A classical real scalar field admits a decomposition $$\phi(x)\sim a_pe^{-ip\cdot x}+a_p^*e^{+ip\cdot x}$$ which tells that at each $x$, there exists a real number i.e., one degree of freedom at each spacetime point $x$.

A classical complex scalar field can be decomposed as $$\phi(x)\sim a_pe^{-ip\cdot x}+b_p^*e^{+ip\cdot x}$$ which implies that it assigns a complex number i.e. two real degrees of freedom to each spacetime point.

Let us now look at a Dirac field. It can be decomposed as $$\Psi(x)\sim \sum\limits_{s=1,2}\Big[a^s_pu^s_pe^{-ip\cdot x}+b^{*s}_pv^s_pe^{+ip\cdot x}\Big].$$

How many real degrees of freedom do I have now at a given spacetime point?

If we pretend for a moment that $u_p^s$ and $v_p^s$ are complex numbers instead of column vectors, it looks like that we have four real degrees of freedom at each $x$: $\Psi(x)$ is made up of two independent complex numbers for each value of $s$.

Is my counting right? Do I make a mistake by treating $u$ and $v$ to be numbers? But I seem to get a problem if I start putting spinor indices.

  • $\begingroup$ You need to perform the full (2nd class constrained system) Hamiltonian analysis to properly the no. of DOF at each spacetime point. $\endgroup$ – DanielC Aug 19 at 20:08

In general, we have

  • Off-shell DOF = # (components) - # (gauge transformations).

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


  1. For a real (complex) scalar field, the number of on- and off-shell DOF is 1 real (complex) DOF, respectively.

  2. For a spinor field, the off-shell and on-shell DOF is $n$ and $\frac{n}{2}$, respectively, where $n$ = #(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.

    The $s$-index on the $u_s$-spinor labels the number of helicity states.

See also my Phys.SE answer here.


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

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