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I am trying to understand the details of the calculations involved in determining the electron's anomalous magnetic moment to second order: the $\alpha/{2\pi}$ term.

There is just one step, where an identity involving gamma matrices is used, that I am not able to prove for myself. It is $\overline{u}(p')[{p\mkern-7.5mu/}{\gamma_\mu} {p\mkern-7.5mu/'}]u(p) = \overline{u}(p')[2m(p_\mu + p'_\mu)]u(p)$, where $p_\mu$ and $p'_\mu$ are both four-momenta, $m$ is the electron mass and I am using the Feynman slash notation: ${p\mkern-7.5mu/} = p_\mu\gamma^\mu$. Nowhere is it explicitly stated that this identity is going to be used, but the derivations results presented are only possible if this is true. I always presumed that it was so obvious that it was not word mentioning.

I have gone through at least a dozen books and papers where the anomalous magnetic moment derivation is explicitly made but in all of them the identity mentioned is used without any justification. I have no doubt it is true. I would just like to be able to prove it.

Can anyone help me out, please?

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closed as off-topic by sammy gerbil, stafusa, AccidentalFourierTransform, Cosmas Zachos, Emilio Pisanty Aug 5 '18 at 16:29

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  • $\begingroup$ Is the changing of argument of the last spinor in the RHS correct? Or just a typo? $\endgroup$ – Francesco Arnaudo Aug 2 '18 at 14:33
  • $\begingroup$ Yes. It is meant to be $p'$ on the left and $p$ on the right. This expression shows up in the nominator of the amplitude. It eventually leads to the transition current from a state with momentum $p$ to a state with momentum $p'$. It is like the Gordon Decomposition Identity. $\endgroup$ – Francisco Alegria Aug 2 '18 at 19:39
  • $\begingroup$ Specifically which paper have you seen this identity? And which other literature has this derivation that you have checked? $\endgroup$ – Triatticus Aug 2 '18 at 20:33
  • $\begingroup$ For example, in Zee's Quantum Field Theory in a Nutshell (2nd edition), page 198, the paragraph after equation (14) where it is stated what eq. (14) becomes. Near the end it says: "the $m^0$ term gives $2m(p' + p)^\mu[-2(1-\alpha)(1-\beta)]$." I have omitted, in my question, the part in square brackets because it is not relevant. $\endgroup$ – Francisco Alegria Aug 2 '18 at 21:27
  • $\begingroup$ Also in Mandl's "Quantum Field Theory" (2nd edition), page 215, eq. (10.62a): $N_0^\mu(p',p) = \gamma^\alpha({p'\mkern-7.5mu/}-{a\mkern-7.5mu/}+m)\gamma^{\mu}({p\mkern-7.5mu/}-{a\mkern-7.5mu/}+m)\gamma_\alpha$. $\endgroup$ – Francisco Alegria Aug 2 '18 at 21:44
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In the current form, the identity does not seem to make any sense. For example, the spinor on the left should be $\overline{u}$, not $u$, otherwise matrix multiplication is ill-defined. Furthermore, I am not sure the arguments of the spinors are correct.

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  • $\begingroup$ Fixed the notation. $\endgroup$ – Francisco Alegria Aug 3 '18 at 9:05
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I managed to get an answer to my question however I should mention that the question itself has some missing relevant information that makes all the difference.

When trying to compute the second order correction to the anomalous magnetic moment of the electron what one usually tries to do is to express the transition current of the electron in the external magnetic field as

$$ \left<p\right|J^\mu\left|p\right>=\overline{u} (p')\left[\gamma^{\mu}F_1(q^2)+\frac{i\sigma^{\mu\nu}q_\nu}{2m}F_2(q^2)\right] u(p). $$ where $q = p'-p$ and $F_1$ and $F_2$ are the form factors. This is eq. 7, on page 196 of Zee's "Quantum Field Theory in a Nutshell" (2nd edition).

It is $F_2$(0) that gives us the anomalous magnetic moment.

One can use the Gordon decomposition identity,

$$ \overline{u} (p')\gamma^{\mu}u(p)=\overline{u} (p')\left[\frac{(p+p')^\mu}{2m}+\frac{i\sigma^{\mu\nu}(p'_\nu-p_\nu)}{2m}\right] u(p), $$

to write

$$ \left<p\right|J^\mu\left|p\right>=\overline{u} (p')\left[\gamma^{\mu}[F_1(q^2)+F_2(q^2)]-\frac{(p'+p)^\mu}{2m}F_2(q^2)\right] u(p). $$

Usually, in the derivations found in the books, one tries to group all terms with $p^\mu$ and $p'^\mu$ to one side and those with $\gamma^\mu$ to the other. The anomalous magnetic moment will thus be the coefficient of the $(p'+p)^\mu$ term.

