We know by the spin-statistics theorem that the real scalar field has to be canonically quantized by commutators. But if we try to use anticommutators, we would expand the field

$$\phi(x)=\int\frac{d^3k}{(2\pi)^3\sqrt{2\omega_k}}\left(a(\mathbf k)e^{-ikx}+a^\dagger(\mathbf k)e^{ikx}\right)$$

Where we have $k^0=\omega_k=\sqrt{m^2+\mathbf k^2}$ and

$$\left\{a(\mathbf k),a(\mathbf k')\right\}=\left\{a^\dagger(\mathbf k),a^\dagger(\mathbf k')\right\}=0,$$

$$\left\{a(\mathbf k),a^\dagger(\mathbf k')\right\}=(2\pi)^3\delta(\mathbf k-\mathbf k').$$

I'm then trying to prove that the microcausality relation for the observable $\phi(x)$ is violated

$$\left[\phi(x),\phi(y)\right]\neq0$$ for $x-y$ spacelike.

However, I can't find a way to write down these commutators in terms of the anticommutators in such a way that it would explicitly be nonzero. How would I proceed?


Expand $\phi$ into the creation and annihilation operators as usual, then use

$$ [a(p), a^{\dagger}(q)] = (2\pi)^3 \delta(p-q) - 2 a^{\dagger}(q) a(p)$$

Where $\delta$ is the delta-function distribution.

You will obtain an operator that is not equal to $0$ on the fermionic Fock space (to show that simply present a “witness” state for which acting on it with the operator in question produces a non-zero state).

  • $\begingroup$ I applied this and worked out the expectation value of the commutator on the state $|\mathbf p\rangle=a^\dagger(\mathbf p)|0\rangle$ and got a nonvanishing $x-y$ dependence. Thanks! $\endgroup$ – Gabriel Golfetti Apr 8 '19 at 12:14

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