General expression for fermionic creation operator subspace

Question

This is perhaps a particular case of the question discussed here.

Given a fermionic Fock space $H$ of dimension $2^N$, that is, with $N$ fermionic modes, let $H_n$ be the subspace of states with $n$ occupied fermions, where $n<N$. Let $\hat{c}^\dagger_i$ be the creation operator for the fermionic mode $i$.

Question: Can all operators $\hat{a}^\dagger$ that map $H_n \rightarrow H_{n+1}$ be written in the form \begin{align} \hat{a}^\dagger &= \sum_{i_1} A_{i_1} \hat{c}^\dagger_{i_1} + \sum_{i_1 i_2 j_1} A_{i_1 i_2 j_1} \hat{c}^\dagger_{i_1} \hat{c}^\dagger_{i_2} \hat{c}_{j_1} + ~ ... ~ + \sum_{i_1...i_{N} j_1 ... j_{N-1}} A_{i_1 i_2 j_1} \hat{c}^\dagger_{i_1} ... \hat{c}^\dagger_{i_N} \hat{c}_{j_1} ... \hat{c}_{j_{N-1}} \text{?} \end{align} You may consider some normal ordering assumption on the coefficients $A_{i_1...i_m j_1...j_{m-1}} \in \mathbb{C}$.

As a bonus question, can we therefore associate such an operator to the quasiparticles of any number conserving fermionic Hamiltonian?

I am looking for a formal proof or a discussion with pointing to a proof in the literature.

Answer 1

Any operator $\hat{a}^\dagger$ that maps $H_n \rightarrow H_{n+1}$ for all $n$, with $H_{N+1} \equiv 0$, is defined by set of matrix elements $$ B^{(n)}_{i_1,\ldots,i_{n+1};j_1,\ldots,j_n} = \langle 0|\hat{c}_{i_{n+1}}\ldots\hat{c}_{i_1}\ \hat{a}^\dagger\ \hat{c}^\dagger_{j_1}\ldots \hat{c}^\dagger_{j_n}|0\rangle $$ Set of vectors $\hat{c}^\dagger_{j_1}\ldots \hat{c}^\dagger_{j_n}|0\rangle$ with $n = 0,1,\ldots,N$, $1\leq j_1<j_2<\ldots < j_n\leq N$ is the orthonormal basis in fermionic Fock space. Hence the operator $\hat{a}^\dagger$ can be written as $$ \hat{a}^\dagger = \sum_{n=0}^{N-1}\sum_{1\leq i_1<\ldots <i_{n+1}\leq N} \sum_{1\leq j_1<\ldots <j_{n}\leq N}B^{(n)}_{i_1,\ldots,i_{n+1};j_1,\ldots,j_n} \hat{c}^\dagger_{i_1}\ldots \hat{c}^\dagger_{i_{n+1}}|0\rangle\langle 0|\hat{c}_{j_{n}}\ldots\hat{c}_{j_1}\tag{a} $$

It is known that the operator of projection on fermionic vacuum $\hat{P}_0 \equiv |0\rangle\langle 0|$ can be written in the form $$ \hat{P}_0 =\ :\!\exp\left(-\sum_{k=1}^N\hat{c}_k^\dagger\hat{c}_k\right)\!:\ = \sum_{l = 0}^N\frac{(-1)^l}{l!}\sum_{k_1,\ldots,k_l=1}^N \hat{c}_{k_1}^\dagger\ldots\hat{c}_{k_l}^\dagger \hat{c}_{k_l}\ldots\hat{c}_{k_1}.\tag{P0} $$ I believe that this formula should be derived in Berezin's book on the method of second quantization. Or one can check it by using following equalities $$ \sum_{k_1,\ldots,k_l=1}^N \hat{c}_{k_1}^\dagger\ldots\hat{c}_{k_l}^\dagger \hat{c}_{k_l}\ldots\hat{c}_{k_1} = \hat{n}_f(\hat{n}_f-1)\ldots(\hat{n}_f-l+1) = \frac{\hat{n}_f!}{(\hat{n}_f-l)!}, $$ where $\hat{n}_f = \sum_{k=1}^N\hat{c}_k^\dagger\hat{c}_k$ is the fermion number operator. It is also easy to check the validity of the following equalities $$ \hat{c}^\dagger_{i_1}\ldots \hat{c}^\dagger_{i_{n+1}}\hat{P}_0 = \hat{c}^\dagger_{i_1}\ldots \hat{c}^\dagger_{i_{n+1}}\sum_{l = 0}^{N-n-1}\frac{(-1)^l}{l!}\sum_{k_1,\ldots,k_l=1}^N \hat{c}_{k_1}^\dagger\ldots\hat{c}_{k_l}^\dagger \hat{c}_{k_l}\ldots\hat{c}_{k_1},\tag{P1}. $$

