In David Tong's lecture notes on quantum field theory, at the top of page 38, we calculate the amplitude for a particle to propagate from $y$ to $x$:

$$\begin{align}\langle0|\phi(x)\phi(y)|0\rangle&=\int\frac{\text d^3p\text d^3p'}{(2\pi)^6}\frac{1}{\sqrt{4E_\vec pE_{\vec p'}}}\langle0|a_{\vec p}a^\dagger_{\vec p'}|0\rangle e^{-ip\cdot x+ip'\cdot y}\\ &=\int\frac{\text d^3p}{(2\pi)^3}\frac{1}{2E_{\vec p}}e^{-ip\cdot(x-y)}\\ &\equiv D(x-y). \tag{2.90} \end{align}$$

However if I take the field operators and expand them by hand I obtain:

$$\langle0|\phi(x)\phi(y)|0\rangle=\langle0|\int\frac{\text d^3p\text d^3p'}{(2\pi)^6}\frac{1}{\sqrt{4E_\vec pE_{\vec p'}}}(a_pe^{-ip\cdot x}a_{p'}e^{-ip'\cdot y} +a_{p}e^{-ip\cdot x}a_{p'}^\dagger e^{ip'\cdot y}\\+a_{p}^\dagger e^{ip\cdot x}a_{p'}e^{-ip'\cdot y}+a_{p}^\dagger e^{ip\cdot x}a_{p'}^\dagger e^{ip'\cdot y})|0\rangle.$$

So in equation 2.90 only the second term from above involving the creation/annihilation operators survives. But I don't see why those other terms vanish in what I've written. I would understand if annihilation operators were annihilating the vacuum and causing terms to vanish but the fourth term does not have an annihilation operator at all.

Also, why in equation 2.90 do we do away with the integral over $p'$ just before obtaining the final result?


1 Answer 1


Your first question can be answered by seeing that a creation operator acting on the bra-vacuum gives 0 because:
\begin{equation} 0= (a|0\rangle)^\dagger = \langle0|a^\dagger \end{equation} That's why we usually use the commutation relation to put all annihilation operators on the right and creation operators on the left.

For your second question, the matrix element basically creates a delta-function for the momentum:
\begin{equation} \langle0|a_p a_{p'}^\dagger|0\rangle = \langle p|p'\rangle = \delta(p-p') \end{equation} Or you could use the commutation relation to put $a_p$ on the right, eliminating the vacuum and obtain the same result.

  • $\begingroup$ Ah of course, thank you! I just tried using the commutation relation to obtain the same result, when we have a term $\langle0|\delta^{(3)}(\vec p-\vec q)|0\rangle$, am I correct to say this is just $\delta^{(3)}(\vec p-\vec q)$, since we pull it out as a "constant" and then use the normalisation of the vaccum to say $\langle0|0\rangle=1$? $\endgroup$
    – Charlie
    Sep 6, 2020 at 22:44
  • 1
    $\begingroup$ Glad I could help! And yes, that is correct since the delta-function is not an operator acting on the states. $\endgroup$
    – Guliano
    Sep 6, 2020 at 22:46
  • $\begingroup$ Great! All problems solved, thanks for your time :) $\endgroup$
    – Charlie
    Sep 6, 2020 at 22:47

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