- #1

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## Homework Statement

Sakurai, problem 1.11

A two-state system is characterized by the Hamiltonian

$$

H = H_{11} | 1 \rangle \langle 1| + H_{22} | 2 \rangle \langle 2| + H_{12} \left[ | 1 \rangle \langle 2| + | 2 \rangle \langle 1| \right]

$$

where ##H_{11}##, ##H_{22}##, and ##H_{12}## are real numbers with the dimension of energy, and ##| 1 \rangle## and ##|2 \rangle## are eigenkets of some observable (##\neq H##). Find the energy eigenkets and corresponding energy eigenvalues. Make sure that your answer makes good sense for ##H_{12} = 0##. (You need to solve the problem from scratch. The following may be used without proof:

$$

( \mathbf{S} \cdot \mathbf{\hat{n}}) | \mathbf{\hat{n}}; + \rangle = \frac{\hbar}{2} | \mathbf{\hat{n}}; + \rangle

$$

with ##| \mathbf{\hat{n}}; + \rangle## given by

$$

| \mathbf{\hat{n}}; + \rangle = \cos \frac{\beta}{2} | + \rangle + e^{i \alpha} \sin \frac{\beta}{2} | - \rangle

$$

where ##\beta## and ##\alpha## are the polar and azimuthal angles, respectively, that characterize ##\mathbf{\hat{n}}##.)

## Homework Equations

Characteristic polynomial

Half-angle relations:

$$

\tan \frac{\beta}{2} = \frac{1 - \cos \beta}{\sin \beta}

$$

$$

cot \frac{\beta}{2} = \frac{1 + \cos \beta}{\sin \beta}

$$

## The Attempt at a Solution

The problem is easy to solve by calculating the eigenvalues of the corresponding matrix, using the characteristic polynomial:

$$

E_{\pm} = \frac{H_{11} + H_{22}}{2} \pm \left[ \left( \frac{H_{11} - H_{22}}{2} \right)^2 + H_{12}^2 \right]^{1/2}

$$

But considering the hint given by Sakurai, I guess that he wants us to solve it in another manner.

So I started like this. From the hint, I posit that one eigenket is akin to ##| \mathbf{\hat{n}}; + \rangle##,

$$

\begin{align*}

H | \mathbf{\hat{n}}; + \rangle &= H_{11} \cos \frac{\beta}{2} | 1 \rangle + H_{22} e^{i \alpha} \sin \frac{\beta}{2} | 2 \rangle + H_{12} \left[ e^{i \alpha} \sin \frac{\beta}{2} | 1 \rangle + \cos \frac{\beta}{2} | 2 \rangle \right] \\

&= E_+ | \mathbf{\hat{n}}; + \rangle \\

&= E_+ \left[ \cos \frac{\beta}{2} | 1 \rangle + e^{i \alpha} \sin \frac{\beta}{2} | 2 \rangle \right]

\end{align*}

$$

which gives me two equations:

$$

\begin{align}

E_+ \cos \frac{\beta}{2} &= H_{11} \cos \frac{\beta}{2} + H_{12} e^{i \alpha} \sin \frac{\beta}{2} \\

E_+ e^{i \alpha} \sin \frac{\beta}{2} &= H_{22} e^{i \alpha} \sin \frac{\beta}{2} + H_{12} \cos \frac{\beta}{2}

\end{align}

$$

From (1), since ##E_+## is real, I set ##\alpha = 0##, so

\begin{align}

E_+ \cos \frac{\beta}{2} &= H_{11} \cos \frac{\beta}{2} + H_{12} \sin \frac{\beta}{2} \\

E_+ \sin \frac{\beta}{2} &= H_{22} \sin \frac{\beta}{2} + H_{12} \cos \frac{\beta}{2}

\end{align}

or

\begin{align}

E_+ &= H_{11} + H_{12} \tan \frac{\beta}{2} \\

E_+ &= H_{22} + H_{12} \cot \frac{\beta}{2}

\end{align}

Taking ##(5)-(6)##, after a bit of algebra, I get

$$

\tan \beta = \frac{H_{12}}{H_{11} - H_{22}}

$$

So far so good.

The problem I face is that I can't go much further than that. I tried getting an expression for ##E_+## by summing equations ##(5)+(6)##, but the best I can get is

$$

\begin{align}

2E_+ &= H_{11} + H_{22} + H_{12} \left( \tan \frac{\beta}{2} + \cot \frac{\beta}{2} \right) \\

E_+ &= \frac{H_{11} + H_{22}}{2} + H_{12} \csc \beta

\end{align}

$$

I've tried many different things, but either I get things in terms of ##\beta / 2## or, as in ##(8)## above, something like ##\csc \beta## which I can't relate to ##\tan \beta##, in order to get something in term of ##H_{11}##, ##H_{22}##, and ##H_{12}##.

If I am going down the right route, some hint would be appreciated. Otherwise, anyone know a good way to get the eigenvalues without the characteristic polynomial?