# Hamiltonian of a Spin in a Magnetic Field

• Hart

## Homework Statement

The hamiltonian of a spin in a magnetic field is given by:

$$\hat{H} = \alpha\left( B_{x}\hat{S_{x}} + B_{y}\hat{S_{y}} + B_{z}\hat{S_{z}}\right)$$

where $\alpha$ and the three components of B all are constants.

Question: Compute the energies and eigenstates of the spin.

## Homework Equations

$$\hat{S_{x}} = \frac{\hbar}{2} \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right]$$

$$\hat{S_{y}} = \frac{\hbar}{2} \left[ \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right]$$

$$\hat{S_{z}} = \frac{\hbar}{2} \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right]$$

## The Attempt at a Solution

.. I don't know how or where to start

Based on my previous experience with helping you, I think you should know where to start. Assemble the 2x2 Hamiltonian matrix, diagonalize it and find the eigenvectors and eigenvalues.

Last edited:
Well, this is actually what I've got so far..

By factoring in the $\frac{\hbar}{2}$ in each matrix:

$$B_{x}\hat{S_{x}} = \left[ \begin{array}{cc} 0 & \left(\frac{B_{x}\hbar^{2}}{2}\right) \\ \left(\frac{B_{x}\hbar^{2}}{2}\right) & 0 \end{array} \right]$$

$$B_{y}\hat{S_{x}} = \left[ \begin{array}{cc} 0 & \left(\frac{-iB_{y}\hbar^{2}}{2}\right) \\ \left(\frac{iB_{y}\hbar^{2}}{2}\right) & 0 \end{array} \right]$$

$$B_{z}\hat{S_{x}} = \left[ \begin{array}{cc} \left( \frac{B_{z}\hbar^{2}}{2}\right) & 0 \\ 0 & \left(\frac{-B_{z}\hbar^{2}}{2}\right) \end{array} \right]$$

Then by factoring in the $\alpha$:

$$B_{x}\hat{S_{x}} = \left[ \begin{array}{cc} 0 & \left(\frac{\alpha B_{x}\hbar^{2}}{2}\right) \\ \left(\frac{\alpha B_{x}\hbar^{2}}{2}\right) & 0 \end{array} \right]$$

$$B_{y}\hat{S_{x}} = \left[ \begin{array}{cc} 0 & \left(\frac{-i\alpha B_{y}\hbar^{2}}{2}\right) \\ \left(\frac{i\alpha B_{y}\hbar^{2}}{2}\right) & 0 \end{array} \right]$$

$$B_{z}\hat{S_{x}} = \left[ \begin{array}{cc} \left( \frac{\alpha B_{z} \hbar^{2}}{2}\right) & 0 \\ 0 & \left(\frac{-\alpha B_{z} \hbar^{2}}{2}\right) \end{array} \right]$$

Therefore the Hamiltonian $\hat{H}$ is the sum of these three matrices:

$$\hat{H} = \left[ \begin{array}{cc} \left(\frac{\alpha B_{z} \hbar^{2}}{2}\right) & \left(\frac{\alpha \hbar^{2}}{2}\left(B_{x} - iB_{y}\right)\right) \\ \left(\frac{\alpha \hbar^{2}}{2}\left(B_{x} + iB_{y}\right)\right) & \left(\frac{-\alpha B_{z} \hbar^{2}}{2}\right) \end{array} \right]$$

Hence my problem with what to do now.

Righto, try again..

By factoring in the $\frac{\hbar}{2}$ in each matrix:

$$B_{x}\hat{S_{x}} = \left[ \begin{array}{cc} 0 & \left(\frac{B_{x}\hbar}{2}\right) \\ \left(\frac{B_{x}\hbar}{2}\right) & 0 \end{array} \right]$$

$$B_{y}\hat{S_{x}} = \left[ \begin{array}{cc} 0 & \left(\frac{-iB_{y}\hbar}{2}\right) \\ \left(\frac{iB_{y}\hbar}{2}\right) & 0 \end{array} \right]$$

$$B_{z}\hat{S_{x}} = \left[ \begin{array}{cc} \left( \frac{B_{z}\hbar}{2}\right) & 0 \\ 0 & \left(\frac{-B_{z}\hbar}{2}\right) \end{array} \right]$$

Then by factoring in the $\alpha$:

$$B_{x}\hat{S_{x}} = \left[ \begin{array}{cc} 0 & \left(\frac{\alpha B_{x}\hbar}{2}\right) \\ \left(\frac{\alpha B_{x}\hbar}{2}\right) & 0 \end{array} \right]$$

$$B_{y}\hat{S_{x}} = \left[ \begin{array}{cc} 0 & \left(\frac{-i\alpha B_{y}\hbar}{2}\right) \\ \left(\frac{i\alpha B_{y}\hbar}{2}\right) & 0 \end{array} \right]$$

$$B_{z}\hat{S_{x}} = \left[ \begin{array}{cc} \left( \frac{\alpha B_{z} \hbar}{2}\right) & 0 \\ 0 & \left(\frac{-\alpha B_{z} \hbar}{2}\right) \end{array} \right]$$

Therefore the Hamiltonian $\hat{H}$ is the sum of these three matrices:

$$\hat{H} = \left[ \begin{array}{cc} \left(\frac{\alpha B_{z} \hbar}{2}\right) & \left(\frac{\alpha \hbar}{2}\left(B_{x} - iB_{y}\right)\right) \\ \left(\frac{\alpha \hbar}{2}\left(B_{x} + iB_{y}\right)\right) & \left(\frac{-\alpha B_{z} \hbar}{2}\right) \end{array} \right]$$

That should be correct now then.

