So, in a question I am given the orthonormal eigenvectors of an operator, $A$, and its non-degenerate eigenvalues - I am not told which eigenvalue corresponds to which eigenvector. The eigenvectors of operator A are expressed as linear combinations of the non-degenerate Hamiltonian, $H$, eigenbasis. I am asked,

"are A and H compatible"?

My question here is, can you say, based on the fact that the eigenvectors of $A$ are expressed as linear combinations of the Hamiltonians eigenbasis, that $A$ and $H$ are compatible? Or must something more be done with the equations I am given, such as proving $[A, H] = 0$?

I think you can, because it implies that A and H share an eigenbasis. But I am uncertain and would highly appreciate if someone could clear this up for me.

Edit: This is the information I am given,

  • $H$ has non-degenerate eigenvalues, $E_1$ and $E_2$, corresponding to eigenvectors $|1>$ and $|2>$ . So its a two-level system.
  • $A$ has orthonormal eigenvectors given by,

$$ |v_1> = a|1> + b|2> $$ $$ |v_1> = b|1> - a|2> $$

  • $A$ has eigenvalues $ \frac{+}{-} 1$.
  • $\begingroup$ Hint: measure AH and HA on a 2x2 example. What do you conclude? $\endgroup$ Dec 9, 2021 at 16:28
  • $\begingroup$ Cheers for the response. I know how to represent H, but I am really stuck for A. I know its eigenvectors and its eigenvalues, but I don't know which eigenvalue belongs to which, so how can I act A on anything? I am going to attach the information I am given to give a better idea of the situation. $\endgroup$
    – imglueck
    Dec 9, 2021 at 17:09
  • 1
    $\begingroup$ Edit has been added. $\endgroup$
    – imglueck
    Dec 9, 2021 at 17:18
  • $\begingroup$ @imglueck Is it given that $|v_1\rangle $ vector corresponds to eigenvalues $+1$? Or you have just written it that way in edit. $\endgroup$ Dec 9, 2021 at 17:22
  • 1
    $\begingroup$ It doesn't matter. Suppose you measure $|1\rangle$ first with H, and then with A. You'll end up in some eigenstate of A. Suppose you measure it first with A, then the outcome with H. You'll end up in some eigenstate of H. What do you conclude? How do you understand "compatible"? $\endgroup$ Dec 9, 2021 at 17:35

1 Answer 1


Given that

$$A|v_1\rangle =aA|1\rangle +bA|2\rangle= v_1|v_1\rangle =v_1a|1\rangle +v_1b|2\rangle $$ $$A|v_2\rangle =bA|1\rangle -aA|2\rangle=v_2|v_2\rangle=v_2b|1\rangle -v_2a|2\rangle $$ where $v_1$ can be $\pm 1$ and $v_2$ can be $\mp 1$. Since it's not given in the problem which is which.

From above, two-equation, it's clear that $$A|1\rangle =\frac{v_1a^2+v_2b^2}{a^2+b^2}|1\rangle +\frac{ab(v_1-v_2)}{a^2+b^2}|2\rangle $$ $$A|2\rangle =\frac{ab(v_1-v_2)}{a^2+b^2}|1\rangle +\frac{v_1 b^2+v_1a^2}{a^2+b^2}|2\rangle $$

It's quite clear from here. For none of the combinations, $|1\rangle $ (or $|2\rangle $) are eigenvector of $A$. So we can conclude that $$[A,H]\not =0$$ Other way to see this, is to see the action of $H$ on $|v_1\rangle $ $$H|v_1\rangle =aE_1|1\rangle +bE_2|2\rangle \not = \lambda_1 |v_1\rangle $$


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