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Two-level quantum system (from Sakurai)

  • #1
DrClaude
Mentor
7,143
3,278

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?
 

Answers and Replies

  • #2
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,270
803
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.
 
  • #3
DrClaude
Mentor
7,143
3,278
Now, "divide" these equations.
I hadn't thought about division. Thanks for the help!
 
  • #4
202
83
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})##.
 
Last edited:
  • #5
DrClaude
Mentor
7,143
3,278
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##
 
  • #6
How do you find the eigenkets in this problem?
 
  • #7
DrClaude
Mentor
7,143
3,278
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.
 
  • #8
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.
 

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