Why the massive spin-1 photon gets more degrees of freedom than massless case; while the massive spin-1/2 electron stays the same as massless case?

Question

Spin 1 field without mass term like photon has 2 real degrees of freedom. The polarization with two states. I think I can denote it as quantum state $|s,s_z> = |1,1>$ and $|1,-1>$.

Spin 1 field with mass term has 3 degrees of freedom, this can be understood from the Goldstone theorem with spin-1 gauge fields. I think I can denote it as quantum state $|s,s_z> = |1,1>$, $|1,0>$, and $|1,-1>$.

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Now Spin 1/2 Dirac fermion field without mass term like $4 \times 2 =8$ real degrees of freedom. (Am I counting the degrees of freedom correct?)

Spin 1/2 Dirac field with mass term also $4 \times 2 =8$ real degrees of freedom. (Am I counting the degrees of freedom correct?)

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This puzzles me: Why the massive spin-1 field gets more degrees of freedom than massless case; while the massive spin-1/2 field stays the same degrees of freedom as massless case?

Perhaps, to rephrase, why the massive photon (spin-1) gets more degrees of freedom than massless case; while the massive electron (spin-1/2) stays the same as massless electron?

At least the dispersion relation for massless to massive case, both cases, spin-1 and spin-1/2, change.

Answer 1

1. To count the on-shell DOF of a spin $s=1$ field we have to solve the EL equations for the Lagrangian density $${\cal L}~=~-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-\frac{1}{2}m^2A_{\mu}A^{\mu}.$$ In the massless case $m=0$, there is a gauge symmetry which removes an additional polarization.

   One may show that in the massive (massless) case the real on-shell DOF are $D-1$ ($D-2$) in $D$ flat spacetime dimensions, respectively, cf. Ref. 1.

2. To count the on-shell DOF of the spin $s=\frac{1}{2}$ Dirac field, we should count the solutions to the Dirac equation.

   One may show that the real on-shell DOF are $2^{[\frac{D}{2}]}$, where $[\cdot]$ denotes the integer part, cf. Ref. 1.

   In particular one may show that the solutions display no discontinuity as $m\to 0$.

3. More generally, there is a (no) discontinuity as $m\to 0$ if $s\geq 1$ ($s\leq \frac{1}{2}$), respectively, cf. e.g. this related Phys.SE post.

References:

1. M.D. Schwartz, QFT & the standard model, 2014; Subsection 8.2.4 + Section 11.2.

Answer 2

The electromagnetic potential has four degrees of freedom. It is its source, the charge-current, that has fewer than four dofs because of charge conservation. Not all four degrees of freedom of the potential are therefore manifest.