11.2: Non-Commuting Operators

In the previous section, we saw that if a particle can be in a definite state for two observables, then the two operators associated with those observables will commute. The converse is therefore also true; if two operators do not commute, then it is not possible for a quantum state to have a definite value of the corresponding two observables at the same time.

We’ve already seen examples of this. A particle can’t have a definite \(x\) spin and a definite \(y\) spin at the same time. If our theory is to be useful, then we would hope that \(\hat{S}_{x}\) and \(\hat{S}_{y}\) would not commute when they operate on a general normalized state \(|\psi\rangle\). Let’s try it first in one order:

\[\begin{aligned}
\hat{S}_{x} \hat{S}_{y}|\psi\rangle &=\frac{\hbar^{2}}{4}\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\left[\begin{array}{cc}
0 & -i \\
i & 0
\end{array}\right]\left[\begin{array}{l}
\psi_{1} \\
\psi_{2}
\end{array}\right] \\
&=\frac{\hbar^{2}}{4}\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\left[\begin{array}{l}
-i \psi_{2} \\
i \psi_{1}
\end{array}\right] \\
&=\frac{\hbar^{2}}{4}\left[\begin{array}{c}
i \psi_{1} \\
-i \psi_{2}
\end{array}\right] \\
&=i \frac{\hbar^{2}}{4}\left[\begin{array}{c}
\psi_{1} \\
-\psi_{2}
\end{array}\right]
\end{aligned}\tag{11.4}\]

Now let’s try it in the other order:

\[\begin{aligned}
\hat{S}_{y} \hat{S}_{x}|\psi\rangle &=\frac{\hbar^{2}}{4}\left[\begin{array}{cc}
0 & -i \\
i & 0
\end{array}\right]\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\left[\begin{array}{l}
\psi_{1} \\
\psi_{2}
\end{array}\right] \\
&=\frac{\hbar^{2}}{4}\left[\begin{array}{cc}
0 & -i \\
i & 0
\end{array}\right]\left[\begin{array}{l}
\psi_{2} \\
\psi_{1}
\end{array}\right] \\
&=\frac{\hbar^{2}}{4}\left[\begin{array}{c}
-i \psi_{1} \\
i \psi_{2}
\end{array}\right] \\
&=-i \frac{\hbar^{2}}{4}\left[\begin{array}{c}
\psi_{1} \\
-\psi_{2}
\end{array}\right]
\end{aligned}\tag{11.5}\]

Clearly the two are not the same; one is the negative of the other. Therefore, \(\hat{S}_{x}\) and \(\hat{S}_{y}\) do not commute when operating on a general state \(\psi\), as expected.

It is interesting to note the effect of \(\hat{S}_{z}\) on this same general state:

\[\begin{aligned}
\hat{S}_{z}|\psi\rangle &=\frac{\hbar}{2}\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right]\left[\begin{array}{l}
\psi_{1} \\
\psi_{2}
\end{array}\right] \\
&=\frac{\hbar}{2}\left[\begin{array}{c}
\psi_{1} \\
-\psi_{2}
\end{array}\right]
\end{aligned}\tag{11.6}\]

Notice that except for the constant out front, the vector produced by \(\hat{S}_{z}\) on this state is the same as the vector produced by \(\hat{S}_{x} \hat{S}_{y}\) and \(\hat{S}_{y} \hat{S}_{x}\). In fact, we can put the two together:

\[\begin{gathered}
\left(\hat{S}_{x} \hat{S}_{y}-\hat{S}_{y} \hat{S}_{x}\right)|\psi\rangle=i \frac{\hbar^{2}}{2}|\psi\rangle \\
{\left[\hat{S}_{x}, \hat{S}_{y}\right]|\psi\rangle=i \hbar \hat{S}_{z}|\psi\rangle}
\end{gathered}\tag{11.7}\]

The term in brackets, \(\left[\hat{S}_{x}, \hat{S}_{y}\right]\) is called the commutator of \(\hat{S}_{x}\) and \(\hat{S}_{y}\). It’s defined by the term in parentheses above it: \(\left(\hat{S}_{x} \hat{S}_{y}-\hat{S}_{y} \hat{S}_{x}\right)\). It works out for the commutators of all three spin angular momentum operators that:

\[\left[\hat{S}_{x}, \hat{S}_{y}\right]=i \hbar \hat{S}_{z}\tag{11.8}\]

\[\begin{aligned}
&{\left[\hat{S}_{y}, \hat{S}_{z}\right]=i \hbar \hat{S}_{x}} \\
&{\left[\hat{S}_{z}, \hat{S}_{x}\right]=i \hbar \hat{S}_{y}}
\end{aligned}\tag{11.9}\]