Determine the energy eigenstates and eigenvectors

Question

Question: Let $\{\vert\psi_1\rangle, \vert\psi_2\rangle \}$ be an orthonormal basis. Define Hamiltonian $\hat{H} = \alpha \left( \vert\psi_1\rangle \langle\psi_2\vert + \vert\psi_2\rangle\langle\psi_1\vert \right)$ for some real, positive $\alpha$. Find the eigenvalues and eigenkets $\{ \vert E_1 \rangle, \vert E_2\rangle \}$ of $\hat{H}$ and express the eigenkets as linear combinations of the $\vert \psi_k \rangle$ basis states.

I set $\vert E_1 \rangle = a_1\vert \psi_1 \rangle + b_1\vert\psi_2\rangle$ and $\vert E_2 \rangle = a_2\vert \psi_1 \rangle + b_2\vert\psi_2\rangle$. and

\begin{align} \hat{H} \vert E_1 \rangle &= \alpha \left(\langle\psi_2\vert E_1\rangle\vert\psi_1\rangle + \langle\psi_1\vert E_1\rangle\vert\psi_2\rangle\right) \\ &= \alpha\left( b_1\vert\psi_1\rangle + a_1\vert\psi_2\rangle \right). \end{align}

Because this needs to be in the form of $E_1\vert E_1\rangle$, I conclude $a_1 = b_1$ and

\begin{equation} \hat{H}\vert E_1\rangle = \alpha \vert E_1\rangle. \end{equation}

Similarly, for $\vert E_2\rangle$, I get

\begin{equation} \hat{H}\vert E_2\rangle = \alpha \vert E_2\rangle. \end{equation}

However, this means the two eigenkets have the same eigenvalue, which does not make sense. I don't know where I did wrong.

Answer 1

Your revised matrix $$ H=\alpha\left[\matrix{0&1\cr 1&0}\right] $$ has eigenvalues $\pm \alpha$. You should be able to find the mutually orthonal eigenvectors now.

Answer 2

Whenever you encounter quadratic equations, even without a linear term, there are always two answers. For example, $$x^2=4$$ has two solutions, $x=2$ and $x=-2.$ You have a similar situation in solving for your eigenvalues.

Also, when finding the eigenkets, even if the eigenvalues are the same (which they aren't here), the eigenkets must be constructed to be orthonormal, i.e., $$\langle E_1|E_2 \rangle = 0 $$

Answer 3

Hint :

This is not precisely an answer but a suggestion to see the answer from a different point of view : that of the Pauli matrices.

So, as @mike stone pointed out in his answer, the representation of your Hamiltonian $\mathrm H$ with respect to the orthonormal basis $\left(\psi_{1},\psi_{2}\right)$ is \begin{equation} \mathrm H\boldsymbol{=}\alpha \begin{bmatrix} 0 & \!\!\hphantom{\boldsymbol{-}}1 \vphantom{\tfrac{a}{b}}\\ 1 & \!\!\hphantom{\boldsymbol{-}}0\vphantom{\tfrac{a}{b}} \end{bmatrix} \boldsymbol{=}\alpha\,\sigma_1 \tag{01}\label{01} \end{equation} where $\sigma_1$ is one of the Pauli matrices \begin{equation} \sigma_1= \begin{bmatrix} 0 & \!\!\hphantom{\boldsymbol{-}}1 \vphantom{\tfrac{a}{b}}\\ 1 & \!\!\hphantom{\boldsymbol{-}}0\vphantom{\tfrac{a}{b}} \end{bmatrix} \quad , \:\:\: \sigma_2= \begin{bmatrix} 0 & \!\!\boldsymbol{-} i \vphantom{\tfrac{a}{b}}\\ i & \!\!\hphantom{\boldsymbol{-}} 0\vphantom{\tfrac{a}{b}} \end{bmatrix} \quad , \:\:\: \sigma_3= \begin{bmatrix} 1 & \!\!\hphantom{\boldsymbol{-}} 0 \vphantom{\frac{a}{b}}\\ 0 & \!\!\boldsymbol{-} 1\vphantom{\frac{a}{b}} \end{bmatrix} \tag{02}\label{02} \end{equation} The eigenstates and eigenvalues problem is equivalent to the diagonalization of $\mathrm H$. The latter one is in turn equivalent to find a similarity transformation usually through a special unitary matrix $U\in$ SU(2) so that \begin{equation} U\mathrm H U^{\boldsymbol{*}}\boldsymbol{=}\mathrm {H_{d}}\,, \qquad \mathrm {H_{d}}\boldsymbol{=}\texttt{diagonal} \tag{03}\label{03} \end{equation} But since the hermiticity and $^{\prime\prime}$tracelessness$^{\prime\prime}$ of $\mathrm H$ are invariant properties under similarity, so the diagonal matrix $\mathrm {H_{d}}$ must be hermitian and traceless. The only candidate for this is the Pauli matrix $\sigma_3$, that is \begin{equation} \mathrm {H_{d}}\boldsymbol{=}\alpha\,\sigma_3\boldsymbol{=}\alpha \begin{bmatrix} 1 & \!\!\hphantom{\boldsymbol{-}} 0 \vphantom{\frac{a}{b}}\\ 0 & \!\!\boldsymbol{-} 1\vphantom{\frac{a}{b}} \end{bmatrix} \tag{04}\label{04} \end{equation} This means that the matrix $U\in$ SU(2) must satisfy the equation \begin{equation} U\mathrm \sigma_1 U^{\boldsymbol{*}}\boldsymbol{=}\sigma_3 \tag{05}\label{05} \end{equation} In a 3-dimensional space $\mathbb{R}^3$ (not the real space) with real coordinates $\left(x_1,x_2,x_3\right)$ the matrix $U$ is a SU(2) representation of a rotation around the unit vector $\mathbf n$ of $x_2-$axis through an angle $\theta$ given by \begin{align} \mathbf{n}&\boldsymbol{=}\left(0,1,0\right) \tag{06a}\label{06a}\\ \theta &\boldsymbol{=}\boldsymbol{-}\dfrac{\pi}{2} \tag{06b}\label{06b} \end{align} so \begin{equation} U\boldsymbol{=} \cos\left(\frac{\theta}{2}\right)\boldsymbol{-}i(\mathbf{n} \boldsymbol{\cdot} \boldsymbol{\sigma})\sin\left(\frac{\theta}{2}\right)\boldsymbol{=}\tfrac{\sqrt{2}}{2}\left(I\boldsymbol{+}i\,\sigma_2\vphantom{\frac{\theta}{2}}\right)\boldsymbol{=}\tfrac{\sqrt{2}}{2} \begin{bmatrix} \hphantom{\boldsymbol{-}}1 & \!\!\hphantom{\boldsymbol{-}}1\:\: \vphantom{\tfrac{a}{b}}\\ \boldsymbol{-}1 & \!\!\hphantom{\boldsymbol{-}}1\:\:\vphantom{\tfrac{a}{b}} \end{bmatrix} \tag{07}\label{07} \end{equation}

$\boldsymbol{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$

Reference 1 : Rotation which diagonalizes the Hamiltonian.

Reference 2 : How does the Hamiltonian changes after rotating the coordinate frame.

Reference 3 : Understanding the Bloch sphere.

$\boldsymbol{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$