Quantum Field Theory – Completeness Relation in Peskin Chapter 7

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In P&S book, in Charpter 7: radiative corrections p. 212 （attached beneath, the completeness relation: (Equation 7.2)
$$\textbf{1}=\left\vert \Omega\right\rangle\left\langle\Omega\right\vert+\sum\limits_{\lambda}\int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_\textbf{p}} \left\vert \textbf{p}\right\rangle\left\langle\textbf{p}\right\vert.\tag{7.2}$$
This is in analogy with Equation 7.1:
$$(\textbf{1})_{1-particle}=\int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_\textbf{p}} \left\vert \textbf{p}\right\rangle\left\langle\textbf{p}\right\vert.\tag{7.1}$$
My question is:what exactly $\sum\limits_{\lambda}$ mean? P&S say: "the sum runs over all zero-momentum states $\left\vert \textbf{p}\right\rangle\\$. I don't really get this, does this mean we use $\lambda$ to denote different particles that are present in scattering processes? Such as: $\lambda=1$ denotes electron and $\lambda=2$ denotes positron?

The original Page 212 of P&S book:

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$|\lambda_0\rangle$ runs over all the single particle states at zero momentum. This will encode spin degrees of freedom -- so for a spin-1/2 particle, you'd sum over $m=1/2$ and $m=-1/2$ states, where $m$ is the spin quantum number (as well as antiparticle states).

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Quantum Field Theory – How to Derive Completeness Relation with Lorentz Invariant Measure

You just need to show that $1_\mathrm{1p}$ acts as the identity on one-particle states: $$ 1_\mathrm{1p}|\boldsymbol q\rangle=\left[\int\frac{d^3\textbf{p}}{(2\pi)^{3}}|\textbf{p}\rangle\frac{1}{2E_\textbf{p}}\langle\textbf{p}|\right]|\boldsymbol q\rangle=\int\frac{d^3\textbf{p}}{(2\pi)^{3}}|\textbf{p}\rangle\frac{1}{2E_\textbf{p}}\overbrace{\langle\textbf{p}|\boldsymbol q\rangle}^{(2\pi)^32E_p\delta(\boldsymbol p-\boldsymbol q)} $$ which indeed equals $|\boldsymbol q\rangle$, as one would expect for the identity operator. Moreover, using $\langle \boldsymbol p|\boldsymbol q_1,\boldsymbol q_2\cdots,\boldsymbol q_n\rangle=0$, we see that $(1)_\mathrm{1p}$ is orthogonal to multiparticle states.

In other words, $1_\mathrm{1p}$ is the identity when acting on one-parcle states, and it annihilates states with two or more particles. Therefore, it is the projector into the one-particle subspace.

In general, if your basis is normalised to $\langle \varphi|\varphi'\rangle=f(\varphi)\delta(\varphi-\varphi')$, then the identity is $$ 1=\int\frac{\mathrm d\varphi}{f(\varphi)}|\varphi\rangle\langle\varphi| $$ as can be seen by acting with $1$ on a general state $|\varphi'\rangle$, and checking that one indeed obtains $1|\varphi'\rangle=|\varphi'\rangle$.
The operator $1_\mathrm{1p}$ projects into the one-particle subspace. You can say that $1\approx 1_\mathrm{1p}$ if and only if you can neglect the multiparticle contribution. For an example, see Källén–Lehmann spectral representation: here, the one-particle subspace generates a pole at $p^2=m^2$. Therefore, as long as $p^2\sim m^2$, the multiparticle contribution is negligible as compared to the one-particle contribution: the latter generates a pole, while the former is regular. In this case, you can approximate $1\approx 1_\mathrm{1p}$. For other cases, there is no general rule: you can neglect the multiparticle contribution if and only if it the one-particle part dominates, which you can only conclude if you explicitly calculate both parts and compare them.
The projector into the $n$ particle subspace is straightforward: $$ 1_n=\frac{1}{n!}\int\frac{d^3\textbf{p}_1}{(2\pi)^{3}2E_\textbf{p}^1}\cdots \frac{d^3\textbf{p}_n}{(2\pi)^{3}2E_\textbf{p}^n}|\textbf{p}_1,\cdots,\boldsymbol p_n\rangle\langle\textbf{p}_1,\cdots\boldsymbol p_n| $$ as can be seen by acting on a general $n$ particle state $|\boldsymbol q_1,\cdots,\boldsymbol q_n\rangle$.

