Isham (in his Lectures on Quantum Theory), in his initial chapters at least, gives examples of only such operators on such wavefunction state-spaces which have only nondegenerate eigenfunctions. I guess that this is due to the reason that it makes the (initial) formalism easier.

Hence for an operator $A$, with eigenvalues $a_1, a_2, \ldots$ with the corresponding (nondegenerate) eigenfunctions $u_1(x), u_2(x), \ldots$, we have $\text{Prob}(A=a_i;\psi)=|c_i|^2$ for a state $\psi = \sum_{i=1}^\infty c_i u_i$.

Question: What if there are two degenerate eigenfunctions corresponding to a single eigenvalue? What will expression of the probability of measuring that degenerate eigenvalue be like?


1 Answer 1


For an operator $A$ with eigenvalues $a_1,a_2,\ldots$ with the corresponding (possibly degenerate) orthonormal eigenfunctions $u_{1,1},\ldots,u_{1,g_1};u_{2,1},\ldots,u_{2,g_2}; \ldots$, such that $Au_{j,k} = a_j u_{j,k}$, measured in a state $$ \psi = \sum_j \sum_{k=1}^{g_j} c_{j,k} u_{j,k}, $$ the probability of obtaining the measurement result $a_i$ is $$ \mathrm{Prob}(A=a_i; \psi) = \sum_{k=1}^{g_i} |c_{i,k}|^2. $$ One can see relatively easily that this is independent of the basis chosen for the $A=a_i$ eigenspace, so long as that basis is orthonormal, and that it coincides with the norm of $\Pi_i \psi$, where $\Pi_i$ is the orthonormal projector onto that subspace.


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