# Two operators commute with the Hamiltonian, but do not commute with each other

I was reading Griffiths, and he made a statement that if two operators commute with the Hamiltonian, but do not commute with each other, then the energy spectrum has to be degenerate. He gave the following reasoning:

If there is not a complete set of simultaneous eigenstates of all three operators, does that mean that there is some state in the Hilbert space that cannot be written as a linear combination of the simultaneous eigenstates? Why do we know from that that "there must be some $$|\psi\rangle$$ such that $$\Lambda|\psi\rangle$$ is distinct from $$|\psi\rangle$$?" (the underlined sentence).

• "[...] cannot be written as a linear combination of the simultaneous eigenstates?" What do you mean by that? Depending on what the $Q$ and $\Lambda$ are, there might not be any states that are simultaneously eigenstates of all three operators.
– noah
Commented Feb 20, 2021 at 20:58
• @noah I was trying to understand the statement "there canot exist a complete set of simultaneous eigenstates of all three operators". I guess what I was trying to say was, does that mean that the simultaneous eigenstates can't form a basis? If there's no simultaneous eigenstate of all three operators, then there is no such basis.
– haha
Commented Feb 20, 2021 at 21:14

## 1 Answer

If $$Q$$ and $$\Lambda$$ don't commute, all we know for certain is that there is no complete basis where each state in that basis is an eigenstate of $$Q$$ and $$\Lambda$$. There might be some states that are eigenstates of both, but that is not guaranteed. So what you called "simultaneous eigenstates" can not span the whole Hilbert space, and may even be an empty set.

The text states that there must be a $$|\psi\rangle$$, such that $$\Lambda|\psi\rangle$$ is distinct from $$|\psi\rangle$$. That is simply the meaning of not having a common complete basis. If $$\Lambda|\psi\rangle \sim |\psi\rangle$$ would hold for all $$|\psi\rangle$$, then $$|\psi\rangle$$ would be an eigenbasis of $$\Lambda$$ (and it is by construction an eigenbasis of $$Q$$), so then $$\Lambda$$ and $$Q$$ would have the same eigenbasis and would commute.