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## Homework Statement

In my quantum class we learned that if two operators commute, we can always find a set of simultaneous eigenvectors for both operators. I'm having trouble proving this for the case of degenerate eigenvalues.

## Homework Equations

Commutator: [itex][A,B]=AB-BA [/itex]

Eigenvalue equation:[itex]A \mid v \rangle = a \mid v \rangle[/itex]

## The Attempt at a Solution

Start off by assuming operators A and B commute so AB=BA.

I think I have the proof for non-degenerate eigenvalues correct:

[itex]A \mid v \rangle = a \mid v \rangle[/itex]

[itex]BA \mid v \rangle = Ba \mid v \rangle[/itex]

[itex]A(B \mid v \rangle) = a(B \mid v \rangle)[/itex]

So [itex] B \mid v \rangle[/itex] is also an eigenvector of A associated with eigenvalue a.

If a is non-degenerate, [itex] B \mid v \rangle[/itex] must be the same eigenvector as [itex] \mid v \rangle[/itex], only multiplied by a scalar.

[itex] B \mid v\rangle=b\mid v \rangle[/itex] which is just the eigenvalue equation for [itex]B[/itex].

For the degenerate case I'm stuck. I can prove that if [itex] \mid v_1 \rangle,\mid v_2 \rangle,...,\mid v_n \rangle[/itex] are eigenvectors of A associated with eigenvalue a, any linear combination of these eigenvectors is also an eigenvector of A with eigenvalue a.

[itex]A (c_1\mid v_1 \rangle + c_2\mid v_2 \rangle) [/itex]

[itex]=c_1A\mid v_1 \rangle + c_2A\mid v_2 \rangle[/itex]

[itex]=c_1a\mid v_1 \rangle + c_2a\mid v_2 \rangle [/itex]

[itex]=a(c_1\mid v_1 \rangle + c_2\mid v_2 \rangle) [/itex]

and so from the fact that [itex] B \mid v_i \rangle[/itex] is an eigenvector of A, [itex] B \mid v_i \rangle[/itex] is a linear combination of the eigenvectors of A associated with eigenvalue a...

[itex] B \mid v_i \rangle= c_1\mid v_1 \rangle + c_2\mid v_2 \rangle + ... + c_n\mid v_n \rangle[/itex]

so I know [itex] B \mid v_i \rangle[/itex] exists within the eigenspace of a, but I'm not sure how to use this to prove that [itex]\mid v_i \rangle[/itex] is an eigenvector of B.

Any hints would be appreciated.