In the derivation that I was looking for I should have said that I was looking to prove that

$$ \overline{u}(p')[{p\mkern-7.5mu/}{\gamma_\mu} {p\mkern-7.5mu/'}]u(p) = \overline{u}(p')[2m(p_\mu + p'_\mu)]u(p) + \text{terms in $\gamma^\mu$}. $$

Now here is my proof. For it I will use the identity

$$ {a\mkern-7.5mu/}\gamma_\mu+\gamma_\mu{a\mkern-7.5mu/}=2a_\mu,\tag{1} $$

which can be obtained from the definition

$$ \gamma_\mu\gamma_\nu+\gamma_\nu\gamma_\mu=2g_{\mu\nu}, $$

doing the product with $a^\nu$ which is a generic four-vector (I will use it with $p$ and $p'$).

Starting from $\overline{u}(p')[{p\mkern-7.5mu/}{\gamma_\mu} {p\mkern-7.5mu/'}]u(p)$ and using (1) to replace ${p\mkern-7.5mu/}\gamma_\mu$ leads to

$$ \overline{u}(p'){p\mkern-7.5mu/}{\gamma_\mu}{p\mkern-7.5mu/'}u(p) = 2p^\mu\overline{u}(p'){p\mkern-7.5mu/'}u(p) -\overline{u}(p')\gamma_\mu{p\mkern-7.5mu/}{p\mkern-7.5mu/'}u(p).\tag{2} $$

Now I am going to use $$ {a\mkern-7.5mu/}{b\mkern-7.5mu/}=-{b\mkern-7.5mu/}{a\mkern-7.5mu/}+2(ab)\tag{3} $$ which is obtained from (1) by doing the product with $b^\mu$. I use (3) to change the order of the product ${p\mkern-7.5mu/}{p\mkern-7.5mu/'}$ in the last term of (2). I thus get

$$ \overline{u}(p'){p\mkern-7.5mu/}{\gamma_\mu}{p\mkern-7.5mu/'}u(p) = 2p_\mu\overline{u}(p'){p\mkern-7.5mu/'}u(p) +\overline{u}(p')\gamma_\mu{p\mkern-7.5mu/'}{p\mkern-7.5mu/}u(p) -2(pp')\overline{u}(p')\gamma_{\mu}u(p).\tag{4} $$

Now I am going to use the Dirac equations ${p\mkern-7.5mu/}u(p)=mu(p)$ and $\overline{u}(p){p\mkern-7.5mu/}=m\overline{u}(p)$ to simplify (4):

$$ \overline{u}(p'){p\mkern-7.5mu/}{\gamma_\mu}{p\mkern-7.5mu/'}u(p) = 2mp_\mu\overline{u}(p')u(p) +m\overline{u}(p')\gamma_\mu{p\mkern-7.5mu/'}u(p) -2(pp')\overline{u}(p')\gamma_{\mu}u(p).\tag{5} $$

Now I am going to use (1) again to switch the order of $\gamma_\mu$ and ${p\mkern-7.5mu/}'$ in the second term of the right hand side of (5):

$$ \overline{u}(p'){p\mkern-7.5mu/}{\gamma_\mu}{p\mkern-7.5mu/'}u(p) = 2mp_\mu\overline{u}(p')u(p) +2mp'_\mu\overline{u}(p')u(p) -m\overline{u}(p'){p\mkern-7.5mu/'}\gamma_{\mu}u(p) -2(pp')\overline{u}(p')\gamma_{\mu}u(p). $$

Again, using the Dirac equation on the third term on the right hand side leads to

$$ \overline{u}(p'){p\mkern-7.5mu/}{\gamma_\mu}{p\mkern-7.5mu/'}u(p) = 2mp_\mu\overline{u}(p')u(p) +2mp'_\mu\overline{u}(p')u(p) -m^2\overline{u}(p')\gamma_{\mu}u(p) -2(pp')\overline{u}(p')\gamma_{\mu}u(p). $$

Finally, grouping the first and second terms as well as the third and forth (on the right hand side) leads to the desired expression:

$$ \overline{u}(p'){p\mkern-7.5mu/}{\gamma_\mu}{p\mkern-7.5mu/'}u(p) = \overline{u}(p')[2m(p_\mu+p'_\mu)]u(p) -[m^2+2(pp')]\overline{u}(p')\gamma_{\mu}u(p). $$

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