Closure Relation For Operator With Degenerate Spectrum Suppose we have an observable represented by a Hermitian operator, $A$. Further, $A$ has at least one eigenvalue that is degenerate. For concreteness lets say $A |\alpha_i> = \alpha | \alpha_i>$ for $i = 1,2, \ldots, n$. When writing the closure relation must we include all the degenerate eigenkets in the sum or is it only necessary to take one?
 A: You must include all of the eigenkets.
To show this, I must create notation for all of the non-degenerate eigenkets. Let's call them $\left|\beta_i\right\rangle$, each of which has a unique eigenvalue $\beta_i$, and $i=1,2,3...,m$. Now, I'll prove by contradiction that leaving out even one of the degenerate eigenkets contradicts the closure relation:
Assume the operator $P$, defined as $\sum_{i=1}^{n-1} \left|\alpha_i\right\rangle \left\langle\alpha_i\right| + \sum_{i=1}^{m} \left|\beta_i\right\rangle \left\langle\beta_i\right|$ is equal to $1$, the identity operator. Then, by definition of the identity, we require $P\left|\alpha_n\right\rangle = \left|\alpha_n\right\rangle$. However, since the eigenkets are orthogonal, $P\left|\alpha_n\right\rangle = \sum_{i=1}^{n-1} \left|\alpha_i\right\rangle \left\langle\alpha_i\right|\left|\alpha_n\right\rangle + \sum_{i=1}^{m} \left|\beta_i\right\rangle \left\langle\beta_i\right|\left|\alpha_n\right\rangle = \left|0\right\rangle$, the null ket. This is a contradiction, hence we must include every degenerate eigenket (in this case $\left|\alpha_n\right\rangle$, the one we left out) in our closure relation. QED
