# Some counting of field degrees of freedom for a classical spin-1/2 Dirac field

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.

• You need to perform the full (2nd class constrained system) Hamiltonian analysis to properly the no. of DOF at each spacetime point. – 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).

Examples:

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.