# Can the necessity of using anti-commutators for Dirac fields and commutator for Klein-Gorden be deduced from the field equations?

We all learned to use the commutator for quantizing the KG field and the anti-commutator for the Dirac field. We are told (which is correct) so that KG-excitations are bosons and Dirac-excitations fermions. But isnt this extra knowledge (we know spin-1/2 is fermionic) we put in.

Can we see this directly from the equations? Would we run into problems (and what kind of) if for example we used commutators to quantize the Dirac equation?

• The spin-statistics theorem essentially says that if you want positive definite energies and you want to protect causality then half-integer spins must anti-commute and integer spins must commute. – gautampk Apr 23 '17 at 15:31

Previous answers and comments stated the main fact that spin-statistics theorem requires to use anti-commutation relations for fermions.

However is worthed (at least in my experience) to try the quantization of Dirac's equation with commutators in the canonical approach.

Let's start writing the Hamiltonian for Dirac equations in $3+1$. Recall the fields carry an spinorial index $\psi_a, (\bar{\psi})^a$. As always, Einstein convention (sumation over repeated indices is used.

$$\mathcal{L}=\bar{\psi}(i\gamma^\mu\partial_\mu-m)\psi \qquad \Pi= \frac{\partial\mathcal{L}}{\partial\dot{\psi}}=i\psi^\dagger\Rightarrow\mathcal{H}=\Pi\dot{\psi}-\mathcal{L} = i\Pi\gamma^0(i\gamma^i\partial_i-m)\psi$$

where i runs over spatial indices ($i\not=0$). A suitable ansatz for the commutation relations:

$$[\psi(x)_a,\Pi(y)^b]=i\delta^3(x-y)\delta_{ab}$$

With those relations and in the same way as KG-field, we can write the fields in term os creation/annihilation operators:

$$\psi(x) = \int d\tilde{p}(a_p^su_s(p)e^{-ipx}+{b_p^s}^\dagger s_s(p)e^{ipx}$$

where $s$ is a spinorial index and $d\tilde{p}$ the usual measure for integration. From the commutation relations we have proposed follows that

$$[a_p^s,{a_q^r}^\dagger]= (2\pi)^3\delta^3(p-q)2p_0 = \mathbf{-}[b_p^s,{b_q^r}^\dagger]$$

The sign on the commutation relations of $b$ operators, which follows from the algebra of the fields is crucial. As for the KG-field, if we try to write the hamiltonian in terms of creation and annihilation operators:

$$\mathcal{H}=\int d\tilde{p} p_0\sum_s({a_p^s}^\dagger a_p^s\mathbf{-}{b_p^s}^\dagger b_p^s)$$

The $b$-modes carry negative energy and the hamiltonian is unbounded from below (the theory at last will not be unitary and will have vacuum instability).

This is, of course, not a proof of the imposibility to quantize the Dirac equation with commutators, but points to the fact that the commutators with fermionic degrees of freedom will carry inconsistencies.

• I think that this could actually be considered as a crude proof for the spin statistics theorem. – gertian Apr 24 '17 at 8:47

In some sense yes.

It can be proven that (under the assumption of unitary and locality) that every particle with integer spin must have and symmetric wave function and that every particle with half integer spin must have an antisymmetric wave function. This theorem is named the Spin statistics theorem.

If we combine this with the fact that commutator brackets create symmetric waves and anticommutator brackets antisymmetric waves we know that every integer spin particle must obey commutation relations !

Finally, the spin can be read of from the Lagrangian since we know that scalars have spin 0, spinors spin 1/2 etc...