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:

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

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