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I'm studying chapter 7 section 1 of Peskin and Schroeder. On page 212, we have the one particle Hilbert space $$\tag{7.1} (1)_{\text{1-particle}}=\int \frac{d^3p}{(2\pi)^3}\frac{1}{2E_p}|p\rangle\langle p|$$

From which Peskin and Schroeder derives the completeness relation for the entire Hilbert space (I guess this is the Fock space of the 1 particle space above?)

$$\tag{7.2} 1=|\Omega\rangle\langle\Omega|+\sum_{\lambda}\int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_p(\lambda)}|\lambda_p\rangle\langle \lambda_p|$$

here $|\lambda_0\rangle$ is an eigenvector of the energy $H$ with momentum zero, $|\lambda_p\rangle$ is the boost of $|\lambda_0\rangle$ with momentum $p$, and assume $|\lambda_p\rangle$ to be relativistically normalized. Let $E_p(\lambda)=\sqrt{|\boldsymbol{p}|^2+m_\lambda^2}$ where $m_\lambda$ is the mass of the state $|\lambda_p\rangle$, that is the energy of the state $|\lambda_0\rangle$

My questions are:

  1. How do we derive 7.2 from 7.1?
  2. By $|\lambda_0\rangle$, do we mean all eigenvectors of $H$? Does it include $|\Omega\rangle$ also? What does it mean for $|\lambda_p\rangle$ to be relativistically normalized?
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  1. How do we derive 7.2 from 7.1?

You can't - it is just there by analogy. 7.2 is the version of 7.1 for the hilbert space in QFT

  1. By $|\lambda_0\rangle$, do we mean all eigenvectors of $H$? Does it include $|\Omega\rangle$ also? What does it mean for $|\lambda_p\rangle$ to be relativistically normalized?

$|\lambda_0 \rangle$ includes all eigenstates of $H$ and $P$, namely the total energy and momentum of all particles in the system. It does not include $|\Omega \rangle \langle \Omega |$ which is pulled out of the sum separately for convenience and to make it clear that it is in fact included.

For the $|\lambda_0 \rangle$ to be "relativistically normalized", the entire identity acting on some $|\lambda'_{p'}\rangle$ should, through a delta function and proper cancellation of the factor including the energy in the integrand, give back the same $|\lambda'_{p'}\rangle$. For this to happen, we need the normalization of $|\lambda_{p}\rangle$ to depend on the energy. The reason it would be called relativistically normalized is that it cancels out with the factor which makes the integrand relativistically covariant, namely

$$\frac{d^3 p}{(2 \pi)^3 2E_p(\lambda)}$$

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