I'm going through Peskin & Schroeder chapter 7, and I am finding it difficult to understand how the branch cut singularity appears in the one-loop correction to the photon propagator. Essentially, the expression coming from the electron loop in a photon propagator is: $$\hat{\Pi}_2(q^2) = - \frac{2 \alpha}{\pi} \int_0^1 \mathrm{d}x\, x(1-x) \log \left[ \frac{m^2}{m^2 - x(1-x)q^2} \right].\tag{7.91}$$

This correction to the photon two-point function is clearly well behaved if $q^2 < 4m^2$, but if $q^2 > 4m^2$ the argument of the logarithm becomes negative for some values of $x$ within the integration range and therefore one has to be careful with logarithmic branch points and all that.

If one defines the value of the logarithm $\log(z)$ in the usual way for $\arg(z) \in (- \pi, \pi)$, then one gets the result Peskin & Schroeder present, which is that for $q^2 > 4m^2$, $$ \hat{\Pi}_2(q^2 + i \epsilon) - \hat{\Pi}_2(q^2 - i \epsilon) = - \frac{2 i \alpha}{3} \sqrt{1 - \frac{4m^2}{q^2}} \left( 1 + \frac{2m^2}{q^2} \right);\tag{7.92}$$ therefore there is a discontinuity in the imaginary part of $\Pi_2(q^2)$ along the real axis of the complex $q^2$ plane, for $q^2 > 4m^2$.

My problem comes from the fact that the definition of the value of the logarithm is somewhat arbitrary. We could have defined $\log(z)$ with a branch cut along, say, the negative imaginary axis. This would produce no difference between $\hat{\Pi}_2(q^2 + i \epsilon)$ and $\hat{\Pi}_2(q^2 - i \epsilon)$, because with this definition $\log (-X + i\epsilon)$ and $\log(-X - i \epsilon)$ are equal for $\epsilon \to 0$ ($X > 0$). We seem not to find a discontinuity in $\Pi_2(q^2)$ along the real axis for $q^2 > 4m^2$ (maybe this definition introduces discontinuities somewhere else, but at least I think you do not have that branch cut).

So, my question is, why do we make that choice for the logarithm branch cut in the first place? Is it that we know, because of some other argument, that the discontinuity in the two point function must be there, and then we force it to appear? Or am I understanding something incorrectly?


1 Answer 1


Peskin & Schroeder write the form for the photon self-energy, eq. (7.71), as $$(-ie)^{2}(-1)\int\frac{d^{4}k}{(2\pi)^{4}}\,{\rm tr}\left[\gamma^{\mu}\frac{i}{\not k-m}\gamma^{\nu}\frac{i}{\not k+\not q-m}\right],$$ and you are correct that, from this expression on its own, you cannot tell where the branch cut in the logarithm that appears in the final answer should lie.

However (and this is one thing that I don’t think Peskin & Schroeder do a good job explaining), there is actually something missing from each of the propagators in the integrand. The expression in square brackets should really be (after we move the Dirac matices to the numerators) $$\left[\gamma^{\mu}\frac{i(\not k+m)}{k^{2}-m^{2}+i\epsilon}\gamma^{\nu}\frac{i(\not k+\not q+m)}{(k+q)^{2}-m^{2}+i\epsilon}\right].$$ If you carry the $i\epsilon$ terms through the whole expression, you will find that the argument of the logarithm has an infinitesimal imaginary part, and the sign of that imaginary part tells you how to place the branch cut in the expression.

  • 1
    $\begingroup$ Many thanks! I will try to do the calculation again carrying the $i\epsilon$ then, I didn't realize P&S were missing that! $\endgroup$
    – Alex V.
    Commented Sep 16, 2020 at 17:50

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