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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 $|1,-1>$.


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?)


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.

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    $\begingroup$ Why should it lose any? The massless particles (of any non-zero spin) have 2 d.o.f., and the spin-1/2 particle has 2 d.o.f. already when massive? The question sounds as if you expect that particles should always lose degrees of freedom when going from massive to massless, but at least for the scalar this can't happen since it has only d.o.f. to begin with, so why expect it for spin-1/2. $\endgroup$
    – ACuriousMind
    Dec 6, 2023 at 23:00
  • $\begingroup$ A good question that deserves a serious answer. $\endgroup$
    – my2cts
    Dec 6, 2023 at 23:05
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    $\begingroup$ Maybe the problem is your first paragraph - there isn't really an "$S_z$" for massless particles, polarization is not the same as spin. See this answer of mine for why massless spin-ful particles end up with 2 d.o.f. generically. $\endgroup$
    – ACuriousMind
    Dec 6, 2023 at 23:20
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    $\begingroup$ But a massless fermion may be Weyl, with half the d.o.f. as Dirac. $\endgroup$ Dec 7, 2023 at 2:14
  • $\begingroup$ Which reference are you using? What is its equation of motion? $\endgroup$
    – my2cts
    Dec 11, 2023 at 8:25

2 Answers 2

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  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.
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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.

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