Proof that $\left \langle 0\mid \hat{S}\mid 0 \right \rangle = e^{\sum (\text{all connected vacuum diagrams})}$

Question

In my book on QFT (Lancaster & Blundell) they state the following equation, without proof: $$\left \langle 0\mid \hat{S}\mid 0 \right \rangle = e^{\sum (\text{all connected vacuum diagrams})}$$ where $\hat{S}$ is the S-matrix. This is what I tried: If we expand the expression above in terms using dyson's series, we get:$$\left \langle 0\mid \hat{S}\mid 0 \right \rangle = \left \langle 0\mid Te^{-i\int d^{4}z\hat{H}_{I}(z)}\mid0 \right \rangle$$ and then maclaurin expanding the exponential gives us the following:$$\left \langle 0\left | T \left [ 1-i\int d^{4}z\hat{H}_I(z)+\frac{(-i)^{2}}{2}\int d^{4}yd^{4}w\hat{H}_I(y)\hat{H}_I(w)+... \right ]\right | 0\right \rangle$$ I am not sure where to go from here, however. Can anyone help me complete this proof?

Answer 1

Note that each term in your MacLaurin$^\dagger$ series represents one or more diagrams. Since they are sandwiched between the vacuum $|0>$, the only nonzero diagrams are those with no external lines. In particular, there will be 'disconnected' diagrams where not all vertices are connected. If you agree that the series generates all the diagrams, then the linked cluster theorem tells you that the exponential of those are the connected ones (where connected means that all vertices in the diagram are connected).

$\dagger$ Most people would call it a Dyson series.

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