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Question: Can anticommuting operators have a simulaneous eigenket? ([1] pr. 1.16)
Two Hermitian operators anticommute
\begin{equation}\label{eqn:anticommutingOperatorWithSimulaneousEigenket:20}
\symmetric{A}{B} = A B + B A = 0.
\end{equation}
Is it possible to have a simultaneous eigenket of \( A \) and \( B \)? Prove or illustrate your assertion.
Answer
Suppose that such a simultaneous non-zero eigenket \( \ket{\alpha} \) exists, then
\begin{equation}\label{eqn:anticommutingOperatorWithSimulaneousEigenket:40}
A \ket{\alpha} = a \ket{\alpha},
\end{equation}
and
\begin{equation}\label{eqn:anticommutingOperatorWithSimulaneousEigenket:60}
B \ket{\alpha} = b \ket{\alpha}
\end{equation}
This gives
\begin{equation}\label{eqn:anticommutingOperatorWithSimulaneousEigenket:80}
\lr{ A B + B A } \ket{\alpha}
=
\lr{A b + B a} \ket{\alpha}
= 2 a b \ket{\alpha}.
\end{equation}
If this is zero, one of the operators must have a zero eigenvalue. Knowing that we can construct an example of such operators. In matrix form, let
\begin{equation}\label{eqn:anticommutingOperatorWithSimulaneousEigenket:120}
A =
\begin{bmatrix}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & a \\
\end{bmatrix}
\end{equation}
\begin{equation}\label{eqn:anticommutingOperatorWithSimulaneousEigenket:140}
B =
\begin{bmatrix}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & b \\
\end{bmatrix}.
\end{equation}
These are both Hermitian, and anticommute provided at least one of \( a, b\) is zero. These have a common eigenket
\begin{equation}\label{eqn:anticommutingOperatorWithSimulaneousEigenket:160}
\ket{\alpha} =
\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}.
\end{equation}
A zero eigenvalue of one of the commuting operators may not be a sufficient condition for such anticommutation.
References
[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.