# Degeneracy and Complete Sets of Commuting Observables

I want to understand how the degeneracy of an operator is related to the existence of a complete set of commuting operators that includes it. I know that if a set of operators commute, they possess a common eigenbasis. My question is, can you always find an operator that commutes with a degenerate operator to form of a complete set and 'lift the degeneracy'. Why / why not?

• May 9, 2020 at 21:59
• How would you know that the spectrum of an operator is degenerate unless you have a commuting operator that lifts the degeneracy?
– user87745
May 9, 2020 at 23:40
• @CosmasZachos I looked at the wikipedia page. The relevant section talked about finding new operators that commute until the degeneracy is lifted, but does not answer why such an operator always exist. May 10, 2020 at 19:19
• Imagine diagonalizing them simultaneously. May 10, 2020 at 19:22

I agree with the answer by Dvij D.C.

Let me add, however, that the answer to your question is trivially yes from a purely mathematical point of view. Consider, for example, the degenerate operator $$A$$: $$A = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix}$$ (i.e., this is the matrix representation of $$A$$ in an eigenbasis). Then the operator $$B = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$ clearly commutes with $$A$$ and lifts the degeneracy.

This construction does not teach us anything about Physics, though.

• Good point. This is a nice explicit demonstration of what OP was asking for 😅
– user87745
May 10, 2020 at 11:04
• Thanks! This is actually what I was hoping for. May 10, 2020 at 19:22

My question is, can you always find an operator that commutes with a degenerate operator to form of a complete set and 'lift the degeneracy'.

I think this is exactly the opposite of the right way to think about this. How would you know that the spectrum of an observable is degenerate? Let's say we want to see if the spectrum of an operator $$\hat{A}$$ is degenerate or not corresponding to the eigenvalue $$a$$. The only way to distinguish between the possible multiple states that constitute the eigensubspace of $$\hat{A}$$ with eigenvalue $$a$$ is to have an operator $$\hat{B}$$ which commutes with $$\hat{A}$$ and thus we can make simultaneous measurement of $$\hat{B}$$ and assign a quantum number apart from $$a$$ to our state. So, the reason we can meaningfully speak of a degenerate spectrum is that we can make simultaneous measurements of two different observables.

In other words, it would be physically empty to speak of degeneracy without having the appropriate commuting observable in the first place.

• Thanks, this lends physical insight to the 'second eigenbasis labelling' that I am learning about now. May 10, 2020 at 19:21
• I didn't understand how simultaneous measurement helps in lifting degeneracy,we still can't distinguish if there are two different states(with same eigenvalue) or we are randomly got same state as before Sep 16, 2021 at 7:30