One-Particle Operators in QFT from the book “Condensed Matter Field Theory”

In "Condensed Matter Field Theory" by Alexander Altland and Ben Simons there is this derivation of one-particle Operators in the formalism of second quantization:

Let us now consider a one-body operator, $\hat{O}_1$, which is diagonal in the one-particle state basis $\{ {\lambda_i} \}$, with $\hat{O}_1 = \sum o_{\lambda_i}|\lambda_i><\lambda_i|$, $o_{\lambda_i}=<\lambda_i|\hat{O}_1|\lambda_i>$. With this definition, one finds that \begin{equation} <n'_{λ_1},n'_{λ_2},...|\hat{O}_1|n_{λ_1},n_{λ_2},...> = \sum_i > o_{\lambda_i} n_{λ_i} <n'_{λ_1},n'_{λ_2},...|n_{λ_1},n_{λ_2},...> = \end{equation} \begin{equation} <n'_{λ_1},n'_{λ_2},...|\sum_i > o_{\lambda_i} \hat{n}_{λ_i}|n_{λ_1},n_{λ_2},...> \end{equation} Since this equality holds for any set of states, one can infer the second quantized representation of the operator $\hat{O}_1$, \begin{equation} \hat{O}_1 = \sum o_{λ_i} \hat{n}_{λ_i} = \sum > o_{λ_i} a^†_{λ_i}a_{λ_i} \end{equation}

Honestly I can't get what they are doing, it seems to me that any inner product of a multiparticle state with a one-particle state like $|\lambda>$ should be equal to zero so I don't understand how does he gets rid of all the $|\lambda_i><\lambda_i|$.

(This is essentially the same question as A question regarding to the one-body operators in N-particle Hilbert space but I tried to write down Altland calculations to help to figure out the answer.)

I believe you are confused by the notation. Here is an example of a multi particle state: $$|\lambda_1, \lambda_1, \lambda_2, \lambda_3, \lambda_3\rangle = |\lambda_1\rangle \otimes |\lambda_1\rangle \otimes |\lambda_2\rangle \otimes |\lambda_3 \rangle \otimes |\lambda_3\rangle\ \ \ \ \ \ (1)$$ In occupation number representation, we write this as $|2,1,2\rangle$. Consider the one-particle operator $\hat{O}$, which acts in Hilbert space:
$$\hat{O} = \sum_{i}O_{\lambda_i}|\lambda_i\rangle\langle\lambda_i|$$
Now, the one body operator (in Fock space) $\mathcal{O}_1$ is given, in terms of $\hat{O}$ by $\mathcal{O}_1 = \sum_i \langle \lambda_i| \hat{O}|\lambda_i\rangle \hat{n}_{\lambda_i}$, thus, when applied to the multi particle state above, we get $$\langle\lambda_3, \lambda_3, \lambda_2, \lambda_1, \lambda_1| \mathcal{O}_1|\lambda_1, \lambda_1, \lambda_2, \lambda_3, \lambda_3\rangle =\langle2,1,2|\mathcal{O}_1|2,1,2\rangle \\= \sum_i \langle \lambda_i| \hat{O}|\lambda_i\rangle \langle2,1,2|\hat{n}_{\lambda_i}|2,1,2\rangle$$
• Altland says that the multi particle version of the operator ($\mathcal{O}_1$) is the sum of the action of the single-particle operator ($\hat{O}$) in each one of the single-particle states of the multi-particle state: $\langle 2,1,2|\mathcal{O}_1|2,1,2\rangle = \sum_i \langle \lambda_i | \hat{O} | \lambda_i \rangle \langle \lambda_3, \lambda_3, \lambda_2, \lambda_1, \lambda_1 | \lambda_i\rangle \langle \lambda_i | \lambda_1, \lambda_1, \lambda_2, \lambda_3, \lambda_3 \rangle$ I think he starts from this supposition. – Patrick Apr 24 '17 at 22:01