# Problem 3.2 from Peskin and Schroeder

This question is actually shockingly easy. It’s one of those questions whose statement is so short and is literally just “prove this identity” that you assume it’s going to take pages of obtuse and demoralising mathematical trickery. Not this time!

The question demands that we prove the Gordon identity, and assures us that this identity will be used in Chapter 6. It goes as follows:

$\bar{u}(p^{\prime})\gamma^{\mu}u(p)=\bar{u}(p^{\prime})\left[\frac{p^{\prime\mu}+p^{\mu}}{2m}+\frac{i \sigma^{\mu\nu}q_{\nu}}{2m}\right]u(p)$

where $q = p^\prime - p$

I’m not going to explicitly go through the calculation of this because it’s trivial once you know the steps and have the necessary information about gamma matrices and whatnot.

• Start by writing $\gamma^{\mu}=\frac{1}{2}(\gamma^{\mu}+\gamma^{\mu})$. This is a trick that dawned on me exactly half way through the solution (when I realised I was clearly missing a term) but the factor of a half in the RHS of the identity should give it away immediately, really.
• What are $u(p)$? We can write down a plane wave solution to the Dirac equation as $\psi(x) = u(p)\exp{(-i p \cdot x)}$ (we can do this since the field also satisfies the Klein-Gordon equation), so $u(p)$ is a column vector with a constraint obtained from plugging this $\psi$ into the Dirac equation[1],

$(i\gamma^{\mu}\partial_{\mu}-m)\psi(x)=0 \rightarrow (\gamma^{\mu}p_{\mu} - m)u(p)=0$

• It’s not terribly important because it doesn’t actually impact the solution, but I feel it relevant to point out that what we have here is a four-dimensional matrix (not a scalar) acting on a column vector. Beside the $m$ term is a sneaky invisible $\mathbb{I}_{4}$, since $\gamma^{\mu} p_{\mu}$ is still a four-dimensional matrix, $p_{\mu}$ being just a number (the $\mu$ component of the momentum four-vector).
• Use the above to write $u(p) = \frac{\gamma^{\nu}p_{\nu}}{m} u(p)$ and substitute this in.
• Recalling that $\bar{u}(p') \equiv u^{\dagger}(p')\gamma^{0}$, combine this with the above to write $\bar{u}(p') = \bar{u}(p')\frac{\gamma^{\nu}p_{\nu}}{m}$. (It is necessary to recall that $(\gamma^{\nu})^{\dagger} = \gamma^0 \gamma^{\nu} \gamma^0$ and $(\gamma^{0})^2 = \mathbb{I}_4$ to get here.)
• After rewriting $u(p)$ and $\bar{u}(p')$ in this way, we end up having some terms like $\gamma^{\mu}\gamma^{\nu}$. This can be rewritten as $g^{\mu\nu} - i \sigma^{\mu \nu}$ using (hint: adding together) the following properties of gamma matrices:

$\{\gamma^{\mu},\gamma^{\nu}\} = 2 g^{\mu \nu}$

$[\gamma^{\mu},\gamma^{\nu}] = -2 i \sigma^{\mu \nu}$

• Noting that $\sigma^{\mu \nu}$ is antisymmetric, we get… the result, as desired.
• Just a note: remember that $p_{\nu}$ is just a number and commutes with everything, so it can be neglected during gamma matrix manipulations.

I hope I have not just committed that wonderful crime students complain about,

“The result obviously follows…”

But… the result does follow, quite obviously!

1. I’m not going to attempt to use Feynman slash notation because I suspect that figuring out how to work it in WordPress will take longer than just writing “\gamma^{\mu}”. This will do, for now.

## 3 thoughts on “Problem 3.2 from Peskin and Schroeder”

1. Mattias says:

I might be missing something obvious, but when you put it all together won’t you end up with two terms having three gamma matrices multiplying each other?

Also, I don’t understand how the result becomes different for the two terms. As I see it you get two identical multiplications (due to how you do the trick with splitting the single gamma matrix up in two terms).

But again, I might be missing something obvious!

2. j says:

Oh wow, a comment! I’m hoping LaTeX works in the comment section, let’s find out…

You should get something like
$\frac{1}{2} \bar{u}(p^{\prime})\left(\gamma^{\mu}\right)u(p) = \frac{1}{2}\left( \bar{u}(p^{\prime}) \gamma^{\mu} u(p) + \bar{u}(p^{\prime}) \gamma^{\mu} u(p)\right)$ for the LHS.

Then you do the transformation trick on the $u(p)$ in one term while leaving the $\bar{u}(p^{\prime})$ alone, which produces a $\frac{p_{\nu}}{m}\bar{u}(p^{\prime})\gamma^{\mu}\gamma^{\nu} u(p)$ and do vice versa on the other, which should produce the same but with $\gamma^{\nu}\gamma^{\mu}$.

Does that make sense?

3. Mattias says:

After I commented here I actually tried by starting from the RHS to work back to the LHS. Then I got to apply the same trick that you now told me, but in reverse, so then I sort of understood how I could do it by starting from the RHS.

Thank you for the answer, I wasn’t sure if you would ever see the comment or not :)