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How to count the number of degrees of freedom of an arbitrary field (vector or tensor)? In other words, what is the mathematical procedure of gauge fixing?

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2 Answers 2

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In this answer, we summarize the results. The analysis itself can be found in textbooks, see e.g. Refs. 1 & 2.

$\downarrow$ Table 1: Massless spin $j$ field in $D$ spacetime dimensions.

$$\begin{array}{ccc} \text{Massless}^1 & \text{Off-shell DOF}^2 & \text{On-shell DOF}^3 \cr j=0 & 1 & 1 \cr j=\frac{1}{2} & n & \frac{n}{2} \cr j=1 & D-1 & D-2 \cr j=\frac{3}{2} & n(D-1) & \frac{n}{2}(D-3) \cr j=2 & \frac{D}{2}(D-1) & \frac{D}{2}(D-3) \cr \vdots &\vdots &\vdots \cr \text{Integer spin }j\in\mathbb{N}_0 & \begin{pmatrix} D+j-2 \cr D-2 \end{pmatrix}+ \begin{pmatrix} D+j-5 \cr D-2 \end{pmatrix}& \begin{pmatrix} D+j-4 \cr D-4 \end{pmatrix}+ \begin{pmatrix} D+j-5 \cr D-4 \end{pmatrix}\cr \text{Integer spin }D=4 & j^2+2 & 2-\delta^j_0 \cr \text{Integer spin }D=5 & \frac{1}{6}(2j+1)(j^2+j+6) & 2j+1\cr \vdots &\vdots &\vdots \cr \text{Half-int. spin }j\in\mathbb{N}_0+\frac{1}{2} & n\begin{pmatrix} D+j-\frac{5}{2} \cr D-2 \end{pmatrix}+ n\begin{pmatrix} D+j-\frac{9}{2} \cr D-2 \end{pmatrix}& \frac{n}{2}\begin{pmatrix} D+j-\frac{9}{2}\cr D-4 \end{pmatrix} \cr \text{Half-int. spin }D=4 &n(j^2+\frac{3}{4}) &\frac{n}{2} \cr \text{Half-int. spin }D=5 &\frac{n}{6}(2j+1)(j^2+j+\frac{9}{4}) &\frac{n}{4}(2j+1) \cr \vdots &\vdots &\vdots \cr \end{array}$$

$^1$For massive multiplets, go up 1 spacetime dimension, i.e. change $D\to D+1$ (without changing the number $n$ of spinor components). E.g. the on-shell DOF for massive 4D fields famously has a factor $2j+1$, cf. the row $D=5$ in Table 1.

$^2$ Off-shell DOF = # (components)- # (gauge transformations).

$^3$ On-shell DOF = # (helicity states)= (Classical DOF)/2, where Classical DOF = #(initial conditions).

$n$=# (spinor components). E.g. a Dirac spinor has $n=2^{[D/2]}$ complex components, while a Majorana spinor has $n=2^{[D/2]}$ real components,

$\downarrow$ Table 2: Antisymmetric $p$-form gauge potential in $D$ spacetime dimensions, $p\in\mathbb{N}_0$.

$$\begin{array}{ccc} p\text{-form gauge potential}& \text{Off-shell DOF} & \text{On-shell DOF} \cr & \begin{pmatrix} D-1 \cr p \end{pmatrix} & \begin{pmatrix} D-2 \cr p \end{pmatrix} \cr \end{array}$$

References:

