$\bullet$ 1. For the one-particle states, the completeness relation is given in Peskin and Schroeder, $$(\mathbb{1})_{1-particle}=\int\frac{d^3\textbf{p}}{(2\pi)^{3}}|\textbf{p}\rangle\frac{1}{2E_\textbf{p}}\langle\textbf{p}|$$ How does one derive this?

$\bullet$ 2. In expanding a general Fock state $|\Psi\rangle$ as $|\Psi\rangle=\mathbb{1}|\Psi\rangle$ can we approximate the identity operator as $$\mathbb{1}\approx(\mathbb{1})_{1-particle}?$$ If yes, when?

$\bullet$ 3. If not, how does one derive similar completeness relations for 2-particle states $|\textbf{p}_1,\textbf{p}_2\rangle$ (or in general, for $N$-particle states $|\textbf{p}_1,\textbf{p}_2...|\textbf{p}_N\rangle$)?

  1. 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 $1|\varphi'\rangle=|\varphi'\rangle$.

  2. 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.

  3. 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 $$ 1=|0\rangle\langle 0|+\sum_{n=1}^\infty 1_n $$

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  • $\begingroup$ Note also that in gauge theories, the sum over all states includes a sum over ghost states too, cf. this PSE post. $\endgroup$ – AccidentalFourierTransform Jan 31 '18 at 15:39
  • $\begingroup$ may I ask what the purpose of the notation for the bold calligraphic momenta is intended to communicate? $\endgroup$ – InertialObserver Apr 12 '19 at 19:15
  • $\begingroup$ @InertialObserver cf. wikipedia on vector notation. $\endgroup$ – AccidentalFourierTransform Apr 18 '19 at 20:06

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