In Peskin and Schroder's QFT book, on page 212, eq.7.2, they use the completeness relation in a derivation involving the two-point correlation function: $$ \mathbf{1}=|\Omega\rangle\langle\Omega|+\sum_{\lambda} \int \frac{d^{3} p}{(2 \pi)^{3}} \frac{1}{2 E_{\mathbf{p}}(\lambda)}\left|\lambda_{\mathbf{p}}\right\rangle\left\langle\lambda_{\mathbf{p}}\right| \ \ \ \ \ \ \ \ \ \ \ \ \ (7.2) $$ I am really troubled about following points:

  1. Why is there a term $|\Omega\rangle\langle\Omega|$? How can the completeness relation of $\mathbf{1}$ have two terms?

  2. Is the summation over $\lambda$ in the second term about summation for different particles with the same mass $m$?

If you have any comment or answer, I would really appreciate it!

  • $\begingroup$ Isn't the Omega part the time part? The integral is the integral for one particle between two points? A wavepacket? $\endgroup$
    – Gerald
    Aug 18, 2022 at 10:34

1 Answer 1


I don't feel that I have a mastery of the equation in your question, but I have some things to offer that can at least help.

In QM, if we resolve the single-particle identity in the momentum basis we see

$$1 = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} |\vec{p} \rangle \langle \vec{p}| d^3p $$

We want something similar in QFT. But the hilbert space in QFT includes states with 0 particles, or 1 particle, or 2 particles, etc. So we need more terms here, which account for states like $|\Omega \rangle$ or $|p_1 p_2 \rangle$. We also want Lorentz covariance, so we need to use a different measure. The general idea is to write something like

$$1 = |\Omega \rangle \langle \Omega | + \int_{-\infty}^{\infty} |p \rangle \langle p| \frac{d^3p}{(2\pi)^3 2E_p} + (\text{2-particle identity}) + ... \tag{1}$$

Then we have covered all states. Because otherwise, the identity acting on for example $|\Omega \rangle$ or $|p_1 p_2 \rangle$ would just give zero! Which is wrong.

The equation you wrote is in the spirit of Eq. (1) above. But, there are deeper reasons why there are more complications than what we have established so far. Namely, an identity operator resolution like (1) uses states $|p \rangle $ constructed from creation/annihilation operators that come from the solution to the equation of motion of the free theory, without interacting terms:

$$a^\dagger(p)|\Omega \rangle = |p \rangle$$

but in an interacting theory, we may not have the momentum eigenstates from the free theory. Generally, the hilbert space of the interacting theory is not the same. For this reason, the use of the label $\lambda_p$ leaves more leeway. Taking a look at the equation you linked,

$$ \mathbf{1}=|\Omega\rangle\langle\Omega|+\sum_{\lambda} \int \frac{d^{3} p}{(2 \pi)^{3}} \frac{1}{2 E_{\mathbf{p}}(\lambda)}\left|\lambda_{\mathbf{p}}\right\rangle\left\langle\lambda_{\mathbf{p}}\right| $$

The integral here is over the total momentum of all particles combined, and $|\lambda_p \rangle$ is a multi-particle state with total momentum $\vec{p}$. The "mysterious" part to me is the sum over $\lambda$, and even the professor with whom I took QFT could not tell me exactly what this sum consisted of. But in principle, it should sum over all states - of any particle number - which have total momentum $\vec{p}$, and for which the total four-momentum operator has eigenvalue

$$P^2 |\lambda_p \rangle = m_\lambda^2 |\lambda_p \rangle$$

Such that, then, $E_p$ is defined as

$$E_p = \sqrt{m_\lambda^2 + \vec{p}^2}$$

For that reason $m_\lambda$ is not the mass of any individual particle but just a defined label for this state in general.

  • $\begingroup$ Thank you very much for your detailed answer! So the identity operator need to be a sum over a complete set of states, and we need to add $|\Omega\rangle\langle\Omega|$ for vacuum state. So in this case identity operator act on them give non-zero. $\endgroup$
    – Daren
    Aug 18, 2022 at 14:59
  • $\begingroup$ Could you please elaborate more why we introduce multiple particle Hilbert space here? I just notice we need to insert this identity operator to the two-point function, how will multiple particle act here? $\endgroup$
    – Daren
    Aug 18, 2022 at 15:02
  • 2
    $\begingroup$ We need a multi-particle hilbert space because QFT is a theory that can accommodate for multiple particles. For example in the scattering of two electrons a huge number of particles can come out depending on the kinetic energy of the electrons. As for the use of this 2pt function, it is just a step in a derivation, I don't know of some physical interpretation. I think he is trying to derive the Källen-Lehmann spectral representation of the 2pt function. $\endgroup$ Aug 18, 2022 at 15:08
  • 1
    $\begingroup$ Thank you very much! $\endgroup$
    – Daren
    Aug 19, 2022 at 0:38
  • $\begingroup$ Glad to help! -- $\endgroup$ Aug 19, 2022 at 2:43

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