s}{u}\right) .
$$

Chai Santi

Numerade Educator

00:49

Problem 19

Repeat the above calculation for an incident virtual photon of mass $k^2 \equiv-Q^2$. Continue to use (6.111). Show that for $\gamma^* \mathrm{e}^{-} \rightarrow$ $\gamma^{-}$(where $\gamma^*$ denotes a virtual photon),
$$
\overline{|\mathscr{R}|^2}=2 e^4\left(-\frac{u}{s}-\frac{s}{u}+\frac{2 Q^2 t}{s u}\right) .
$$

We shall make use of this result in Chapter 10 .

Salamat Ali

Numerade Educator

02:29

Problem 20

Restore the mass $m$ of the electron and show that at high energy, $s \rightarrow \infty$, the integrated cross section for Compton scattering is
$$
\sigma=\frac{1}{64 \pi^2 s} \int \sqrt{|\Omega|^2} d \Omega \rightarrow \frac{2 \pi \alpha^2}{s} \log \left(\frac{s}{m^2}\right) .
$$

Note that at high energy the dominant contribution comes from $9 \pi_2$, via a glancing collision in which the $u$-channel electron is almost on mass shell.

Mayukh Banik

Numerade Educator

04:23

Problem 21

Show, by using particle helicities, that high-energy Compton scattering via the first diagram of Fig. 6.12 is, in the center-of-mass frame, given by
$$
\begin{aligned}
& \left.\left|\overline{\left.R_1\right|^2} \propto\right| d_{++}^{1 / 2}(\theta)\right|^2+\left|d^{1 / 2}(\theta)\right|^2 \\
& =(1+\cos \theta) \simeq-\frac{u}{2 s}, \\
&
\end{aligned}
$$
in agreement with (6.112). An example of this type of calculation is described in Section 6.6.

Arpit Gupta

Numerade Educator

03:33

Problem 22

Draw the lowest-order Feynman diagrams for the pair annihilation process
$$
\mathrm{e}^{+}\left(p_1, s_1\right)+\mathrm{e}^{-}\left(p_2, s_2\right) \rightarrow \gamma\left(k_1, \varepsilon_1\right)+\gamma\left(k_2, \varepsilon_2\right)
$$
Check your answer with Fig. 6.13. Use the Feynman rules to show that
$$
-i \mathscr{K}=(i e)^2 \bar{v}^{\left(s_1\right)}\left(p_1\right)\left(k_1^* \frac{i}{p_1-k_1-m} k_2^*+\xi_2^* \frac{i}{p_2-k_1-m} \xi_1^*\right) u^{\left(s_2\right)}\left(p_2\right) .
$$

Verify that, in the high-energy limit, the spin-averaged rate is given by
$$
\overline{|\mathscr{R}|^2}=2 e^4\left(\frac{u}{t}+\frac{t}{u}\right) \text {. }
$$

The $\mathrm{e}^{+} \mathrm{e}^{-} \rightarrow \gamma \gamma$ cross section has both forward and backward peaks, corresponding to the $t$ - and $u$-channel exchanged electrons being almost on mass shell. Result $(6.117)$ can also be obtained by crossing the amplitude for Compton scattering. To go from $\gamma \mathrm{e}^{-} \rightarrow \gamma \mathrm{e}^{-}$to $\mathrm{e}^{+} \mathrm{e}^{-} \rightarrow \gamma \gamma$, we simple "cross" the ingoing photon with the outgoing electron,
$$
\begin{aligned}
& k, \varepsilon \rightarrow-k_2, \varepsilon_2^* \\
& p^{\prime} \rightarrow-p_1 \quad \text { and } \quad \bar{u}^{\left(s^{\prime}\right)}\left(p^{\prime}\right) \rightarrow \bar{v}^{\left(s_1\right)}\left(p_1\right) .
\end{aligned}
$$

The initial electron and outgoing photon are unaltered:
$$
u^{(s)}(p) \equiv u^{\left(s_2\right)}\left(p_2\right), \quad k^{\prime}, \varepsilon^{\prime *} \rightarrow k_1, \varepsilon_1^* .
$$

Making these substitutions in (6.103) and (6.104) gives the pair annihilation amplitude (6.117).

Chai Santi

Numerade Educator

01:04

Problem 23

Use the Feynman rules to evaluate $\mathscr{R}\left(\gamma \mathrm{e}^{-} \rightarrow \gamma \mathrm{e}^{-}\right)$corresponding to the two Feynman diagrams of Fig. 6.12 with the electron taken to have spin 0 . Show that the result is not invariant under the gauge transformation (6.66). Demonstrate that gauge invariance is restored if diagram 6.16 is included with a vertex factor $2 i e^2 \mathrm{~g}^{\mu \nu}$.

Chai Santi

Numerade Educator

01:04

Problem 23

Chai Santi

Numerade Educator