  1. D.Z. Freedman & A. Van Proeyen, SUGRA, 2012.

  2. H. Nastase, Intro to SUGRA, arXiv:1112.3502; chapter 5.

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  • $\begingroup$ Are you suggesting that a spin $j=1/2$ particle has $2^{[D/2]}$ spin states, while a spin $j=1$ has $D-1$ spin states? Surely that cannot be correct: how can a particle with less spin $1/2<1$ have more spin states $2^{[D/2]}>D-1$?! (Or is this answer about fields only, with no relevance to particles?) $\endgroup$ Commented Jan 23, 2018 at 17:27
  • $\begingroup$ Yes, this is for fields and their corresponding particles. $\endgroup$
    – Qmechanic
    Commented Jan 23, 2018 at 17:41
  • $\begingroup$ Notes for later: The stabilizer subgroup/isotropy group/little group of the Lorentz group $O(D\!-\!1,1)$ for (1) massive and (2) massless particles are (the double cover) of $O(D\!-\!1)$ and $E(D\!-\!2)\cong O(D\!-\!2)\ltimes\mathbb{R}^{D-2},$ respectively. Sketched proof: $[P_{\lambda},M_{\mu\nu}]\propto \eta_{\lambda[\mu}P_{\nu]}$. Consider fixed $P_{\lambda}$ where only one coordinate (1) $P_0$ or (2) $P_+$ is non-zero, respectively. Let's call the corresponding coordinate-index $\mu_0$. Then conservation of $P_{\lambda}$ means that $M_{\mu\nu}$ cannot contain a $\mu_0$-coordinate.$\Box$ $\endgroup$
    – Qmechanic
    Commented Jan 15, 2019 at 10:23
  • $\begingroup$ Related: physics.stackexchange.com/q/457269/2451 $\endgroup$
    – Qmechanic
    Commented Jan 28, 2019 at 22:04
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Maybe, it will be particularly the answer on your question.

It's convenient to classificate the fields due to Wigner classification of the Poincare group representation. First assume only massless case. In this case there aren't mass Casimir operator $\hat {P}^{2}$ and spin Casimir operator $\hat {W}^{2}$, but the Pauli-Lubanski operator is proportional to 4-impulse operator with factor $\hat {h} = \frac{(\hat {\mathbf S} \cdot \hat {\mathbf p})}{|\mathbf p|}$ which is called helicity. It is invariant operator (for massless fields), so we may classificate fields by it. The state with fixed helicity may be connected only with state with opposite one; it is possible if theory is invariant under spatial inversions. So there is only two (maximum) degrees of freedom for massless field of an arbitrary spin. With this interpretation you can understand gauge fixing procedure only as the criterion of irreducibility (mass-zero) of the (Poincare group) representation of the field, and the answer on your question about counting the degrees of freedom is always two.

For better understanding of how gauge fixing decreases the number of freedom let's assume the massive spin-$s$ case. Let's use the irreducibility conditions $(1)-(4)$ for it. They leaves only $2s + 1$ degrees of freedom (in the beginning there were $4^{s}$ components). After that let's set mass $m$ in $(1)$ to zero. Then condition $(2)$ will cut an additional degrees of freedom.

For example, spin-one field: $$ \tag 5 (\partial^{2} + m^{2})A_{\mu} = 0, \quad \partial_{\mu}A^{\mu} = 0, $$ so there is three degrees of freedom.

Let's set $m$ to zero. Then implies additional gauge freedom, and we can set $u^{\mu}A_{\mu} = 0$ for an arbitrary timelike 4-vector $u_{\mu}$. It decrease the number of degrees of freedom by one. It is possible because there is transformations $A_{\mu} \to A_{\mu} + \partial_{\mu} f$, which may satisfy the first equation of $(5)$. So we get two degrees of freedom.

Spin-two field: $$ \tag 6 (\partial^{2} + m^{2})A^{\mu \nu} = 0, \quad A^{\mu \nu} = A^{\nu \mu}, \quad \partial_{\nu}A^{\mu \nu} = 0, \quad A_{\mu}^{\mu} = 0. $$ The second leaves 10 degrees of freedom, the third leaves 6 degrees and the last one leaves 5 degrees.

Let's set $m$ to zero. Then we can set $u_{\mu}A^{\mu \nu} =0$, which will cut an additional 3 degrees of freedoms (the fouth is equal to one of $\partial_{\mu}A^{\mu \nu} = 0$, so we again get 2 degrees. It is possible because there are gauge transformations $A_{\mu \nu} \to A_{\mu \nu} + \partial_{\mu} f_{\nu} + \partial_{\nu} f_{\mu}$ which may satisfy the first equation of $(5)$.

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