Projection operator in second quantization

Question

Let our system is defined on some complete orthogonal basis $\{|n\rangle \}$. I know that in first quantization a projection operator is defined as $P_n=|n\rangle \langle n|$ which when applied on some arbitrary state $|\alpha\rangle=\sum_n a_n|n\rangle$ projects out the contribution of $|n\rangle$ to $|\alpha \rangle$. For example, let some operator, say velocity $\hat{v}$ is written as $\hat{v}=\sum_{nm} v_{nm} |n\rangle \langle m|$ with $ v_{nm}=\langle n|\hat{v}|m\rangle$. When we apply $P_n$ we get contribution of vector $|n\rangle$ in $\hat{v}$.

How we translate all this in second quantization formalism where we have field operators? Basically, I have an operator $\hat{v}$ given as in k-space $$ \hat{v} = \sum_k v_{aa} \hat{a}_k^+\hat{a}_k + v_{ab} \hat{a}_k^+\hat{b}_k + v_{ba} \hat{b}_k^+\hat{a}_k + v_{bb} \hat{b}_k^+\hat{b}_k $$ with $v_{ij}$ are some numbers. Now, I want to calculate the contribution of the $a_k^+$ operator only. So, I want to built a projection operator (similar to first quantizationformalism). How do I do that in second quantization?

Answer 1

You can get the 2nd quantization form (that means, the generalization of a single particle operator to the multiparticle system) using the following rule

$$A^{(mult)}=\sum_{nm} \langle n|A^{(single)}|m\rangle a^{\dagger}_n a_m $$

Therefore, the projection operator into a state $k$ that you're looking for is just $a^{\dagger}_k a_k$. Keep in mind that it's action on a symmetrized manybody state composed of single particle states $\psi_n$

$$ |\Psi \rangle=S\big(|\psi_1\rangle |\psi_2\rangle|\psi_3\rangle...\big)$$

is given by

$$ a^{\dagger}_k a_k |\Psi \rangle= S\big(\langle k|\psi_1\rangle |k\rangle |\psi_2\rangle|\psi_3\rangle...\big)+S\big(|\psi_1\rangle \langle k|\psi_2\rangle|k\rangle |\psi_3 \rangle...\big)+...$$