规范固定后标量QED理论中有质量自由度的计数分析

Alright, let's break down this problem step by step, starting with the baseline from the given Lagrangian:
$$\mathcal{L} = - \frac14 F^{\mu\nu} F_{\mu\nu} + (D^\mu \phi)^* (D_\mu \phi) - m^2 \phi^* \phi$$

First, let's recap the ungauged physical degrees of freedom (d.o.f.) to set a reference:

• The photon field $A_\mu$: it's a 4-component vector, but U(1) gauge symmetry eliminates 2 redundant d.o.f., leaving 2 massless real d.o.f. (the transverse polarizations we observe for real photons).
• The complex scalar $\phi$: as a complex field, it has 2 massive real d.o.f. (its independent real and imaginary parts, both physical before any gauge fixing).
  Total physical d.o.f.: $2 + 2 = 4$.

Now let's analyze what happens when we switch to unitary gauge, where we fix the gauge by setting $\phi$ to a real field: $\phi = \frac{1}{\sqrt{2}} \rho(x)$, with $\rho(x)$ a real scalar field.

1. The Real Scalar Field $\rho(x)$ #

By forcing $\phi$ to be real, we've completely eliminated the phase degree of freedom of the original complex scalar. All that's left is the modulus degree of freedom, encoded in $\rho(x)$. This is a single real scalar field with a kinetic term and a mass term $- \frac{1}{2} m^2 \rho^2$, so it contributes 1 massive real d.o.f.

2. The Vector Field $A_\mu(x)$ #

Here's the critical shift: when we fix unitary gauge, we've fully fixed the U(1) gauge symmetry—there's no remaining gauge transformation that can keep $\phi$ real (any phase rotation would turn $\phi$ back into a complex field, violating the gauge condition).

A 4-component vector field with no gauge redundancy might seem like it has 4 d.o.f., but we still have a Hamiltonian constraint: the time component $A_0$ has no kinetic term, leading to a primary constraint that removes 1 d.o.f. This leaves 3 massive real d.o.f. for $A_\mu$.

Even without spontaneous symmetry breaking (so the vacuum expectation value $\langle \rho \rangle = 0$), the Lagrangian now includes an interaction term $\frac{1}{2} e^2 \rho^2 A^\mu A_\mu$. While this isn't a constant mass term, it still means the longitudinal polarization of $A_\mu$ is no longer a gauge artifact—it's a physical, dynamic degree of freedom, just coupled to the scalar field.

Total Physical Degrees of Freedom #

Adding these up: $1 + 3 = 4$, which exactly matches the original count. This makes perfect sense—gauge fixing never creates or destroys physical d.o.f., it just rearranges how we count them. The phase d.o.f. from the original complex scalar gets "converted" into the longitudinal polarization of the vector field, turning the massless photon (2 d.o.f.) into a massive vector boson (3 d.o.f.), while the scalar loses one d.o.f. (the phase) to become a single real scalar (1 d.o.f.).

内容的提问来源于stack exchange，提问作者knzhou