Operators and their Eigenstates in Quantum Mechanics

Question

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?

Answer 1

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.

Answer 2

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}