# Weird quantum linear operator [closed]

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?

• $\psi_1$ 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 $\hat Q$. Jan 14, 2021 at 10:07
• 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. Jan 16, 2021 at 0:51

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.
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?