I have tried several textbooks to study quantum field theory. Below is a quick summary over them.

Maggiore's A Modern Introduction to Quantum Field Theory is a great starting point, especially for those who have already studied group theory. Unfortunately, there are many omissions in its derivations, as well as some serious (yet hard to detect) errors starting from Chapter 4. I would suggest that the reader should complete Chapter 2 and Chapter 3, and then move on to Peskin and Schroeder.

Weinberg's The Quantum Theory of Fields is great for those who already know quite a bit about quantum field theory. But it contains too much information and will only overwhelm the uninitiated, although one should keep a copy of it for reference purposes.

Peskin and Schroeder's An Introduction to Quantum Field Theory is probably the best book overall. Certain pedagogical choices are in my opinion suboptimal. For example, complex scalar fields and Majorana fields are only included in exercises, although some key insight could be gained by treating real scalar fields as a special case of complex scalar fields, and Majorana fields as a special case of Dirac fields, and drawing an analogy between the two. Also, the discussion on C-, P-, and T-symmetries has some issues and only includes Dirac fields. Despite these imperfections, it is probably still the best choice for novices. One can always turn to Weinberg if they find certain parts unclear or confusing in Peskin and Schroeder.

The rest of this post will focus on Peskin and Schroeder. I will include notes that I believe are useful for first-time learners. It is impossible to gain a real understanding of quantum field theory without doing its exercises. There are solutions available online posted by various individuals, including Xianyu, Bourjaily, and Pokraka. Unfortunately, all of them have serious problems here and there. I will point to the best solution to each problem, often with some typo corrections. In the rare cases where none of them is satisfactory, I will post my own solution.

Here is a link to the official Errata.

Chapter 1

The reader can safely skip it during the first read. It is probably a good idea for those who have the habit of trying to understand every single sentence.

Chapter 2

Notes

Page 25

In the derivation of (2.45), we use the fact that \([\phi(\mathbf{x}, t), \nabla \phi(\mathbf{x'}, t)] = 0\). Note that \([\phi(\mathbf{x}, t), \nabla_{} \phi(\mathbf{x'}, t)]\) really means \([\phi(\mathbf{x}, t), \nabla_{\mathbf{x'}} \phi(\mathbf{x'}, t)]\) and therefore it is equal to \(\nabla_{\mathbf{x'}}[\phi(\mathbf{x}, t), \phi(\mathbf{x'}, t)] = 0\).

Alternatively, we can calculate \([\phi(\mathbf{x}, t), \nabla \phi(\mathbf{x'}, t)]\) from (2.25), and get something like \(\int \frac{d^3p}{(2\pi)^3} \frac{\mathbf{p}}{\omega_\mathbf{p}} (e^{i \mathbf{p} \cdot (\mathbf{x}-\mathbf{x'})} + e^{i \mathbf{p} \cdot (\mathbf{x'}-\mathbf{x})}) \), which vanishes because the integrand is odd.

Page 27

To get the right sign for the denominator in (2.52), we need to rotate \(p\) continuously on the complex plane from a real number to an imaginary number.

Problems

2.1

Xianyu's solution is good, except for an obvious typo in (2.5), where the last \(F_{\mu\lambda}\) should be \(F_{\lambda\mu}\).

It should be stressed that when we calculate energy and momentum densities, we are free to add or subtract any 4-divergence terms, because their integral leads to surface terms, which we take to vanish at infinity. The implication is that any densities are well defined only in a smeared sense, i.e. as an average over a big enough area.

2.2

Xianyu's solution is good, except for the following typos or minor issues.

In the derivation of (2.7), \(b_\mathbf{q} a_\mathbf{q}\) should be \(b_\mathbf{p} a_\mathbf{q}\) before the fourth equal sign. Also, \(d^x\) should be \(d^3 \frac{p}{(2 \pi)^3}\) in the last two lines, .

Regarding (2.19), there should be a minus sign if we use (2.12). But as usual, it is a matter of definition so we don't really care. Also, the order of \(\phi^*\) and \(\dot{\phi}\) does not matter, since its choice will only introduce an infinite constant term that we will throw away in the end. However, this particular order does make it more complicated for (d). \(e^{i(E_\mathbf{p} + E_\mathbf{q} t)}\) after the 5th equal sign in the derivation should be \(e^{i(E_\mathbf{p} + E_\mathbf{q})t}\).

In (2.23), the last \(\tau_{ij}\) should be \(\tau_{ij}^*\).

In (2.24), the last \(\tau^\alpha_{ij}\) is correct, because

\[\dot{\Phi}_i \tau_{ij}^* \Phi^*_j = \Phi^*_j \tau_{ij}^* \dot{\Phi}_i = \Phi^*_i \tau_{ji}^* \dot{\Phi}_j = \Phi^*_i \tau_{ij} \dot{\Phi}_j.\]

The first equality is up to an infinite constant term, while the last one is because \(\tau^* = \tau^T\).

In (2.25), the integrand should be \(a^\dagger_{i \mathbf{p}} \tau^a_{ij} a_{j \mathbf{p}} - b^\dagger_{j \mathbf{p}} \tau^a_{ij} b_{i \mathbf{p}}\), or equivalently, \(a^\dagger_{i \mathbf{p}} \tau^a_{ij} a_{j \mathbf{p}} - b^\dagger_{i \mathbf{p}} {\tau^a}^*_{ij} b_{j \mathbf{p}}\). I will privilege the second notation, which makes the rest of the problem easier. It is easy to verify that \(Q^a = {Q^a}^\dagger\), so that the charge is real.

In the derivation of (2.26), \(+(a\rightarrow b)\) should be \(-(a\rightarrow b)\) but it does not affect the result.

In the derivation of (2.27), all three \(t\)s sandwidged between \(b^\dagger\) and \(b\) should be \(t^*\)s. Also, \(a\rightarrow b\) needs to reflect such complex conjugations and this is what causes the change of sign between the second line and the third line. The end result is however correct.

It is worth pointing out that in the thoery of free fields, a complex scalar field is simply made of two real fields of the same mass. This can be clearly seen when we write \(\phi = \phi_R + i \phi_I\), where both \(\phi_R\) and \(\phi_I\) are real. The Langragian of \(\phi\) is simply the sum of the Langragians of the two real fields. Complex scalar fields become more interesting when we introduce higher-order terms into the Langragian, e.g. \( (\phi^* \phi)^2 = \phi_R^4+\phi_I^4+2\phi_R^2 \phi_I^2\), effectively coupling the two otherwise independent real fields.

The Langragian in this problem does not include such higher-order terms, and \(n\) identical complex scalar fields are equivalent to \(2n\) identical real scalar fields. Therefore we can replace \(SU(n)\) by \(SO(2n)\) in the solution and the number of generators (and thus conserved charges) increases from \(n^2-1\) to \(n(2n-1)\), without counting the regular charge corresponding to \(U(1)\).

2.3

Xianyu's solution is good. For the definition of \(K_1\), refer to this Wikipedia page.

Chapter 3

Notes

Page 51

The explanation on the Lorentz invariance of (3.77) may be a bit cryptic for those less proficient in group theory. Here is some elaboration.

The upper half of (3.43) can be written as \(m(\psi_L)_\alpha=i(\sigma^\mu)_{\alpha\beta} \partial_\mu (\psi_R)_\beta\), where \((\psi_L)_\alpha\), \((\psi_R)_\beta\), and \(\partial_\mu\) transform according to \((\frac{1}{2},0)\), \((0,\frac{1}{2})\), and (covariant) vector representations of the Lorentz group. (See Problem 3.1 for more explanation on \((\frac{1}{2},0)\) and \((0,\frac{1}{2})\).)

This implies that the matching \(\alpha\), \(\beta\), and \(\mu\) transform according to \((\frac{1}{2},0)\), \((0,\frac{1}{2})\), and (contravariant) vector representations of the Lorentz group.

With \(\mu\) contracted, the LHS of (3.77) is a tensor with four indices which transform as described in the highlighted text.

On the RHS, \(\epsilon\) is an invariant tensor in both \((\frac{1}{2},0)\) and \((0,\frac{1}{2})\) representations.

This makes (3.77) a tensor equation. So we only need to verify it for one of the 16 components and the rest are automatically guaranteed because these components can be rotated into one another.

Page 52

To show the equality between the first line and the second line of (3.82), we re-write (3.80) as

\[\epsilon_{\beta\alpha}(\sigma^\mu)_{\beta\gamma}=(\bar{\sigma}^\mu)_{\beta\alpha}\epsilon_{\gamma\beta},\]

, apply it twice to \(\epsilon_{\beta\delta}(\sigma^\nu)_{\beta\rho}(\bar{\sigma}^\lambda)_{\rho\sigma}(u_{2L})_\sigma\), and finally re-lable a few dummy variables.

Page 53

We should read the LHS of (3.89) as \([\psi(x)_a, \psi^\dagger(y)_b]\), and the \(1_{4x4}\) on the RHS as \(\delta_{ab}\).

Page 59

The last sentence on this page should not be surprising at all. If we sandwich the LHS of (3.110) between \(\langle\psi|\) and \(|\psi\rangle\), we are rotating the state \(|\psi\rangle\) by \(U(Λ^{-1})\).

Page 60

After the discussion on the spin, we could also apply Noether's theorem to boosts. We will similarly get two terms.

• The one corresponding to orbital angular momentum is the relativistic version of Newton's first law.
• The one corresponding to spin gives us 3-momentum conservation. (We simply replace \(\Sigma^3\) by (3.26).)

For more detail, see this post.

For the derivation of \(j^0\), we use the fact that \(\mathcal{J}^0\) as in (2.10) is 0, because \(\partial_t\) does not appear in \(\delta\mathcal{L} = - \theta (x \partial_y - y \partial_x) \mathcal{L}\). (Note that \(\mathcal{L}\) is a scaler.)

Page 62

The symmetry corresponding to \(Q\) is \(\Psi \rightarrow e^{i\theta}\Psi\).

It becomes more interesting when we couple the Dirac field to an electromagnetic field through minimal coupling. See Section 3.5.4 in Maggiore.

Regarding the polarization of \(\psi_\alpha(x)|0\rangle\), we can apply the number operator \(N\), which is similar to \(H\) as in (3.104), and conclude that \(\psi_\alpha(x)|0\rangle\) contains one positron. It is however not an eigenstate of \(S\), and so it has mixed polarization.

Page 65

3.6 is the most problematic section in this book. Throughout this section \(P\ldots P\), \(T\ldots T\), and \(C\ldots C\) should be \(P\ldots P^{-1}\), \(T\ldots T^{-1}\), and \(C\ldots C^{-1}\). Note that \(P\) really stands for \(U(P)\) (and similarly for \(T\) and \(C\)). So although \(PP=TT=CC=1\) in the Minkowsky space, there is no guarantee that \(U(P)U(P)=U(T)U(T)=U(C)U(C)=1\) in the Hilbert space. In fact, as pointed out in the Errata, \(U(T)U(T)=-1\) in some cases. For \(U(P)U(P)\) and \(U(C)U(C)\), we can later set them to one by choice. But it is probably not a good idea to assume it implicitly from the very beginning.

In fact, if we consistently use \(T\psi(x)T^{-1}\) in the book, the correction in the Errata (p.69) will be unnecessary.

Also notice that any observable can be written in the form \(\langle 0| O |0\rangle\). When we do a parity transformation to the vacuum, it becomes \(\langle 0|P^\dagger| O |P|0\rangle\). So the phase of \(P|0\rangle\) is canceled any way and we may as well set \(P|0\rangle = |0\rangle\).

The discussion on intrinsic parity \(\eta_a^2,\eta_b^2=\pm 1\) is valid only if we are dealing with particles who are their own antiparticles, since we may have \(aa\) (or \(a^\dagger a^\dagger\)) in observables. But otherwise, \(\psi\) will always appear in pairs with \(\bar{\psi}\) (or equivalently \(\psi^\dagger\)) in any observable, allowing \(\eta\) to be any complex number of unit magnitude.

In (3.124), we are using the fact that \(U(P)\) is unitary. See Weinberg for a full discussion on this topic.

Page 68

Notation-wise, it might be a good idea to write \(\xi^{\updownarrow s}\) instead of \(\xi^{-s}\) to avoid weird situations like \(\xi^{-(-s)} = -\xi^s\).

Page 69

In (3.139), we are using the fact that \(T\) is unitary (or strictly speaking, anti-unitary) so that \(T^{-1} = T^\dagger\), where the \(\dagger\) sign denotes anti-adjoint rather than adjoint.

Problems

3.1

Pokraka's solution is very detailed and looks quite good.

3.2

Both Xianyu's and Pokraka's solutions are fine.

3.3

Note that the only purpose of \(k_1\) is to define unambiguously \(u_{R0}\). We can redefine \(k_1\) as \(k_1 - x k_0\) such that \(k_1\) has no time component. It will still give us \(k_0 \cdot k_1 = 0\) and won't change \(u_{R0}\), since we are talking about massless particles. We may therefore demand from the very beginning that \(k_1\) has no time component.

Similarly, the purpose of \(k_0\) is to specify a direction. Its 3-momentum, which is equal to its energy, has no consequence for \(u_L(p)\) or \(u_R(p)\), both of which are always properly normalized.

Both Xianyu's and Pokraka's solutions are fine.

3.4

Note that the equation in part (a) is nothing other than (3.39) by having \(\psi_L = \chi\) and \(\psi_R = i \sigma^2 \chi^*\), and that the action in part (b) is nothing but half of (3.34) by having \(\psi_L = \chi\) and \(\psi_R = i \sigma^2 \chi^*\). In fact, part (c) shows that a Majorana field is a special case of Dirac fields, when we set \(\chi_1=\chi_2\). This is very similar to a real scalar field, which is a special case of complex scalar fields, with the extra constraint that \(\phi=\phi^*\). In both cases, the extra constraint makes a particle its own antiparticle.

It is useful to know the following equalities for matrices made of Grassmann numbers.

• \((A B)^T = - B^T A^T\)
• \((A B)^\dagger = B^\dagger A^\dagger\), just as for matrices made of c-numbers.

Xianyu's solution to part (d) has a serious error. Pokraka's solution looks quite good.

3.5

Xianyu's solution is good, except for the following typos (and a note).

In the derivation of \(\delta(\chi^\dagger i \bar{\sigma}^\mu \partial_\mu \chi)\) before (3.39), \(i F^* \epsilon^\dagger \bar{\sigma}^2 \partial_\mu \chi\) (appearing twice) should be \(i F^* \epsilon^\dagger \bar{\sigma}^\mu \partial_\mu \chi\).

In (3.39), \(\sigma^\mu \sigma^\nu \sigma_\nu\) should be \(\sigma^\mu \bar{\sigma}^\nu \partial_\nu\).

For the four equations immediately after (3.41), note that both \(\epsilon\) and \(\chi\) are made of Grassmann numbers.

In (3.48), \(i\phi\chi^T\sigma^2\chi\) should be \(ig\phi\chi^T\sigma^2\chi\).

In (3.49), we are assuming that \(g\) is real. Otherwise \(g\) should be \(g^*\).

3.6

For part (a), Xianyu's solution is good.

For part (b), one should instead consult this, which also has a few typos, of which the most important is that \(G_{abcd}\) should be \(M_{abcd}\).

For part (c), Xianyu's solution is good. Note that in (3.58), \(\gamma^\mu\) either commutes or anticommutes with \(\Gamma^D\).

3.7

Just as in the main text, \(P\ldots P\), \(T\ldots T\), and \(C\ldots C\) should be \(P\ldots P^{-1}\), \(T\ldots T^{-1}\), and \(C\ldots C^{-1}\).

Xianyu's solution is good for parts (a) and (b), except that in the equation right after (3.61), a factor of \(-\frac{i}(2)\) should be removed.

For part (c), see my comment below.

It should be stressed that we are not required to consider cases with \(\partial_\mu\). Such cases are non-trivial. It is common to see arguments (in some other textbooks!) that a \(\partial_\mu\) is always matched with either another \(\partial^\mu\) or a bilinear \(\bar{\psi} \gamma^\mu \psi\) and in both cases \(CPT = 1\).

This is a bad argument because this same logic would lead us to the conclusion that a Dirac field would be antisymmetric under \(C\) alone, at least when \(m=0\). (See the table on Page 71.) But it is clearly not true. The reason is that an extra minus sign can arise through integration by part, such as in the case of \(\bar{\psi}i\gamma^\mu\partial_\mu\psi\) in the Dirac Langragian.

3.8

For part (a), Pokraka is in general good, except for the following.

In the Fournier transform in (1.3), \(e^{i\mathbf{k}\cdot\mathbf{r}}\) should be \(e^{-i\mathbf{k}\cdot\mathbf{r}}\), although it does not really matter because the purpose of (1.3) is to show that parity of \(\psi_{n l m_l}(\mathbf{k})\) is the same as that of \(\psi_{n l m_l}(\mathbf{x})\).

Also notice that when we calculate \( P|\Psi_{n l m_l;s,m_s} \rangle\) as in (1.2), \(P\) passes right through \(\psi_{nlm_l}(\mathbf{k})\), instead of changing it to \(\psi_{nlm_l}(-\mathbf{k})\). To see why, simply consider \(\psi(\mathbf{k})=\delta^3(\mathbf{k}-\mathbf{k_0})\).

The discussion around \(T\) is however wrong. \(T\) not only flips \(m_l\), but also changes \(\psi_{nlm_l}(\mathbf{k})\) to \(\psi_{nlm_l}^*(\mathbf{k})\). Therefore spin eigenstates do not have a well-defined quantum number under \(T\).

For part (b), to ensure that electrodynamics be invariant to \(P\) and \(C\), we need to make \(\Delta H\) invariant, and this means \(A^\mu\) must have the same \(P\) and \(C\) parity as \(j^\mu\).

