# I Does a field operator always commute with itself?

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1. Jul 13, 2017

### Frank Castle

In quantum field theory (QFT), the requirement that physics is always causal is implemented by the microcausality condition on commutators of observables $\mathcal{O}(x)$ and $\mathcal{O}'(y)$, $$\left[\mathcal{O}(x),\mathcal{O}'(y)\right]=0$$ for spacelike separations. Intuitively, I've always understood this as the requirement that a measurement of the observable $\mathcal{O}(x)$ at the point $x^{\mu}$ cannot influence a measurement of the observable $\mathcal{O}'(y)$ at the point $y^{\mu}$ is the two points are spacelike separated.

Now, consider a scalar field $\phi(x)$. Is it true that it always commutes with itself regardless of the spacetime separation? i.e. is $$\left[\phi(x),\phi(y)\right]=0$$ true $\forall\;x^{\mu},\,y^{\mu}$?

If so, how does one interpret this physically? Is it analogous to ordinary quantum mechanics (QM) in which the different position operators $\hat{x}^{i}$ all commute among themselves (since a particle can be in a simultaneous eigenstate of its 3-position)?

2. Jul 13, 2017

### hilbert2

The commutator of the Heisenberg picture scalar field operators can be written with propagators as

$[\phi (x) , \phi (y)] = D(x-y) - D(y-x)$.

For spacelike separations this vanishes as shown in Peskin&Schroeder and other sources. It's not always zero.

3. Jul 13, 2017

### Frank Castle

What is the physical interpretation of this though?

In which cases does the commutator of a field with itself not vanish? Is it only the case when one has anti-commuting fields such as spinors?

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