# Operators and their Eigenstates in Quantum Mechanics

Suppose that a quantum state $$|\Psi\rangle$$ is given by $$|\Psi\rangle = \Sigma_{i=1}^{n}\alpha_{i}|\psi_{i}\rangle$$, where each $$|\psi_{i}\rangle$$ is an eigenstate of a Hermitian operator $$M$$ and $$\sum_i\vert \alpha_i\vert^2=1$$.

Generally, $$|\Psi\rangle$$ would not be an eigenstate of $$M$$ for $$n>1$$. Then can we know (in some systematic way) if there is a Hermitian operator, say, $$M^*$$, of which $$|\Psi\rangle$$ is an eigenstate? And is there any interesting relationship between $$M$$ and $$M^*$$ in that case?

• Is'nt $M^*$ necessarily any complex number times the identity (in that basis) ? Jun 29 at 23:04
• Any quantum state $|\psi\rangle$ is an eigenstate of the projector $|\psi\rangle\langle\psi|$ with eigenvalue 1. Jun 29 at 23:06

Then how do you find a linear operator M* of which $$|\Psi\rangle$$ is an eigenstate?

Assuming M is a "normal" operator (more on that below), we don't know anything about what $$|\Psi \rangle$$ is right now, so just solve the equation

$$M^* |\Psi \rangle = \lambda |\Psi \rangle$$

as usual. The fact that you can write $$\Psi$$ in terms of the eigenstates of M doesn't mean anything about $$\Psi$$, because the eigenstates of a normal operator form a basis for all vectors. So we could write any vector like that.

I put this in bold just because it is at the core of what I think you are missing. And it is also for that reason that there would be no reason for a relationship between $$M$$ and $$M^*$$.

On normal operators (edit): An operator is "normal" if has $$[M, M^{\dagger}] = 0$$. Just about every operator we use in QM is normal in this sense:

• Any Hermitian operator is normal
• Any unitary operator is normal
• Any diagonalizable operator is normal (in fact, normal $$\iff$$ diagonalizable)

It looks like you are talking about a normal operator since you have written what looks like a sum of basis vectors, in which case $$M$$ would be diagonalizable. But in case you weren't - there are occasional non-normal operators in QM, like the creation/annihilation operators. The answer I gave above would not apply to these. That being said, for non-normal operators, I don't think there is any useful relationship between their eigenstates ($$\psi_i$$) and an operator $$M$$ which has as an eigenstate some vector $$\Psi$$ which is in the span of their eigenstates of the $$\psi_i$$ (phew... what a mouthful). The relationship is just too fuzzy.

• Thanks for the clarification! Yes, I think I was asking specifically about normal operators. I was wondering if a system is in a superposition of eigenstates of an observable M it is usually in an eigenstate of another observable M*. This will be true for spin observables. But I wonder if it can hold in other cases. For instance, if a system is in a superposition of multiple eigenstates of the momentum operator, is there any observable for which the state is an eigenstate, no matter how complicated or arbitrary the observable may be? Jun 30 at 17:57
• I infer from your answer that there might be no systematic or interesting answer to my question. Thanks a lot! Jun 30 at 17:57
• From the comment I understand a bit better how you are thinking. For 2d spinors specifically, since each pure state can be mapped one-to-one to the bloch sphere's surface at some point $\hat{n}$, there is a spin operator $\sigma \cdot \pm \hat{n}$ for which that spinor is an eigenvector. Any other eigenbasis of $\sigma \cdot \hat{m}$ with $n \neq m$ would express the same state as a superposition. Jun 30 at 20:23
• In general, if a system is in a superposition of eigenstates of an observable M, yes there would be another observable M* for which the system is an eigenstate. However generally in QM there are tons of observables M* for which this is true. Only in 2d are your options so constrained that you can always assign something tangible to M*, namely a direction along the bloch sphere. I thought about it, and in 3d or more, it would not be the case that every state is an eigenvector of some $\sigma \cdot \hat{n}$. That is because $\hat{n}$ still only has 2 degrees of freedom, while psi now has 6-2=4. Jun 30 at 20:31
• To your last question: For instance, if a system is in a superposition of multiple eigenstates of the momentum operator, is there any observable for which the state is an eigenstate, no matter how complicated or arbitrary the observable may be? The answer is yes, there is certainly some M* for which the state is an eigenstate. In fact there would be infinitely many such operators in the case of momentum eigenstates: letting the momentum superposition state be called $|\phi \rangle$, any operator in the form $\lambda |\phi \rangle \langle \phi | + c_1|\phi_1 \rangle \langle \phi_1| + ...$ where Jun 30 at 20:35

Like doublefelix says, the fact that $$\vert \Psi\rangle$$ is a superposition of $$\vert \psi_i\rangle$$ doesn't mean anything. If $$M$$ was a normal operator, any vector could be written that way.

Also, note that $$M^*$$ isn't even uniquely defined by the condition of having $$\vert \Psi\rangle$$ as an eigenstate. E.g. $$M^*_1 = \lambda \vert \Psi \rangle \langle \Psi\vert$$ is a solution for any $$\lambda \in \mathbb{C}$$. Another would be $$M^*_2 = \lambda \vert \Psi \rangle \langle \Psi\vert + \lambda' \vert \Psi^\perp\rangle \langle \Psi^\perp\vert$$, where $$\vert \Psi^\perp\rangle$$ is any vector orthogonal to $$\vert\Psi\rangle$$. In fact, any operator of the form $$M^* = \lambda \vert \Psi \rangle \langle \Psi\vert +P,$$ would be a valid choice for $$M^*$$, for any operator $$P$$ that has $$\vert\Psi\rangle$$ in its kernel. You can easily see this by calculating $$M^*\vert \Psi\rangle$$: \begin{align} M^*\vert\Psi\rangle &= \big(\lambda \vert \Psi \rangle \langle \Psi\vert +P\big)\vert\Psi\rangle\\ &= \lambda \vert \Psi \rangle+P\vert\Psi\rangle\\ &= \lambda \vert \Psi \rangle. \end{align}

• Thanks! I think I should have added that both M and M* are (at least) Hermitian, since I'm only interested in operators representing observables. Jun 30 at 18:00