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Consider the theory of scalar QED with the Lagrangian $$\mathcal{L} = - \frac14 F^{\mu\nu} F_{\mu\nu} + (D^\mu \phi)^* (D_\mu \phi) - m^2 \phi^* \phi \tag{1}$$ where $\phi$ is a complex scalar field with mass $m$. Counting the degrees of freedom, we have

  • two massless real degrees of freedom from $A_\mu$
  • two massive real degrees of freedom from $\phi$

Now, even though there's no symmetry breaking going on here we can still choose to go to unitary gauge, i.e. fixing the gauge so that $\phi$ is real. We now have the gauge-fixed Lagrangian $$\mathcal{L} = - \frac14 F^{\mu\nu} F_{\mu\nu} + (D^\mu \varphi) (D_\mu \varphi) - \frac12 m^2 \varphi^2\tag{2}$$ where $\varphi$ is a canonically normalized real scalar field, and there is no gauge symmetry. Then we have

  • three real degrees of freedom from $A_\mu$
  • one massive real degree of freedom from $\phi$

where I know there are three real degrees of freedom in $A_\mu$, because gauge fixing always removes one and we have no gauge fixing here.

What confuses me is that there must be two massive degrees of freedom, just as there were in the original theory. So that somehow means that one of the degrees of freedom in $A_\mu$ is massive while the other two aren't -- but how can that be? There's no mass term for $A_\mu$ in sight.

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  • $\begingroup$ Note: I've seen this question which has some overlap, though I don't think the answer addresses my question. In particular I don't understand how the mass term for the longitudinal part of $A_\mu$ comes about. $\endgroup$ – knzhou Feb 17 '18 at 20:15
  • $\begingroup$ related: Why can't a real scalar couple to the electromagnetic field?. In your second Lagrangian, the coupling between $A$ and $\varphi$ is not through a conserved current, so the longitudinal mode is not decoupled, and the counting of d.o.f. is not straightforward. $\endgroup$ – AccidentalFourierTransform Feb 17 '18 at 22:31
  • $\begingroup$ @AccidentalFourierTransform There’s indeed a longitudinal mode, but I’m having a hard time seeing why it has mass $m$! $\endgroup$ – knzhou Feb 17 '18 at 23:18
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What you can remove with a field redefinition (of the form of a $U(1)$ gauge transformation) is the phase of $\phi$. But the total number of degrees of freedom (dof's) is not going to change since the resulting lagrangian is not longer gauge invariant. The new counting is $3+1$ where the $3$ dof's come from $A_\mu$ (with a covariant constraint, see below, so that really $3=4-1$), and the 1 dof comes instead from the scalar radial mode of $\phi$.

More explicitly, write $$\phi(x)=e^{ie\pi(x)}\frac{h(x)}{\sqrt{2}}$$ (with both $\pi$ and $h(x)$ real scalar fields) and the covariant derivative $D_\mu \phi=(\partial_\mu-i e A_\mu)\phi$ becomes $$ D_\mu \phi=e^{ie\pi}\left[ie(\partial_\mu\pi-A_\mu)+\frac{1}{\sqrt{2}}\partial_\mu h\right] $$ so that the lagrangian reads $$ \mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\frac{e^2}{2}h^2(\partial_\mu\pi-A_\mu)^2+\frac{1}{2}(\partial_\mu h)^2-\frac{m^2}{2}h^2\,. $$ Now, defining the new variable $A_\mu^\prime=A_\mu-\partial_\mu\pi$ (which is the gauge invariant combination), the lagrangian becomes $$ \mathcal{L}=-\frac{1}{4}F^\prime_{\mu\nu}F^{\prime \mu\nu}+\frac{e^2}{2}h^2 A_\mu^{\prime 2}+\frac{1}{2}(\partial_\mu h)^2-\frac{m^2}{2}h^2 $$ in full analogy to the abelian Higgs mechanism except that $h$ has vanishing vacuum expectation value. Now, this last lagrangian depends on the fields $A_\mu^\prime$ and $h$: how many dof's are there? Well, the $h(x)$ certainly counts 1. The $A_\mu^\prime$ on the other hand counts $3$ and neither $4$ (even though $\mu=0,1,2,3$) nor $2$. Indeed, from the equations of motion $\partial_\mu F^{^\prime\mu\nu}=-e^2h^2 A^{\prime \nu}$ we see the covariant constrain $$ \partial_\mu (A^{\prime\mu} h^2)=0 $$ which sends us from $4$ to $3$. But notice that there is no gauge invariance for $A_\mu^\prime$ in its lagrangian above (the $e^2 h^2 A_\mu^{\prime 2}$-term breaks it) and no longitudinal mode can be therefore removed in that way: $A_\mu^\prime$ is a different physical configuration than say $A^\prime_\mu-\partial_\mu \Omega$ (in particular one solves the equation of motion, the other doesn't). So the counting ends here and it matches that for massive spin-1 plus a real scalar, that is $3+1$ as expected. Notice however, that the spin-1 particle is not massive and all this gymnastic is just artificial as you wanted to move around one scalar dof $\pi$ inside $A_\mu$.

