A function of an operator is defined by its power series: \(F(\hat A)=\sum_na_n\hat A^n\); e.g. \(\exp(a\hat A)=\displaystyle\sum_n\frac{a^n}{n!}\hat A^n\).
\([F(\hat A),G(\hat A)]=0\); if \([\hat A,\hat B]=0\) then \([\hat B,F(\hat A)]=0\).
\([F(\hat A)]^\dagger=F^*(\hat A^\dagger)\); e.g. \([\exp(ia\hat A)]^\dagger=\exp(-ia^*\hat A^\dagger)\).
Baker–Campbell–Hausdorff (when \([\hat A,\hat B]\) commutes with \(\hat A\) and \(\hat B\)): \[e^{\hat A}e^{\hat B}=e^{\hat A+\hat B}\,e^{[\hat A,\hat B]/2}.\]
General expansion: \[e^{\hat A}\hat Be^{-\hat A}=\hat B+[\hat A,\hat B]+\frac1{2!}[\hat A,[\hat A,\hat B]]+\frac1{3!}[\hat A,[\hat A,[\hat A,\hat B]]]+\cdots\]
Useful: \([\hat x,F(\hat p)]=i\hbar F'(\hat p)\) and \([\hat p,F(\hat x)]=-i\hbar F'(\hat x)\).
Example 4.1 Position–momentum commutators
\(\hat P_x=-i\hbar\,\partial_x\), \(\hat X\psi=x\psi\). Find (a) \([\hat X,\hat P_x]\); (b) \([\hat X^2,\hat P_x]\) and \([\hat X,\hat P_x^2]\); (c) \([\hat X^2,\hat P_x^2]\).
Solution
(a) \([\hat X,\hat P_x]\psi=x(-i\hbar\psi')+i\hbar(x\psi)'=i\hbar\psi\) \(\Rightarrow[\hat X,\hat P_x]=i\hbar\).
(b) \([\hat X^2,\hat P_x]=\hat X[\hat X,\hat P_x]+[\hat X,\hat P_x]\hat X=2i\hbar\hat X\); \([\hat X,\hat P_x^2]=\hat P_x[\hat X,\hat P_x]+[\hat X,\hat P_x]\hat P_x=2i\hbar\hat P_x\).
(c) \([\hat X^2,\hat P_x^2]=\hat X[\hat X,\hat P_x^2]+[\hat X,\hat P_x^2]\hat X=2i\hbar(\hat X\hat P_x+\hat P_x\hat X)\).
Example 4.2 Commuting matrices
\(\hat A=a\begin{pmatrix}1&0\\0&1\end{pmatrix}\), \(\hat B=b\begin{pmatrix}0&1\\1&0\end{pmatrix}\). (a) Find \([\hat A,\hat B]\). (b) Write the eigenvectors of \(\hat B\) in the eigenbasis of \(\hat A\).
Solution
(a) \(\hat A\hat B-\hat B\hat A=ab\begin{pmatrix}0&1\\1&0\end{pmatrix}-ba\begin{pmatrix}0&1\\1&0\end{pmatrix}=0\): they commute.
(b) \(\hat A\): \(\left|a_1\right\rangle=\begin{pmatrix}1\\0\end{pmatrix}\), \(\left|a_2\right\rangle=\begin{pmatrix}0\\1\end{pmatrix}\), both eigenvalue \(a\) (degenerate).
\(\hat B\): \(\left|b_1\right\rangle=\frac1{\sqrt2}\begin{pmatrix}1\\1\end{pmatrix}\) (\(+b\)), \(\left|b_2\right\rangle=\frac1{\sqrt2}\begin{pmatrix}1\\-1\end{pmatrix}\) (\(-b\)).
\(\left|b_1\right\rangle=\frac1{\sqrt2}(\left|a_1\right\rangle+\left|a_2\right\rangle)\), \(\left|b_2\right\rangle=\frac1{\sqrt2}(\left|a_1\right\rangle-\left|a_2\right\rangle)\) — common eigenvectors of both.
Example 4.3 Parity commutes with a symmetric Hamiltonian
For \(V(-x)=V(x)\), prove \([\hat H,\hat\pi]=0\), where \(\hat H=-\dfrac{\hbar^2}{2m}\partial_x^2+V(x)\).
Solution
\(\hat H\hat\pi\psi(x)=\hat H\psi(-x)=-\dfrac{\hbar^2}{2m}\dfrac{\partial^2\psi(-x)}{\partial x^2}+V(x)\psi(-x)\).
\(\hat\pi\hat H\psi(x)=-\dfrac{\hbar^2}{2m}\dfrac{\partial^2\psi(-x)}{\partial(-x)^2}+V(-x)\psi(-x)\)
