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I Does a field operator always commute with itself?

  1. Jul 13, 2017 #1
    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)?
     
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  3. Jul 13, 2017 #2

    hilbert2

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    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.
     
  4. Jul 13, 2017 #3
    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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