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How should I interpret the left-hand side of this expression

$$ \langle \phi(k)\phi(-k) \rangle ~=~ \frac{\mathrm{i}}{k^2 -m^2},$$

which appears on pg. 3 of Matt Strassler's TASI 2001 notes: http://arxiv.org/abs/hep-th/0309149

This equation is presented in a discussion of the classical field theory of a massive complex scalar. What are the brackets indicating? An average? An expectation value?

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1 Answer 1

In real space, $\langle \phi(y)\phi^\dagger(x)\rangle = \langle 0 | \phi(y)\phi^\dagger(x)|0\rangle$, which reads the amplitude you create a particle at $x$ and annihilate it at $y$. The brackets means expectation value, in this case the vacuum average (it could be thermal, if you are doing Euclidean field theory).

The formula you wrote is in momentum representation, if you do an inverse Fourier transform, namely integrate $k$, you get the real space correlation.

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Maybe I'm being foolish, but surely your answer only works in a quantum field theory. I was under the impression that Strassler was talking about an analogous quantity in classical field theory. –  AClassicalCaseOfConfusion Sep 21 '12 at 16:00
    
Hi Classical Case of Confusion, a simple way to see that you (and/or Matt) are foolish is to restore the normal units and see that the right hand side should really have an extra factor of $\hbar$ in the conventional units (because $\phi^2$ has units of action over $m^2$, think of a mass term, and the RHS is $1/m^2$ only as you wrote it), so it's surely no average in classical field theory. It's the 2-point function in quantum field theory - but this 2-point function may be evaluated by solving a problem (Green's function problem) in classical field theory. –  Luboš Motl Sep 21 '12 at 16:16
    
Incidentally, the expectation value of the product in the classical vacuum is clearly zero because $\phi(k)=0$ in the vacuum for all values of $k$. The nonzero value is proportional to $\hbar$ because it's a quantum effect, due to the uncertainty principle sort of. Also, if you considered the expectation value in a classical thermal state, the result would clearly be different and temperature-dependent. –  Luboš Motl Sep 21 '12 at 16:25
    
Thanks, Luboš. That's cleared it up nicely. –  user12418 Sep 21 '12 at 19:06
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