# What is the probability in case of degeneracy?

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?

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.