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>

So, we want to find eigenfunctions of <math>\vec{L} \cdot \vec{S}</math>, and in some situations, we may also want to find <math>\vec{S_1} \cdot \vec{S_2}</math> eigenfunctions.

Well, if we let <math>\vec{J} = \vec{L}+\vec{S}</math>, we'll have the following… <math>[J^2,\vec{L} \cdot \vec{S}] = 0</math> and <math>[J_z,\vec{L} \cdot \vec{S}] = 0</math>, and then we can use <math>\vec{J}</math> to play with <math>\vec{L} \cdot \vec{S}</math> in a simple useful way that will come up in Chapter 6.

In short, let's just trust that this addition of angular momentum technique we develop in this chapter will be very useful later, and move on to the technique itself.

The basic idea of the following section is to define <math>J^2</math> and then show how to find the eigenvalues of <math>J^2</math> for the basic spin-up and spin-down eigenfunctions. In the lecture, and in what follows, we just found the eigenvalues for the combined state <math>\uparrow\,_1\uparrow\,_2</math>. At the end, Yuichi just wrote out a matrix that included the other eigenvalues for the other combined spin states. They can be found in a similar manner to what is shown below. I hope my writeup makes some sense.

First of all, we know that <math>\left|S_1\right|^2=\frac{1}{2}(\frac{1}{2}+1)\hbar^2 = \frac{3}{4}\hbar^2</math> and by the same method <math>\left|S_2\right|^2 = \frac{3}{4}\hbar^2</math>.

For each of the two Hilbert spaces (one for S1 and one for S2) we have a pair of spin-up and spin-downs represented as follows: <math>\chi_1^+ = \begin{pmatrix} 1
0\end{pmatrix}_1 \:\:</math> <math>\chi_1^- = \begin{pmatrix} 0
1\end{pmatrix}_1</math>
<math>\chi_2^+ = \begin{pmatrix} 1
0\end{pmatrix}_2 \:\:</math> <math>\chi_2^- = \begin{pmatrix} 0
1\end{pmatrix}_2</math>
Remember these can also be represented with up and down arrows, which will be done below.

We have to be careful with these since each pair is associated with a different Hilbert space. If we want to be able to represent combined states of S1 and S2, one way to do it is to use a column 4-vector and define it like this…
<math>\uparrow\uparrow = \begin{pmatrix} 1
0 \\0\\0\end{pmatrix} \:\: \uparrow\downarrow = \begin{pmatrix} 0
1 \\0\\0\end{pmatrix} \:\: \downarrow\uparrow = \begin{pmatrix} 0
0 \\1\\0\end{pmatrix} \:\: \downarrow\downarrow = \begin{pmatrix} 0
0 \\0\\1\end{pmatrix}</math>

Now, let's try adding <math>\vec{S_1}</math> and <math>\vec{S_2}</math> when they are both in the spin up state….
<math>\vec{J}=\vec{S_1} + \vec{S_2}</math> so
<math>J^2\uparrow\uparrow=(\vec{S_1} + \vec{S_2})^2\uparrow\uparrow = (S_1^2+S_2^2+2S_1 \cdot S_2)\uparrow\uparrow</math> (This last bit works because S1 and S2 commute.)

Now, how do we find this sum?
Well, we already know that
<math>\vec{S_1}^2 \uparrow\;_1\uparrow\:_2 = (\frac{3}{4}\hbar^2\uparrow\;_1)\uparrow\;_2</math> and, similarly,
<math>\vec{S_2}^2 \uparrow\;_1\uparrow\:_2 = (\frac{3}{4}\hbar^2\uparrow\;_2)\uparrow\;_1</math>

but what about <math>2\vec{S_1}\cdot\vec{S_2}\uparrow\,_1\uparrow\,_2</math> ??

Well,
<math>2\vec{S_1}\cdot\vec{S_2}\uparrow\,_1\uparrow\,_2 = 2(\vec{S_{1x}}\,\vec{S_{2x}}+\vec{S_{1y}}\,\vec{S_{2y}}+\vec{S_{1z}}\,\vec{S_{2z}})\uparrow\,_1\uparrow\,_2</math>

Now, we know all of the spin matrices in the above expression, so we can just plug them all in and solve, and we will get
<math>J^2\uparrow\uparrow = \frac{1}{2}\hbar^2\uparrow\uparrow</math>.
Yuichi I mis-stated here. It should have been:<math>2\vec{S_1}\cdot\vec{S_2}\uparrow\uparrow = \frac{1}{2}\hbar^2\uparrow\uparrow</math>.

Using a similar technique, we can also find (with corrections)…

<math>2\vec{S_1}\cdot\vec{S_2}\uparrow\downarrow = -\frac{1}{2}\hbar^2\uparrow\downarrow + \hbar^2\uparrow\downarrow
2\vec{S_1}\cdot\vec{S_2}\downarrow\uparrow = -\frac{1}{2}\hbar^2\downarrow\uparrow + \hbar^2\downarrow\uparrow
2\vec{S_1}\cdot\vec{S_2} \downarrow\downarrow = \frac{1}{2}\hbar^2\downarrow\downarrow \\</math>

We can use our 4 element column vector notation described above to represent <math>J^2</math> using a single matrix, which is
<math>J^2 = \frac{1}{2}\hbar^2 \begin{pmatrix} 1 & 0 & 0 & 0
0 & -1 & 2 & 0
0 & 2 & -1 & 0
0 & 0 & 0 & 1 \end{pmatrix} </math>

Yuichi and this should have been <math>2\vec{S_1}\cdot\vec{S_2} = \frac{1}{2}\hbar^2 \begin{pmatrix} 1 & 0 & 0 & 0
0 & -1 & 2 & 0
0 & 2 & -1 & 0
0 & 0 & 0 & 1 \end{pmatrix} </math>

Happy Thanksgiving!

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