Question: The operators A and B both have a denumerable number of eigenstates. Of these, the single eigenstate is known to be common to both.
The operators A and B both have a denumerable number of eigenstates. Of these, the single eigenstate is known to be common to both. That is, Aq=ap Bo=bo (a) What can be said about the commutability of A and ? (b) Suppose that it is further know that all the eigenstates of A and B are degenerate. Dose this additional information in any way change your answer to part (a)?
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ANSWER a From the simultaneous eigenfunctions of the commuting operator... View full answer
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