Responsible party: ice IX, Daniel Faraday

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Please try to include the following

• main points understood, and expand them - what is your understanding of what the points were.
  • expand these points by including many of the details the class discussed.
• main points which are not clear. - describe what you have understood and what the remain questions surrounding the point(s).
  • Other classmates can step in and clarify the points, and expand them.
• How the main points fit with the big picture of QM. Or what is not clear about how today's points fit in in a big picture.
• wonderful tricks which were used in the lecture.
  

This lecture had two main sections. In the first section, we talked about the expectation values of Sx, Sy, Sz, and S^2

In the second section, we began to talk about the addition of angular momenta.

We know that <math><s^2> = \frac{3}{4}\hbar^2</math> and we know that <math><s_y>^2 = (\frac{1}{2}\hbar)^2</math>

so we can say that <math><s^2> \; > \; (\frac{1}{2}\hbar)^2</math>

Another way to look at this is to note that <math><s^2>=<s_x^2>+<s_y^2>+<s_z^2> </math>

and <math>\sigma_s_x = <s_x^2> - <s_x>^2</math>

but we also know that <math>\sigma_s_x > 0</math>. We know that because we consider <math><s_z></math> to be a 'special' axis in which we know the spin, which makes the spin of the other two axes have some nonzero uncertainty.

Therefore, we can say that <math><s_x^2> \:>\: <s_x>^2</math>.

In other news, let's look at <math>s_z</math>. Remember that <math>\uparrow</math> and <math>\chi_+</math> mean the same thing. We can also represent this spin up state as the column vector <math>$\begin{pmatrix} 1\\ 0\end{pmatrix}$</math>

So, therefore, <math> <s_z> = (1 \: 0) s_z \left( \begin{array}{c} 1
0 \end{array} \right) = \frac{1}{2}\hbar </math>

Squaring this, we have <math> <s_z>^2 = (\frac{1}{2}\hbar)^2 </math>

We also know that <math> <s_z>^2 = <s_z^2> </math> because <math>s_z</math> is an eigenstate with 0 uncertainty. Also, we have <math> <s_x> = <s_y> = 0</math>

Yuichi at this point asked a good question: How can we have <math><s_x> = <s_y> = 0</math> but <math><s_x^2> = <s_y^2> = (\frac{1}{2}\hbar)^2</math>? That seems a little strange when you first look at it.

Here's the answer: If you measure <math>s_x</math> you will always get <math>\pm\frac{1}{2}\hbar</math>. If you don't square your results, the average will be zero. But if you do square the results, you will always get <math>(\frac{1}{2}\hbar)^2</math>. The same thing goes for <math>s_y</math> and <math>s_y^2</math>.

On the subject of what you can measure when, we briefly discussed a Stern-Gerlach thought experiment: If you send a beam of 1/2 spin particles through a S-G apparatus and separate particles in the z-direction, you have separated your particles into two groups, one with spin up in the z-direction and one with spin down in the z-direction.

Now, take the z-spin up particles, and run them through a second S-G apparatus which is aligned in the x direction. Now, you have two groups of particles, one with spin up in the x-direction and one with spin down in the x-direction, all of which have z-spin up, right?

Wrong! When you separate the particles based on their x-direction spin, you lose all your knowledge of the z-direction spin. In fact, you cause the z-direction spin to become indeterminate. This is because x-direction spin and z-direction spin are incompatible observables.

Example: when looking at fine structure, <math>\vec{L}</math> and <math>\vec{S}</math> will be important and useful. In a magnetic field, the hamiltonian will have an extra term, so that <math>\hat{H} = \alpha \vec{L} \cdot \vec{S} + (\frac{p^2}{2m} + V)</math> , and <math>\alpha</math> can be determined from E&M, which can be seen in chapter 6.

Notice that we are going to ignore the <math>(\frac{p^2}{2m} + V)</math> term of the hamiltonian since that part isn't changing from the magnetic field; we're not interested in that part of it right now.

Anyway, if the magnetic field we apply is in the <math>\hat{z}</math> direction, then we have <math>\vec{B}=B_z</math> and <math>\hat{H}=-\alpha B_z S_z</math> In that case, which is associated with the Zeeman Effect, <math>S_z</math> is an eigenfunction of <math>\hat{H}</math>

If the magnetic field is not on, then the <math>\vec{L} \cdot \vec{S}</math> becomes important, and <math>[\vec{L} \cdot \vec{S},S_z] \neq 0</math>

Me or the other guy will finish the notes later; happy Thanksgiving!

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