# Homework Help: Spin 1/2 particle in B field

1. Nov 4, 2007

### ultimateguy

1. The problem statement, all variables and given/known data
Consider a spin 1/2 particle placed in a magnetic field $$\vec{B_0}$$ with components:

$$B_x = \frac{1}{\sqrt{2}} B_0$$
$$B_y = 0$$
$$B_z = \frac{1}{\sqrt{2}} B_0$$

a) Calculate the matrix representing, in the {| + >, | - >} basis, the operator H, the Hamiltonian of the system.
b) Calculate the eigenvalues and the eigenvectors of H.
c) The system at time t = 0 is in the state | - >. What values can be found if the energy is measured, and with what probabilities?

2. Relevant equations
$$\omega_0 = - \gamma B_0$$
$$H = \omega_0 S_z$$
$$S_z = \frac{\hbar}{2} $\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right)$$$

3. The attempt at a solution

I'm stuck on part a).

My initial instinct is to do this:

$$H = \omega_0 S_z$$
$$H = - \gamma \vec{B_0} S_z$$
$$H = - \gamma \vec{B_0} \frac{\hbar}{2} $\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right)$$$

But $$\vec{B_0}$$ is a 3D column vector, and I can't multiply that into a 2x2 matrix. And I have to somehow express that with | + > and | - >... I have a feeling I'm on the wrong track.

2. Nov 4, 2007

### malawi_glenn

The basical hamiltonian is on the form:

$$H = \vec{B} \cdot \vec{S} = B_x \cdot S_x + B_y \cdot S_y + B_z \cdot S_z$$

And the $$S_x = \frac{1}{2} \sigma _x$$ pauli matrix, etc

(I use natural units, so dont bother)¨

I hope my hint helped you anyway to solve a)

Last edited: Nov 4, 2007
3. Nov 4, 2007

### ultimateguy

So with this I get:

$$H = \frac{1}{\sqrt{2}} B_0 \frac{\hbar}{2} $\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array} \right)$$$
$$+ \frac{1}{\sqrt{2}} B_0 \frac{\hbar}{2} $\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right)$$$

$$H = B_0 \hbar $\left( \begin{array}{cc} \frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} \\ \frac{1}{2\sqrt{2}} & \frac{-1}{2\sqrt{2}} \\ \end{array} \right)$$$

Right?

4. Nov 4, 2007

### malawi_glenn

Yes, that looks correct. Now find eigenvalue and eigenvector of this matrix.

5. Nov 4, 2007

### ultimateguy

For the eigenvalue:

$$(\frac{B_0 \hbar}{2\sqrt2})^2 (1-\lambda)(1-\lambda) - (\frac{B_0 \hbar}{2\sqrt2})^2 = 0$$

$$(1-\lambda)(1-\lambda) = 0$$
$$\lambda = 1$$

And the eigenvector:

$$\frac{B_0 \hbar} {2\sqrt2} $\left( \begin{array}{cc} 0 & 1 \\ 1 & -2 \\ \end{array} \right)$ \times $\left(\begin{array}{c} c_1 \\ c_2 \\ \end{array} \right)$ = 0$$

This gives me $$c_2 = 0$$ and $$c_1 = \frac{B_0 \hbar} {2\sqrt2}$$. I think I did something wrong.

6. Nov 4, 2007

### malawi_glenn

for matrix:

$$A = $\left( \begin{array}{cc} a & a\\ a & -a \\ \end{array} \right)$$$

The secular eq is $$(a- \lambda )(-a- \lambda ) - a^2 = 0$$

if lambda is the eigenvalue.

7. Nov 4, 2007

### ultimateguy

Thanks, I found the problem, I should have had $$(1-\lambda)(-1-\lambda) = 1$$

8. Nov 4, 2007

### ultimateguy

Ok so I get the following system of equations for the case of the eigenvalue $$+\sqrt2$$

$$(1-\sqrt2)c_1 + c_2 = 0$$
$$c_1 + (-1-\sqrt2)c_2 = 0$$

which according to myself and my calculator has no solution....

9. Nov 4, 2007

### malawi_glenn

If you can get eigenvalues to a matrix, then there exists corresponding eigenvectors.

Eigen vectors are, by using Matlab:

for sqrt2 = (-0.92388,-0.38268)
for -sqrt2 = (0.38268,-0.92388)

Last edited: Nov 4, 2007
10. Nov 4, 2007

### ultimateguy

Ok I found the two eigenvectors using Matlab. I'm not sure how to write them when applied to the system so that it makes sense.

$$|\psi(t)> = 0.92388 | + > + 0.382683 | - >$$
$$|\psi(t)> = 0.382683 | + > - 0.92388 | - >$$

Is this correct? And for part c), which one do I use to find the probability?

11. Nov 4, 2007

### malawi_glenn

But in order to get them analytically, just do substituion
$$c_1 = (1+\sqrt2)c_2$$
From your second equation and put in into the first one and solve for c_2

Now how does a state evolve with time? Ever heard of "Time evolution operator" or similar?

Time evolution of a ket is
$$|a(t) \rangle = \exp (-i E_a t/\hbar)|a(0) \rangle$$
where E_a is the energy eigenvalue of that ket.

$$|\psi +> = 0.92388 | + > + 0.382683 | - >$$
$$|\psi -> = 0.382683 | + > - 0.92388 | - >$$

Dont use time, as you did, it is not correct.
The psi + has eigenvalue +sqrt2 etc.
Now I have helped you very much.

Last edited: Nov 4, 2007
12. Nov 4, 2007

### ultimateguy

Yes I've seen it. So the eigenvectors are just for $$|\psi(0)>$$ ?

13. Nov 4, 2007

### malawi_glenn

no, just for the hamiltonian.

at time = 0; the state is in |->

Then you must find out what just |-> is in superposition of the eigenvectors to the hamiltonian, in order to get the time evolution.

i.e you should first write |-> = a|phi + > + b|phi ->