I'm sure this is really simple, and I might be right; it's just that I'm not sure. I'm asked to prove that the Klein-Gordon equation it's invariant under global gauge transformations.

In Greiner's book "Relativistic Quantum Mechanics" there's a section where it explains how the K-G equations it's invariant under gauge transformations, "1.10 Gauge invariance of the Coupling", where he uses the most general form of a gauge transformation $$A'_{\mu}(x) = A_{\mu}(x) + \frac{\partial \chi(x)}{\partial x^{\mu}}$$ and the wave function transformed it's $$\psi' = \exp \left(\frac{ie}{\hbar c}\chi\right)\psi$$ now it's of my understanding that a global gauge transformation when the $\chi \neq \chi(x)$, in other words, the $\chi$ it's just a constant, doesn't depend on the coordinates.

In a local gauge transformation $\chi = \chi (x)$, it does depend on where you are standing in the space, so here Greiner uses a local gauge transformation because $\chi$ depends on the coordinates, and when substituting on the K-G equation the form of the equation its preserved, and because of the derivative of the wave function, now as I said earlier, I'm asked to prove that the Klein-Gordon equation it's invariant under global gauge transformations. My question is wouldn't it be trivial? Because if $\chi$ doesn't depend on $x$ then: $$A'_{\mu}(x) = A_{\mu}(x) + \frac{\partial \chi}{\partial x^{\mu}} = A_{\mu}(x) + 0 = A_{\mu}(x)$$ and $$\psi' = \exp \left(\frac{ie}{\hbar c}\chi\right)\psi$$ would only be a constant phase which doesn't really affect the equation so it's global invariant? I suppose it's not that easy and I'm doing something wrong, please help.

  • $\begingroup$ Does the question as about a specific type of global transformation, i.e. U(1) or SO(2), or do you have to show it for a general transformation? $\endgroup$ – Ollie113 May 17 '19 at 10:08
  • $\begingroup$ I't doesn't specify a type, i have to show it for a general transformation $\endgroup$ – Luis.Alberto May 17 '19 at 16:29

In case of a required global gauge transformation a gauge field $A_\mu$ is not needed since the K-G equation is already invariant under a coordinate-independent phase transformation of which a global gauge transformation is a special case. It can be easily checked that $\psi'$ and $\psi$ fulfill the same equation:

$$(\partial^\mu \partial_\mu + m^2)\psi =0$$

since the a coordinate-independent phase transformation commutes with the derivative. Only in case of a required local gauge transformation invariance one would use the more general equation:

$$(D^\mu D_\mu + m^2)\psi =0$$

with $D_\mu = \partial_\mu + i e A_\mu$ and $A'_\mu = A_\mu +\partial_\mu\chi$. Of course the more general equation is also invariant under global gauge transformations, but this observation is considered of little interest in general.

  • $\begingroup$ Ok, but how do i prove that the K-G equation it's invariant under a coordinate-independent phase transformation? $\endgroup$ – Luis.Alberto May 17 '19 at 16:34
  • $\begingroup$ The K-G equation is linear. So if $\psi$ is a solution, $C\psi$ is also a solution with C=constant. So you can choose $C=exp(\frac{ie}{\hbar c}\chi)$. $\chi$ is just a number, it does not depend on anything. $C$ is actually a complex number, but as said constant. You can also multiply the original equation for $\psi$ on the left side with $e^{i\chi}$, ($e/\hbar/c\equiv 1$ for simplicity) then commutate this factor with $\partial^\mu\partial_\mu$ and with $m^2$ and you get $e^{i\chi}\psi = \psi'$ on the right side. So $\psi'$ does what $\psi$ does, fulfilling the K-G equation. $\endgroup$ – Frederic Thomas May 17 '19 at 17:48
  • $\begingroup$ Ok thanks a lot, now it's more clear $\endgroup$ – Luis.Alberto May 17 '19 at 17:51

You correctly observe that there is no such thing as a global gauge transformation. That should be your answer. There are of course global coordinate transformations giving the well known conserved Noether currents. Gauge transformations are not only not global but also are not coordinate transformations.

  • $\begingroup$ Yeah that's what i thought, thanks a lot! $\endgroup$ – Luis.Alberto May 17 '19 at 17:52

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