How to calculate probabilities of eigenvalue of observable with multiple eigenvectors (degeneracy) I was calculating the probability that for a two spin system, prepared in a state $\mid rr \rangle$ (which can be expressed as a combination of basis vectors $\mid u \rangle$ and $\mid d \rangle$), a given spin would be +1. I calculated the state $\mid\Psi(t)\rangle$ with no difficulty. My thinking was that the observable corresponding to one spin would be $\sigma_z \otimes I$. I knew that the eigenvalues would be +1 and -1, as these are the possible spin values. However, each eigenvalue had two eigenvectors. I have never dealt with degeneracy before and didn't know quite how to handle it. When I calculated the probability $P(t)_{+1}=\langle\lambda_1\mid\Psi(t)\rangle\langle\Psi(t)\mid\lambda_1\rangle$ for each eigenvector with eigenvalue 1, I got $\frac{1}{4}$. 
I think, since the probability the spin is 1 is $\frac{1}{2}$, it would be logical to add the probabilities of each eigenvector with a shared eigenvalue together to obtain the probability for the eigenvalue. In the basis I was using, the eigenvectors for eigenvalue 1 were equivalent to $\mid uu \rangle$ and $\mid ud \rangle$. In both of these vectors, the spin being measured is in a state $\mid u \rangle$, or +1. I think that the probabilities for $\mid uu \rangle$ and $\mid ud \rangle$ were both $\frac{1}{4}$, which is what I was actualy calculating. However, the probability for the eigenvalue, 1, is the sum of these probabilities, $\frac{1}{2}$. This would agree with experiment and my math.
Keep in mind the entire paragraph above is my thinking and may be wrong. However, it seems correct to me. Is is correct that, in a state of degeneracy, the probability that and observable measures a specific eigenvalue is equal to the sum of the probabilities of the eigenvectors of the eigenvalue?
 A: To calculate the probability of an eigenvalue of an observable being measured, you must calculate the probability $P_\lambda(t) = \langle\lambda\mid\Psi(t)\rangle\langle\Psi(t)\mid\lambda\rangle$ for each eigenvector of the eigenvalue. For non-degenerate states, there will be one and this will be the probability. In cases of degeneracy, the probability to measure an eigenvalue is the sum of the probabilities of the eigenvectors. In mathematical format: $$P_\lambda(t) = \sum_i{\langle\lambda_i\mid\Psi(t)\rangle\langle\Psi(t)\mid\lambda_i\rangle},$$ where $\lambda_i$ is all eigenvectors of the observable being measured with eigenvalue $\lambda$.
This equation works with your math. Because the probability of each eigenvector of an eigenvalue was $\frac{1}{4}$, the sum of which is $\frac{1}{2}$, it agrees with experimental data and mathematical theory. Lastly, it is in line with the principles of quantum mechanics: The probability of an event is the sum the probabilities of all possible ways in which the event could happen.