Since \(j^\mu\) has \(C\) parity equal to \(-1\), the same must be true for \(A^\mu\). Knowing that photon is its own antiparticle, we can safely assume that the quantization of \(A^\mu\) should look like

\[A^{\mu}(x) = \int \frac{d^3 p}{(2\pi)^3)} \frac{1}{\sqrt{2E_p}} \sum_s (a^s_\mathbf{p} \epsilon^s(p) e^{-i p x} + {a^s}^\dagger_\mathbf{p} {\epsilon^s}^*(p) e^{i p x}), \]

and

\[C a^s_\mathbf{p} C^{-1} = - a^s_\mathbf{p}. \]

An (unnormalized) two-photon state in the center-of-mass frame can be expressed as

\[ \int \frac{d^3 k}{(2\pi)^3)} \sum_{s,s'} \psi_{s,s'}(\mathbf{p}) {a^s}^\dagger_\mathbf{p} {a^{s'}}^\dagger_{-\mathbf{p}} | 0 \rangle. \]

Its quantum number under \(C\) is \(+1\), the same as the spin-0 ground state of positronium (\(l=0\), \(s=0\)), or any positronium state with even \(l+s\).

The same argument can be made for a 3-photon state and the spin-1 ground state (\(l=0\), \(s=1\)), or any positronium state with odd \(l+s\).

For 1-photon transitions between positronium levels, \(l+s\) of the positronium must change from odd to even or from even to odd, before and after the transition.

Chapter 4

Notes

Page 86

Note that in (4.23), \(t_0\) is still very much present. It is hidden inside the definition of \(H_I\).

Page 103

The explicit factor \(e^{-i\mathbf{b}\cdot\mathbf{k}_\mathcal{B}}\) is essential in the following calculation. Also, its interpretation is completely logical and the statement that suggests it's merely a convention is rather misleading.

\(\int \frac{d^3 k_\mathcal{B}}{(2\pi)^3} \phi_\mathcal{B}(k_\mathcal{B}) |k_\mathcal{B}\rangle\) describes one incoming particle of \(\mathcal{B}\), localized at a certain point in space. Applying \(e^{-i P \cdot \mathbf{b}}\), we get \(e^{-i \mathbf{b} \cdot \mathbf{k}_\mathcal{B}}\) inside the integral, which is simply the original incoming particle translated by \(\mathbf{b}\).

In the calculation leading to (4.79), there are really three things over which we need to average.

• all possible forms of wave packets of \(\mathcal{A}\)
• all possible forms of wave packets of \(\mathcal{B}\)
• spatial shift of \(\mathcal{B}\) by \(\mathbf{b}\).

In the most general case, \(\phi_\mathcal{B}(\mathbf{k}_\mathcal{B})\) in (4.68) should be written as \(\phi_\mathcal{B}(\mathbf{k}_\mathcal{B}, \mathbf{b})\), to allow the possibility that the distribution of all possible wave packet forms of \(\mathcal{B}\) may depend on \(\mathbf{b}\). If, however, we assume such dependency does not exist, \(\phi_\mathcal{B}(\mathbf{k}_\mathcal{B}, \mathbf{b})\) becomes \(\phi_\mathcal{B}(\mathbf{k}_\mathcal{B}) e^{-i \mathbf{b} \cdot \mathbf{k}_\mathcal{B}}\). This also allows us to decouple the integrand and first integrate over \(\mathbf{b}\), keeping the same \(\phi_\mathcal{B}(\mathbf{k}_\mathcal{B})\), and then integrate over different forms of wave packets \(\phi_\mathcal{B}(\mathbf{k}_\mathcal{B})\) and \(\phi_\mathcal{A}(\mathbf{k}_\mathcal{A})\). The latter integrals turn out to be unnecessary if we assume that \(\phi_\mathcal{B}(\mathbf{k}_\mathcal{B})\) and \(\phi_\mathcal{A}(\mathbf{k}_\mathcal{A})\) are sufficiently peaked, as explained in the text.

Page 104

Note that \(S\) and \(T\) are (infinite-dimensional) matrices, whose rows are labeled by states like \(\langle p_1 p_2|\) and columns by states like \(|k_\mathcal{A} k_\mathcal{B}\rangle\). Obviously, those states, though orthorgonal to each other, are not eigenstates of \(S\), because otherwise \(S\) would be a diagonal matrix. The split of \(S\) into an identity matrix and \(iT\) is rather arbitrary. \(iT\) also has non-zero diagonal elements. Also, note that the \(H\) in the \(S\) matrix is the full interactive theory Hamiltonian. Thus the \(1\) in (4.72) is not the same as the 0-th Taylor expansion of the right-hand-side of (4.89).

\(i\mathcal{M}\), on the other hand, is just an entry in the \(S\) matrix. \((2\pi)^4 \delta^{(4)}\) is from our state normalization convention and thus outside the \(S\) matrix.

Page 105

It is not entirely clear how (4.76) leads to (4.78). What is missing is somewhere in the derivation factors of \(\delta^{(4)}(k_\mathcal{A}-\bar{k}_\mathcal{A})\) and \(\delta^{(4)}(k_\mathcal{B}-\bar{k}_\mathcal{B})\). Below are the steps to reach such factors.

Note that \(\delta^{(4)}(k_\mathcal{A}+k_\mathcal{B}-\sum p_f)\) and \(\delta^{(4)}(\bar{k}_\mathcal{A}+\bar{k}_\mathcal{B}-\sum p_f)\), put together, are equivalent to \(\delta^{(4)}(k_\mathcal{A}+k_\mathcal{B}-\bar{k}_\mathcal{A}-\bar{k}_\mathcal{B})\) and \(\delta^{(4)}(k_\mathcal{A}+k_\mathcal{B}-\sum p_f)\).

The first one, combined with \(\delta^{(2)}(k^\perp_\mathcal{B}-\bar{k}^\perp_\mathcal{B})\), gives us also \(\delta^{(2)}(k^\perp_\mathcal{A}-\bar{k}^\perp_\mathcal{A}).\)

(Note that \(\perp\) denotes the components of \(k\) that lie in the plane where \(\mathcal{B}\) particles are created. It is perpendicular to the path of collision in direction \(z\).)

Further, from \(\delta(E_\mathcal{A} + E_\mathcal{B} - \bar{E}_\mathcal{A} - \bar{E}_\mathcal{B})\) and \( \delta(k^z_\mathcal{A} + k^z_\mathcal{B} - \bar{k}^z_\mathcal{A} - \bar{k}^z_\mathcal{B})\), we can get, assuming \(\mathcal{A}\) and \(\mathcal{B}\) have different masses, \(\delta(k^z_\mathcal{A}-\bar{k}^z_\mathcal{A}) \delta(k^z_\mathcal{B}-\bar{k}^z_\mathcal{B})\).

If \(\mathcal{A}\) and \(\mathcal{B}\) have the same mass, however, we also get an extra term \(\delta(k^z_\mathcal{A}-\bar{k}^z_\mathcal{B}) \delta(k^z_\mathcal{B}-\bar{k}^z_\mathcal{A})\).

Fortunately, this would give us \(\phi_\mathcal{A}(\mathbf{k}_\mathcal{A}) \phi_\mathcal{B}(\mathbf{k}_\mathcal{A})\) in the integral. We can safely assume this to be zero, because the wave packets of \(\mathcal{A}\) and \(\mathcal{B}\) do not overlap.

Page 106

For an explicit proof that (4.80) is Lorentz invariant, refer to this link.

Page 119

Regarding the assignment of the momentum to the direction of particle-number flow, the same argument can be made for complex scalar fields, whose particles and antiparticles are not the same. But it does not matter in the end because the propagator is an even function.

For Majorana fields, it is a lot more complicated. Refer to these slides or this paper.

Problems

4.1

The overall approach in Pokraka's solution is correct, despite the following errors.

Part (b)

The argument that starts with since \(p^0>0\) at the end is wrong. We integrate over \(p^0\) from \(-\infty\) to \(\infty\). There are two poles at \(p^0=E_\mathbf{p}-i\epsilon\) and at \(p^0=-E_\mathbf{p}+i\epsilon\). The lack of an explicit term in the form of \(e^{i\cdot p \cdot c}\), where c is either a positive or a negative real number, implies that we have to assume that \(|\tilde{j}(p)|^2\), as am analytical function on the complex plane, vanishes fast enough as \(|p|\rightarrow\infty\). With this assumption, we can freely choose to do the contour integration in either the upper or the lower plane. Both choices lead to the same result. (Although the choice does affect \(p^0\), and therefore \(\tilde{j}(p)\), it has no consequence after the integration.)

Part (c)

The explanation on the symmetry factor is unfortunatley completely off. Below is how it should be calculated.

A Feynman diagram with \(n\) propagators and \(2n\) gives us \((-\lambda)^n\), mupltiplied by a factor of \(\frac{1}{(2n)!}\) (from Taylor expansion), as well as a factor of \((2n-1)!!\). (The first vertex can choose among \((2n-1)\) vertices, the next one among \((2n-3)\) vertices, etc.) The total amplitude of the process is therefore

\[\sum_{n=0}^\infty \frac{(2n-1)!!}{(2n)!} (-\lambda)^n = \sum_{n=0}^\infty \frac{1}{2^n n!} (-\lambda)^n = e^{-\lambda/2}. \]

The total probability is thus simply \( e^{-\lambda} \).

Part (d)

Instead of \(\langle 0|a_\mathbf{k}^n\), we should have \(\langle 0|a_{\mathbf{k}_1} a_{\mathbf{k}_2} \dots a_{\mathbf{k}_n}\). Within the total amplitude, there is a factor of \(n!\) from the \(n!\) different ways to contract \(\phi(x_i)\) with \(a_{\mathbf{k}_j}\), which is then cancelled by \(\frac{1}{n!}\) from Taylor's expansion.

To calculate the probability, we need to integrate over \(\mathbf{k}_1, \dots, \mathbf{k}_n\), and divide by \(\frac{1}{n!}\) because those \(n\) particles are identical.

4.2

Pokraka's solution shows how to calculate \(i\mathcal{M}\), while Xianyu did a better job calculating \(\Gamma\).

Note that we should take \(\mathbf{P}_\Phi=0\) because decay rates, by definition, should be calculated in the COM frame. Also, the calculation becomes simpler if one remembers \(\int \frac{dp}{(2\pi)^3} p = \int \frac{dE_\mathbf{p}}{(2\pi)^3} E_\mathbf{p} \).

4.3

Part (a)

Pokraka's solution is detailed very well motivated, despite the following error.

\(E_{CM}\) is not equal to \(2m\) unless in the nonrelativistic limit.

Part (b)

Pokraka's solution is detailed very well motivated, despite the following errors.

The statement \(\Phi^i_{\mathrm{min}}=0\) is wrong.

In the super long derivation of \(V(\Phi^2)\), the third-to-the-last and the last equal signs should be removed.

In the propagator of the massive \(\sigma\) field, it should be \(m=\sqrt{2}\mu\).

Part (c)

Xianyu's solution is good, although it uses Mandelstam variables, which are to be introduced in a later chapter. Note that in the first three diagrams, \(\frac{1}{2}\) from the second-order taylor expansion is cancelled by \(2\) from the symmetry between the two vertices. Also, all appreances of \(\mathcal{M}\) really should be \(i\mathcal{M}\).

Part (d)

Xianyu's solution is good, except that in (4.36), \(\lambda \Phi^i \Phi^i\) should be written as \(\lambda \Phi^j \Phi^j\) to avoid any confusion.

To get the new \(V(\Phi^2)\), simply replace \(\sigma\) by \(\sigma+\frac{a}{2\mu^2}\) in Pokraka's solution to part (b). The last two terms in the original derivation give now to the pions an extra mass term

\[(\boldsymbol{\pi} \cdot \boldsymbol{\pi}) \left(\sqrt{\lambda}\mu \frac{a}{2\mu^2} + O(a^2)\right),\]

corresponding to a mass of \(a^{1/2}\lambda^{1/4}\mu^{-1/2}\).

Further, this replacement does not affect the coefficient of the \((\boldsymbol{\pi} \cdot \boldsymbol{\pi})^2\) term, but it does change the coefficient of the \((\boldsymbol{\pi} \cdot \boldsymbol{\pi})\sigma\) term from \(\sqrt{\lambda}\mu\) to \(\sqrt{\lambda}\mu + \lambda \frac{a}{2\mu^2}\)

Compared to (4.34), now we have an extra term \(2i \frac{\lambda^{3/2}a}{\mu^3}\left(\frac{s+t+u}{2\mu^2}+3\right)\), and the result is simply (4.39). Also note that now \(s+t+u=4m^2=\frac{4\alpha\sqrt{\lambda}}{\mu}\) are proportional to \(a\). Therefore we have, at threshold,

\[i\mathcal{M}=\frac{10i\lambda^{3/2}a}{\mu^3} + O(a^2).\]

4.4

Xianyu's solution is very good, except for the following.

• In (4.58), there should be an overall minus sign. Also, \(u(p)\) should be \(u(k)\).
• In (4.61), it should be \(|\mathbf{k}|^2\) in the denominator.

Chapter 5

Notes

Page 137

To understand the one but last paragraph on this page, note that the curve in Figure 5.1 is \(\sigma \cdot E^2_{cm}\) instead of \(\sigma\).

Page 144

About Figure 5.4, note that if we measure the spins of the muons along the \(z\)-direction, as well as the momenta, total spin-\(z\) will always be conserved. However, if we measure the helicity (i.e. spin in the direction of momentum) we will get a distribution.

If we only measure the spins of the muons, but not the momenta, things are more complicated, because it is angular momentum, rather than total spin, that is conserved. (See (3.111).) Since there is angular dependency in the wave function of the two muons, orbital angular momentum is non-zero, and therefore total spin itself is not conserved. The muon wave function is basically a mixture of \(1s\) and \(2p\) (with \(L_z = 0\)). Since \(J_z = L_z + S_z\), \(S_z\) is still conserved.

There is thus no contradiction with Section 5.3.

Page 149

In (5.42), there should be a minus sign according Notations and Conventions.

Page 150

Here is how to derive (5.51).

From (5.34) and (5.36), we have

\[\Gamma(k) = -i \sqrt{2} e^2 \sigma^i \epsilon^i.\]

Then we have, from (5.50)

\[ \begin{eqnarray} && \mathrm{tr} \left( \frac{\Gamma(k)\dot {\mathbf{n}^*}^j \sigma^j}{\sqrt{2}i} \right) \\ &=& - e^2 \mathrm{tr} \left( \mathbf{\epsilon}^i {\mathbf{n}^*}^j \sigma^i \sigma^j \right) \\ &=& - e^2 \mathrm{tr} \left( \mathbf{\epsilon}^i {\mathbf{n}^*}^j (\delta^i_j + i \epsilon_{ijk} \sigma^k) \right) \\ &=& - e^2 \mathrm{tr} \left( \mathbf{\epsilon}^i {\mathbf{n}^*}^i + i \epsilon_{ijk} \mathbf{\epsilon}^i {\mathbf{n}^*}^j \sigma^k) \right) \end{eqnarray} \]

Since the trace of all sigma matrices is 0, the trace of the second part is 0 and we have

\[\mathrm{tr} \left( \frac{\Gamma(k)\dot {\mathbf{n}^*}^j \sigma^j}{\sqrt{2}i} \right) = - 2e^2 \mathbf{\epsilon}^i {\mathbf{n}^*}^i.\]

Page 152

It is interesting that we multiply Equation (5.57) by a color factor of 3 for the case of a spin-1 \(q\bar{q}\) decay. We are summing instead of averaging over the three colours.

The reason is that a quarkonium is always in the state of \( \frac{1}{\sqrt{3}} (r\bar{r} + g\bar{g} + b\bar{b})\), instead of having three eigenstates \(r\bar{r}\), \(g\bar{g}\), and \(b\bar{b}\).

Page 160

To derive (5.79), integrate by parts on the RHS of (5.78) and assume the integrand goes to 0 at infinity.

Page 161

To derive (5.81), note if there are multiple gamma matrices between \(\bar{u}\) and \(u\), their order needs to be reversed when we take the complex conjugate. Also see Page 132 for reference.

Page 167

To derive the un-numbered equation, we should also calculate \(2p\cdot k\) besides \(2k\cdot k'\) in the center-of-mass frame and in the lab frame.

Problems

5.1

There is a minor correction in the Errata.

Xianyu's solution is good except for the following.

• In (5.5), \(|\mathbf{k}|^4\) should be \(|\mathbf{k}|^2\).
• In (5.8), a factor of \(\frac{1}{4}\) is missing in the intermediate step.
• Immediately above (5.12), \(4k^2\sin^2\frac{\theta}{2}\) should have a minus sign.
• In (5.12), it should be \(\sin^4\frac{\theta}{2}\) in the denominator.

5.2

Xianyu's solution is good except for the following.

• In (5.19), \(E_{CM}\) should be \(E_{CM}^2\).

5.3

In Problem 3.3, \(u_{R0}\) was defined using another vector \(k_1\). But once \(u_{R0}\) was defined, \(k_1\) was no longer needed anywhere. So we may as well pick an \(u_{R0}\) from the very beginning.

Part (a)

Xianyu's solution is good except for the following.

• In (5.28), the last \((1-\gamma^5)\) should be \((1+\gamma^5)\).

Part (b)

Xianyu's solution is good except for the following.

• Immediately below (5.32), \(C \mathrm{tr}(\not{p_1}\gamma^\mu\not{p_2})\) should be \(C \mathrm{tr}(\not{p_1}\gamma^\mu\not{p_2}\frac{1+\gamma^5}{2}\not{k_0})\).

Part (c)

Another way to argue about the form of \([M]\) is as follows.

Any 4-by-4 matrix can be written as a linear combination of the 16 \(\Gamma\) matrices, of which 8 are block-wise diagonal (1, \(\gamma^5\), \(\gamma^\mu \gamma^\nu\)), and 8 are block-wise off-diagonal (\(\gamma^\mu\), \(\gamma^\mu \gamma^5\)). The RHS, being itself block-wise off-diagonal, must be a linear combination of the 8 block-wise off-diagonal \(\Gamma\) matrices.

Xianyu's solution is good except for the following.

• Between (5.33) and (5.34), \(\gamma^\mu M=-M\gamma^5\) should be \(\gamma^5 M=-M\gamma^5\).

Part (d)

Xianyu's solution is good except for the following.

• In (5.36), \((-i e^2)\) should be \((-i e)^2\).
• In (5.39), \(E_{CM}\) should be \(E_{CM}^2 = s\). Idem for the rest of the solution.
• In (5.40), a factor of \(\frac{4}{s^2}\) is missing in the intermediate step.

