Quantum Mechanics for Engineers 

© Leon van Dommelen 

D.18 Eigenfunctions of commuting operators
The fact that two operators that commute have a common set of
eigenfunctions can be seen as follows: assume that is an
eigenfunction of with eigenvalue . Then since and
commute, .
Comparing start and end, must be an eigenfunction of
with eigenvalue just like itself is. If there is no
degeneracy of the eigenvalue, that must mean that equals
or is at least proportional to it. That is the same as
saying that is an eigenfunction of too. (In the special
case that is zero, is still an eigenfunction of
, with eigenvalue zero.)
If there is degeneracy, the eigenfunctions of are not unique and
you can mess with them until they all do become eigenfunctions of
too. That can be shown assuming that the problem has been
approximated by a finitedimensional one. Then and become matrices
and the eigenfunction become eigenvectors. Consider each eigenvalue
of in turn. There will be more than one eigenvector corresponding
to a degenerate eigenvalue . Now by completeness, any
eigenvector can be written as a combination of the
eigenvectors of , and more particularly as
where is a combination of the eigenvectors
of with eigenvalue and a combination of the
eigenvectors of with other eigenvalues.
The vectors and separately are still eigenvectors
of if nonzero, since as noted above, converts eigenvectors
of into eigenvectors with the same eigenvalue or zero. (For
example, if was not , would have
to make up the difference, and can only produce
combinations of eigenvectors of that do not have eigenvalue
.) Now replace the eigenvector by either
or , whichever is independent of the other eigenvectors
of . Doing this for all eigenvectors of you achieve that
the replacement eigenvectors of are either combinations of the
eigenvectors of with eigenvalue or of the other eigenvectors
of . The set of new eigenvectors of that are
combinations of the eigenvectors of with eigenvalue can now be
taken as the replacement eigenvectors of with eigenvalue
. They are also eigenvectors of . Repeat for all
eigenvalues of .
Similar arguments can be used recursively to show that more generally,
a set of operators that all commute have a common set of eigenvectors.
The operators do not really have to be Hermitian, just
diagonalizable
: they must have a complete set of
eigenfunctions.
The above derivation assumed that the problem was finitedimensional,
or discretized some way into a finitedimensional one like you do in
numerical solutions. The latter is open to some suspicion, because
even the most accurate numerical approximation is never truly exact.
Unfortunately, in the infinitedimensional case the derivation
gets much trickier. However, as the hydrogen atom and harmonic
oscillator eigenfunction examples indicate, typical infinite systems
in nature do satisfy the relationship anyway.