Now, substituting (P1) into (a) gives the equality $$ \hat{a}^\dagger = \sum_{n=0}^{N-1}\ \sum_{l = 0}^{N-n-1}\frac{(-1)^l}{l!}\sum_{1\leq i_1<\ldots <i_{n+1}\leq N} \sum_{1\leq j_1<\ldots <j_{n}\leq N}\sum_{k_1,\ldots,k_l=1}^N B^{(n)}_{i_1,\ldots,i_{n+1};j_1,\ldots,j_n} $$ $$ \hat{c}^\dagger_{i_1}\ldots \hat{c}^\dagger_{i_{n+1}}\hat{c}_{k_1}^\dagger\ldots\hat{c}_{k_l}^\dagger \hat{c}_{k_l}\ldots\hat{c}_{k_1}\hat{c}_{j_{n}}\ldots\hat{c}_{j_1} $$ that has exactly the desired form.

Answer 2

Yes, you can. To prove this, let us first define what we mean by a general fermionic creation operator.

Let $f_n: \mathcal{H}_n \to \mathcal{H}_{n+1}$ be a linear map from the Hilbert space of $n$ fermions to $n+1$ fermions. A general fermion creation operator would be defined as $$ f = f_0 \oplus f_1 \oplus f_2 \oplus \dots \oplus f_{N-1} $$ for arbitrary choices of $f_n$.

Next, we want to prove that any such $f$ can be expressed in the form you described, that is, $$ f = \sum_{i_1} A_{i_1} c^\dagger_{i_1} + \sum_{i_1,i_2,j_1} A_{i_1,i_2,j_1} c^\dagger_{i_1}c^\dagger_{i_2}c_{j_1} + \dots $$ I will just sketch the proof.

1. Look a general operator of the form $\sum_{i_1,\dots ,i_{n+1},j_1,\dots, j_n} A_{i_1\dots i_{n+1}j_1\dots j_n} c^\dagger_{i_1}\dots c^\dagger_{i_{n+1}}c_{j_1}\dots c_{j_n} $. Show that for any $f_n$, there is a choice for the matrix $A$ such that the action of that operator on $\mathcal{H}_n$ is the same as $f_n$. You can show this constructively by deriving the required matrix elements of $A$ by looking at the matrix elements of $f_n$. Let us call this operator $g_n$. Note that $g_n \neq f_n$ as it also has nontrivial action on $\mathcal{H}_k, k>n$.

2. Let $k>n$. The action of $g_n$ on $\mathcal{H}_k$ can be cancelled by adding a counterterm $$ g_{k,n}(g_n)=-\sum_{i_1,\dots ,i_{n+1},j_1,\dots, j_n} A^{(g_n)}_{i_1\dots i_{n+1}j_1\dots j_n} c^\dagger_{i_1}\dots c^\dagger_{i_{n+1}} \bigg(\sum_{m_1,\dots,m_{k-n}}c^\dagger_{m_1}\dots c^\dagger_{m_{k-n}}c_{m_{k-n}}\dots c_{m_1}\bigg)c_{j_1}\dots c_{j_n} $$

3. Define $h_0 = g_0, h_n = g_n + \sum_{k=0}^{n-1}g_{n,k}(h_k)$.

4. Show that $f = \sum_{n=0}^{N-1}h_n$. Each of the $h_n$ are manifestly in the form you require.

This constructively shows that for any $f_n$, there exists a choice of $A_{i_1}, A_{i_1,i_2,j_1},\dots$ such that $$ f = f_0 \oplus f_1 \oplus f_2 \oplus \dots \oplus f_{N-1} = \sum_{i_1} A_{i_1} c^\dagger_{i_1} + \sum_{i_1,i_2,j_1} A_{i_1,i_2,j_1} c^\dagger_{i_1}c^\dagger_{i_2}c_{j_1} + \dots. $$

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Explanation of the direct sum notation:

I will give an explicit explanation of the direct sum notation for $N=2$. The Fock space is spanned by $$ \vert{00}\rangle,\vert{01}\rangle,\vert{10}\rangle,\vert{11}\rangle $$ Let us choose $f_0(\vert{00}\rangle) = a \vert{01}\rangle + b \vert{10}\rangle$ and $f_1(\vert{01}\rangle) = c\vert{11}\rangle $, $f_1(\vert{10}\rangle) = d\vert{11}\rangle$. The operator $f_0 \oplus f_1: \mathcal{H}_0\oplus \mathcal{H}_1 \to \mathcal{H}_1\oplus \mathcal{H}_2$ can be naturally extended as an operator $f:\mathcal{H} \to \mathcal{H}$ by giving it zero action on the $\mathcal{H}_2$ sector. In matrix form, we write $$ f = \begin{pmatrix}0 & 0& 0 & 0 \\ a & 0& 0 & 0 \\ b & 0& 0 & 0 \\ 0 & c& d & 0 \end{pmatrix} $$ in the occupation number basis listed earlier. Thanks to @hft to helping me clear up the sloppy notation.