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It its very often stated that in QFT, we don't actually deal with operator valued functions (assign a field operator to each point in space time), but instead with operator valued distributions (in general that means that we have to smear over some test region before we can take expectation values of operators).

I'd like to know where this becomes apparant in scalar real QFT (the most simply case one could imagine). I tried to write the the CCR and see if I can say anything interesting about the non-smeared eigenvalues of the field $\Phi$, but I'm stuck.

Qmechanic
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Quantumwhisp
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    How are you "writing the CCR" without already having to say that the fields are distributions? The r.h.s. of $[\phi(x,t),\pi(y,t)] = \delta(x-y)$ is clearly a distribution, is it not? – ACuriousMind Aug 28 '22 at 15:08
  • @ACuriousMind yes, the rhs is a distribution, but I wasn't sure wether this also means the operators itself must be distributions. – Quantumwhisp Aug 28 '22 at 16:37
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    $a_{p}|p\rangle$ is not a part of the Fock space because it's equal to $a_pa^{\dagger}_p|0\rangle=[a_p, a^{\dagger}_p]|0\rangle=\delta(0)|0\rangle$. This means that $a_p$ is not an operator on the Fock space because it produces delta function coefficients when operated on $|p\rangle$. But I'm not sure how this argument carries over to space dependent operators – Ryder Rude Aug 28 '22 at 16:45

1 Answers1

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Since the commutation relations are $$ [\phi(x,t),\pi(y,t)] = \mathrm{i}\delta(x-y)$$ at least one of $\phi$ and $\pi$ must be a distribution, too, since functions are closed under multiplication and subtraction: $\phi\pi - \pi\phi$ when $\pi$ and $\phi$ are both ordinary functions could not result in a $\delta$-distribution on the r.h.s.

ACuriousMind
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    Put a Poisson bracket on the LHS and classical fields on spacetime are also distributions. – DanielC Aug 28 '22 at 21:11
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    @DanielC but the Poisson bracket is not simply $\phi \pi - \pi \phi$, so not just a multiplication and subtraction. In fact it will involve functional derivatives which can generate delta functions. – Gold Aug 28 '22 at 21:25