| dbp:proof
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- Let be a set of orthonormal states that form a complete eigenbasis for each of the two compatible observables and represented by the self-adjoint operators and with corresponding eigenvalues and , respectively. This implies that
:
for each mutual eigenstate . Because the eigenbasis is complete, we can expand an arbitrary state according to
:
where . The above results imply that
:
for any state . Thus, , meaning that the two operators commute. (en)
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When has non-degenerate eigenvalues:
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Let be a complete set of orthonormal eigenkets of the self-adjoint operator corresponding to the set of real-valued eigenvalues .
If the self-adjoint operators and commute, we can write
:
So, if , we can say that is an eigenket of corresponding to the eigenvalue . Since both and are eigenkets associated with the same non-degenerate eigenvalue , they can differ at most by a multiplicative constant. We call this constant . So,
: ,
which means is an eigenket of , and thus of and simultaneously. In the case of , the non-zero vector is an eigenket of with the eigenvalue .
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When has degenerate eigenvalues:
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Suppose each is -fold degenerate. Let the corresponding orthonormal eigenkets be .
Since , we reason as above to find that is an eigenket of corresponding to the degenerate eigenvalue . So, we can expand in the basis of the degenerate eigenkets of :
:
The are the expansion coefficients. The coefficients form a self-adjoint matrix, since . Next step would be to diagonalize the matrix . To do so, we sum over all with constants . So,
:
So, will be an eigenket of with the eigenvalue if we have
:
This constitutes a system of linear equations for the constants . A non-trivial solution exists if
:
This is an equation of order in , and has roots. For each root we have a non-trivial solution , say, . Due to the self-adjoint of , all solutions are linearly independent. Therefore they form the new basis
:
is simultaneously an eigenket of and with eigenvalues and respectively. (en)
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