# Representation of a general unitary operator acting on all electrons in terms of creation and annihilation operators

My goal is to find the representation of a general unitary operator acting on all electrons in terms of creation and annihilation operators. Suppose a set of single-particle basis functions $\{\psi_i:i=1,2,...\}$ is connected to another basis functions $\{\phi_i:i=1,2,...\}$ by a unitary transformation $$|\psi_i\rangle = e^A|\phi_i\rangle = \sum_j (e^A)_{ji} |\phi_j\rangle$$ where $A$ is antihermitian to ensure the unitarity and $(e^A)_{ji} = \langle \phi_j | e^A |\phi_i\rangle$. A Slater determinant $|\psi_1\psi_2\ldots\psi_N\rangle$ for $N$ electrons is then obtained from the determinant $|\phi_1\phi_2\ldots\phi_N\rangle$ by $$|\psi_1\psi_2\ldots\psi_N\rangle = \prod_{k=1}^N e^{A(k)}|\phi_1\phi_2\ldots\phi_N\rangle$$ where $e^{A(k)}$ acts on the k-th particle. My goal is to find the representation for $\prod_{k=1}^N e^{A(k)}$ in terms creation and annihilation operators. My proposal is the following, let's take an example of two-electron Slater determinant $|\psi_a\psi_2\rangle = e^{A(1)}e^{A(2)}|\phi_a\phi_b\rangle$, then my hypothesis is $$e^{A(1)}e^{A(2)} = \frac{1}{2} \sum_{ij} \langle \phi_i | e^{A(1)} |\phi_j\rangle c_i^\dagger c_j \sum_{kl} \langle \phi_k | e^{A(2)} |\phi_l\rangle c_k^\dagger c_l \hspace{4cm}(1)$$ where $c_i^\dagger$ and $c_i$ are creation and annihilation operators for the set $\{\phi_i:j=1,2,...\}$. For the set $\{\psi_i:i=1,2,...\}$, $a_i^\dagger$ and $a_i$ are used.

PROOF:

Each of $e^{A(k)}$ is single-particle by nature, therefore its second quantization representation is $e^{A(k)} = \sum_{ij} \langle \phi_i | e^{A(k)} |\phi_j\rangle c_i^\dagger c_j$. Using this for $e^{A(1)}$ and $e^{A(2)}$ and let $e^{A(1)}e^{A(2)}$ acts on an arbitrary determinant $|\phi_a\phi_b\rangle$ one obtains $$\sum_{j} \left(\sum_{i} \langle \phi_i | e^{A(1)} |\phi_j\rangle c_i^\dagger \right) c_j \sum_{l} \left( \sum_{k} \langle \phi_k | e^{A(2)} |\phi_l\rangle c_k^\dagger \right) c_l |\phi_a\phi_b\rangle$$ Now by the definition of $e^A$, the terms in the first and second parentheses are $a_j^\dagger$ and $a_l^\dagger$ respectively. So that we have \begin{aligned} \sum_{j} a_j^\dagger c_j \sum_{l} a_l^\dagger c_l |\phi_a\phi_b\rangle &= \sum_{j} a_j^\dagger c_j (a_a^\dagger c_a + a_b^\dagger c_b) |\phi_a\phi_b\rangle \\ &= \sum_{j} a_j^\dagger c_j (|\psi_a\phi_b\rangle + |\phi_a\psi_b\rangle)\\ &= 2|\psi_a\psi_b\rangle \end{aligned} Since there is a factor of two in the end, we must add 1/2 in the representation of our total unitary operator giving rise to eq. (1). It can be seen that the above proof extends to a general $N$ electrons for which the representation is $$\prod_{k=1}^N e^{A(k)} = \frac{1}{N!} \sum_{ij} \langle \phi_i | e^{A(1)} |\phi_j\rangle c_i^\dagger c_j \ldots \sum_{kl} \langle \phi_k | e^{A(N)} |\phi_l\rangle c_k^\dagger c_l$$ Can someone verify my proof or whether my hypothesis is wrong to begin with? Regardless of whether I am right or wrong I am thankful if you can also give me reference that shows this representation because I have not been able to find such references. Moreover, the all-particle unitary operator is of multiplicative type (i.e. not that of additive type whose second quantization representation are most commonly considered in textbooks), does my proof also extend to a general multiplicative operators such as projection operator?