I'm trying to understand the derivation of the Kallen-Lehmann representation given in Peskin & Schroeder (pages 211-214). I would really appreciate if anyone on here could answer a few questions I have about this derivation.

At one point they insert the identity into a propagator. The one-particle identity is given by \begin{equation} \mathbf{1} = \int \frac{d^3p}{(2\pi)^3}\frac{1}{2E_{\mathbf{p}}}|\mathbf{p}\rangle \langle \mathbf{p}|. \end{equation} So, I think in principle we could write the interacting identity as \begin{equation} \mathbf{1} = \sum_{n=0}^{\infty}\int \frac{d^3p}{(2\pi)^3}\frac{1}{2E}|\mathbf{p}_1 ... \mathbf{p}_n\rangle \langle \mathbf{p}_1 ... \mathbf{p}_n|, \end{equation} where $E$ is the total system's energy. Correct me if this is wrong.

However, Peskin & Schroeder write this in terms of the states $|\lambda_{\mathbf{p}}\rangle$, which are Lorentz boosts of the state $|\lambda_{0}\rangle$, such that $\mathbf{P} |\lambda_{0}\rangle = 0$.

My questions are: is $|\lambda_{0}\rangle$ a state with arbitrarily many particles in it?

And why introduce $|\lambda_{0}\rangle$ in the first place? Is it just to simplify the expression for the spectral function? Or does it have some physical meaning?



The notation $|\lambda_0\rangle$ is a bit misleading. It simply means a generic eigenstate of the full interacting Hamiltonian, with zero spatial momentum $\vec p$. All other eigenvalues are lumped into a single label $\lambda$, so you may as well imagine $\lambda=\{n_1,n_2,\ldots\}$ where $n_i$ are indices labelling distinct states/eigenvalues. The only eigenvalues being made explicit are the eigenvalues of $\vec P$, i.e. the state $|\lambda_0\rangle$ has eigenvalue $\vec P=0$, and the state $|\lambda_{\mathbf{p}}\rangle$ has eigenvalue $\vec P = \mathbf p$. This all leads to figure 7.2 on page 214 of P&S.

Your multi-particle identity operator doesn't quite work because it only works for the non-interacting case. The states you wrote are only eigenstates of the non-interacting Hamiltonian.

The point of introducing notation for an arbitrary state having total spatial momentum, $|\lambda_{\mathbf{p}}\rangle$, is that the Källén–Lehmann spectral representation specifically selects those states that have zero spatial-momentum $\mathbf p = 0$. The authors are utilizing the (trivial) fact that, any state at all can be transformed by Lorentz boost into a state with zero spatial momentum.

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    $\begingroup$ > The authors are using the (trivial) fact that, any state at all can be transformed by Lorentz boost into a state with zero spatial momentum. At risk of being very pedantic, I should mention that this is not true for "any state at all". It's true for momentum eigenstates. A state without definite momentum cannot be transformed into a state with zero momentum by a Lorentz boost. $\endgroup$
    – Zack
    Jul 6 at 4:58
  • $\begingroup$ @Zack Fair point. I meant "any state at all having definite momentum $\vec p$ ...", but I see how that could be missed. $\endgroup$ Jul 7 at 3:11

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