In (5.37), note that \(\bar{v}_R(p_1)\gamma_\mu v_R(p_2)\) can be written as \([\bar{v}_R(p_1)]_a [\gamma_\mu]_{ab} [v_R(p_2)]_b\). Also, \(\bar{v}_R(p_1) u_R(k_1)\) vanishes and this is most obvious in the chiral representation.

Part (e)

Xianyu's solution is a bit more messed up here.

By simple visual inspection in the chiral representation, we can see that the \(t\)-channel term in (5.41) is 0. A similar observation leads to (5.45), which contains a sign error.

(5.43), (5.44), and (5.46) can be easily obtained by replacing variables.

• (5.43) can be obtained from (5.42) by \(p_1 \leftrightarrow k_2\).
• (5.44) can be obtained from (5.41) by \(k_1 \leftrightarrow p_1\) and \(k_2 \leftrightarrow p_2\).
• (5.46) can be obtained from (5.45) by \(k_1 \leftrightarrow p_2\) and \(k_2 \leftrightarrow p_1\).

(5.46) has the same sign error as in (5.45).

It should be \(s\) instead of \(2s\) in the denominator in (5.47), (5.48), and (5.49).

Sum all possibilities up, divided it by four, and we get the correct answer (5.50).

5.4

Part (a)

Xianyu's solution is overall correct, except for the following.

• For the triplet states, we should set \(\xi \xi'^\dagger\) to \(\frac{1}{2}\sigma^1+\frac{i}{2}\sigma^2\), \(-\frac{1}{2}\sigma^1+\frac{i}{2}\sigma^2\), and \(-\frac{1}{\sqrt{2}}\sigma^3\). The sign in (5.61) is actually correct, despite the error corrected in Errata. Here we have \(\xi \xi’^\dagger\) instead of \(\xi’ \xi^\dagger\).
• \(2m\) should be \(m\) in the denominator in (5.63).
• \(\delta(m-E_1)\) should be \(\delta(2m-2E_1)\) in (5.64). This error miraculously cancels the previous error and eventually gives us the correct result.

Part (b)

The answer provided by Xianyu is not wrong (except for some minor sign errors). But it takes the direction of \(\mathbf{k}\) to be the \(x^3\)-axis, and thus implicitly uses the same direction to label spin states. This will present a problem later in part (d), where we need to allow for an arbitrary angle between the direction of \(\mathbf{k}\) and the \(z\)-direction that is used to label both the spin states, and various spatial wavefunctions. Also, we will distinguish the \(x\)-axis and \(y\)-axis explicitly, because that is what we will do with the spatial wavefunctions.

Below is the necessary corrections to Xianyu's solution.

In (5.65), we need instead the following. (Note that the relative angle between \(p_1\) and \(k_1\) here is \(-\theta\) instead of \(\theta\). But it does not matter in the end.)

\[ \begin{eqnarray} k_1^\mu=(E, k \sin\theta\cos\phi, k \sin\theta\sin\phi, k \cos\theta) & \qquad\qquad\qquad & k_2^\mu=(E, -k \sin\theta\cos\phi, -k \sin\theta\sin\phi, -k \cos\theta) \\ p_1^\mu=(E, 0, 0, E) & \qquad\qquad\qquad & p_2^\mu=(E, 0, 0, -E) \\ \epsilon_\pm^\mu(p_1)=\frac{1}{\sqrt{2}}(0, \mp 1, -i, 0) & \qquad\qquad\qquad & \epsilon_\pm^\mu(p_2)=\frac{1}{\sqrt{2}}(0, \pm 1, -i, 0) \end{eqnarray} \]

correspondingly, we have the following.

\[ \begin{eqnarray} \sqrt{k_1 \cdot \sigma} = \sqrt{k_2 \cdot \bar{\sigma}} = \sqrt{m} - \frac{k\sin\theta\cos\phi}{2\sqrt{m}} \sigma^1 - \frac{k\sin\theta\sin\phi}{2\sqrt{m}} \sigma^2 - \frac{k\cos\theta}{2\sqrt{m}} \sigma^3 \\ \sqrt{k_2 \cdot \sigma} = \sqrt{k_1 \cdot \bar{\sigma}} = \sqrt{m} + \frac{k\sin\theta\cos\phi}{2\sqrt{m}} \sigma^1 + \frac{k\sin\theta\sin\phi}{2\sqrt{m}} \sigma^2 + \frac{k\cos\theta}{2\sqrt{m}} \sigma^3 \end{eqnarray} \]

(The other two equations remain the same.)

After some tedious calculation, we have the following.

\[ \begin{eqnarray} \Gamma^{\mu\nu}_t &=& 2m^2\sigma^\nu\sigma^3\sigma^\mu - mk(\sigma_\theta\sigma^\nu\sigma^\mu+\sigma^\nu\sigma^\mu\sigma_\theta+2\sigma^\nu\sigma_\theta\sigma^\mu)+O(k^2) \\ \Gamma^{\mu\nu}_u &=& -2m^2\sigma^\mu\sigma^3\sigma^\nu - mk(\sigma_\theta\sigma^\mu\sigma^\nu+\sigma^\mu\sigma^\nu\sigma_\theta+2\sigma^\mu\sigma_\theta\sigma^\nu)+O(k^2), \end{eqnarray} \]

where \(\sigma_\theta = \sigma^1 \sin\theta \cos\phi + \sigma^2 \sin\theta \sin\phi + \sigma^3 \cos\theta\).

The part between the brackets in (5.66) thus becomes

\[-2\cos\theta\sigma^\mu\sigma^3\sigma^\nu-2\cos\theta\sigma^\nu\sigma^3\sigma^\mu+(\sigma_\theta\sigma^\mu\sigma^\nu+\sigma^\mu\sigma^\nu\sigma_\theta+2\sigma^\mu\sigma_\theta\sigma^\nu)+(\sigma_\theta\sigma^\nu\sigma^\mu+\sigma^\nu\sigma^\mu\sigma_\theta+2\sigma^\nu\sigma_\theta\sigma^\mu).\]

Now we have something quite different to (5.67).

\[ \begin{eqnarray} i\hat{\mathcal{M}}^{S=1,S_z=-1}_{\pm\pm} = -2i e^{-i\phi} \sin\theta \frac{e^2k}{m} & \qquad\qquad & i\hat{\mathcal{M}}^{S=1,S_z=-1}_{+-} = 0 & \qquad\qquad & i\hat{\mathcal{M}}^{S=1,S_z=-1}_{-+} = 4i e^{i\phi} \sin\theta \frac{e^2k}{m} \\ i\hat{\mathcal{M}}^{S=1,S_z=0}_{\pm\pm} = -2\sqrt{2}i \cos\theta \frac{e^2k}{m} & \qquad\qquad & i\hat{\mathcal{M}}^{S=1,S_z=0}_{\pm\mp} = 0 \\ i\hat{\mathcal{M}}^{S=1,S_z=1}_{\pm\pm} = 2i e^{i\phi} \sin\theta \frac{e^2k}{m} & \qquad\qquad & i\hat{\mathcal{M}}^{S=1,S_z=1}_{+-} = -4i e^{-i\phi} \sin\theta \frac{e^2k}{m} & \qquad\qquad & i\hat{\mathcal{M}}^{S=1,S_z=1}_{-+} = 0 \\ i\hat{\mathcal{M}}^{S=0}_{\pm\pm} = 0 & \qquad\qquad\qquad & i\hat{\mathcal{M}}^{S=0}_{\pm\mp} = 0 \end{eqnarray} \]

Note that we are ignoring terms proportional to zero powers of \(\mathbf{k}\), because they don’t matter for \(L=1\), where the spatial wavefunction is odd and such terms lead to 0.

Part (c)

This problem has serious issues.

In the definition of \(|B(\mathbf{k})\rangle\), \(a^\dagger\) should be understood as a row vector \((a^\dagger_\uparrow, a^\dagger_\downarrow)\) while \(b^\dagger\) as a column vector \((b^\dagger_\downarrow, -b^\dagger_\uparrow)^T\). (The parameter \(\mathbf{k}\) does not add any value and I will leave it out from now on.)

With this definition, we can easiy verify the following for a spin-1 electron-positron pair.

\[ \begin{eqnarray} |S_z=1\rangle &=& |\uparrow\uparrow\rangle = a^\dagger \left( -\frac{1}{2}(\sigma^1+i\sigma^2) \right) b^\dagger |0\rangle \\ |S_z=0\rangle &=& \frac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle) = a^\dagger \left( \frac{1}{\sqrt{2}}\sigma^3 \right) b^\dagger |0\rangle \\ |S_z=-1\rangle &=& |\downarrow\downarrow\rangle = a^\dagger \left( \frac{1}{2}(\sigma^1-i\sigma^2) \right) b^\dagger |0\rangle \end{eqnarray} \]

We see that the matrix between \(a^\dagger\) and \(b^\dagger\) is simply \(\frac{1}{\sqrt{2}}\mathbf{n}\cdot\mathbf{\sigma}\), where \(\mathbf{n}\) is defined on Page 150 (and corrected in the Errata).

Similarly for orbital angular momentum \(L=1\), we have the following.

\[ \begin{eqnarray} |L_z=1\rangle &=& -\frac{1}{\sqrt{2}}(\psi_1 + i \psi_2) \\ |L_z=0\rangle &=& \psi_3 \\ |L_z=-1\rangle &=& \frac{1}{\sqrt{2}}(\psi_1 - i \psi_2) \end{eqnarray} \]

With this in mind, we can write the \(|J=2,J_z=2\rangle\) state as follows.

\[ \begin{eqnarray} && |J=2,J_z=2\rangle \\ &=& |L_z=1,S_z=1\rangle \\ &=& \sqrt{2M} \int \frac{d^3 p}{(2\pi)^3} \left(-\frac{1}{\sqrt{2}}(\psi_1 + i \psi_2)\right) a^\dagger \left( -\frac{1}{2}(\sigma^1+i\sigma^2) \right) b^\dagger |0\rangle \\ &=& \sqrt{2M} \int \frac{d^3 p}{(2\pi)^3} a^\dagger \left( \psi_1 \left(\frac{1}{2\sqrt{2}}\sigma^1 + \frac{i}{2\sqrt{2}}\sigma^2\right) + \psi_2 \left(\frac{i}{2\sqrt{2}}\sigma^1 - \frac{1}{2\sqrt{2}}\sigma^2\right) \right) b^\dagger |0\rangle \\ \end{eqnarray} \]

Comparing it to the definition of \(|B\rangle\), we have \(\Sigma^1 = \frac{1}{2\sqrt{2}}\sigma^1 + \frac{i}{2\sqrt{2}}\sigma^2\), \(\Sigma^2 = \frac{i}{2\sqrt{2}}\sigma^1 - \frac{1}{2\sqrt{2}}\sigma^2\) and \(\Sigma^3 = 0\), or equivalently, \(h^{ij}=\begin{pmatrix}\frac{\sqrt{3}}{2\sqrt{2}} & \frac{\sqrt{3}i}{2\sqrt{2}} & 0 \\ \frac{\sqrt{3}i}{2\sqrt{2}} & -\frac{\sqrt{3}}{2\sqrt{2}} & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}\).

If we adopt these Clebsch–Gordan coefficients to decompose \(J\) into \(L\) and \(S\) (e.g. \(|J=1,J_z=0\rangle=\frac{1}{\sqrt{2}}|L_z=1,S_z=-1\rangle-\frac{1}{\sqrt{2}}|L_z=-1,S_z=1\rangle\)), we have the following \(h^{ij}\) for \(J=2\),

\[ \begin{eqnarray} J_z=2&:& h^{ij}=\begin{pmatrix} \frac{\sqrt{3}}{2\sqrt{2}} & \frac{\sqrt{3}i}{2\sqrt{2}} & 0 \\ \frac{\sqrt{3}i}{2\sqrt{2}} & -\frac{\sqrt{3}}{2\sqrt{2}} & 0 \\ 0 & 0 & 0 \end{pmatrix} \\ J_z=1&:& h^{ij}=\begin{pmatrix} 0 & 0 & -\frac{\sqrt{3}}{2\sqrt{2}} \\ 0 & 0 & -\frac{\sqrt{3}i}{2\sqrt{2}} \\ -\frac{\sqrt{3}}{2\sqrt{2}} & -\frac{\sqrt{3}i}{2\sqrt{2}} & 0 \end{pmatrix} \\ J_z=0&:& h^{ij}=\begin{pmatrix} -\frac{1}{2} & 0 & 0 \\ 0 & -\frac{1}{2} & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} \\ J_z=-1&:& h^{ij}=\begin{pmatrix} 0 & 0 & \frac{\sqrt{3}}{2\sqrt{2}} \\ 0 & 0 & -\frac{\sqrt{3}i}{2\sqrt{2}} \\ \frac{\sqrt{3}}{2\sqrt{2}} & -\frac{\sqrt{3}i}{2\sqrt{2}} & 0 \end{pmatrix} \\ J_z=-2&:& h^{ij}=\begin{pmatrix} \frac{\sqrt{3}}{2\sqrt{2}} & -\frac{\sqrt{3}i}{2\sqrt{2}} & 0 \\ -\frac{\sqrt{3}i}{2\sqrt{2}} & -\frac{\sqrt{3}}{2\sqrt{2}} & 0 \\ 0 & 0 & 0 \\ \end{pmatrix} \\ \end{eqnarray} \]

and the following for \(J=1\).

\[ \begin{eqnarray} J_z=1&:& \mathbf{n}=(\frac{i}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0) \\ J_z=0&:& \mathbf{n}=(0, 0, -i) \\ J_z=-1&:& \mathbf{n}=(-\frac{i}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0) \end{eqnarray} \]

For \(J=0\), we can verify that it should be \(\Sigma^i=-\frac{1}{\sqrt{6}}\sigma^i\), with an extra sign compared to the one given in the original text.

Since all \(\Sigma\) matrices are expressed in terms of \(\sigma\) matrices. It is worthwhile to see explicitly what each \(\sigma\) matrix means. \(\sigma^1\) corresponds to \(|S_z=-1\rangle - |S_z=1\rangle\) (\(a^\dagger_\downarrow b^\dagger_\downarrow - a^\dagger_\uparrow b^\dagger_\uparrow\)), \(\sigma^2\) corresponds to \(i |S_z=-1\rangle + i|S_z=1\rangle\) (\(i a^\dagger_\downarrow b^\dagger_\downarrow + i a^\dagger_\uparrow b^\dagger_\uparrow\)), and \(\sigma^3\) corresponds to \(\sqrt{2}|S_z=0\rangle\) (\(a^\dagger_\uparrow b^\dagger_\downarrow + a^\dagger_\downarrow b^\dagger_\uparrow\)).

Part (d)

Combining the results from Part (b) and Part (c), we have the following.

\[ \begin{eqnarray} \sigma^1_{\pm\pm} \rightarrow -4i \frac{e^2}{m} k^x & \qquad\qquad & \sigma^1_{+-} \rightarrow 4i \frac{e^2}{m} (k^x - i k^y) & \qquad\qquad & \sigma^1_{-+} \rightarrow 4i \frac{e^2}{m} (k^x + i k^y) \\ \sigma^2_{\pm\pm} \rightarrow -4i \frac{e^2}{m} k^y & \qquad\qquad & \sigma^2_{+-} \rightarrow 4 \frac{e^2}{m} (k^x - i k^y) & \qquad\qquad & \sigma^2_{-+} \rightarrow -4 \frac{e^2}{m} (k^x + i k^y) \\ \sigma^3_{\pm\pm} \rightarrow -4i \frac{e^2}{m} k^z & \qquad\qquad & \sigma^3_{\pm\mp} \rightarrow 0 \end{eqnarray} \]

Also note that there is a sign error in the formula given in Part (b) in the original textbook. We also use the following for the positronium spatial wavefunction with \(L=1\) and electron mass \(m\).

\[\psi_{2,1,i}(\mathbf{x}) = \sqrt{\frac{(\alpha m)^5}{\pi}} \frac{x^i}{32} e^{-\alpha m r/4}, \]

which gives us \(f(0) = \frac{\alpha^{5/2} m^{5/2}}{32 \sqrt{\pi}} \).

Let us start with \(J=0\).

\[ \begin{eqnarray} && i\hat{\mathcal{M}}(P_{J=0,J_z=0}\rightarrow \gamma_\pm\gamma_\pm) \\ &=& \frac{1}{\sqrt{m}} \int \frac{d^3k}{(2\pi)^3} \frac{-1}{\sqrt{6}} (-4i \frac{e^2}{m}) \left( \psi_1(\mathbf{k}) k^x + \psi_2(\mathbf{k}) k^y + \psi_3(\mathbf{k}) k^z \right) \\ &=& \frac{4i e^2}{\sqrt{6}m^{3/2}} (-i) \left( \frac{\partial \psi_1(\mathbf{x})}{\partial x} + \frac{\partial \psi_2(\mathbf{x})}{\partial y} + \frac{\partial \psi_3(\mathbf{x})}{\partial z} \right) \Bigg |_{\mathbf{x}=0} \\ &=& \frac{12 e^2}{\sqrt{6}m^{3/2}} f(0) \\ &=& \sqrt{\frac{3\pi\alpha^7}{8}} m \end{eqnarray} \]

Following exactly the same procedure, we get \( i\hat{\mathcal{M}}(P_{J=0,J_z=0}\rightarrow \gamma_\pm\gamma_\mp) = 0\), which should not be a surprise given the conservation of \(J_z\).

By summing up \(\gamma_+\gamma_+\) and \(\gamma_-\gamma_-\), and taking into account that the two photons are indistinguishable, we have \(\Gamma(P_{J=0}\rightarrow 2\gamma) = \frac{3}{128}\alpha^7 m\).

For \(J=1\), we have \( i\hat{\mathcal{M}}(P_{J=1,J_z=0,\pm 1}\rightarrow \gamma_\pm\gamma_\pm) = i\hat{\mathcal{M}}(P_{J=1,J_z=0,\pm 1}\rightarrow \gamma_\pm\gamma_\mp) = 0\), by following the same calculation.

