\prob{50}
The state of a quantum mechanical system is described by a wave function  . Consider two physical observables that have discrete eigenvalues: observable A with eigenvalues  , and observable B with eigenvalues  . Under what circumstances can all wave functions be expanded in a set of basis states, each of which is a simultaneous eigenfunction of both A and B?
- Only if the values and are nondegenerate
- Only if A and B commute
- Only if A commutes with the Hamiltonian of the system
- Only if B commutes with the Hamiltonian of the sytem
- Under all circumstances
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Quantum Mechanics }Simultaneous Eigenstates
QM in verse...
Two operators, both alike in state functions,
In fair bases, where we lay our scene,
From ancient grudge break new mutiny...
Two operators unlike in eigenvalues
Yet star-crossed lovers commute.
So anyway, the problem gives  and  . That is, both A and B share the same eigenstate  .
Consider ![\langle a^{'} | [A,B]| a^{''} \rangle =\langle a^{'}| AB-BA| a^{''} \rangle = \langle a^{'} |(A\beta^{'}-B\alpha^{''}) | a^{''} \rangle = \langle a^{'} |(\beta^{'}A-\alpha^{''}B) | a^{''} \rangle = \langle a^{'} |(\alpha^{''}\beta^{''}-\beta^{''}\alpha^{''}) | a^{''} \rangle = (\alpha^{''}\beta^{''}-\beta^{''}\alpha^{''}) \langle a^{'} | a^{''} \rangle](/cgi-bin/mimetex.cgi?\langle a^{'} | [A,B]| a^{''} \rangle =\langle a^{'}| AB-BA| a^{''} \rangle = \langle a^{'} |(A\beta^{'}-B\alpha^{''}) | a^{''} \rangle = \langle a^{'} |(\beta^{'}A-\alpha^{''}B) | a^{''} \rangle = \langle a^{'} |(\alpha^{''}\beta^{''}-\beta^{''}\alpha^{''}) | a^{''} \rangle = (\alpha^{''}\beta^{''}-\beta^{''}\alpha^{''}) \langle a^{'} | a^{''} \rangle) .
But, the scalar term is 0. This implies that in order for both operators to have the same eigenstate, ![[A,B]=0](/cgi-bin/mimetex.cgi?[A,B]=0) .
(Also, one knows that  by definition of orthonormal eigenfunctions.)
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