quantum mechanics - $[A_1, H] =[A_2, H] = 0$ but $[A_1, A_2] neq 0$?

I am having a difficult time understanding this problem.

Suppose $[A_1, A_2] \ne 0,$ $[A_1, H] = 0,$ $[A_2, H] = 0.$ Show that the energy eigenstates of $H$ are in general degenerate.

It seems to me that if $[A_1, H] = 0$ and $[A_2, H] = 0,$ then $H$ shares eigenstates with both $A_1$ and $A_2$. Hence, any eigenstate of $H$ is also one of $A_1$ and $A_2\Rightarrow[A_1, A_2]=0$. Clearly, this is not so, since the question forces me to assume the opposite. What did I do wrong?

Answer

If $H$ commutes with $A_1$, then it will indeed share an eigenbasis with it. Your mistake is in supposing that it will share the same eigenbasis with both $A_i$s.

Examples are easy to provide:

• On the trivial side, if $H=E_0\mathbf 1$ is trivial, then it shares an eigenbasis with $A_1=x$ and it shares an eigenbasis $A_2=p$, but it cannot share an eigenbasis with both.


• On the more physically meaningful side, the total angular momentum (as well as spherically-symmetric hamiltonians like the hydrogenic one) shares an eigenbasis with $L_z$ and ditto for $L_x$, but it can't do so with both. To make a shared eigenbasis you need to break the spherical symmetry and choose a quantization axis, and you will get different bases depending on which axis you choose.

In every degenerate eigenspace of $H$ you can always find non-commuting observables of this sort. This is easy to prove: if $\newcommand{\ket}[1]{|#1\rangle}\newcommand{\bra}[1]{\langle #1|}\ket 1$ and $\ket 2$ share an energy eigenvalue, then you can choose $$ A_1=\ket 1\bra1-\ket 2\bra 2\quad\text{and}\quad A_2=\ket1\bra 2+\ket2\bra1, $$ or in matrix language $$ A_1=\begin{pmatrix}1&0\\0&-1\end{pmatrix} \quad\text{and}\quad A_2=\begin{pmatrix}0&1\\1&0\end{pmatrix} $$ in the $\{\ket 1,\ket 2\}$ basis. You can easily check that these matrices don't commute. To be more explicit, here are eigenstates of $H$ which are not eigenstates of $A_2$, thought there is indeed some (other) basis for this eigenspace which $H$ shares with $A_2$. What is this other shared eigenbasis? What are the matrix representations of the two operators in the two eigenbases?

The only way you can avoid this is by giving the observables absolutely no breathing room for this sort of monkey business, and that is by having one-dimensional, nondegenerate energy eigenspaces. If this is the case, then for each eigenspace where $H$ and $A_1$ share a basis $\{\ket\phi\}$, $A_1$ needs to share with $H$ a basis of $\mathrm{span}\,\{\ket\phi\}$, which can differ from $\ket\phi$ by, at most, a trivial phase. The shared eigenbases are then the same, and $A_1$ must commute with $A_2$.