For \(J=2\), most \(i\hat{\mathcal{M}}\) terms vanish, excecpt for

\[ \begin{eqnarray} && i\hat{\mathcal{M}}(P_{J=2,J_z=2}\rightarrow \gamma_+\gamma_-) \\ &=& \frac{1}{\sqrt{m}} \int \frac{d^3k}{(2\pi)^3} \frac{1}{2\sqrt{2}} (\psi_1\sigma^1_{+-}+i\psi_1\sigma^2_{+-}+i\psi_2\sigma^1_{+-}-\psi_2\sigma^2_{+-}) \\ &=& \frac{1}{\sqrt{m}} \int \frac{d^3k}{(2\pi)^3} \frac{1}{2\sqrt{2}} (4i \frac{e^2}{m}) \left( 2\psi_1(\mathbf{k}) k^x + 2\psi_2(\mathbf{k}) k^y \right) \\ &=& \sqrt{\frac{\pi\alpha^7}{2}} m, \end{eqnarray} \]

and

\[ \begin{eqnarray} && i\hat{\mathcal{M}}(P_{J=2,J_z=-2}\rightarrow \gamma_-\gamma_+) \\ &=& \frac{1}{\sqrt{m}} \int \frac{d^3k}{(2\pi)^3} \frac{1}{2\sqrt{2}} (\psi_1\sigma^1_{-+}-i\psi_1\sigma^2_{-+}-i\psi_2\sigma^1_{-+}-\psi_2\sigma^2_{-+}) \\ &=& \frac{1}{\sqrt{m}} \int \frac{d^3k}{(2\pi)^3} \frac{1}{2\sqrt{2}} (4i \frac{e^2}{m}) \left( 2\psi_1(\mathbf{k}) k^x + 2\psi_2(\mathbf{k}) k^y \right) \\ &=& \sqrt{\frac{\pi\alpha^7}{2}} m. \end{eqnarray} \]

By summing up \(\gamma_+\gamma_-\) and \(\gamma_-\gamma_+\) and averaging over the five \(J_z\)s, we have \(\Gamma(P_{J=2}\rightarrow 2\gamma) = \frac{1}{160}\alpha^7 m\).

It is very nice to see that \(i\hat{\mathcal{M}}(P \rightarrow 2\gamma)\) is non-zero only when \(J_z\) is conserved.

5.5

To derive \(\sum_i \epsilon_\mu^{(i)} \epsilon_\nu^{(i)*} \rightarrow -g_{\mu\nu}\), simply set \({\epsilon^{(3)}}^\mu = (k/m,0,0,E/m)\) and \(k^\mu = (E,0,0,k)\). To same derivation on Page 160 gives us this result.

Xianyu's solution is generally correct, with a few exceptions.

Part (a)

In (5.80), \(ig\) should be \(-ig\), although it has no further consequence.

In (5.84), \(dp\) should be \(dE_p\) in the last line.

Part (b)

In (5.86), \(\frac{i}{\not{k}_1-\not{p}_1}\) really means \(\frac{i(\not{k}_1-\not{p}_1)}{(k_1-p_1)^2}\). It's a notation that is introduced in Chapter 7.

Part (c)

In the textbook we have Show that the denominator of the propagator then never becomes smaller than \(O(m_e^2/s)\). This sentence is very misleading. When we choose not to ignore electron mass, the denominator of the propagator is \(2 x E(k \cos\theta - E)\). Note that \(x\) is fully determined by \(E\), \(m_e\), \(m_B\) and \(\theta\). The minimal value of its absolute value is reached when \(\theta=0\). Also with \(E-k \approx m_e^2/2E\), the minimal absolute value is \(x m_e^2\).

Energy conservation gives us \(x = 1 - m_B^2/s\). So the minimal absolute value is \((s-m_B^2)m_e^2/s\). If we compare this to \(t\) in the derivation in Xianyu's solution, where \(m_e\) is set to 0 in (5.89), we have

\[t = -(s-m_B^2)m_e^2/s = (m_B^2-s) \sin^2 (\theta/2),\]

giving us \(\theta^2 = 4 m_e^2/s\).

So in (5.92), the upper limit of the integral should really be \(1 - 2 m_e^2/\) instead of \(1 - m_e^2/\). But such a constant term does not matter when only the leading logarithmic term, proportional to \(\log(s/m_e^2)\), is retained. Also, if we choose to keep the \(\cos^2\theta\) from (5.91) inside the integral, instead of just setting it to 1, we get the same end result to the leading logarithmic term proportional to \(\log(s/m_e^2)\).

5.6

Part (a)

The definitions of \(\epsilon_+^\mu\) and \(\epsilon_-^\mu\) should be swapped. Since both sides of the definitions are Lorentz vectors, we only need to show it is valid in a particular frame where \(k^\mu=(k,0,0,k)\), \(p^\mu=(p,p\sin\theta,0,p\cos\theta)\), \(u_R(k)=\sqrt{2k} (0,0,1,0)^T\) and \(u_R(p)=\sqrt{2p} (0,0,\cos(\theta/2),\sin(\theta/2))^T\). Then we have \(\epsilon_-^\mu=\frac{1}{\sqrt{2}} (\cot(\theta/2),1,-i,\cot(\theta/2))\). We get similar results for \(\epsilon_+^\mu\).

Also, the statement The second term on the right gives zero when dotted with any photon emission amplitude \(\mathcal{M}^\mu\) is inaccurate. \(\mathcal{M}^\mu\) should be replaced by \(\mathcal{M}^\mu(k) {\mathcal{M}^\nu}^*(k)\).

Xianyu's solution is good. Note that in the derivation of (5.95), we can freely add terms like \(\bar{u}_R(k) \gamma^\mu u_L(p) \bar{u}_L(p) \gamma^\nu u_R(k)\), because terms like \(\bar{u}_R(k) \gamma^\mu u_L(p)\) vanish as it can be easily seen in the chiral representation.

Part (b)

(5.97) in Xianyu's solution should be re-derived accordingly as follows.

\[ \begin{eqnarray} && i\hat{\mathcal{M}}(e_L^- e_R^+ \rightarrow \gamma_L \gamma_R) \\ &=& -(ie)^2 {\epsilon_-}_\mu^*(p_1) {\epsilon_+}_\nu^*(p_2) \bar{u}_L(k_2) \left( \gamma^\nu \frac{i(\not{k_1}-\not{p_1})}{(k_1-p_1)^2} \gamma^\mu + \gamma^\mu \frac{i(\not{k_1}-\not{p_2})}{(k_1-p_2)^2} \gamma^\nu \right) u_L(k_1) \\ &=& -ie^2 \frac{[k_1\gamma_\mu p_1\rangle\langle k_2\gamma_\nu p_2]}{4\sqrt{(k_1\cdot p_1)(k_2\cdot p_2)}} \left(\frac{\langle k_2 \gamma^\nu (\not{k_1}-\not{p_1}) \gamma^\mu k_1]}{t} + \frac{\langle k_2 \gamma^\mu (\not{k_1}-\not{p_2}) \gamma^\nu k_1]}{u} \right) \\ &=& -ie^2 \frac{\langle p_1 \gamma_\mu k_1]\langle k_2\gamma_\nu p_2]}{2t} \left(\frac{\langle k_2 \gamma^\nu k_1]\langle k_1 \gamma^\mu k_1] - \langle k_2 \gamma^\nu p_1]\langle p_1 \gamma^\mu k_1]}{t} + \frac{\langle k_2 \gamma^\mu k_1]\langle k_1 \gamma^\nu k_1] - \langle k_2 \gamma^\mu p_2]\langle p_2 \gamma^\nu k_1]}{u} \right) \\ &=& \frac{-2ie^2}{t} \left(\frac{\langle k_1 p_1 \rangle [k_1 k_1] \langle k_2 k_2 \rangle [k_1 p_2] - \langle p_1 p_1 \rangle [k_1 k_1] \langle k_2 k_2 \rangle [p_1 p_2]}{t} + \frac{\langle k_2 p_1 \rangle [k_1 k_1] \langle k_2 k_1 \rangle [k_1 p_2] - \langle k_2 p_1 \rangle [k_1 p_2] \langle k_2 p_2 \rangle [k_1 p_2]}{u} \right) \\ &=& \frac{2ie^2}{t} \frac{\langle k_2 p_1 \rangle [k_1 p_2] \langle k_2 p_2 \rangle [k_1 p_2]}{u} \end{eqnarray} \]

This is simply (5.99) in Xianyu's solution.

Likewise, (5.97) in Xianyu's solution actually gives us \(i\hat{\mathcal{M}}(e_L^- e_R^+ \rightarrow \gamma_R \gamma_L)\).

Therefore, \(\frac{t}{u}\) and \(\frac{u}{t}\) in (5.100) and (5.101) in Xianyu's solution should be swapped and this change gives us the same result (5.102).

Chapter 6

Notes

Page 178

The placement of the poles at \(k^0=\pm|\mathbf{k}|\) below the real axis might seem a bit arbitrary. The point is that this is the only way to ensure that \(A^\mu(x)\) with \(t<0\) be the Coulomb potential of an unaccelerated charge.

The un-numbered equation at the bottom of the page is a bit sloppy. The \(k\) in \(k^2\) in the denominator is a 4-momentum with \(k^0\) set to \(\mathbf{k}\cdot\mathbf{p}/p^0\). It should have been written similar to Eq. (6.6).

For some more details on the electromagnetic field, one can refer to Shankar's Principles of Quantum Mechanics, Section 18.5.

Page 182

Technically, the last equality in (6.17) should be approximate equality.

Page 187

Regarding the Born approximation of scattering from a potential, refer to the discussion around (4.123) and (4.124).

Immediately below (6.35), by vanishes at \(\mathbf{q}=0\), the author means that radiative corrections vanish.

Page 188

There is a sign error in the equation/defintion of \(\tilde{B}^k(\mathbf{q})\).

Page 190

For the statement the denominator depends only on \(l^2\) to make sense, we need to assume that \(p\) corresponds to an external (thus non-virtual) electron, and thus \(p^2=m_e^2\).

Page 191

For a detailed derivation of (6.44), see this post.

To see why \(q^2<0\), note that we can always find a frame where the incoming and outgoing electrons have the same energy.

Page 192

For a detailed derivation of the numerator, see this post.

Page 193

The statement the Wick rotation cannot be justified does not seem to make sense. Wick rotation should be fine up to \(l^4\) in the numerator when \(m=3\).

Page 195

The statement the divergence appears in the worst possible place... means that an infinite \(\Lambda\) leads to an infinite \(\log\frac{z\Lambda^2}{\Delta}\).

Page 208

The "correction" to Eq. (6.86) in the official Errata actually makes things worse. The original text is perfectly correct. Only its interpretation requires some explanation.

(6.86) is a conditional probability, i.e., given that we observe those hard photons, with the threshold \(E_l\) being \(E_+\), what is the probability that there are exactly \(n\) soft photons with energy between \(E_-\) and \(E_+\), and whatever number of soft photos with energy less than \(E_-\). This can be clearly seen when we can multiply it to (6.84) as follows.

\[ \begin{eqnarray} && \left(\frac{d\sigma}{d\Omega}\right)_0 \times \exp\left[-\frac{\alpha}{\pi}f_{\mathrm{IR}}(q^2)\log\left(\frac{-q^2}{E_+^2}\right)\right] \times \frac{1}{n!}\left[\frac{\alpha}{\pi}f_{\mathrm{IR}}(q^2)\log\left(\frac{E_+^2}{E_-^2}\right)\right]^n \times \exp\left[-\frac{\alpha}{\pi}f_{\mathrm{IR}}(q^2)\log\left(\frac{E_+^2}{E_-^2}\right)\right] \\ &=& \left(\frac{d\sigma}{d\Omega}\right)_0 \times \exp\left[-\frac{\alpha}{\pi}f_{\mathrm{IR}}(q^2)\log\left(\frac{-q^2}{E_-^2}\right)\right] \times \frac{1}{n!}\left[\frac{\alpha}{\pi}f_{\mathrm{IR}}(q^2)\log\left(\frac{E_+^2}{E_-^2}\right)\right]^n \end{eqnarray} \]

Problems

6.1

Xianyu's solution is probably not wrong. But there are many gaps in the derivations. So it is better to refer to Bourjaily's solution.

Below is a list of the minor errors.

• \((k' \cdot p)(k \cdot p)\) should be \((k' \cdot p)(k \cdot p')\) at the bottom of Page 1. This error is propagated to the top of Page 3, but corrected thereafter.
• It should be \(p' = (E-E'+m,E\hat{z}-\vec{k'})\) on Page 2.
• It should be \(k' = (E',0,E'\sin\theta,E'\cos\theta)\) on Page 2.
• It should be \(k^2=0=k'^2+2q\cdot k'+q^2=2p'\cdot k'-2p\cdot k'+q^2=2p'\cdot k'-2Em\) on Page 2.
• \(F_s\) should be \(F_2\) on Page 3.
• It should be \(\frac{q^2}{4}=-\frac{2E^2m^2}{1+\frac{2E}{m}\sin^2\frac{\theta}{2}}\frac{q^2}{2m^2}\sin^2\frac{\theta}{2}\) on Page 3.
• In the third line of the derivation of \(\sigma\) on Page 4, \(dE\) should be \(dE'\). A more serious conceptual error is that \(\bar{|\mathcal{M}|^2}\) should be inside the integral throughout the derivation. Basically we should move the integral sign to the leftmost on every single line.

6.2

Xianyu's solution is good.

Part (a)

Note that the expansion is simply decomposing a 4-vector into four linearly independent basis vectors.

Part (b)

To see how to reach (6.19) from the three preceding equations in Xianyu's solution, simply notice that when we throw away the time dimension in the equation in Part (a) in the textbook where \(A\), \(B\), \(C\), and \(D\) are defined, \(\mathbf{\epsilon}_1\), \(\mathbf{\epsilon}_2\), and \(\mathbf{q}\) are orthorgonal to each other. We immediately get (6.19) by multiplying \(\mathbf{\epsilon}_1\) or \(\mathbf{\epsilon}_2\) on both sides of the equation.

Part (c)

In (6.20) and (6.21), \(\hat{\mathcal{M}}_\nu(q)\) should be \(\hat{\mathcal{M}}_\nu^*(q)\).

There are a few issues with (6.22).

First, Xianyu is already using an assumption from Part (e), that the target cross sections are independent of the photon polarization, which means \(\epsilon_1^\mu{\epsilon_1^\nu}^*\) and \(\epsilon_2^\mu{\epsilon_2^\nu}^*\) are equivalent to each other, as long as they are dotted into \(\hat{\mathcal{M}}_\mu(q)\hat{\mathcal{M}}_\nu^*(q)\). Eq. (5.75) in the textbook then implies that we should have \(\epsilon_1^\mu{\epsilon_1^\nu}^*\rightarrow-\frac{1}{2}g^{\mu\nu}\) and \(\epsilon_2^\mu{\epsilon_2^\nu}^*\rightarrow-\frac{1}{2}g^{\mu\nu}\). Note that the factor should be \(-\frac{1}{2}\) instead of \(-\frac{1}{3}\), because there are in fact also \(\epsilon_0\) and \(\epsilon_3\) that cancel each other. So the factor of \(\frac{1}{3}\) in the second line of (6.22) should be \(\frac{1}{2}\) instead. This factor somehow got lost between the second and the third line of (6.22).

(If we postpone this assumption, we simply deal with the two polarizations separately by splitting (6.22) into two equations. We will reach the same conclusion nonetheless.)

Second, the minus sign in the third line should not be there. I guess Xianyu got it when he replaced the integral variable \(p'\) (or equivalently \(E'\)) by \(x\). But we should then also swap the upper and lower limits of the integral over \(x\). What those values are exactly does not really matter in the end, as long as the lower limit is less than the upper limit, because the \(\delta^4\) term will fix it. The only integral in (6.22) that is not fixed by the \(\delta^4\) term is that over \(\theta\).

Third, \(|\hat{\mathcal{M}}^\mu|^2\) is effectively defined as \(-\hat{\mathcal{M}}^\mu \hat{\mathcal{M}}_\mu\) with a minus sign coming from averaging over photon polarizations.

In (6.23), \((1-\cos\theta)^4\) should be \((1-\cos\theta)^2\).

Part (d)

The left-hand side of (6.24) means that the lower limit of the integral over \(\theta\) is now effectively \(\frac{x}{1-x}\frac{m}{E}\). Factors like \(\frac{x}{1-x}\), or the upper bound of \(\theta\), do not matter compared to the dominant term \(\log\frac{E^2}{m^2}\). The minus sign on the right-hand side, however, should be removed. (This error cancels the previous sign error in (6.22).)

Part (e)

In (6.25), a factor of \(\frac{1}{2}\) should be added, due to the error in Part (c). Note that \(Ex\) is simply \(E_\gamma\), the energy of the equivalent photon, so we can write, instead of (6.25),

\[d\sigma=\frac{1}{2} dx \frac{1}{2E_\gamma2M_t}\frac{\alpha}{2\pi} \frac{1+(1-x)^2}{x}\log\frac{E^2}{m^2} \int\frac{d^3p_t}{(2\pi)^3 2E_t}|\hat{\mathcal{M}}^\mu(q)|^2 (2\pi)^4 \delta^{(4)}(\sum p_i).\]

Note that this is exactly what we would write down if we pretend it is an incoming photon hitting the target, and replace the regular expression \(\int \frac{d^3p_\gamma}{(2\pi)^3 2E_\gamma}\ldots\) by \(\int N_\gamma dx\). The factor of \(\frac{1}{2}\) is then simply averaging over the photon helicity.

6.3

Xianyu's solution is probably not wrong. But there are many gaps in the derivations. So it is better to refer to Bourjaily's solution.

Part (a)

Note that we don't care about terms that only involve a \(\gamma^\mu\) because they affect \(F_1\) but for \(g\) we only need \(F_1(0)\), which is always set to 1.

There are the following errors on Page 2.

• In the first line of the calculation of the numerator, \(u(p)\) is missing at the end.
• It should be \(\textbf{(iv)} \rightarrow 2mp'^\mu\).
• \((y-x)(z-1)\) should be \((x-y)(1+z)\) in (a.4).

Part (b)

I have nothing to add to Bourjaily's solution.

Part (c)

It should be rather obvious that the \(a\) should be \(\lambda\)'s subscript rather a standalone parameter in the textbook.

There are the following errors on Page 4.

