Suppose I have a fermionic Fock space $H$ of dimension $2^n$. If I fix an operator $O$ acting on $H$ that commutes with the number operator $N$, I typically make an assumption internally that such an operator can be written as a normal ordered polynomial of the creation/annihilation operators in the following way: $$O = k + \sum_{ij} c_{ij} a_i^\dagger a_j + \sum_{ijkl} c_{ijkl}a_i^\dagger a_j^\dagger a_k a_l + \cdots + \sum_{i_1\ldots i_n j_i\ldots j_n} c_{i_1\cdots i_n j_1 \cdots j_n} a_{i_1}^\dagger\cdots a_{i_n}^\dagger a_{j_1}\cdots a_{j_n}.$$ For an arbitrary (not necessarily number conserving operator), I also assume that it can be written in the above form, but with number broken terms included (so, terms like $a_i$ and $a_i^\dagger a_j a_k$ would be included also). For example, fermionic Hamiltonians and generators for unitaries for systems of physical interest (like: the electronic Hamiltonian, Hubbard model, Jaynes-Cummings model, ...), are always presented in the above form as far as I can tell.

I am wondering if my assumption that all Fock space operators can be written in this form is a correct assumption. This leads to two questions I have regarding the correspondance between linear operators $O \in \mathcal{L}(H)$, and formal normal ordered polynomials of the creation/annhiliation operators.

  1. Given an arbitary operator $O$, does there exist a collection of coefficients such that $O$ can be written as a normal ordered polynomial of fermionic operators (including potentially the number breaking terms if $[O, N] \not=0$).

  2. Is the association of operators to normal ordered polynomials well-defined? What I mean is: is there a "canonical" way to associate, to each operator, a normal ordered polynomial?

My thoughts on these questions are as follows. For question 1, it is sufficient to answer the question with the "normal ordered" condition dropped, since any polynomial can be brought to normal ordered form by successive applications of the canonical anticommutation relations. Moreover, there seems to be an implicit assumption that once the second quantization procedure has been done, any operator can be written as a polynomial of fermionic operators, but I am not entirely sure why that is the case.

For question 2, I have seen it written in passing that there is a unique normal ordered polynomial associated to each operator, but I have no idea why this is the case. There are also technical issues regarding the fact that certain coefficients with repeated indices in the polynomial don't affect the operator associated to a polynomial, because of the fermionic anticommutation relations. For example any choice of $c_{iikl}$ will not affect the operator since $a_i^\dagger a_i^\dagger a_k a_l = 0$. If it is possible to associate each operator to a unique or "canonical" polynomial, how does one take into account potential issues like this one?


I will try to answer your question 1. It will also give an (extremely ineffective) solution to your question 2.

Let me start from the following observation: For any finite dimensional Hilbert space $H$, one can show that $\mathcal{L}(H)$ is also a Hilbert space of dimension $\text{dim}(H)^2$. Suppose $\{|{\alpha}\rangle\}$ is an orthonormal basis for $H$, then a basis for $\mathcal{L}(H)$ is just $|\alpha\rangle\langle \beta|$.

You can already see that the counting works in the Fock space case: for each fermionic mode, there are four independent operators: $1, a_i, a_i^\dagger, a_i^\dagger a_i$. Taking the product of such operators from different modes gives $4^n$ linearly-independent operators, which is the expected dimension of $\mathcal{L}(H)=2^{2n}$. Then any such product can be expressed as a sum of normal-ordered monomials of $a$'s and $a^\dagger$'s, using canonical commutation relation.

Now I can give a more formal proof. We can work with the occupation number basis of the Hilbert space, i.e $|m_1,m_2,\dots, m_n\rangle$, which will be abbreviated as $|\{m\}\rangle$. Then a complete basis for the operators in this Hilbert space is $|\{m\}\rangle\langle \{m'\}\rangle$. This can be written as a bunch of creation/annihilation operators multiplying (in some proper order) on the projector $|\{0\}\rangle\langle \{0\}|=\prod_i a_ia_i^\dagger$. So this operator $|\{m\}\rangle\langle \{m'\}\rangle$ is a polynomial in the $a$'s and $a^\dagger$'s. This shows that the entire $\mathcal{L}(H)$ is just a polynomial of $a$ and $a^\dagger$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.