Two-level quantum system (from Sakurai)

[DrClaude]

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?

 

[George Jones]

You need to solve the problem from scratch.

should be "You need not solve the problem from scratch."

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}

Rewrite these equations as
$$\begin{align}
\left( E_+ - H_{11} \right) \cos \frac{\beta}{2} &= H_{12} \sin \frac{\beta}{2} \\
H_{12} \cos \frac{\beta}{2} &= \left( E_+ - H_{22} \right) \sin \frac{\beta}{2}
\end{align}$$

Now, "divide" these equations.

 

[DrClaude]

Now, "divide" these equations.

I hadn't thought about division. Thanks for the help!

 

[Fred Wright]

Starting from scratch the Hamiltonian for a quantum particle in a steady magnetic field is,$$H=\mathbf \mu \cdot \mathbf B$$
where $\mathbf \mu = \gamma \mathbf s$ is the magnetic moment of the particle, $\gamma$ is the gyromagnetic ratio, $\mathbf s$ is the spin, $\mathbf B= B \hat n$ is the magnetic field and $\hat n$ is the normal vector pointing in the direction of the steady field in the unit sphere. The normal in the sphere is $\hat n=(cos(\alpha)sin(\frac {\beta}{2}),sin(\alpha)sin(\frac {\beta}{2}),cos(\frac {\beta}{2}))$. The $\frac {\beta}{2}$ comes in because the domain of the polar angle is half a period. Now write the spin in terms of the Pauli matrices, $\mathbf s = \frac {\hbar}{2} \mathbf \sigma$ and take the inner product,
$$H=\frac {\hbar}{2} \gamma B \mathbf \sigma \cdot \hat n =\frac {\hbar}{2} \gamma B \left ( cos(\alpha)sin(\frac {\beta}{2}) \sigma_x + sin(\alpha)sin(\frac {\beta}{2}) \sigma_y + cos(\frac {\beta}{2}) \sigma_z \right )$$We find,$$ H=\frac {\hbar}{2} \gamma B
\begin{pmatrix}
cos(\frac {\beta}{2}) & e^{-i\alpha} sin(\frac {\beta}{2}) \\
e^{i\alpha} sin(\frac {\beta}{2}) & -cos(\frac {\beta}{2})
\end{pmatrix}$$
By inspection we see that the eigenvalues are $\frac {\hbar}{2} \gamma B$ and $-\frac {\hbar}{2} \gamma B$ and the eigenvectors are $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$
Because $H_{12}=H_{21}$ is real, $\alpha = 0$ and we get $H_{22}=\frac {\hbar}{2} \gamma B cos(\frac {\beta}{2}), H_{11}=-\frac {\hbar}{2} \gamma B cos(\frac {\beta}{2})$ and $H_{12}= H_{21}=\frac {\hbar}{2} \gamma B sin(\frac {\beta}{2})$.

 

[DrClaude]

Starting from scratch the Hamiltonian for a quantum particle in a steady magnetic field is,$$H=\mathbf \mu \cdot \mathbf B$$
where $\mathbf \mu = \gamma \mathbf s$ is the magnetic moment of the particle, $\gamma$ is the gyromagnetic ratio, $\mathbf s$ is the spin, $\mathbf B= B \hat n$ is the magnetic field and $\hat n$ is the normal vector pointing in the direction of the steady field in the unit sphere. The normal in the sphere is $\hat n=(cos(\alpha)sin(\frac {\beta}{2}),sin(\alpha)sin(\frac {\beta}{2}),cos(\frac {\beta}{2}))$. The $\frac {\beta}{2}$ comes in because the domain of the polar angle is half a period. Now write the spin in terms of the Pauli matrices, $\mathbf s = \frac {\hbar}{2} \mathbf \sigma$ and take the inner product,
$$H=\frac {\hbar}{2} \gamma B \mathbf \sigma \cdot \hat n =\frac {\hbar}{2} \gamma B \left ( cos(\alpha)sin(\frac {\beta}{2}) \sigma_x + sin(\alpha)sin(\frac {\beta}{2}) \sigma_y + cos(\frac {\beta}{2}) \sigma_z \right )$$

My question was about the clever way the equations need to be manipulated to reach the solution, not the physics of the problem.

We find,$$ H=\frac {\hbar}{2} \gamma B
\begin{pmatrix}
cos(\frac {\beta}{2}) & e^{-i\alpha} sin(\frac {\beta}{2}) \\
e^{i\alpha} sin(\frac {\beta}{2}) & -cos(\frac {\beta}{2})
\end{pmatrix}$$
By inspection we see that the eigenvalues are $\frac {\hbar}{2} \gamma B$ and $-\frac {\hbar}{2} \gamma B$ and the eigenvectors are $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$

These are only the eigenvectors for $\beta = 0, 2\pi, \ldots$

 

[Silicon-Based]

How do you find the eigenkets in this problem?

 

[DrClaude]

How do you find the eigenkets in this problem?

Once you have the eigenvalues, you put them back in the eigenequation and solve for the coefficients of the eigenvectors $(c_1, c_2)^T$. Note that the solution is not unique, so you can either solve simultaneously the normalization equation $|c_1|^2 + |c_2|^2 = 1$ or take arbitrarily $c_1 = 1$, find $c_2$, and renormalize.

 

[Silicon-Based]

Once you have the eigenvalues, you put them back in the eigenequation and solve for the coefficients of the eigenvectors $(c_1, c_2)^T$. Note that the solution is not unique, so you can either solve simultaneously the normalization equation $|c_1|^2 + |c_2|^2 = 1$ or take arbitrarily $c_1 = 1$, find $c_2$, and renormalize.

Is that really the way it was meant to be done? The standard procedure is supposed to be quite messy, and I thought there would be a way to make use of the hint to find the eigenkets. Using the hint to find the eigenvalues is not much quicker than just evaluating the Hamiltonian and solving det(H-EI)=0.