# A naive question on the eigenvalues of fermionic operators?

Let $A$ be a fermionic operator which is a product of odd number of fermion operators or a summation of them, say

$A=C_{i_1}^{\dagger}\cdot \cdot\cdot C_{i_m}^{\dagger}C_{j_1}\cdot \cdot\cdot C_{j_n}$ or $A=\sum w(i_1\cdot \cdot\cdot i_mj_1\cdot \cdot\cdot j_n) C_{i_1}^{\dagger}\cdot \cdot\cdot C_{i_m}^{\dagger}C_{j_1}\cdot \cdot\cdot C_{j_n}$

where $C_i,C_j^{\dagger}$ are fermion operators satisfying the standard anticommutaion relations, and $m+n$ is an odd number. [$w(i_1\cdot \cdot\cdot i_mj_1\cdot \cdot\cdot j_n)$ are the coefficients.]

My question is: If $\lambda(\neq0)$ is an eigenvalue of $A$, then is $-\lambda$ also an eigenvalue of $A$ ?

If the above is true, how to prove it？If it is wrong, what is the counterexample?

(i) Define an unitary operator : $\mathcal{U}(\chi)=e_{}^{-i\chi {\underset{k}\sum_{}^{}}C_{k}^{\dagger}C_{k}^{}}$ for $\chi \in \mathbb{R}$.

(ii) Notice : $\mathcal{U}(\chi)_{}^{\dagger}C_{k}^{\dagger}\mathcal{U}(\chi)_{}^{}=e_{}^{i\chi}C_{k}^{\dagger}$ and $\mathcal{U}(\chi)_{}^{\dagger}C_{k}^{}\mathcal{U}(\chi)_{}^{}=e_{}^{-i\chi}C_{k}^{}$.

(iii) So, for the operator defined in the question : $A=\underset{\{s_{p}^{}\},\{t_{q}^{}\}}{\sum_{}^{}}w[\{s_{p}^{}\},\{t_{q}^{}\}]\underset{\text{ordered products}}{\tilde\prod_{p=1}^{m}\tilde \prod_{q=1}^{n}}C_{s_{p}^{}}^{\dagger}C_{t_{q}^{}}^{}$, we have : $\mathcal{U}(\chi)_{}^{\dagger}A\mathcal{U}(\chi)_{}^{}=e_{}^{i\chi(m-n)}A$.

(iv) For odd number $m+n$ and $\chi=\pi$ we get : $A\mathcal{U}(\pi)_{}^{}=-\mathcal{U}(\pi)_{}^{}A$.

(v) Now, if $A|\lambda\rangle=\lambda|\lambda\rangle$ then $A\left[\mathcal{U}(\pi)_{}^{}|\lambda\rangle\right]=-\lambda\left[\mathcal{U}(\pi)_{}^{}|\lambda\rangle\right]$.

(vi) More generally we have : $A|\lambda\rangle=\lambda|\lambda\rangle$ $\Rightarrow$ $A\left[\mathcal{U}(\chi)_{}^{}|\lambda\rangle\right]=\left[e_{}^{i\chi(m-n)}\lambda\right]\left[\mathcal{U}(\chi)_{}^{}|\lambda\rangle\right]$.

• I see, thank u very much. I like your last conclusion (vi), from this, it seems that the eigenvalue of $A$ must be zero if $m\neq n$, right? – Kai Li Aug 5 '18 at 14:18
• Because if it is nonzero, there would be infinite number of eigenvalues of $A$, this seems impossible. – Kai Li Aug 5 '18 at 14:22
• @KaiLi Identity given in (iii) provides information of non-zero matrix elements of $A$ in the Fock space. Regarding your second comment, I have to think more about Fermion case, for bosonic Fock space (same analysis holds), consider a single mode and $A=a$, then it is clear that $A$ has uncountable eigenvalues and eigenstates corresponding to $|\alpha| e_{}^{i\chi}$ can be generated through the unitary transformation $\mathcal{U}(\chi)=e_{}^{i\chi a_{}^{\dagger}a_{}^{}}$ from the eigenstate corresponding to $|\alpha|$. – Sunyam Aug 5 '18 at 18:35