# Mystery of the lost degree of freedom in a gauge theory

1. Feb 29, 2012

### TriTertButoxy

Something is totally not making sense. In a complex scalar field theory, I have two field degrees of freedom, which I parametrize in polar field coordinates: $\phi = \rho e^{i\theta}/\sqrt{2}$, where $\rho$ and $\theta$ are real-valued; and its Lagrangian takes the form:
$$\mathcal{L} = \frac{1}{2}\partial_\mu\rho\partial^\mu\rho-V(\rho)+\frac{1}{2}\rho^2(\partial_\mu\theta)^2.$$
I have written this Lagrangian so that it is has a global U(1) symmetry (under $\rho\rightarrow\rho$ and $\theta\rightarrow\theta+\alpha$ with $\alpha$ being a spacetime-independent parameter). And I have a conserved charge, allowing me to get positively- and negatively- charged particles.

All is good so far; but now when I gauge the U(1), something funny happens. Suddenly $\theta$ becomes an "unphysical degree of freedom." If the potential $V(\phi)$ is such there is no spontaneous symmetry breaking, i.e. no vacuum expectation value for $\rho$, the Lagrangian reads:
$$\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\frac{1}{2}{ \partial }_\mu\rho{\partial}^\mu\rho-V(\rho)+\frac{1}{2}\rho^2({\partial}_\mu\theta)^2+e\rho A_\mu\partial^\mu\theta+e^2A^2\rho^2.$$
And now, the gauge transformation rule for the fields are:
$$A_\mu\rightarrow A_\mu-\frac{1}{e}{\partial_\mu}\alpha$$$$\rho\rightarrow\rho$$$$\theta\rightarrow\theta+\alpha,$$where now the gauge parameter $\alpha$ is spacetime-dependent.
The first term is the gauge-invariant kinetic term for the gauge field, and the next two terms involve the gauge invariant field $\rho$. The last three terms are individually not gauge-invariant, but added together, are gauge-invariant.

I shall now make a particular choice of gauge such that $\theta=0$, by appropriately choosing $\alpha$. This is analogous to the unitary gauge when spontaneous symmetry breaking does occur (but again, my potential is designed so this doesn't happen). With this choice of gauge my lagrangian reads:
$$\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\frac{1}{2}{ \partial }_\mu\rho{\partial}^\mu\rho-V(\rho)+e^2A^2\rho^2.$$
And all the dependence on $\theta$ is gone. Given that no symmetry breaking has occurred, the gauge field remains massless, and only has two physical degrees of freedom (the transverse polarizations) and since $\rho$ is a real-valued field, it represents only one field degree of freedom. Yet, this theory is supposed to describe the interaction of charged particles with the gauge field. How is this describing the interaction between charged particles when I have only one real scalar field??? What the heck is going on???

Bottom line: How can real-valued $\rho$ accommodate particle and antiparticle?

Last edited: Feb 29, 2012
2. Mar 1, 2012

### Demystifier

One of your charge degrees of freedom vanishes only when you fix the gauge in one particular way (the way which eliminates the phase of the scalar field). But when you fix the gauge, then you no longer have the gauge symmetry. And when you don't have gauge symmetry, then the vector field has 3 physical degrees of freedom (not only 2). So in essence, by fixing the gauge you transform one degree of freedom of the scalar field into one new degree of freedom of the vector field. The total number of physical degrees of freedom (equal to 4) has not changed.

3. Mar 1, 2012

### TriTertButoxy

This can't be true without qualifiers. In quantum electrodynamics, the (covariant) Lorentz gauge fixing condition is used by particle physicists. By fixing the gauge using that gauge condition doesn't leave the vector field with 3 physical degrees of freedom. There are still only two (transverse) physical degrees of freedom. Please explain.

Secondly; if what you said were true in the above example, the third vector physical degree of freedom (the spatial longitudinal mode) is totally decoupled from the theory (c.f. when symmetry breaking does occur, the coupling to the longitudinal mode is proportional to the mass of the gauge field). So this means I only have three overall degrees of freedom that are interacting with each other (the rho along with the the two transverse vector ones). Yet, the theory is supposed to describe four interacting degrees of freedom: the positively- and negatively- charged scalars, along with the two transverse vector degrees of freedom. Again, please elaborate.

4. Mar 1, 2012

### vanhees71

But in your unitary gauge Lagrangian you have a kind of mass term from the coupling to the "radial" dof. of the scalar field. Even if this has no vev, it still requires three physical degrees of freedom rather than two for the vector field.

5. Mar 1, 2012

### TriTertButoxy

I don't buy that argument even in the slightest. If I had used the standard parameterization of the fields, using $\phi$ and $\phi^\dagger$, the Lagrangian would read
$$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+|\partial_\mu\phi|^2-V(\phi)-ieA_\mu(\phi\partial^\mu\phi^\dagger-\phi^\dagger\partial^\mu\phi)+e^2A^2|\phi|^2.$$
And in the standard Lorentz gauge condition, the last term $e^2A^2|\phi|^2$ which in your words "a kind of mass term" would still be there. However, in the absence of spontaneous symmetry breaking, the vector field has no mass. Bad argument! Please explain... this issue seems to have been overlooked...

6. Mar 1, 2012

### Ben Niehoff

The discrepancy is related to exactly what happens when you take a gauge fixing condition and plug it back into the Lagrangian.

The simple fact is that this Lagrangian:

$$\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\frac{1}{2}{ \partial }_\mu\rho{\partial}^\mu\rho-V(\rho)+e^2A^2\rho^2.$$
does not have a gauge symmetry, and hence the vector field $A_\mu$ has 3 degrees of freedom.

However, if you assume the Lorentz gauge condition

$$\partial_\mu A^\mu = 0$$
there is still residual gauge freedom under

$$A_\mu \rightarrow A_\mu + \partial_\mu \lambda, \qquad \partial^\mu \partial_\mu \lambda = 0$$

7. Mar 2, 2012

### Demystifier

As explained by Ben Niehoff above, the Lorentz gauge condition is not really a gauge fixing condition, in the sense that it does not fix the gauge completely. (A complete gauge fixing is provided, e.g., by the Coulomb gauge.) Thus, with the Lorentz gauge condition you still have a vector field with 3 degrees of freedom.