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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.

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