An observable in quantum mechanics is represented by a hermitian matrix. When an observation is made the state collapses to one of the eigenstates of the matrix. What happens if some eigenvalues are degenerate, how is the state from the degenerate subspace chosen? Say for two-level system that would mean that the observable is proportional to the identity matrix. Does the corresponding measurement leave the state unchanged or does is rotate it randomly? Also, if the observable is nearly-degenerate, should one expect its behaviour to be close to the strictly degenerate observable? It seems to me that continuity here is not possible, which is physically puzzling.

  • $\begingroup$ You might be interested in this: physics.stackexchange.com/questions/220071/… $\endgroup$
    – Jakob
    Feb 5 at 13:37
  • $\begingroup$ It will collapse to the subspace spanned by the degenerate eigenstates. In the case of the identity operator, this is the whole space. $\endgroup$ Feb 5 at 15:42
  • $\begingroup$ @RahulArvind is this an independent postulate or it somehow follows from others? And what about if the operator is nearly degenerate -- then the result could only be one of the two orthogonal vectors, but for very-very small level spacing should not one expect some kind of continuity with the degenerate case? $\endgroup$ Feb 5 at 17:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.