With this, the identity over the full Hilbert space is $$ 1=|0\rangle\langle 0|+\sum_{n=1}^\infty 1_n $$

[Physics] Expectation values in a quantum field theory

The idea is exactly the same as in single particle QM. Take the expectation value $\langle 0 | \hat\phi(x) | 0 \rangle$ and insert a complete set of field eigenstates $|\varphi\rangle$; these are the states that satisfy $\hat\phi(x) |\varphi\rangle = \varphi(x)|\varphi\rangle$. We do this with a path integral:

$$\langle 0 | \hat\phi(x) | 0 \rangle = \int \mathcal{D}\varphi\ \langle 0 | \hat\phi(x)|\varphi\rangle \langle\varphi | 0 \rangle = \int \mathcal{D}\varphi\ \langle 0 |\varphi\rangle \langle\varphi | 0 \rangle \varphi(x).$$

So your probability distribution is $p[\varphi] = |\langle\varphi | 0 \rangle|^2$, just like in regular QM. I'm not sure yet how to calculate this, but by analogy with the harmonic oscillator it would probably be along the lines of $\exp(-\varphi^2)$.

Edit: I think I've managed to calculate it, though I make no guarantees of this being correct. In particular I'm pretty sure there should be some $2\pi$'s around.

We proceed by analogy with the harmonic oscillator. You can find the ground state wavefunction by using the fact that $\langle x | \hat a | 0 \rangle = 0$; this gives you a differential equation for $\langle x | 0 \rangle = 0$. In much the same way, we write $\langle \varphi | \hat a_\mathbf{p} | 0 \rangle = 0$. We then express $\hat a_\mathbf{p}$ in terms of $\hat \phi$ and $\hat \pi$, and use the fact that acting on a wavefunctional $\hat \pi$ is $-i \delta / \delta \varphi(x)$. I did it following Peskin and Schroeder's conventions for normalizations and working in the Schrödinger picture, and arrived at the following equation (kx is the dot product $\mathbf{k} \cdot \mathbf{x}$):

$$ \int \mathrm{d}^3x\ e^{-ikx} \left( \omega_k \varphi(x) \langle\varphi | 0 \rangle + \frac{\delta}{\delta \varphi(x)}\langle\varphi | 0 \rangle \right) = 0$$

We will eventually express everything in terms of the Fourier transform $\hat \varphi$ of $\varphi$ (who knew, right?), so we will need to use the fact that

$$ \int \mathrm{d}^3x\ e^{-ikx} \frac{\delta}{\delta \varphi(x)}\langle\varphi | 0 \rangle = \frac{\delta}{\delta \hat \varphi} \langle\varphi | 0 \rangle \bigg|_{-k}$$

so our equation becomes

$$\frac{\delta}{\delta \hat \varphi (k)} \langle\varphi | 0 \rangle = - \omega_k \hat \varphi (-k) \langle\varphi | 0 \rangle$$

and we can guess the solution

$$\langle\varphi | 0 \rangle = \exp \left( -\frac12 \int \mathrm{d}^3k\ \omega_k \hat \varphi(k) \hat \varphi(-k) \right).$$

You could rewrite this in position space but the Fourier transform of $\omega_k$ involves Bessel functions so I don't know if it's worth it.

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Quantum Field Theory – How to Derive Completeness Relation with Lorentz Invariant Measure

[Physics] Expectation values in a quantum field theory

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