# Kallen-Lehmann representation derivation

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 $$$$\mathbf{1} = \int \frac{d^3p}{(2\pi)^3}\frac{1}{2E_{\mathbf{p}}}|\mathbf{p}\rangle \langle \mathbf{p}|.$$$$ So, I think in principle we could write the interacting identity as $$$$\mathbf{1} = \sum_{n=0}^{\infty}\int \frac{d^3p}{(2\pi)^3}\frac{1}{2E}|\mathbf{p}_1 ... \mathbf{p}_n\rangle \langle \mathbf{p}_1 ... \mathbf{p}_n|,$$$$ where $$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?

Thanks!

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.
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.
• @Zack Fair point. I meant "any state at all having definite momentum $\vec p$ ...", but I see how that could be missed. – Arturo don Juan Jul 7 at 3:11