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chrisd
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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.