Question

Show that two noncommuting operators cannot have a complete set of common eigenfunctions. Hint: Show that if $\hat{\mathbf{P}}$and $\hat{\mathbf{Q}}$have a complete set of common eigenfunctions, then$\mathbf{[}\hat{\mathbf{P}}\mathbf{·}\hat{\mathbf{Q}}\mathbf{]}\mathit{f}\mathbf{=}\mathbf{0}$ for any function in Hilbert space.

Short answer

A whole set of common eigenfunctions cannot be shared by two noncommuting operators.

Step-by-step solution

Concept used

The same complete set of common eigenfunctions:

$\hat{\mathbf{P}}{\mathit{f}}_{\mathbf{n}}\mathbf{=}{\mathit{\lambda }}_{\mathbf{n}}{\mathit{f}}_{\mathbf{n}}\mathbf{}\mathbf{\text{and}}\mathbf{}\hat{\mathbf{Q}}{\mathit{f}}_{\mathbf{n}}\mathbf{=}{\mathit{\mu }}_{\mathbf{n}}{\mathit{f}}_{\mathbf{n}}$

Calculation

Assuming the operators $\hat{\mathbf{P}}$and $\hat{\mathbf{Q}}$have the same complete set of common eigenfunctions, that is:

$\hat{\mathrm{P}}{f}_n={\lambda }_n{f}_n\text{and}\hat{Q}{f}_n={\mu }_n{f}_n$

And suppose the set $f_n$is complete, so that any function in Hilbert space$f\left(x\right)$can be expressed as a linear combination, that is:

$f=\sum c_n{f}_n$

Solving for the above function,

$$\begin{gathered} [\hat{\mathrm{P}},\hat{Q}]\mathrm{f}=(\hat{\mathrm{P}}\hat{\mathrm{Q}}-\hat{\mathrm{Q}}\hat{\mathrm{P}})\sum c_n{f}_n \\ =\hat{\mathrm{P}}\left(\sum c_n{\mu }_n{f}_n\right)-\hat{Q}\left(\sum c_n{\lambda }_n{f}_n\right) \\ =\sum c_n{\mu }_n{\lambda }_n{f}_n-\sum c_n{\lambda }_n{\mu }_n{f}_n \\ =0 \end{gathered}$$

If two operators have the same set of eigenfunctions, the commutator will be zero.