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For a problem sheet at uni, I need to find eigenvalues and normalised eigenstates of a linear operator. This operator is $\hat{Q}$ and is defined by its action on the normalised eigenstates of the system Hamiltonian: $\hat{Q}\psi_1=\psi_2$, $\hat{Q}\psi_2=\psi_1$, $\hat{Q}\psi_n=0$ for $n>2$. I thought the way operators work is that they always solve the equation $\hat{Q}\psi_1=(eigenvalue) \psi_1$ so I do not really understand how the form of the operator that I am given is even possible. Can anyone explain?

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    $\begingroup$ $\psi_1$ is an eigenstate of the Hamiltonian not of $\hat Q$. Why not try writing down a general linear combination of the $\psi_n$ and find a condition for it to be an eigenstate of $\hat Q$. $\endgroup$
    – kaylimekay
    Jan 14, 2021 at 10:07
  • $\begingroup$ At a first glance $\phi_1\boldsymbol{\equiv}\left(\psi_1\boldsymbol{+}\psi_2\right)/\sqrt{2}$, $\phi_2\boldsymbol{\equiv}\left(\psi_1\boldsymbol{-}\psi_2\right)/\sqrt{2}$ are normalized eigenstates of eigenvalues $\boldsymbol{+}1$,$\boldsymbol{-}1$ respectively and all $\phi_n\boldsymbol{\equiv}\psi_n,n\boldsymbol{>}2$ are normalized eigenstates of eigenvalue $0$. Thus the operator $\hat{Q}$ is completely determined. $\endgroup$
    – Frobenius
    Jan 16, 2021 at 0:51

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The states $\psi_n$ are not eigenstates of the operator $\hat{Q}$, only the Hamiltonian. Thus, while $\psi_n$ satisfy $$\hat{H}\psi_n = E_n \psi_n,$$ in general $$\hat{Q}\psi_n \not \propto \psi_n.$$

Your job is to find a bunch of states $\phi_n$ which are eigenstates of the $\hat{Q}$ operator. i.e., a bunch of states $\phi_n$ that satisfy: $$\hat{Q} \phi_n = q_n \phi_n,\tag{1}\label{1}$$ for some numbers $q_n$. You can do this by remembering that you can express any arbitrary state $\phi$ as a linear combination of the energy eigenstates $\psi_m$ (since they form a basis). You can start off by assuming the following ansatz $$\phi = \sum_m c_m \psi_m,$$ and you now need to find the $c_m$s which allow you to satisfy Equation (\ref{1}). If you plug your ansatz into the equation, and you'd get: $$\hat{Q}\phi = q \phi \quad \quad \implies \quad \quad \sum_m c_m\hat{Q}\psi_m = q \sum_m c_m \psi_m.$$

In other words, if you know how $\hat{Q}$ acts on the energy eigenbasis $\psi_m$, you can solve the above equation and find the different values of $q$ that satisfy the above equation (there may -- and in this case, will -- be more than one!). From this, you can easily find the different $c_m$s as well. This should be straightforward.

Hint: Remember that the states $\psi_m$ are linearly independent.

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A way to look at the problem that may clarify your confusion is to consider it in matrix form.

It's not difficult to see that in the basis of the hamiltonian eigenstates $\psi_i$ i.e. $\psi_1=(1,0,0,\dots)^T$, $\psi_2=(0,1,0,\dots)^T$, $\psi_3=(0,0,1,\dots)^T$ and so on, the operator $Q$ is represented by the matrix $$ [Q] = \begin{pmatrix} 0 & 1 & 0 & 0 & \dots \\ 1 & 0 & 0 & 0 & \dots \\ 0 & 0 & 0 & 0 & \dots \\ 0 & 0 & 0 & 0 & \dots \\ \vdots & \vdots & \vdots & \vdots & \ddots & \end{pmatrix} $$ Do you have any idea on how to find eigenvectors and eigenvalues of this matrix?

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