• It should be \(\Delta=-xyq^2+(1-z)^2m_e^2+zm_a^2\).
• There is an overall sign error in \(i\mathcal{M}\).
• \((y-x)(z-1)\) should be \((x-y)(1+z)\) in (a.4).

Chapter 7

Notes

Page 212

For a better understanding of the following sentence, refer to this post.

This term is usually zero by symmetry; for higher-spin fields, it is zero by Lorentz invariance.

Page 216

A review of the text leading to (4.57) is recommended before one studies An Example: The Electron Self-Energy.

Page 221

This link provides a partial explanation to the claim below. However, it is still unclear why we would need linearity in \(m_0\).

The mass shift must therefore be proportional to \(m_0\), and so, by dimensional analysis, it can depend only logarithmically on \(\Lambda\).

Page 224

The following statement is meant to be valid only at the singularity.

The denominator is just that of Eq. (7.5): p2 — ra2.

Page 227

The following statement requires some qualification.

By a similar analysis, it is easy to confirm that disconnected diagrams should be disregarded because they do not have the singularity structure, with a product of four poles, indicated on the right-hand side of (7.42).

For a 2-particle→2-particle scattering, a disconnected diagram would lead to two poles. But for a 3-particle→3-particle scattering (\(\mathrm{a,b,c}\rightarrow\mathrm{1,2,3}\)), we can have a disconnected diagram with six poles with a and b connected to 1 and c connected to 2 and 3.

Also, remember that the Feynman propagators for the two external lines in each diagram in the 1PI must not be included in the expression for \(-iM^2(p^2)\). In particular, see (7.16) and the discussion below (7.21).

Page 231

If one pays close attention to the derivation of (7.49), one should notice that there is nothing preventing \(f\) from being the same as \(a\) or \(b\).

Page 233

Regarding the phrase in the physical region \(k^0 > 2m\), here we are already assuming the center-of-mass frame. In general, \(i\mathcal{M}\) is a function of \(k^2\) instead of \(k^0\).

Page 234

The paragraph above (7.53) deserves some elaboration.

We certainly could close the integration contour upward and pick up the residues of the poles in the upper half-plane. We would, however, get exactly the same result.

In (7.54), we only considered the pole at \(q^0=-(1/2)k^0+E_\mathbf{q}\). If we also include the other pole \(q^0=(1/2)k^0+E_\mathbf{q}\), we will get an extra term similar to the second line of (7.54), but with \((k^0-2E_\mathbf{q})\) in the denominator replaced by \((k^0+2E_\mathbf{q})\). This will not lead to any pole in the integrand and therefore will only contribute to the real part of \(\delta M\). Note that strictly speaking, the equality in (7.54) is incorrect because it ignores this irrelevant but nonetheless non-zero contribution.

To see why (7.53) is true, notice the following.

\[ \begin{eqnarray} && \delta\left( (k/2+q)^2-m^2 \right) \\ &=& \delta\left( (k^0/2+q^0)^2-E_\mathbf{q}^2 \right) \\ &=& \delta\left( (k^0/2+q^0+E_\mathbf{q})(k^0/2+q^0-E_\mathbf{q}) \right) \\ &=& \delta\left(q^0 - (-k^0/2-E_\mathbf{q}) \right) \frac{1}{|k^0/2+q^0-E_\mathbf{q}|} + \delta\left(q^0 - (-k^0/2+E_\mathbf{q}) \right) \frac{1}{|k^0/2+q^0+E_\mathbf{q}|} \\ \end{eqnarray} \]

Subtitute this into (7.52), we have

\[i\delta \mathcal{M} = -2\pi i \frac{\lambda^2}{2} \int \frac{d^3 q}{(2\pi)^4} \left( \frac{1}{(k^0+E_\mathbf{q})^2-E_\mathbf{q}^2} \frac{1}{2E_\mathbf{q}} + \frac{1}{(k^0-E_\mathbf{q})^2-E_\mathbf{q}^2} \frac{1}{2E_\mathbf{q}} \right),\]

which is exactly the same as (7.54) if we had included both poles.

Regarding the principle value \(P\), refer to A.4 in Shankar's Principles of Quantum Mechanics.

To verify the claim about (7.55), we substitue (7.53) and (7.55) into (7.52) and have

\[ \begin{eqnarray} && i\delta \mathcal{M} \\ &\rightarrow& \frac{\lambda^2}{2} \int \frac{d^4 q}{(2\pi)^4} (-2i\pi)^2 \delta\left( (k/2+q)^2-m^2 \right) \delta\left( (k/2-q)^2-m^2 \right) \\ &=& - \frac{1}{(2\pi)^2} \frac{\lambda^2}{2} \int d^4 q \left( \delta\left(q^0 - (-k^0/2-E_\mathbf{q}) \right) \frac{1}{|k^0/2+q^0-E_\mathbf{q}|} + \delta\left(q^0 - (-k^0/2+E_\mathbf{q}) \right) \frac{1}{|k^0/2+q^0+E_\mathbf{q}|} \right) \left( \delta\left(q^0 - (k^0/2+E_\mathbf{q}) \right) \frac{1}{|k^0/2-q^0-E_\mathbf{q}|} + \delta\left(q^0 - (k^0/2-E_\mathbf{q}) \right) \frac{1}{|k^0/2-q^0+E_\mathbf{q}|} \right) \\ &=& - \frac{1}{(2\pi)^2} \frac{\lambda^2}{2} \int d^3 q \left( \delta(k^0+2E_\mathbf{q}) \frac{1}{2E_\mathbf{q}k_0} + \delta(-k^0) \frac{1}{2E_\mathbf{q}(k_0+2E_\mathbf{q})} + \delta(-k^0) \frac{1}{2E_\mathbf{q}(k_0-2E_\mathbf{q})} + \delta(-k^0+2E_\mathbf{q}) \frac{1}{2E_\mathbf{q}k_0} \right) \\ &=& - \frac{1}{\pi} \frac{\lambda^2}{2} \int_m^\infty dE_{\mathrm{q}} E_{\mathrm{q}} |q| \left( \delta(k^0+2E_\mathbf{q}) \frac{1}{2E_\mathbf{q}k_0} + \delta(-k^0) \frac{1}{2E_\mathbf{q}(k_0+2E_\mathbf{q})} + \delta(-k^0) \frac{1}{2E_\mathbf{q}(k_0-2E_\mathbf{q})} + \delta(-k^0+2E_\mathbf{q}) \frac{1}{2E_\mathbf{q}k_0} \right) \\ &=& - \frac{1}{\pi} \frac{\lambda^2}{2} \int_m^\infty dE_{\mathrm{q}} |q| \delta(-k^0+2E_\mathbf{q}) \frac{1}{2k_0}, \end{eqnarray} \]

which is indeed the discontinuity \(i\delta \mathcal{M}(k^2+i\epsilon) - i\delta \mathcal{M}(k^2-i\epsilon)\).

Page 235

To make the discussion around (7.56) a bit more explicit, we have, starting from (7.52) and making the substitutions up to and including (7.56),

\[ \begin{eqnarray} && \mathrm{Disc} (i \mathcal{M}) \\ &=& \frac{\lambda^2}{2} \int \frac{d^4 p_1}{(2\pi)^4} \int \frac{d^4 p_2}{(2\pi)^4} (2\pi)^4 \delta^{(4)}(p_1+p_2-k) (-2i\pi)^2 \delta(p_1^2-m^2) \delta(p_2^2-m^2) \\ &=& -\frac{\lambda^2}{2} \int \frac{d^4 p_1}{(2\pi)^3} \int \frac{d^4 p_2}{(2\pi)^3} (2\pi)^4 \delta^{(4)}(p_1+p_2-k) \delta\left((p_1^0)^2-E_\mathrm{p_1}\right) \delta\left((p_2^0)^2-E_\mathrm{p_2}\right) \\ &=& -\frac{\lambda^2}{2} \int \frac{d^3 p_1}{(2\pi)^3} \int \frac{d^3 p_2}{(2\pi)^3} (2\pi)^4 \delta^{(4)}(p_1+p_2-k) \frac{1}{2E_\mathrm{p_1}} \frac{1}{2E_\mathrm{p_2}} \\ &=& -\frac{1}{2} \int \frac{d^3 p_1}{(2\pi)^3} \frac{1}{2E_\mathrm{p_1}} \int \frac{d^3 p_2}{(2\pi)^3} \frac{1}{2E_\mathrm{p_2}} |\mathcal{M}(k)|^2 (2\pi)^4 \delta^{(4)}(p_1+p_2-k), \end{eqnarray} \]

where \(\mathcal{M}(k)=-\lambda+O(\lambda^2)\) as can be seen in the discussion under (4.98).

Page 236

Note that for the order-\(\alpha^2\) Bhabha scattering diagrams shown in Fig. 7.6, the factor of \(\frac{1}{2}\) is absent since it is about the scattering of an electron and a positron, which are not identical.

The validity of (7.57) cannot be established by the arguments given in the textbook for the following reasons.

• The claim in the paragraph above (7.57) that \(-iM^2(p^2)\) can be thought of as the sum of all amputated diagrams is dubious. If we refer back to (4.54), the first sub-diagram cannot be expressed as something conforming to Figure 7.4; the second one is unambiguous; the third however presents a problem, as the suggested "amputated diagram" can contain the first loop, the second loop, or both, and this will lead to multiple counting of the same sub-diagram in the complete, un-amputated diagram.
• Even if we accept the previous claim any way, now we rely on the un-numbered equation at the top of Page 229. It is however \(S\), instead of \(iT\), on the left-hand side. So the left-hand side will give us an extra term \(\langle p|1|p\rangle\) in addition to just \(i\mathcal{M} (2\pi)^4\delta^{(4)}(0)\).

In fact, one could argue that we should not be allowed to use the LSZ formula at all, because we derived the LSZ formula by assuming asymptotic states.

There are probably some other deeper reasons why (7.57) is valid.

For 1-particle → 1-particle “scattering” where the particle is stable, The right-hand side of (7.49) vanishes, because \(\mathcal{M}\) is the matrix element of \(T\), which is in turn defined as \(S=1+iT\). \(\mathcal{M}\) must vanish since \(\langle p|S|p\rangle=(2\pi)^4\delta^{(4)}(0)\).

To get a better understanding of unstable particles, I found the following two sources quite helpful.

• Quantum Field Theory by Professor Lowell S. Brown. Section 6.3. Note that certain conventions used in the book are different. For (6.3.10), refer to (6.1.48) and notice that \(G^{-1}(z)\) itself has a \(-\pi i \sigma(z)\) for \(z\) under the real axis.
• Master thesis Resonances and poles in the second Riemann sheet by Thomas Wolkanowski.

Unlike a "real" particle, which manifests itself as a delta function in Figure 7.3, an unstable particle is a pole on the "second" Riemmann sheet right below the real axis through analytical continuation - actually also on a "zeroth" sheet by symmetry right above the real axis - and would manifest itself as a bump in Figure 7.3. In a sense, its presense is "sensed" on the branch cut (since everything is through analytical continuation and therefore continuous).

Page 241

There is a sign error in the two equations on this page.

Page 246

I am not sure how to prove the claim If, for example, \(\Pi^{\mu\nu}(q)\) contained a term \(M^2g^{\mu\nu}\) (with no compensating \(q^\mu q^\nu\) term), the photon mass would be shifted to \(M\).

Page 254

As for \(q=-Q\), we would be integrating downward on the left side of the cut, and upward on the right side of the cut with the original \(Q\). Now with \(q\), which is \(Q\) rotated by \(\frac{\pi}{2}\) clockwise, it would be twice the integral below the real axis from \(2m\) to infinity.

Page 255

To see where \(\frac{5}{3}\) is from in the un-numbered equation, use this calculator.

Problems

7.1

Most solutions floating online use dimensional regularization. For the sake of variety, I will use the good old Pauli-Villars procedure instead. (Feel free to consult Xianyu's solution, just note that a factor of \(\frac{1}{2}\) is missing in his (7.1). I will make further no mention of his solution to this problem.)

One starts with (7.52). Note its similarity with (7.16). Following the same procedure with minimal modifications (e.g., replacing the numerator by a constant, setting \(\mu=m_0\), etc.), we have

\[ \begin{eqnarray} && i\delta\mathcal{M} \\ &=& \frac{\lambda^2}{2} \int_0^1 \frac{i}{(4\pi)^2}\log\left(\frac{\Delta_\Lambda}{\Delta}\right) dx \\ &=& \frac{\lambda^2}{2} \int_0^1 \frac{i}{(4\pi)^2}\log\left(\frac{\Lambda^2-x(1-x)k^2}{m_0^2-x(1-x)k^2}\right) dx \\ &=& \frac{\lambda^2}{2} \int_0^1 \frac{i}{(4\pi)^2}\log\left(\frac{\Lambda^2}{m_0^2-x(1-x)k^2}\right) dx, \end{eqnarray} \]

where as usual, we set \(\Lambda\rightarrow\infty\).

Since \(x(1-x) \le \frac{1}{4}\), nothing funny happens within the logarithm and \(\delta\mathcal{M}\) is real as long as \(k^2 < 4 m_0^2\). When \(k^2 \ge 4 m_0^2\), we calculate \(\log\left(\frac{\Lambda^2}{m_0^2-x(1-x)k^2}\right)\) through analytic continuation. We have

\[ \begin{eqnarray} && \mathrm{Im}\log\left(\frac{\Lambda^2}{m_0^2-x(1-x)(k^2+i\epsilon)}\right) \\ &=& \mathrm{Im}\log\left(-\frac{\Lambda^2}{x(1-x)k^2-m_0^2} + i\epsilon \right) \\ &=& \pi, \end{eqnarray} \]

when \(\frac{1}{2}-\frac{1}{2}\sqrt{1-\frac{4m_0^2}{k^2}} < x < \frac{1}{2}+\frac{1}{2}\sqrt{1-\frac{4m_0^2}{k^2}}\).

Therefore we have

\[ \begin{eqnarray} && \mathrm{Im} \delta\mathcal{M} \\ &=& \frac{\lambda^2}{2(4\pi)^2} \int_{\frac{1}{2}-\frac{1}{2}\sqrt{1-\frac{4m_0^2}{k^2}}}^{\frac{1}{2}+\frac{1}{2}\sqrt{1-\frac{4m_0^2}{k^2}}} \mathrm{Im}\log\left(\frac{\Lambda^2}{m_0^2-x(1-x)k^2}\right) dx \\ &=& \frac{\lambda^2}{32\pi} \sqrt{1-\frac{4m_0^2}{k^2}} \\ \end{eqnarray} \]

7.2

Part (a)

Calculation leading to \(Z_1\) starts with (6.38). With the regularization required for this problem, we will have to introduce photon mass \(\mu\). This leads to the following modifications.

• (6.38): \((k-p)^2+i\epsilon\) has to be replace by \((k-p)^2-\mu^2+i\epsilon\);
• (6.43): an extra term \(-z\mu^2\) has to be added;
• (6.44): an extra term \(z\mu^2\) has to be added to \(\Delta\).

If we cut off \(l_E\) at \(\Lambda\), (6.49) becomes

\[\frac{i}{2(4\pi)^2}\left(\frac{\Lambda^2}{(\Lambda^2+\Delta)^2}+\frac{1}{\Lambda^2+\Delta}-\frac{1}{\Delta}\right)=\frac{-i}{2(4\pi)^2}\frac{\Lambda^4}{(\Lambda^2+\Delta)^2\Delta},\]

and (6.50) becomes

\[\frac{-i}{2(4\pi)^2}\left(\frac{\Lambda^4}{(\Lambda^2+\Delta)^2}+\frac{2\Lambda^2}{\Lambda^2+\Delta}-2\log(\frac{\Lambda^2+\Delta}{\Delta})\right).\]

Setting \(q=0\), we have from (6.47)

\[ \begin{eqnarray} && \delta\Gamma^\mu(q=0) \\ &=& \gamma^\mu 2ie^2\int_0^1dx dy dz \delta(x+y+z-1) \left[(-1) \frac{-i}{2(4\pi)^2}\left(\frac{\Lambda^4}{(\Lambda^2+\Delta)^2}+\frac{2\Lambda^2}{\Lambda^2+\Delta}-2\log(\frac{\Lambda^2+\Delta}{\Delta})\right)+2(1-4z+z^2)m^2(\frac{-i}{2(4\pi)^2}\frac{\Lambda^4}{(\Lambda^2+\Delta)^2\Delta})\right], \end{eqnarray} \]

where \(\Delta=(1-z)^2m^2+z\mu^2\) since we have set \(q=0\).

Then from (7.47) we have

\[ \begin{eqnarray} && \delta Z_1 \\ &=& -2ie^2\int_0^1dx dy dz \delta(x+y+z-1) \left[(-1) \frac{-i}{2(4\pi)^2}\left(\frac{\Lambda^4}{(\Lambda^2+\Delta)^2}+\frac{2\Lambda^2}{\Lambda^2+\Delta}-2\log(\frac{\Lambda^2+\Delta}{\Delta})\right)+2(1-4z+z^2)m^2(\frac{-i}{2(4\pi)^2}\frac{\Lambda^4}{(\Lambda^2+\Delta)^2\Delta})\right] \\ &=& -2ie^2\int_0^1 dz (1-z) \left[\frac{i}{2(4\pi)^2}\left(\frac{\Lambda^4}{(\Lambda^2+\Delta)^2}+\frac{2\Lambda^2}{\Lambda^2+\Delta}-2\log(\frac{\Lambda^2+\Delta}{\Delta})\right)+2(1-4z+z^2)m^2(\frac{-i}{2(4\pi)^2}\frac{\Lambda^4}{(\Lambda^2+\Delta)^2\Delta})\right] \\ &=& \frac{e^2}{(4\pi)^2} \int_0^1 dz (1-z) \left[\left(\frac{\Lambda^4}{(\Lambda^2+\Delta)^2}+\frac{2\Lambda^2}{\Lambda^2+\Delta}-2\log(\frac{\Lambda^2+\Delta}{\Delta})\right)-2(1-4z+z^2)m^2\frac{\Lambda^4}{(\Lambda^2+\Delta)^2\Delta}\right]. \end{eqnarray} \]

Setting \(\Lambda\rightarrow\infty\), we have

\[\delta Z_1 = \frac{e^2}{(4\pi)^2} \int_0^1 dz (1-z) \left[\left(3-2\log(\frac{\Lambda^2+\Delta}{\Delta})\right)-2(1-4z+z^2)\frac{m^2}{\Delta}\right].\]

Now we turn to \(Z_2\). With the same procedure, (7.18) becomes

\[\frac{i}{(4\pi)^2}\left(\log(\frac{\Lambda^2+\Delta}{\Delta})-\frac{\Lambda^2}{\Lambda^2+\Delta}\right),\]

where \(\Delta=-x(1-x)p^2+x\mu^2+(1-x)m^2\).

