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I am studying Mandl and Shaw's book on QFT and I am trying to understand the different definitions of the propagator functions, or $\Delta$-functions. One $\Delta$-function is defined (and derived) in one section as

$$ i\hbar c \Delta^+(x-y) = [\phi^+(x), \phi^-(y)] = \frac{ic}{2(2\pi)^3} \int \frac{d^3\textbf{k}}{\omega_{\textbf{k}}} e^{-ik(x-y)}, $$

where $\phi^{\pm}(x)$ are the positive and negative frequency parts of $\phi(x)$ which is a complex scalar field. This derivation is somewhat straightforward and I think I understand it. The problem for me is when the same function is later presented as

$$ i\hbar c \Delta^+(x-y) = \langle 0 | [\phi^+(x), \phi^-(y)] | 0 \rangle. $$

This seems to imply that

$$ \langle 0 | [\phi^+(x), \phi^-(y)] | 0 \rangle = [\phi^+(x), \phi^-(y)], $$

which looks a bit strange. This definition with the vacuum expectation value seems very important for further chapters but is barely motivated at all. Am I missing something obvious here?

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    $\begingroup$ This might be of interest. $\endgroup$ Sep 20, 2022 at 14:06

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It's simply because that commutator is proportional to identity operator, as your result imply. Considering a normalized vacuum state then $$ \left[ \hat{\phi}^+(x), \hat{\phi}^-(y) \right] = \text{const} \, \hat{I} \\ \langle 0 | \left[ \hat{\phi}^+(x), \hat{\phi}^-(y) \right] | 0 \rangle = \text{const}$$

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    $\begingroup$ And, by that, it is a (common) abuse of notation. $\endgroup$ Sep 20, 2022 at 14:01
  • $\begingroup$ @JasonFunderberker Why? $\endgroup$
    – Rob Tan
    Sep 20, 2022 at 14:10
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    $\begingroup$ Are we taking the commutation relation outside of the expectation value because it is a constant and using <0|0> = 1? $\endgroup$
    – Tjommen
    Sep 20, 2022 at 14:17
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    $\begingroup$ @Tjommen I don't really understand the "outside". The commutator of those two operators is an operator too, but very simple, proportional to identity operator on your space of states. When you consider the expectation value of this operator respect to any normalized state, in particular the vacuum one, you have $\langle 0 | \text{const} \cdot \hat{I} | 0 \rangle = \text{const}$ and that's all $\endgroup$
    – Rob Tan
    Sep 21, 2022 at 9:26
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    $\begingroup$ @Tjommen Yes, that is a basic fact / defintion of the inner product (with the convention that it is linear in the second and anti-linear in the first argument): For $c\in \mathbb C$ and $\psi,\phi \in H$ we have $ \langle \psi,c \phi\rangle = c\,\langle \psi,\phi\rangle$ and $\langle c\psi,\phi\rangle = \bar c \, \langle \psi,\phi\rangle$. $\endgroup$ Sep 23, 2022 at 12:27

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