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  • $\begingroup$ But why is the new component of $A_\mu$ massive? I don’t see a mass term for it anywhere, but we should conserve the number of massive degrees of freedom. $\endgroup$ – knzhou Feb 22 '18 at 10:12
  • $\begingroup$ @knzhou Please read carefully what I have written and all the explicit details I have provided. Again, the spin-1 boson is not massive even though the counting of degrees of freedom goes like if it was, since gauge invariance in the new variables ($A_\mu^\prime$ and $h$) is lost, precisely as it happens for a massive spin-1. But again, there is no massive spin-1 particle here, and the number of degrees of freedom is conserved, 4 . $\endgroup$ – TwoBs Feb 22 '18 at 14:23
  • $\begingroup$ I haven't said that all three components of $A_\mu$ are massive. The issue is that in the original Lagrangian, there are clearly two massive degrees of freedom, the two components of the $\phi$. If you claim all three components of $A_\mu$ are massless, then we have lost one massive degree of freedom. $\endgroup$ – knzhou Feb 22 '18 at 14:38
  • $\begingroup$ @knzhou The counting of degrees of freedom is 1+1+2=4 at the beginning and it can be rewritten after some manipulation as 1+3=4. But the number and composition of massless and massive particles does not change whatever manipulation you do. In the original lagrangian there were two massless spin-0 particles (hence 2 dofs) and one massless spin-1 particle (hence 2 more dofs). In the new lagrangian there are still 4 dofs; they are mixed in the new fields in a twisted way. Yet they correspond to the three massless particles: the two spin-0 massless particles and the one spin-1 massless particle. $\endgroup$ – TwoBs Feb 22 '18 at 15:27
  • $\begingroup$ "In the original lagrangian there were two massless spin-0 particles (hence 2 dofs)." No, the $\phi$ field has a mass. The whole point of my question is where its two massive degrees of freedom go. $\endgroup$ – knzhou Feb 22 '18 at 15:37
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Before we can compare with the second Lagrangian (2), the first Lagrangian (1) should include a gauge-fixing term ${\cal L}_{\rm gf}$, e.g. ${\cal L}_{\rm gf} = \lambda~ {\rm Im}(\phi),$ where $\lambda$ is a Lagrange multiplier. After integrating out $\lambda$ and ${\rm Im}(\phi)$ the Lagrangian (1) becomes the Lagrangian (2).

Why we need to consider gauge-fixed (rather than un-gauge-fixed) Lagrangians is e.g. discussed in my Phys.SE answer here.

For both Lagrangians, the scalar field effectively induces a mass term for the $A_{\mu}$-field.

$\downarrow$ Table 1: Real DOF of OP's Lagrangians.

$$\begin{array}{ccc} \text{Lagrangian}& \text{Off-shell DOF}^1 & \text{On-shell DOF}^2 \cr (1)& 2+4-1=5 &2+3-1=4 \cr (2)& 1+4-0=5 &1+3-0=4 \end{array}$$

$^1$ Off-shell DOF = # (components)- # (gauge transformations).

$^2$ On-shell DOF = # (helicity states)= (Classical DOF)/2, where Classical DOF = #(initial conditions).

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