From (7.17), we have

\[ \begin{eqnarray} && \Sigma_2(p) \\ &=& \frac{e^2}{(4\pi)^2} \int_0^1 dx (-2x\not{p}+4m) \left(\log(\frac{\Lambda^2+\Delta}{\Delta})-\frac{\Lambda^2}{\Lambda^2+\Delta}\right). \end{eqnarray} \]

From (7.26), we have

\[ \begin{eqnarray} && \delta Z_2 \\ &=& \frac{d\Sigma}{d\not{p}} \Bigg|_{\not{p}=m} \\ &=& \frac{e^2}{(4\pi)^2} \int_0^1 dx \left[(-2x) \left(\log(\frac{\Lambda^2+\Delta}{\Delta})-\frac{\Lambda^2}{\Lambda^2+\Delta}\right) + (-2x\not{p}+4m) \left(\frac{1}{\Lambda^2+\Delta}-\frac{1}{\Delta}+\frac{\Lambda^2}{(\Lambda^2+\Delta)^2}\right)(-2x)(1-x)\not{p} \right] \Bigg|_{\not{p}=m} \\ &=& \frac{e^2}{(4\pi)^2} \int_0^1 dx \left[(-2x) \left(\log(\frac{\Lambda^2+\Delta}{\Delta})-\frac{\Lambda^2}{\Lambda^2+\Delta}\right) - 4x(2-x)(1-x)m^2 \left(\frac{1}{\Lambda^2+\Delta}-\frac{1}{\Delta}+\frac{\Lambda^2}{(\Lambda^2+\Delta)^2}\right) \right] \\ &=& \frac{e^2}{(4\pi)^2} \int_0^1 dx \left[(-2x) \left(\log(\frac{\Lambda^2+\Delta}{\Delta})-\frac{\Lambda^2}{\Lambda^2+\Delta}\right) + 4x(2-x)(1-x)m^2 \frac{\Lambda^4}{(\Lambda^2+\Delta)^2\Delta} \right], \end{eqnarray} \]

with now \(\Delta=(1-x)^2m^2+x\mu^2\).

Setting \(\Lambda\rightarrow\infty\), we have

\[\delta Z_2 = \frac{e^2}{(4\pi)^2} \int_0^1 dx \left[(-2x) \left(\log(\frac{\Lambda^2+\Delta}{\Delta})-1\right) + 4x(2-x)(1-x)\frac{m^2}{\Delta} \right].\]

Notice that in both cases, we have the same \(\Delta\) when we replace \(z\) by \(x\) in \(Z_1\).

Clearly \(\delta Z_1 \neq \delta Z_2\).

Part (b)

We start with (6.38). The calculation can be greatly simplified since we are only interested in the case where \(q=0\). (See (7.47).) Therefore we may as well set \(q=0\) (and thus \(p=p'\) and \(k=k'\)) from the very beginning. Then (6.38) becomes

\[ \begin{eqnarray} && \bar{u}(p) \delta\Gamma^\mu(p,p) u(p) \\ &=& \int \frac{d^dk}{(2\pi)^d} \frac{-ig_{\nu\rho}}{(k-p)^2-\mu^2+i\epsilon} \bar{u}(p) (-ie\gamma^\nu) \frac{i(\not{k}+m)}{{k}^2-m^2+i\epsilon} \gamma^\mu \frac{i(\not{k}+m)}{{k}^2-m^2+i\epsilon} (-ie\gamma^\rho) u(p) \\ &=& -ie^2 \int \frac{d^dk}{(2\pi)^d} \frac{\bar{u}(p) [(2-d)\not{k}\gamma^\mu\not{k} + (2-d)m^2\gamma^\mu + 2dmk^\mu ]u(p)}{((k-p)^2-\mu^2+i\epsilon) ({k}^2-m^2+i\epsilon) ({k}^2-m^2+i\epsilon)} \\ &=& -ie^2 \int \frac{d^dl}{(2\pi)^d} \int_0^1 dxdydz \delta(x+y+z-1) \frac{2 \bar{u}(p) [(2-d)\frac{2-d}{d}l^2\gamma^\mu + (2-d)(1+z^2)m^2\gamma^\mu + dmz(p^\mu+p^\mu) ]u(p)}{D^3} \\ &=& -ie^2 \int \frac{d^dl}{(2\pi)^d} \int_0^1 dxdydz \delta(x+y+z-1) \frac{2 \bar{u}(p) \gamma^\mu[(2-d)\frac{2-d}{d}l^2 + (2-d)(1+z^2)m^2 + 2dm^2z]u(p)}{D^3}, \end{eqnarray} \]

where we have made exactly the same substitutions as those leading to (6.47), except that now \(\Delta=(1-z)^2m^2+z\mu^2\). Note that our result is consistent with (6.47).

From this we can directly read

\[ \begin{eqnarray} && \delta\Gamma^\mu(q=0) \\ &=& \gamma^\mu (2ie^2) \int \frac{d^dl}{(2\pi)^d} \int_0^1 dxdydz \delta(x+y+z-1) \frac{(d-2)\frac{2-d}{d}l^2 + (d-2)(1+z^2)m^2 - 2dm^2z}{D^3} \\ &=& \gamma^\mu (-2e^2) \int_0^1 dz (1-z) \left[(d-2)\frac{2-d}{d}\frac{1}{(4\pi)^{d/2}}\frac{d}{2}\frac{\Gamma(2-\frac{d}{2})}{\Gamma(3)}(\frac{1}{\Delta})^{2-\frac{d}{2}} - ((d-2)(1+z^2) - 2dz)m^2 \frac{1}{(4\pi)^{d/2}}\frac{\Gamma(3-\frac{d}{2})}{\Gamma(3)}(\frac{1}{\Delta})^{3-\frac{d}{2}}\right] \\ &=& \gamma^\mu \frac{2e^2}{(4\pi)^{d/2}} \int_0^1 dz (1-z) \left[\frac{(d-2)^2}{2}\frac{\Gamma(2-\frac{d}{2})}{\Gamma(3)}(\frac{1}{\Delta})^{2-\frac{d}{2}} + ((d-2)(1+z^2) - 2dz)m^2 \frac{\Gamma(3-\frac{d}{2})}{\Gamma(3)}(\frac{1}{\Delta})^{3-\frac{d}{2}}\right] \\ &=& \gamma^\mu \frac{2e^2}{(4\pi)^2} \int_0^1 dz (1-z) \left[\frac{4-4\epsilon}{4}\left(\frac{2}{\epsilon}-\log\Delta-\gamma+\log(4\pi)\right) + 2(1-4z+z^2)m^2 \frac{1}{2}(\frac{1}{\Delta})\right] \\ &=& \gamma^\mu \frac{2e^2}{(4\pi)^2} \int_0^1 dz (1-z) \left[\frac{2}{\epsilon}-\log\Delta-\gamma+\log(4\pi)-2 + \frac{(1-4z+z^2)m^2}{\Delta}\right]. \end{eqnarray} \]

Equivalently, we have

\[\delta Z_1 = -\frac{2e^2}{(4\pi)^2} \int_0^1 dz (1-z) \left[\frac{2}{\epsilon}-\log\Delta-\gamma+\log(4\pi)-2 + \frac{(1-4z+z^2)m^2}{\Delta}\right].\]

For \(Z_2\). we start with (7.16), which now becomes

\[ \begin{eqnarray} && \Sigma_2(p) \\ &=& -ie^2 \int \frac{d^dk}{(2\pi)^d} \frac{\gamma^\mu(\not{k}+m)\gamma^\mu}{(k^2-m^2+i\epsilon)((p-k)^2-\mu^2+i\epsilon)} \\ &=& -ie^2 \int \frac{d^dk}{(2\pi)^d} \frac{-(d-2)\not{k}+dm}{(k^2-m^2+i\epsilon)((p-k)^2-\mu^2+i\epsilon)} \\ &=& -ie^2 \int_0^1 dx \int \frac{d^dl}{(2\pi)^d} \frac{-(d-2)x\not{p}+dm}{(l^2-\Delta+i\epsilon)^2} \\ &=& \frac{e^2}{(4\pi)^{d/2}} \int_0^1 dx (-(d-2)x\not{p}+dm) \frac{\Gamma(2-\frac{d}{2})}{\Gamma(2)} (\frac{1}{\Delta})^{2-\frac{d}{2}} \\ &=& \frac{e^2}{(4\pi)^2} \int_0^1 dx \left(-(2-\epsilon)x\not{p}+(4-\epsilon)m\right) \left(\frac{2}{\epsilon}-\log\Delta-\gamma+\log(4\pi)\right). \end{eqnarray} \]

From (7.26), we have

\[ \begin{eqnarray} && \delta Z_2 \\ &=& \frac{d\Sigma}{d\not{p}} \Bigg|_{\not{p}=m} \\ &=& \frac{e^2}{(4\pi)^2} \int_0^1 dx \left[-(2-\epsilon)x \left(\frac{2}{\epsilon}-\log\Delta-\gamma+\log(4\pi)\right) + \left(-(2-\epsilon)xm+(4-\epsilon)m\right) (-\frac{1}{\Delta}) (-x)(1-x)2m \right] \\ &=& -\frac{2e^2}{(4\pi)^2} \int_0^1 dx x \left[\frac{2}{\epsilon}-\log\Delta-\gamma+\log(4\pi)-1 - \frac{2(2-x)(1-x)m^2}{\Delta} \right]. \end{eqnarray} \]

Again, notice that we have in both cases the same \(\Delta\) when we replace \(z\) by \(x\) in \(Z_1\). Also notice that \(\int_0^1 x dx = \int_0^1 (1-x) dx\). So we have (with \(\Delta=(1-x)^2m^2+x\mu^2\))

\[ \begin{eqnarray} && \delta Z_1 - \delta Z_2 \\ &=& -\frac{2e^2}{(4\pi)^2} \left[ \int_0^1 dx \left(x-(1-x)\right) \log\Delta + \int_0^1 dx (1-x) \left(-1 + \frac{(1-4x+x^2)m^2}{\Delta} + \frac{2x(2-x)m^2}{\Delta}\right) \right] \\ &=& -\frac{2e^2}{(4\pi)^2} \left[ \int_0^1 d(x^2-x) \log\Delta + \int_0^1 dx (1-x) \frac{(1-x^2)m^2-\Delta}{\Delta} \right] \\ &=& -\frac{2e^2}{(4\pi)^2} \left[ -\int_0^1 dx \frac{x^2-x}{\Delta} \left(2(x-1)m^2+\mu^2\right) + \int_0^1 dx (1-x) \frac{2x(1-x)m^2-x\mu^2}{\Delta} \right] \\ &=& -\frac{2e^2}{(4\pi)^2} \int_0^1 dx (1-x) \frac{2x(x-1)m^2 + x\mu^2 + 2x(1-x)m^2-x\mu^2}{\Delta} \\ &=& 0. \end{eqnarray} \]

7.3

Part (a)

Let us go back to (6.38). We only need to replce the virtual photon by a virtual \(\phi\). As far as \(\delta Z_1\) is concerned, we could simply set \(q=0\) as we did in 7.2. But we will keep \(q\) this time so that we can verify directly the Ward identity. (From the proof of the Ward identity, it should be clear that the presence of \(\phi\) should change absolutely nothing. But it is still nice to verify it here.)

We have (with the same or similar substitutions as in the textbook)

\[ \begin{eqnarray} && \bar{u}(p') \delta\Gamma^\mu(p',p) u(p) \\ &=& \int \frac{d^dk}{(2\pi)^d} \frac{i}{(k-p)^2-m_\phi^2+i\epsilon} \bar{u}(p') (-i\frac{\lambda}{\sqrt{2}}) \frac{i(\not{k'}+m)}{{k'}^2-m^2+i\epsilon} \gamma^\mu \frac{i(\not{k}+m)}{{k}^2-m^2+i\epsilon} (-i\frac{\lambda}{\sqrt{2}}) u(p) \\ &=& \frac{i\lambda^2}{2} \int \frac{d^dk}{(2\pi)^d} \frac{\bar{u}(p') [\not{k'}\gamma^\mu\not{k} + m^2\gamma^\mu + m(\not{k'}\gamma^\mu+\gamma^\mu\not{k})]u(p)}{((k-p)^2-\mu^2+i\epsilon) ({k}^2-m^2+i\epsilon) ({k}^2-m^2+i\epsilon)} \\ &=& i\lambda^2 \int \frac{d^dl}{(2\pi)^d} \int_0^1 dxdydz \delta(x+y+z-1) \frac{\bar{u}(p') N u(p)}{D^3}, \end{eqnarray} \]

where \( N \) is

\[ \begin{eqnarray} &=& \not{l}\gamma^\mu\not{l} + ((1-y)\not{q}+z\not{p})\gamma^\mu(-y\not{q}+z\not{p}) + m^2\gamma^\mu + m(\not{q}\gamma^\mu+2(l^\mu-yq^\mu+zp^\mu)) \\ &=& \frac{2-d}{d}l^2\gamma^\mu + \left((1-y)m + (y+z-1)\not{p}\right)\gamma^\mu\left(-y\not{p'}+(y+z)m\right) + m^2\gamma^\mu + m(m\gamma^\mu-\not{p}\gamma^\mu+2(l^\mu-yq^\mu+zp^\mu)) \\ &=& \frac{2-d}{d}l^2\gamma^\mu - y(1-y)m\gamma^\mu\not{p'} - y(y+z-1)\not{p}\gamma^\mu\not{p'} + [(y+z-1)(y+z) - 1]m\not{p}\gamma^\mu + [(1-y)(y+z) + 1 + 1 ]m^2\gamma^\mu - 2ymq^\mu + 2zmp^\mu. \end{eqnarray} \]

Note that the equality signs above, as well as often the case in the following derivations, are valid only when everything is sandwiched between \(\bar{u}(p')\) and \(u(p)\) and inside an integral over \(l\).

Using the following equality

\[ \begin{eqnarray} && \not{p} \gamma^\mu \not{p'} \\ &=& (2p^\mu-\gamma^\mu\not{p}) \not{p'} \\ &=& 2m p^\mu - \gamma^\mu (2p\cdot p' - \not{p'}\not{p}) \\ &=& 2m p^\mu - \gamma^\mu (2p\cdot p') + 2m {p'}^\mu - m\not{p'}\gamma^\mu \\ &=& 2m (p^\mu + {p'}^\mu) + \gamma^\mu ((p-p')^2 - 2m^2) - m^2 \gamma^\mu \\ &=& 2m (p^\mu+{p'}^\mu) + \gamma^\mu (q^2 - 3m^2), \end{eqnarray} \]

as well as \(\gamma^\mu\not{p'}=2{p'}^\mu-m\gamma^\mu\) and \(\not{p}\gamma^\mu=2{p}^\mu-m\gamma^\mu\), we have

\[ \begin{eqnarray} && N \\ &=& \gamma^\mu\frac{2-d}{d}l^2 + m^2\gamma^\mu\left[y(1-y)+1+(1-y-z)(y+z)-3y(1-y-z)+(1-y)(y+z)+2\right] + q^2\gamma^\mu y(1-y-z) - 2mp^\mu\left[(1-y-z)(y+z)+1\right] - 2m{p'}^\mu y(1-y) + 2my(1-y-z)(p^\mu+{p'}^\mu) - 2myq^\mu \\ &=& \gamma^\mu\frac{2-d}{d}l^2 + m^2\gamma^\mu(3+2z-z^2) + q^2\gamma^\mu y(1-y-z) - m(p^\mu+{p'}^\mu)(1-z^2) - mq^\mu(2y+z^2+2yz-1). \end{eqnarray} \]

Now using the Gordon identity (6.32) and \(x+y+z=1\), we have

\[ \begin{eqnarray} && N \\ &=& \gamma^\mu\frac{2-d}{d}l^2 + m^2\gamma^\mu(3+2z-z^2) + q^2\gamma^\mu y(1-y-z) - m(2m\gamma^\mu-i\sigma^{\mu\nu}q_\nu)(1-z^2) + mq^\mu(x-y)(1+z) \\ &=& \gamma^\mu\frac{2-d}{d}l^2 + m^2\gamma^\mu(1+2z+z^2) + q^2\gamma^\mu y(1-y-z) +im\sigma^{\mu\nu}q_\nu(1-z^2) + mq^\mu(x-y)(1+z) \\ &=& \gamma^\mu\left[\frac{2-d}{d}l^2 + m^2(1+z)^2 + q^2y(1-y-z)\right] +im\sigma^{\mu\nu}q_\nu(1-z^2) + mq^\mu(x-y)(1+z). \end{eqnarray} \]

From the result above, the Ward identity is clearly satisfied. (See the discussion leading to (6.47).)

