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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?

Qmechanic
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Tjommen
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1 Answers1

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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 $$ \left[ \hat{\phi}^+(x), \hat{\phi}^-(y) \right] = \text{const} \, \hat{I} \\ \langle 0 | \left[ \hat{\phi}^+(x), \hat{\phi}^-(y) \right] | 0 \rangle = \text{const}$$

Rob Tan
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    And, by that, it is a (common) abuse of notation. – Tobias Fünke Sep 20 '22 at 14:01
  • @JasonFunderberker Why? – Rob Tan Sep 20 '22 at 14:10
  • Well, if you define, which the book does as far as I understand the question, both of your equations as $\Delta$, it is an abuse of notation. On the one hand it denotes an operator, on the other hand a $c$-number. So strictly speaking, one should really use two different symbols. – Tobias Fünke Sep 20 '22 at 14:12
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    Are we taking the commutation relation outside of the expectation value because it is a constant and using <0|0> = 1? – Tjommen Sep 20 '22 at 14:17
  • @JasonFunderberker I never defined $\Delta$ here. Anyway sometimes to use notation freely makes things less cumbersome :) – Rob Tan Sep 21 '22 at 09:23
  • I obviously meant the book here... And you're right, abuse of notation is fine (and can be helpful) if everyone knows what is meant. – Tobias Fünke Sep 21 '22 at 09:24
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    @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 – Rob Tan Sep 21 '22 at 09:26
  • What I meant was $\langle 0| const \cdot \hat{I} |0\rangle = const \langle0| \hat{I} |0 \rangle = $ const. Does it make sense to bring out the constant like this? – Tjommen Sep 22 '22 at 14:09
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    @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 $\langle c\psi,\phi\rangle = \bar c , \langle \psi,\phi\rangle$. – Tobias Fünke Sep 23 '22 at 12:27