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Let $\psi(x)$ be the field of the electron. Its Fourier transformed two-point function reads $$ \langle\psi\bar\psi\rangle=\frac{1}{\not p-m-\Sigma(\not p)}. $$

If we calculate $\Sigma(\not p)$, we observe that it depends on the gauge parameter $\xi$, which in principle is not a problem because $\Sigma(\not p)$ is not observable by itself.

But if we think of a gauge transformation as taking $\psi\to\mathrm e^{i\alpha(x)}\psi(x)$, then the two-point function should satisfy $$ \langle\psi\bar\psi\rangle\to \langle\psi\mathrm e^{i\alpha(x)}\mathrm e^{-i\alpha(x)}\bar\psi\rangle=\langle\psi\bar\psi\rangle $$

Therefore, one would naïvely expect $\Sigma(\not p)$ to be gauge invariant, and therefore it shouldn't depend on $\xi$. What is the solution to this contradiction? Why do our expectations fail?

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  • $\begingroup$ The catch, I suspect, lies in the calculation of $\Sigma$. Strictly speaking it is the summation of an infinite number of orders, but in calculations one necessarily have to truncate it at some finite order. This truncation introduces the gauge dependence. If one were able to compute it to all orders, it would be gauge independent, I think. $\endgroup$ – flippiefanus Dec 7 '16 at 11:37
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    $\begingroup$ @flippiefanus if that works it would certainly be satisfying. But Im not 100% sure that that works: after all, when computing $S$ matrix elements, we ask for $\xi$-independence order by order in perturbation theory, right? $\endgroup$ – AccidentalFourierTransform Dec 7 '16 at 11:41
  • $\begingroup$ How does the gauge transformation $\mathrm{e}^{\mathrm{i}\alpha(x)}$ change $\xi$? In a theory with $\xi$, you have abandoned gauge invariance because you fixed the gauge, so I don't see how the requirement that $\langle\psi\bar\psi\rangle$ be "gauge invariant" is consistent with having fixed the gauge by introducing $\xi$. $\endgroup$ – ACuriousMind Dec 7 '16 at 14:51
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    $\begingroup$ In any case, the two point function is non-local $\psi(0)\bar\psi(x)$, so it is not gauge invariant. $\endgroup$ – Thomas Dec 7 '16 at 16:05
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    $\begingroup$ You need $S(x,y)=\langle\psi(x)\bar\psi(y)\rangle$. By Lorentz invariance $S(x,y)$ only depends on $x-y$, and $S(p)$ is the FT in $x-y$. $\endgroup$ – Thomas Dec 7 '16 at 16:21
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The propagator $S(p)$ is the Fourier transform of the two-point function $S(x,y)=\langle\psi(x)\bar\psi(y)\rangle$, $$ S(p) = \int \frac{d^p}{(2\pi)^4} \, \exp(-ip\cdot(x-y)) \, S(x,y)\, . $$ Note that because of Lorentz invariance $S(x,y)$ does not depend on $x+y$. Clearly, the two-point function is non-local and not gauge invariant.

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As an alternative to Thomas answer, we note that if we write the transformation law explicitly, we get $$ \langle\psi(x)\bar\psi(y)\rangle\to \langle\psi(x)\mathrm e^{i\alpha({\color{red}x})}\mathrm e^{-i\alpha({\color{red}y})}\bar\psi(y)\rangle=\mathrm e^{i(\alpha(x)-\alpha(y))}\langle\psi(x)\bar\psi(y)\rangle $$

We see that the two-point function fails to be gauge invariant because the fields are evaluated at different points and thus the local phases don't cancel off each other. This wasn't evident in the OP because I didn't write the space-time labels explicitly. Silly me.

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  • $\begingroup$ In this sense the Dirac field $\psi(x)$ is also gauge non-invariant since it changes. The more important thing is that the variable change is not obliged to preserve the form of equations, if it helps solve the equations. $\endgroup$ – Vladimir Kalitvianski Dec 7 '16 at 17:48

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