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I have two points not clear in Peskin & Schroeder's QFT on page 166.

  1. On Figure 5.6, "Since helicity is conserved, a unit of spin angular momentum is converted to orbital angular momentum", I am really puzzled on this sentence. In my understanding, in this Center-of-Mass frame, the initial total spin is zero, also the finial total spin is zero. For the orbital angular momentum, I see that the "back-scattering" probability is large, so their need to have some orbital angular momentum in principle. But I am really puzzled on how this one unit of spin angular momentum converted to orbital angular momentum. By the wave, the book mentioned the final states is a "p-wave", also what is this "p-wave"?

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  1. I also troubled with the sign in eq.(5.101), I thought their need to have a "minus" sign, the reason as follows: for the equation above (5.101), I thought it should be $$ \bar{\sigma}\cdot (p-k^{\prime})\simeq -\bar{\sigma}^{1}(p-k^{\prime})^{1}=\sigma^{1}\cdot (-\omega\chi) $$ so it has a "minus" sign difference with book, and we have used the approximation $$ p-k^{\prime}\simeq (0,-\omega \chi,0,0) $$

If you have any comments on above questions, I am really appreciate it.

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  • $\begingroup$ I answered your first question. For your second question in the second inequality the minus sign is already coming from the fact that the spatial 1 component of the 4-vector $\bar{\sigma}$ is negative so that is why P and S already don't have a bar in the second line (unlike in your question). $\endgroup$
    – octonion
    Commented Jul 17, 2022 at 5:26
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    $\begingroup$ @octonion Thanks for your detailed explanation about the first question. While for the second question, we know the definition of $\bar{\sigma}=\{1,-\sigma^{1},-\sigma^{2},-\sigma^{3}\}$, so their have a minus sign, and another minus sign is from $g^{\mu \nu}$, so I thought it should be $\sigma^{1}\cdot (p-k^{\prime})^{1}$. Is their still something that I missing? $\endgroup$
    – Daren
    Commented Jul 17, 2022 at 9:03
  • $\begingroup$ Good point, perhaps they really do have a sign error in their expression. They end up squaring the amplitude to compare with the cross section so they might not have noticed. $\endgroup$
    – octonion
    Commented Jul 17, 2022 at 12:11
  • $\begingroup$ @octonion Thanks！Yes when they squared the amplitude, the differential cross section is invariant. $\endgroup$
    – Daren
    Commented Jul 17, 2022 at 13:51

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The total spin about the z-axis in the initial state is +1/2 (photons have spin 1), and in the final state it is -1/2 (if $\chi=0$, to be exact), so there must be some additional +1 orbital angular momentum about the z-axis in the final state compared to whatever was there in the initial state.

The amplitude is calculated in terms of well defined initial 3-momentum pointing along the z-axis, and final momentum pointing in a direction specified by the angle $\chi$. But to really talk about orbital angular momentum we ought to expand in terms of spherical harmonics. Consider a table of spherical harmonics. Notice that all the spherical harmonics vanish at $\theta=0$ unless $m=0$, so even though there is not a well-defined initial total orbital angular momentum $l$, there is definitely initially zero z-component.

So we know the $z$ component in the final state should be $m=1$, and if you again take a look at that table you'll notice all of the $m=1$ spherical harmonics are proportional to $-\sin\theta\approx \chi$, which is consistent with Peskin and Schroeder's calculation. By the way ``p-wave" just means the $l=1$ orbital angular momentum state, and it is a slight abuse of terminology in my opinion since what is really well-defined is that $m=1$, not the $l$ value.

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