Eigenbasis of Anti-Commuting Operators

Suppose that we have two operators $A$ and $B$ such that:

$$ AB=-BA. $$

In other words, they anti-commute. What can we say about their eigenbasis? It is well known that for commuting operators, they share an eigenbasis, i.e. their eigenkets are identical.

First, let’s look at the representation of $B$ in the basis defined by the eigenbasis of $A$. Let $A=\sum_{i} a_i\ket{a_i}\bra{a_i}$ be the spectral decomposition of $A$. Then,

$$ \bra{a_i}(AB+BA)\ket{a_j} = (a_i+a_j)\bra{a_i}B\ket{a_j}=0 \tag{1} $$

This tells us immediately that if $a_i\neq-a_j$, then $\bra{a_i}B\ket{a_j}=0$. One immediate consequence is that if all of the eigenvalues of $A$ are nonzero, then the diagonal of $B$ in the $A$ basis is 0, since the diagonal is where $a_i=a_j$ and $a_i=-a_j$ is only met if $a_i=0$. Then sum of the diagonal elements of $B$ is 0 as well – and since the trace is basis-independent, this is the trace of $B$. That is to say, if $A$ is invertible (has no zero-eigenvalues) then any operator which anti-commutes with $A$ will have to have 0 trace.

Another consequence of Equation (1) is that the matrix elements of $B$ in the $A$ basis are only nonzero for entries where $a_i=-a_j$. This tells us that unless an invertible operator has at least one set of eigenvalues which are each others negation, that operator can only anti-commute with the zero operator.

If $A$ is invertible, has distinct eigenvalues, and has an odd dimension (e.g. 3x3) then $B$ is singular.