11.2: Non-Commuting Operators
- Page ID
- 56868
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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:
\[\left[\hat{S}_{x}, \hat{S}_{y}\right]=i \hbar \hat{S}_{z}\tag{11.8}\]
\[\begin{aligned}
&{\left[\hat{S}_{y}, \hat{S}_{z}\right]=i \hbar \hat{S}_{x}} \\
&{\left[\hat{S}_{z}, \hat{S}_{x}\right]=i \hbar \hat{S}_{y}}
\end{aligned}\tag{11.9}\]