We can also read off the extra \(\delta Z_1\) due to the virtual \(\phi\).

\[ \begin{eqnarray} && \delta Z_1 \\ &=& -i\lambda^2 \int \frac{d^dl}{(2\pi)^d} \int_0^1 dxdydz \delta(x+y+z-1) \frac{\frac{2-d}{d}l^2 + m^2(1+z)^2}{(l^2-\Delta)^3} \\ &=& -i\lambda^2 \frac{i}{(4\pi)^{d/2}} \int_0^1 dxdydz \delta(x+y+z-1) \left[ \frac{2-d}{d} \frac{d}{2}\frac{\Gamma(2-\frac{d}{2})}{\Gamma(3)}(\frac{1}{\Delta})^{2-\frac{d}{2}} - m^2(1+z)^2\frac{\Gamma(3-\frac{d}{2})}{\Gamma(3)}(\frac{1}{\Delta})^{3-\frac{d}{2}} \right] \\ &=& \frac{\lambda^2}{2(4\pi)^2} \int_0^1 dz (1-z) \left[ (\frac{\epsilon}{2}-1) (\frac{2}{\epsilon}-\log\Delta-\gamma+\log(4\pi)) - \frac{m^2(1+z)^2}{\Delta} \right] \\ &=& -\frac{\lambda^2}{2(4\pi)^2} \int_0^1 dz (1-z) \left[ \frac{2}{\epsilon}-\log\Delta-\gamma+\log(4\pi)-1 + \frac{m^2(1+z)^2}{\Delta} \right]. \end{eqnarray} \]

For \(Z_2\). we start with (7.16), which now becomes

\[ \begin{eqnarray} && \Sigma_2(p) \\ &=& \frac{i\lambda^2}{2} \int \frac{d^dk}{(2\pi)^d} \frac{\not{k}+m}{(k^2-m^2+i\epsilon)((p-k)^2-m_\phi^2+i\epsilon)} \\ &=& \frac{i\lambda^2}{2} \int_0^1 dx \int \frac{d^dl}{(2\pi)^d} \frac{x\not{p}+m}{(l^2-\Delta+i\epsilon)^2} \\ &=& -\frac{\lambda^2}{2(4\pi)^{d/2}} \int_0^1 dx (x\not{p}+m) \frac{\Gamma(2-\frac{d}{2})}{\Gamma(2)} (\frac{1}{\Delta})^{2-\frac{d}{2}} \\ &=& -\frac{\lambda^2}{2(4\pi)^2} \int_0^1 dx \left(x\not{p}+m\right) \left(\frac{2}{\epsilon}-\log\Delta-\gamma+\log(4\pi)\right). \end{eqnarray} \]

From (7.26), we have

\[ \begin{eqnarray} && \delta Z_2 \\ &=& \frac{d\Sigma}{d\not{p}} \Bigg|_{\not{p}=m} \\ &=& -\frac{\lambda^2}{2(4\pi)^2} \int_0^1 dx \left[x \left(\frac{2}{\epsilon}-\log\Delta-\gamma+\log(4\pi)\right) + \left(xm+m\right) (-\frac{1}{\Delta}) (-x)(1-x)2m \right] \\ &=& -\frac{\lambda^2}{2(4\pi)^2} \int_0^1 dx x \left[\frac{2}{\epsilon}-\log\Delta-\gamma+\log(4\pi) + \frac{2(1+x)(1-x)m^2}{\Delta} \right]. \end{eqnarray} \]

Again, notice that we have in both cases the same \(\Delta\) when we replace \(z\) by \(x\) in \(Z_1\). Also notice that \(\int_0^1 x dx = \int_0^1 (1-x) dx\). So we have (with \(\Delta=(1-x)^2m^2+xm_\phi^2\))

\[ \begin{eqnarray} && \delta Z_1 - \delta Z_2 \\ &=& -\frac{\lambda^2}{2(4\pi)^2} \left[ \int_0^1 dx \left(x-(1-x)\right) \log\Delta + \int_0^1 dx (1-x) \left(-1 + \frac{(1+x)^2m^2}{\Delta} - \frac{2x(1+x)m^2}{\Delta}\right) \right] \\ &=& -\frac{\lambda^2}{2(4\pi)^2} \left[ \int_0^1 d(x^2-x) \log\Delta + \int_0^1 dx (1-x) \frac{(1-x^2)m^2-\Delta}{\Delta} \right] \\ &=& -\frac{\lambda^2}{2(4\pi)^2} \left[ -\int_0^1 dx \frac{x^2-x}{\Delta} \left(2(x-1)m^2+m_\phi^2\right) + \int_0^1 dx (1-x) \frac{2x(1-x)m^2-xm_\phi^2}{\Delta} \right] \\ &=& -\frac{\lambda^2}{2(4\pi)^2} \int_0^1 dx (1-x) \frac{2x(x-1)m^2 + xm_\phi^2 + 2x(1-x)m^2-xm_\phi^2}{\Delta} \\ &=& 0. \end{eqnarray} \]

Part (b)

In Part (a), we have shown that \(\delta Z_1 = \delta Z_2\) for fermion-photon interactions. What this equality means physically is that in the limit that \(q\rightarrow0\), the observed coupling constant for the fermion-photon-fermion vertex is equal to \(e=\sqrt{Z_3}e_0\), where \(e_0\) is the "bare" charge in the Langragian.

Now we want to verify whether the same is true for the fermion-\(\phi\)-fermion vertex.

To calculate its corresponding \(Z_1\), we again starts with (6.38). We need to calculate the \(\delta Z_1\) contribution from photon and that from \(\phi\). From now on, \(q\) denotes the momentum of the \(phi\) particle. As before, we can simply set it to \(0\) since we are only interested in this limit.

\[ \begin{eqnarray} && \bar{u}(p) \delta\Lambda_\gamma(p,p) u(p) \\ &=& \int \frac{d^dk}{(2\pi)^d} \frac{-ig_{\nu\rho}}{(k-p)^2-\mu^2+i\epsilon} \bar{u}(p) (-ie\gamma^\nu) \frac{i(\not{k}+m)}{{k}^2-m^2+i\epsilon} \frac{i(\not{k}+m)}{{k}^2-m^2+i\epsilon} (-ie\gamma^\rho) u(p) \\ &=& -ie^2 \int \frac{d^dk}{(2\pi)^d} \frac{\bar{u}(p) [dk^2 + dm^2 - 2(d-2)m\not{k} ]u(p)}{((k-p)^2-\mu^2+i\epsilon) ({k}^2-m^2+i\epsilon) ({k}^2-m^2+i\epsilon)} \\ &=& -ie^2 \int \frac{d^dl}{(2\pi)^d} \int_0^1 dxdydz \delta(x+y+z-1) \frac{2 \bar{u}(p) [dl^2 + d(zp)^2 + dm^2 -2(d-2)mz\not{p} ]u(p)}{D^3} \\ &=& -ie^2 \int \frac{d^dl}{(2\pi)^d} \int_0^1 dxdydz \delta(x+y+z-1) \frac{2 \bar{u}(p) [dl^2 + (d(z^2-2z+1)+4z)m^2]u(p)}{D^3}. \end{eqnarray} \]

Therefore we have

\[ \begin{eqnarray} && \delta Z_{1,\gamma} \\ &=& -\delta\Lambda_\gamma(q=0) \\ &=& 2ie^2 \int \frac{d^dl}{(2\pi)^d} \int_0^1 dxdydz \delta(x+y+z-1) \frac{dl^2 + (d(z^2-2z+1)+4z)m^2}{D^3} \\ &=& 2ie^2 \frac{i}{(4\pi)^{d/2}} \int_0^1 dz(1-z) \left[ \frac{d^2}{2}\frac{\Gamma(2-\frac{d}{2})}{\Gamma(3)}(\frac{1}{\Delta})^{2-\frac{d}{2}} - (d(z^2-2z+1)+4z)m^2 \frac{\Gamma(3-\frac{d}{2})}{\Gamma(3)}(\frac{1}{\Delta})^{3-\frac{d}{2}} \right] \\ &=& 2ie^2 \frac{i}{(4\pi)^2} \int_0^1 dz(1-z) \left[ (4-2\epsilon)(\frac{2}{\epsilon}-\log\Delta-\gamma+\log(4\pi)) - \frac{2(z^2-z+1)m^2 }{\Delta} \right] \\ &\approx& -\frac{8e^2}{(4\pi)^2\epsilon} \end{eqnarray} \]

Similarly, we have

\[ \begin{eqnarray} && \bar{u}(p) \delta\Lambda_\phi(p,p) u(p) \\ &=& \int \frac{d^dk}{(2\pi)^d} \frac{i}{(k-p)^2-m_\phi^2+i\epsilon} \bar{u}(p) \frac{-i\lambda}{\sqrt{2}} \frac{i(\not{k}+m)}{{k}^2-m^2+i\epsilon} \frac{i(\not{k}+m)}{{k}^2-m^2+i\epsilon} \frac{-i\lambda}{\sqrt{2}} u(p) \\ &=& \frac{i\lambda^2}{2} \int \frac{d^dk}{(2\pi)^d} \frac{\bar{u}(p) [k^2 + m^2 + 2m\not{k} ]u(p)}{((k-p)^2-m_\phi^2+i\epsilon) ({k}^2-m^2+i\epsilon) ({k}^2-m^2+i\epsilon)} \\ &=& \frac{i\lambda^2}{2} \int \frac{d^dl}{(2\pi)^d} \int_0^1 dxdydz \delta(x+y+z-1) \frac{2 \bar{u}(p) [l^2 + (zp)^2 + m^2 + 2mz\not{p} ]u(p)}{D^3} \\ &=& \frac{i\lambda^2}{2} \int \frac{d^dl}{(2\pi)^d} \int_0^1 dxdydz \delta(x+y+z-1) \frac{2 \bar{u}(p) [l^2 + (z^2+2z+1) m^2 ]u(p)}{D^3}. \end{eqnarray} \]

Therefore we have

\[ \begin{eqnarray} && \delta Z_{1,\phi} \\ &=& -\delta\Lambda_\phi(q=0) \\ &=& -i\lambda^2 \int \frac{d^dl}{(2\pi)^d} \int_0^1 dxdydz \delta(x+y+z-1) \frac{l^2 + (z^2+2z+1)m^2}{D^3} \\ &=& -i\lambda^2 \frac{i}{(4\pi)^{d/2}} \int_0^1 dz(1-z) \left[ \frac{d}{2}\frac{\Gamma(2-\frac{d}{2})}{\Gamma(3)}(\frac{1}{\Delta})^{2-\frac{d}{2}} - (z^2+2z+1)m^2 \frac{\Gamma(3-\frac{d}{2})}{\Gamma(3)}(\frac{1}{\Delta})^{3-\frac{d}{2}} \right] \\ &=& -i\lambda^2 \frac{i}{(4\pi)^2} \int_0^1 dz(1-z) \left[ (1-\frac{\epsilon}{4})(\frac{2}{\epsilon}-\log\Delta-\gamma+\log(4\pi)) - \frac{(z^2+2z+1)m^2 }{2\Delta} \right] \\ &\approx& \frac{\lambda^2}{(4\pi)^2\epsilon} \end{eqnarray} \]

Note that \(\delta Z_2\) is about the so-called electron self-energy, which we already have from Problem 7.2(b) and from Part (a). If we retain only the parts proportional to \(\frac{1}{\epsilon}\), we have

\[ \begin{eqnarray} \delta Z_{2,\gamma} &\approx& -\frac{2e^2}{(4\pi)^2\epsilon} \\ \delta Z_{2,\phi} &\approx& -\frac{\lambda^2}{2(4\pi)^2\epsilon}. \end{eqnarray} \]

Clearly \(\delta Z_1 - \delta Z_2 \approx -\frac{6e^2}{(4\pi)^2\epsilon} + \frac{3\lambda^2}{2(4\pi)^2\epsilon} \) does not vanish. So the rescaling of this vertex at \(q^2 = 0\) is not canceled by the correction to \(Z_2\).

Final Project

Before jumping into calculations, we should notice the similarity between parts (a) and (d) and the electron vertex function (Section 6.3), between parts (c) and (d) and soft Bremsstralung (Section 6.1), and finally between part (e) and how the infrared divergences from the two phenomena cancel each other (Section 6.5).

Comparing \(\Delta H\) to (4.129), we can easily infer that the Feynman rule for \(\bar{\psi}_{fi}\gamma^\mu\psi_{fi}\) (without summation) should be \(-ig\gamma^\mu\), while the gluon propagator is formally the same as that of photons. Also refer to my comment on quarkonium, which explains the color factor of 3, and equally applies to the gluon-quark-antiquark interaction.

Part (a)

Treating all particles as massless except the gluon, and remembering that on internal fermion lines (propagators), the momentum must be assigned in the direction of particle-number flow, we have

\[ \begin{eqnarray} && i\delta \mathcal{M} \\ &=& \int \frac{d^4k}{(2\pi)^4} \frac{-ig_{\rho\sigma}}{(k-k_1)^2-\mu^2+i\epsilon} \bar{u}(k_1) (-ig\gamma^\sigma) \frac{i\not{k}}{{k}^2+i\epsilon} (i e Q_f \gamma^\nu) \frac{i(\not{k}-\not{q})}{{(k-q)}^2+i\epsilon} (-ig\gamma^\rho) v(k_2) \frac{-ig_{\mu\nu}}{q^2+i\epsilon} \bar{v}(p_2) (-ie\gamma^\mu) u(p_1) \\ &=& \bar{u}(k_1) (-ie) \left[ \int \frac{d^4k}{(2\pi)^4} \frac{ 2i g^2 (-Q_f) (\not{k}-\not{q}) \gamma^\nu \not{k} }{ ((k-k_1)^2-\mu^2+i\epsilon) ({k}^2+i\epsilon) ((k-q)^2+i\epsilon) } \right] v(k_2) \frac{-ig_{\mu\nu}}{q^2+i\epsilon} \bar{v}(p_2) (-ie\gamma^\mu) u(p_1). \end{eqnarray} \]

We can focus on the integral between the brackets, which is equivalent to \(\gamma^\nu\) in the tree diagram. Denoting it as \(\delta F_1(q^2) \gamma^\nu\), we have

\[ \delta F_1(q^2) \gamma^\nu = -2i g^2 Q_f \int \frac{d^4k}{(2\pi)^4} \frac{ (\not{k}-\not{q}) \gamma^\nu \not{k} }{ ((k-k_1)^2-\mu^2+i\epsilon) ({k}^2+i\epsilon) ((k-q)^2+i\epsilon) } \]

Instead of evaluating it from scratch, we could take advantage of its similarity to (6.38) and crossing symmetry (5.67). The correspondence between (6.38) and ours is the following. (There is actually an extra complication with the sign. But it affects the correction and the tree diagram in the same way so we may ignore it.)

• \( e^2 \leftrightarrow -g^2 Q_f \)
• \( p \leftrightarrow -k_2 \)
• \( u(p) \leftrightarrow v(-k_2) \)
• \( p' \leftrightarrow k_1 \)
• \( k \leftrightarrow k - q \)

Another difference is that we have set the fermion mass to zero and the gluon mass to \(\mu\). The net effect is that

• (6.44) becomes \(\Delta = -xyq^2 + z\mu^2\);
• (6.54) remains essentially the same, with all necessary modifications listed above.

(6.56) then becomes

\[ \delta F_1(q^2) = -\frac{\alpha_g Q_f}{2\pi} \int_0^1 dx dy dz \delta(x+y+z-1) \left[\log\frac{\mu^2 z}{\mu^2 z - q^2 xy} + \frac{q^2(1-x)(1-y)}{\mu^2 z-q^2 xy} \right] + O(\alpha_g^2). \]

There are a few points worth mentioning.

• Within the logarithm, it makes more sense to retain the quark mass, instead of that of the virtual gluon. But it will not affect the result when we take the limit where both masses tend to zero.
• Close to the end of the calculation, \(q^2\) should be replaced by \(q^2+i\epsilon\) when we carry out analytical continuation of the logarithm. The sign of \(i\epsilon\) can be easily read from the equations below (6.47), and this will determine the sign of the imaginary part of \(\delta F_1(q^2)\), although it does not really matter when we calculate the cross section. (In fact, even the magnitude of the imaginary part does not matter if we restrict to the first order of \(\alpha_g\).)
• Strictly speaking, \(F_1(q^2=0) = -Q_f\). But the sign obviously does not matter when we are calculating the cross section.

Part (b)

For (i) and (ii), Xianyu's solution is quite clear.

For (iii), we have the following. (Remember that quarks and antiquarks are treated as massless and therefore \(E_1=|\mathbf{k}_1|\) and \(E_2=|\mathbf{k}_2|\).)

\[ \begin{eqnarray} && \int d\Pi_3 \\ &=& \int \frac{d^3\mathbf{k}_1}{(2\pi)^3} \frac{d^3\mathbf{k}_2}{(2\pi)^3} \frac{d^3\mathbf{k}_3}{(2\pi)^3} \frac{1}{(2E_1)(2E_2)(2E_3)} (2\pi)^4 \delta^{(4)} (q-k_1-k_2-k_3) \\ &=& \int \frac{d^3\mathbf{k}_1}{(2\pi)^3} \frac{d^3\mathbf{k}_2}{(2\pi)^3} \frac{1}{(2E_1)(2E_2)(2E_3)} (2\pi) \delta (E_q-E_1-E_2-E_3) \Big|_{\mathbf{k}_3=-\mathbf{k}_1-\mathbf{k}_2;E_3=\sqrt{{k_3}^2+\mu^2}} \\ &=& \int \frac{d E_1 E_1^2 d\cos\theta_1 d\phi_1}{(2\pi)^3} \frac{d E_2 E_2^2 d\cos\theta_{12} d\phi_2}{(2\pi)^3} \frac{1}{(2E_1)(2E_2)(2E_3)} (2\pi) \delta \left(E_q-E_1-E_2-\sqrt{E_1^2+E_2^3+2E_1E_2\cos\theta_{12}+\mu^2}\right) \Big|_{\mathbf{k}_3=-\mathbf{k}_1-\mathbf{k}_2;E_3=\sqrt{{k_3}^2+\mu^2}} \\ &=& \frac{1}{{32\pi^3}} \int d E_1 d E_2 d\cos\theta_{12} \frac{E_1 E_2}{E_3} \frac{E_3}{E_1 E_2} \delta \left(\cos\theta_{12} - \frac{E_q^2-2E_q(E_1+E_2)+2E_1E_2-\mu^2}{2E_1E_2} \right) \Big|_{\mathbf{k}_3=-\mathbf{k}_1-\mathbf{k}_2;E_3=\sqrt{{k_3}^2+\mu^2}} \\ &=& \frac{1}{{32\pi^3}} \int d E_1 d E_2 \Big|_{\cos\theta_{12} = \frac{E_q^2-2E_q(E_1+E_2)+2E_1E_2-\mu^2}{2E_1E_2}} \\ &=& \frac{q^2}{{128\pi^3}} \int d x_1 d x_2 \Big|_{\cos\theta_{12} = \frac{1-(x_1+x_2)+x_1x_2/2-\mu^2/q^2}{x_1x_2/2}}. \end{eqnarray} \]

