Källén–Lehmann spectral representation

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The following derivation employs the mostly-minus metric signature.

In order to derive a spectral representation for the propagator of a field ${\displaystyle \Phi (x)}$, one considers a complete set of states ${\displaystyle \{|n\rangle \}}$ so that, for the two-point function one can write

${\displaystyle \langle 0|\Phi (x)\Phi ^{\dagger }(y)|0\rangle =\sum _{n}\langle 0|\Phi (x)|n\rangle \langle n|\Phi ^{\dagger }(y)|0\rangle .}$

We can now use Poincaré invariance of the vacuum to write down

${\displaystyle \langle 0|\Phi (x)\Phi ^{\dagger }(y)|0\rangle =\sum _{n}e^{-ip_{n}\cdot (x-y)}|\langle 0|\Phi (0)|n\rangle |^{2}.}$

Next we introduce the spectral density function

${\displaystyle \rho (p^{2})\theta (p_{0})(2\pi )^{-3}=\sum _{n}\delta ^{4}(p-p_{n})|\langle 0|\Phi (0)|n\rangle |^{2}}$.

Where we have used the fact that our two-point function, being a function of ${\displaystyle p_{\mu }}$, can only depend on ${\displaystyle p^{2}}$. Besides, all the intermediate states have ${\displaystyle p^{2}\geq 0}$ and ${\displaystyle p_{0}>0}$. It is immediate to realize that the spectral density function is real and positive. So, one can write

${\displaystyle \langle 0|\Phi (x)\Phi ^{\dagger }(y)|0\rangle =\int {\frac {d^{4}p}{(2\pi )^{3}}}\int _{0}^{\infty }d\mu ^{2}e^{-ip\cdot (x-y)}\rho (\mu ^{2})\theta (p_{0})\delta (p^{2}-\mu ^{2})}$

and we freely interchange the integration, this should be done carefully from a mathematical standpoint but here we ignore this, and write this expression as

${\displaystyle \langle 0|\Phi (x)\Phi ^{\dagger }(y)|0\rangle =\int _{0}^{\infty }d\mu ^{2}\rho (\mu ^{2})\Delta '(x-y;\mu ^{2})}$

where

${\displaystyle \Delta '(x-y;\mu ^{2})=\int {\frac {d^{4}p}{(2\pi )^{3}}}e^{-ip\cdot (x-y)}\theta (p_{0})\delta (p^{2}-\mu ^{2})}$.

From the CPT theorem we also know that an identical expression holds for ${\displaystyle \langle 0|\Phi ^{\dagger }(x)\Phi (y)|0\rangle }$ and so we arrive at the expression for the time-ordered product of fields

${\displaystyle \langle 0|T\Phi (x)\Phi ^{\dagger }(y)|0\rangle =\int _{0}^{\infty }d\mu ^{2}\rho (\mu ^{2})\Delta (x-y;\mu ^{2})}$

where now

${\displaystyle \Delta (p;\mu ^{2})={\frac {1}{p^{2}-\mu ^{2}+i\epsilon }}}$

a free particle propagator. Now, as we have the exact propagator given by the time-ordered two-point function, we have obtained the spectral decomposition.