As far as then diagonalizing the Hamiltonian matrix derived, I have read the info on that Wiki page, and a few other places online, but I still don't get how to diagonalize it.. or atleast, not in a way that doesn't get horribly messy.

I've got an idea, but to ease on latex code inclusion on a method that could be completely wrong, I've simplified it..

So say for a matrix:

$$H = \left[ \begin{array}{cc} 5 & 4 \\ (-4) & 2 \end{array} \right]$$

I take these steps:

$$det(H) = det\left[ \begin{array}{cc} (5-\lambda) & 4 \\ (-4) & (2-\lambda) \end{array} \right]=0$$

$$\left[\left(5 - \lambda\right)\left(2-\lambda\right)\right] - \left[(4)(-4)\right] = 0$$

$$\left[10 - 7\lambda + \lambda^{2} + 16\right] = \lambda^{2} - 7\lambda + 26 = 0$$

And then the Eigenvalues are the values of $\lambda$ for which that equation holds. Obviously the numbers I chose were not very good as to give a nice answer, but hopefully the jist is clear.

.. do I do it at all something like this?!?

You do it exactly like this, but with the Hamiltonian 2x2 matrix that you assembled.

Good good, right I've had a go at that and got to this stage:

$$\lambda^{2} - \left(\left(\frac{\alpha^{2}\hbar^{2}}{4}\right)\left(\hat{B_{x}^{2}} + \hat{B_{y}^{2}} + \hat{B_{z}^{2}}\right)\right) = \lambda^{2} - \left(\left(\frac{\alpha^{2}\hbar^{2}}{4}\right)\left(B\right)\right) = \lambda^{2} - \left(\frac{\alpha^{2}\hbar^{2}B}{4}\right)\right) = 0$$

Then the eigenvalues of $\hat{H}$ are the valid values of $\lambda$:

$$\lambda = \sqrt{\left(\frac{\alpha^{2}\hbar^{2}B}{4}\right)\right)} = \frac{\alpha \hbar \sqrt{B}}{2}\right)$$

.. hopefully this is heading towards a correct solution?

First off Bx2+By2+Bz2=B2

Also, the equation λ2-A2 = 0 has two solutions. What are they?

.. so it should be:

$$\lambda = \sqrt{\left(\frac{\alpha^{2}\hbar^{2}B^{2}}{4}\right)\ \right)} = \pm \left( \frac{\alpha \hbar {B}}{2}\right)\right)$$

??

Hart said:
.. so it should be:

$$\lambda = \sqrt{\left(\frac{\alpha^{2}\hbar^{2}B^{2}}{4}\right)\ \right)} = \pm \left( \frac{\alpha \hbar {B}}{2}\right)\right)$$

??

You have correctly found the eigenvalues. Now proceed to find the eigenvectors as outlined in the wikipedia example.

Right, I've looked at the Wiki example but got stuck at this point, where I have two linear equations:

$$\frac{\alpha \hbar}{2}\left(xB_{z}+y\left(B_{x} + iB_{y}\right)\left) = \frac{\alpha \hbar}{2}\left(Bx\right)$$

$$\frac{\alpha \hbar}{2}\left(x\left(B_{x}+iB_{y}\right)-2yB\right) = \frac{\alpha \hbar}{2}\left(By\right)$$

I know I have to somehow equate these, then set/choose eigenvectors, something like that.

.. just can't see where to go with it now at the moment though.

You need to select one eigenvalue, say λ1 and write down
(H11 - λ1)x + H12y = 0
where Hij are the matrix elements of the Hamiltonian. The equation above and the normalization condition
x2 + y2 = 1

should give you the eigenvector for λ1. Repeat with λ2.

Note: I will be signing off for the next few days, so if you need more help, perhaps someone else may be able to do it.

Right from what you said I've got this:

Chosen eigenvalue:

$$\lambda_{1} = \left( \frac{\alpha \hbar {B}}{2}\right)\right)$$

Then use:

(H11 - λ1)x + (H12)y = 0

where Hij are the matrix elements of the Hamiltonian.

Hence:

$$\left(\left(\frac{\alpha B_{z} \hbar}{2}\right) - \left( \frac{\alpha \hbar {B}}{2}\right)\right)\right) x + \left(\frac{\alpha \hbar}{2}\left(B_{x} - iB_{y}\right)\right) y = 0$$

Then I need to find what the values of x and y are, which are the respective components of the eigenvector associated to this eigenvalue.

You said to use $x^{2} + y^{2}=1$ and I should be able to use this to solve the above equation to get the eigenvector, but I just can't see what to do. I'm sure it's not too difficult, but hopefully I'm on the right track so far.

Note: Thanks for letting me know you won't be online for a while now, hopefully someone else will comment on this thread with some help.