Since \(-1 \le \cos\theta_{12} \le 1\), we have

\[ \begin{eqnarray} && -x_1x_2/2 \le 1-(x_1+x_2)+x_1x_2/2-\mu^2/q^2 \le x_1x_2/2 \\ &\iff& -x_1x_2 \le 1-(x_1+x_2)-\mu^2/q^2 \le 0 \\ &\iff& 1 - x_1 - \mu^2/q^2 \le x_2 \le 1 - \frac{\mu^2/q^2}{1 - x_1}, \end{eqnarray} \]

and therefore

\[ \int d\Pi_3 = \frac{q^2}{{128\pi^3}} \int_0^{1-\mu^2/q^2} d x_1 \int_{1 - x_1 - \mu^2/q^2}^{1 - \frac{\mu^2/q^2}{1 - x_1}} d x_2. \]

Part (c)

For pedagogical purposes, I will keep the quark mass and the electron mass for now. We have

\[ \begin{eqnarray} && i\mathcal{M} \\ &=& \epsilon_\rho^*(k_3) \bar{u}(k_1) \left[ (-ig\gamma^\rho) \frac{i(\not{k}_1+\not{k}_3+m_q)}{(k_1+k_3)^2-m_q^2+i\epsilon} (i e Q_f \gamma^\nu) + (i e Q_f \gamma^\nu) \frac{i(-\not{k}_2-\not{k}_3+m_q)}{(-k_2-k_3)^2-m_q^2+i\epsilon} (-ig\gamma^\rho) \right] v(k_2) \frac{-ig_{\mu\nu}}{q^2+i\epsilon} \bar{v}(p_2) (-ie\gamma^\mu) u(p_1) \\ &=& -i g e^2 Q_f \epsilon_\rho^*(k_3) \bar{u}(k_1) \left[ \gamma^\rho \frac{\not{k}_1+\not{k}_3+m_q}{(k_1+k_3)^2-m_q^2+i\epsilon} \gamma^\mu + \gamma^\mu \frac{-\not{k}_2-\not{k}_3+m_q}{(-k_2-k_3)^2-m_q^2+i\epsilon} \gamma^\rho \right] v(k_2) \frac{1}{q^2+i\epsilon} \bar{v}(p_2) \gamma^\mu u(p_1). \end{eqnarray} \]

The part between \( \bar{u}(k_1)\) and \(v(k_2)\), multiplied with its complex conjugate, will form the main ingredient of what is callled \(H_{\mu\nu}\) in the textbook. An argument based on general principles for

\[q^\mu H_{\mu\nu} = 0\]

is essentially the same as the Ward identity. To verify it explicitly, we only need to focus on the part between \( \bar{u}(k_1)\) and \(v(k_2)\) in \(i\mathcal{M}\). Also remember that for fermions we have \( \bar{u}(k) \not{k} = \bar{u}(k) m \) and \( \not{k} v(k) = -m v(k) \).

\[ \begin{eqnarray} && \bar{u}(k_1) \left[ \gamma^\rho \frac{\not{k}_1+\not{k}_3+m_q}{(k_1+k_3)^2-m_q^2+i\epsilon} \not{q} + \not{q} \frac{-\not{k}_2-\not{k}_3+m_q}{(-k_2-k_3)^2-m_q^2+i\epsilon} \gamma^\rho \right] v(k_2) \\ &=& \bar{u}(k_1) \left[ \gamma^\rho \frac{\not{k}_1+\not{k}_3+m_q}{(k_1+k_3)^2-m_q^2+i\epsilon} (\not{k}_1+\not{k}_2+\not{k}_3) + (\not{k}_1+\not{k}_2+\not{k}_3) \frac{-\not{k}_2-\not{k}_3+m_q}{(-k_2-k_3)^2-m_q^2+i\epsilon} \gamma^\rho \right] v(k_2) \\ &=& \bar{u}(k_1) \left[ \gamma^\rho \frac{\not{k}_1+\not{k}_3+m_q}{(k_1+k_3)^2-m_q^2+i\epsilon} (\not{k}_1+\not{k}_3-m_q) + (\not{k}_2+\not{k}_3+m_q) \frac{-\not{k}_2-\not{k}_3+m_q}{(-k_2-k_3)^2-m_q^2+i\epsilon} \gamma^\rho \right] v(k_2) \\ &=& 0. \end{eqnarray} \]

Multiplying by \(g_{\mu\nu}\) on both sides of

\[\int d\Pi_3 H^{\mu\nu} = (g^{\mu\nu} - \frac{q^\mu q^\nu}{q^2}) \cdot H\]

we have

\[\int d\Pi_3 g^{\mu\nu} H_{\mu\nu} = 3 \cdot H.\]

Now we have, keeping in mind that \(L_{\mu\nu} q^\mu = 0\) due to the Ward identity,

\[L_{\mu\nu} \int d\Pi_3 H^{\mu\nu} = g^{\mu\nu}L_{\mu\nu} H = \frac{1}{3} (g^{\mu\nu}L_{\mu\nu}) \int d\Pi_3 g^{\rho\sigma} H_{\rho\sigma}.\]

For the rest, one can refer to Xianyu's solution. The calculation of (7.55) is rather tedious but no special techinques are required except the numerator algebra listed in A.3 of the textbook. For the factor of 3 due to color charges, refer to my previous comments.

Part (d)

In Xianyu's solution, note that we are already throwing away \(O\left(\mu^4/q^4\right)\) terms in (7.59).

Denoting \(\frac{\mu^2}{q^2}\) simply as \(\epsilon\), we need to calculate the following integral.

\[\int_0^{1-\epsilon} \int_{1-x_1-\epsilon}^{1-\frac{\epsilon}{1-x_1}} F\left(x_1, x_2,\epsilon\right) \mathrm{d}x_2 \mathrm{d}x_1,\]

where

\[F(x_1, x_2,\epsilon) = \frac{2(x_1+x_2-1+\epsilon)(1+\epsilon)}{(1-x_1)(1-x_2)} + \left(\frac{1}{(1-x_1)^2} + \frac{1}{(1-x_2)^2}\right)\left((1-x_1)(1-x_2)-\epsilon\right).\]

First, noticing that \(x_1\) and \(x_2\) are completely symmetric in both \(F(x_1, x_2,\epsilon)\) and the integration area, we can replace \(F(x_1, x_2,\epsilon)\) by the following.

\[ F(x_1, x_2,\epsilon) \rightarrow \frac{-4(1+\epsilon)}{1-x_1} + \frac{2(1+\epsilon)^2}{(1-x_1)(1-x_2)} + 2 \frac{1-x_2}{1-x_1} - 2 \frac{\epsilon}{(1-x_1)^2} \]

Only the integral of the second term is tricky. We have

\[ \begin{eqnarray} && \int_0^{1-\epsilon} \int_{1-x_1\epsilon}^{1-\frac{\epsilon}{1-x_1}} \frac{2(1+\epsilon)}{(1-x_1)(1-x_2)}\mathrm{d}x_2 \mathrm{d}x_1 \\ &=& \left(2+O(\epsilon)\right) \int_0^{1-\epsilon} \frac{ \log\left(x_1+\epsilon\right)-\log\left(\frac{\epsilon}{1-x_1}\right)}{1-x_1} \mathrm{d}x_1 \\ &=& \left(2+O(\epsilon)\right) \left(\int_0^1 \frac{\log(x_1)}{1-x_1} \mathrm{d}x_1 + o(1)|_{\epsilon\rightarrow 0} \right) - \left(2+O(\mu^2)\right) \int_0^{1-\epsilon} \frac{\log(\epsilon)-\log(1-x_1)}{1-x_1} \mathrm{d}x_1 \\ &=& -\frac{\pi^2}{3} - \left(2+O(\epsilon)\right) \left(-\log^2(\epsilon) + \frac{1}{2} \log^2(\epsilon) \right) \\ &=& -\frac{\pi^2}{3} + \log^2(\epsilon). \end{eqnarray} \]

(We can safely throw away not only \(O(\epsilon)\) terms, but also \(O(\epsilon)\log(\epsilon)\) and \(O(\epsilon)\log^2(\epsilon)\) terms, because they all tend to zero when \(\epsilon\) does.)

For the other integrals, it is easier to use a computer. Below is a piece of Python code that calculates the integral of the last term of \(F(x_1,x_2,\epsilon)\).

from sympy import *
init_printing()
x_1, x_2, epsilon = symbols('x_1 x_2 epsilon', positive=True)

x_2_upper = 1 - epsilon/(1-x_1)
x_2_lower = 1 - x_1 - epsilon

integrand = epsilon/((1-x_1)**2)
inner = integrate(integrand, (x_2,x_2_lower,x_2_upper))
outer = integrate(inner,(x_1,0,1-epsilon))
simplify(outer)

Therefore the end result is

\[ \begin{eqnarray} && \int_0^{1-\epsilon} \int_{1-x_1\epsilon}^{1-\frac{\epsilon}{1-x_1}} F(x_1,x_2,\epsilon) \\ &=& -4\left(-\log(\epsilon)-2\right) + \left(-\frac{\pi^2}{3}+\log^2(\epsilon)\right) + 2\left(-\frac{1}{2}\log(\epsilon)-1\right) - 2\left(\frac{1}{2}\right) \\ &=& \log^2(\frac{\mu^2}{q^2}) + 3 \log(\frac{\mu^2}{q^2}) + 5 -\frac{\pi^2}{3}. \end{eqnarray} \]

And we have

\[ \sigma(e^+e^-\rightarrow \bar{q}qg) = \frac{4\pi\alpha^2}{3s} \cdot 3Q_f^2 \cdot \frac{\alpha}{2\pi} \left[ \log^2\left(\frac{\mu^2}{q^2}\right) + 3 \log\left(\frac{\mu^2}{q^2}\right) + 5 -\frac{\pi^2}{3} \right]. \]

Parts (e) and (f)

Focusing on the integral from part (a), and substituting \(Q^2=-q^2\), we have

\[ \begin{eqnarray} && \int dxdydz \delta(x+y+z-1) \left[ \log\frac{\mu^2 z}{\mu^2 z + Q^2 x y} - \frac{Q^2 (1-x)(1-y)}{\mu^2 z + Q^2 x y} \right] \\ &=& \int_0^1 dz \int_0^{1-z} \left[ \log\frac{\mu^2 z}{\mu^2 z + Q^2 (1-y-z) y} - \frac{Q^2 (y+z)(1-y)}{\mu^2 z + Q^2 (1-y-z) y} \right] dy. \end{eqnarray} \]

But we have

\[ \begin{eqnarray} && \int_0^{1-z} \log\frac{\mu^2 z}{\mu^2 z + Q^2 (1-y-z) y} dy \\ &=& \int_0^{1-z} \Big[ \log\left(\mu^2 z\right) - \log\left(\mu^2 z + Q^2 (1-y-z) y\right) \Big] dy \\ &=& (1-z)\log\left(\mu^2 z\right) - (1-z)\log\left(\mu^2 z\right) + \int_0^{1-z} \frac{Q^2 y(1-2y-z)}{\mu^2 z + Q^2 (1-y-z) y} dy. \end{eqnarray} \]

Therefore we have

\[ \begin{eqnarray} && \int dxdydz \delta(x+y+z-1) \left[ \log\frac{\mu^2 z}{\mu^2 z + Q^2 x y} - \frac{Q^2 (1-x)(1-y)}{\mu^2 z + Q^2 x y} \right] \\ &=& -\int_0^1 dz \int_0^{1-z} \frac{y^2+z}{\frac{\mu^2}{Q^2} z + (1-y-z) y} dy \\ &=& -\int_0^1 \left[\frac{1-y}{\frac{\mu^2}{Q^2} - y} + \left(y^2-\frac{y(1-y)}{\frac{\mu^2}{Q^2}-y}\right)\int_0^{1-y} \frac{1}{\frac{\mu^2}{Q^2} z + (1-y-z) y} dz \right] dy \\ &=& -\int_0^1 \left[\frac{1-y}{\frac{\mu^2}{Q^2} - y} + \left( \frac{y^2}{\frac{\mu^2}{Q^2}-y} - \frac{y(1-y)}{(\frac{\mu^2}{Q^2}-y)^2} \right) \log\left(\frac{\mu^2}{Q^2} z + (1-y-z) y\right)\Big|_{z=0}^{1-y} \right] dy \\ &=& -\int_0^1 \left[\frac{1-y}{\frac{\mu^2}{Q^2} - y} + \left( \frac{y^2}{\frac{\mu^2}{Q^2}-y} - \frac{y(1-y)}{(\frac{\mu^2}{Q^2}-y)^2} \right) \left(\log\left(\frac{\mu^2}{Q^2} (1-y)\right) - \log\left((1-y) y\right)\right) \right] dy \\ &=& -\int_0^1 \left[1 - \frac{1}{y+\frac{\mu^2}{q^2}} + O(\mu^2) + \left( -y + 1 - \frac{1+O(\mu^2)}{y+\frac{\mu^2}{q^2}} + \frac{\frac{\mu^2}{q^2}+O(\mu^4)}{(y+\frac{\mu^2}{q^2})^2} + O(\mu^2) \right) \left(\log\left(\frac{\mu^2}{Q^2}\right) - \log(y)\right) \right] dy \\ &=& -1 - \log\left(\frac{\mu^2}{q^2}\right) + \left(-\frac{1}{2} - \log\left(\frac{\mu^2}{q^2}\right) - 1 \right) \log\left(\frac{\mu^2}{Q^2}\right) - \int_0^1 \left[\left( y - 1 + \frac{1+O(\mu^2)}{y+\frac{\mu^2}{q^2}} - \frac{\frac{\mu^2}{q^2}+O(\mu^4)}{(y+\frac{\mu^2}{q^2})^2} + O(\mu^2) \right) \log(y) \right] dy \\ &=& -1 - \log\left(\frac{\mu^2}{q^2}\right) + \left(-\frac{3}{2} - \log\left(\frac{\mu^2}{q^2}\right) \right) \log\left(\frac{\mu^2}{-q^2-i\epsilon}\right) - \left(-\frac{1}{4} + 1 - \log\left(\frac{\mu^2}{q^2}\right) \right) - \left(1+O(\mu^2)\right) \int_0^1 \frac{\log(y)}{y+\frac{\mu^2}{q^2}} dy \\ &=& -\frac{7}{4} - \left(\frac{3}{2} + \log\left(\frac{\mu^2}{q^2}\right) \right) \left( \log\left(\frac{\mu^2}{q^2}\right) +i\pi \right) - \left(1+O(\mu^2)\right) \left(-\frac{1}{2}\log^2\left(\frac{\mu^2}{q^2}\right)-\frac{\pi^2}{6}\right) \\ &=& -\frac{1}{2}\log^2\left(\frac{\mu^2}{q^2}\right) - \frac{3}{2}\log\left(\frac{\mu^2}{q^2}\right) - \frac{7}{4} + \frac{\pi^2}{6} - i\pi \left(\log\left(\frac{\mu^2}{q^2}\right)+\frac{3}{2}\right). \end{eqnarray} \]

Most integrals in the intermediate steps of the derivation above can be better done by a computer, except the following.

\[ \begin{eqnarray} && \int_0^1 \frac{\log(x)}{x+\epsilon} \mathrm{d}x \\ &=& \int_0^1 \frac{\log(x+\epsilon-\epsilon)}{x+\epsilon} \mathrm{d}x \\ &=& \int_0^1 \frac{1}{x+\epsilon}\left(\log(x+\epsilon) - \sum_{n=1}^\infty \frac{1}{n} \left(\frac{\epsilon}{x+\epsilon}\right)^n \right) \mathrm{d}x \\ &=& \frac{1}{2}\log^2(x+\epsilon) \Bigg|_0^1 - \int_0^1 \left( \sum_{n=1}^\infty \frac{1}{n}\frac{\epsilon^n}{(x+\epsilon)^{n+1}}\right) \mathrm{d}x \\ &=& -\frac{1}{2}\log^2(\epsilon) + O(\epsilon) + \sum_{n=1}^\infty \left( \frac{1}{n^2}\frac{\epsilon^n}{ (x+\epsilon)^n} \Bigg|_0^1 \right) \\ &=& -\frac{1}{2}\log^2(\epsilon) + O(\epsilon) + \sum_{n=1}^\infty \left( -\frac{1}{n^2} + O(\epsilon^n) \right) \\ &=& -\frac{1}{2}\log^2(\epsilon) - \frac{\pi^2}{6} + O(\epsilon) \end{eqnarray} \]

Omitting all \(O(\alpha_g^2)\) terms, we have

\[ \begin{eqnarray} && F_1(q^2) \\ &=& -Q_f \left[1 + \frac{\alpha_g}{2\pi} \int_0^1 dx dy dz \delta(x+y+z-1) \left(\log\frac{\mu^2 z}{\mu^2 z - q^2 xy} + \frac{q^2(1-x)(1-y)}{\mu^2 z-q^2 xy} \right) \right] \\ &=& -Q_f \left[1 - \frac{\alpha_g}{4\pi} \left( \log^2\left(\frac{\mu^2}{q^2}\right) + 3\log\left(\frac{\mu^2}{q^2}\right) + \frac{7}{2} - \frac{\pi^2}{3} + i\pi \left(2\log\left(\frac{\mu^2}{q^2}\right)+3\right) \right) \right]. \end{eqnarray} \]

It is mathematically problematic to treat the coefficient of \(\alpha_g\) as if it were a small value despite of the infrared divergence, but let us pretend it is okay to do so. Then we have

\[ \begin{eqnarray} && \sigma(e^+e^-\rightarrow\bar{q}q) \\ &=& \frac{4\pi\alpha^2}{3s} \cdot 3 F_1(q^2) F_1^*(q^2) \\ &=& \frac{4\pi\alpha^2}{3s} \cdot 3 Q_f^2 \left[1 - \frac{\alpha_g}{2\pi} \left( \log^2\left(\frac{\mu^2}{q^2}\right) + 3\log\left(\frac{\mu^2}{q^2}\right) + \frac{7}{2} - \frac{\pi^2}{3} \right) \right]. \end{eqnarray} \]

Combining this with the result from part (d), we have

\[ \sigma(e^+e^-\rightarrow\bar{q}q \:\mathrm{or}\: \bar{q}qg) = \frac{4\pi\alpha^2}{3s} \cdot 3 Q_f^2 \cdot \left(1 + \frac{3\alpha_g}{4\pi} \right). \]