Given an observable that has a partially discrete and partially continuous spectrum of eigenvalues associated to it with the order of the spectrum's degeneracy being greater than 1, how would you derive an expression for the probability of finding a particular eigenvalue upon measuring this observable?


For a particular nondegenerate eigenvalue of an operator $O$, the probability that the system in state $|\psi\rangle$ will be measured to be in the eigenstate $|v\rangle$ associated with that eigenvalue is given by

$$P=|\langle v |O|\psi\rangle|^2$$

For a particular eigenvalue with degeneracy $n$, we can freely choose an $n$-dimensional orthonormal basis of the eigenspace $\{|v_i\rangle\}_{i=1}^n$. The probability that it will be measured to be in that particular eigenstate is equal to the probability that it will be in any of the eigenstates $\{|v_i\rangle\}_{i=1}^n$ or any superposition thereof. Therefore, we can extend the nondegenerate case to the following:

$$P=\sum_{i=1}^n|\langle v_i|O|\psi\rangle|^2$$

Note that this holds whether the spectrum is continuous or discrete, as long as you remember that this is the probability that you obtain exactly this eigenvalue given the exact state of the system $|\psi\rangle$.

  • $\begingroup$ You might care to add a bit about the case for the continuous spectrum: this would transform your answer from good to excellent. $\endgroup$ – ZeroTheHero May 19 '18 at 14:40
  • $\begingroup$ The answer I wrote above applies to either case; in the continuous case, you get Dirac delta functions in the sum (from e.g. some constant multiplied by $\langle x|x' \rangle$). It's also not entirely clear what degeneracy would mean in the continuous case. Is the entire continuous part of the spectrum degenerate? Is only part of it? What happens if that part is uncountable and of measure zero? $\endgroup$ – probably_someone May 20 '18 at 5:30

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