===== Nov 23 (Mon) Spin operators, probabilities, Larmor precession, S-G experiment =====
** Responsible party: vinc0053, ice IX ** 

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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.\\


In this lecture the spin operators and their expectation values were briefly covered (see 11/25 lecture note for a more complete description), and we discussed the probabilities of observing specific spin states with respect to different axes. Also, we went over the concepts of Larmor precession and ended with a little Stern-Gerlach food for thought.

==== I. Probabilities ====
The general spin quantum state (for spin 1/2) can be described by a linear combination of the eigenstates in the familiar way:\\
<math>\chi= \begin{pmatrix} a\\b \end{pmatrix}=a\chi_+\:+\:b\chi_-,</math>\\
where <math>\chi_+</math> and <math>\chi_-</math> are the spin-up and -down basis vectors (spinors?) with the eigenvalues <math>\pm\frac{\hbar}{2}</math>, respectively. From previous experience, the probability of observing a particle in a specific state upon measurement of some observable is the square of the coefficient that comes from normalization; the probabilities must add up to unity: <math>\left| a \right|^2+\left| b \right|^2=1</math>.

So if one were to measure <math>S_z</math>, the probability of finding <math>\frac{+\hbar}{2}</math> would be <math>\left| a \right|^2</math>. This is simple, but if the desired observable is not in the z-direction, perhaps the interest lies the x-direction instead, one needs to figure out the eigenvalues (also <math>\pm\frac{\hbar}{2}</math>) and eigenvectors characteristic of said direction. The state in the z-direction is related to the x-direction:

<math>\begin{pmatrix} a\\b \end{pmatrix}_z=\alpha \chi_+ ^{(x)}+\beta\chi_- ^{(x)}</math>,
where <math>\chi_+ ^{(x)}=\frac{1}{sqrt{2}}\begin{pmatrix} 1\\1 \end{pmatrix}</math> and <math>\chi_- ^{(x)}=\frac{1}{sqrt{2}}\begin{pmatrix} 1\\-1 \end{pmatrix}</math>; these are just eigenvectors found in the normal manner. One can see that <math>\alpha=\frac{a+b}{sqrt{2}}</math> and <math>\beta=\frac{a-b}{sqrt{2}}</math> from <math>\chi_{\pm}</math>. Thus, the probability of obtaining <math>\frac{\hbar}{2}</math> in the +x-direction would be:

<math>\left|\alpha\right|^2=\left|\frac{a+b}{sqrt{2}}\right|^2=\frac{\left|a\right|^2+\left|b\right|^2+2Reab*}{2}=\frac{1+2Reab*}{2}</math>

The utility of these coefficients is seen in calculating the expectation values of the <math>S_x,\:S_y,\:S_z</math> and <math>S^2</math> spin operators:

  * <math><S_z>=\frac{1}{2}\hbar(\left|a\right|^2-\left|b\right|^2)</math>

  * <math><S_x>=\chi^{\dag}S_x\chi=\frac{\hbar}{2}(2Reab*)</math>

  * <math><S_y>=\frac{\hbar}{2}(2Imab*)</math>

More stuff can be done with these expectation values, as will be seen below and in the next lecture note.

==== II. Larmor Precession ====
Larmor precession occurs when a charged particle with angular momentum, more specifically a particle with a magnetic dipole moment <math>\mu</math>, is acted on by an external magnetic field , say in the z-direction <math>B=B_o\hat{k}</math>. The B-field exerts a torque equal to <math>\mu\times B</math> on the particle, but since <math>\mu=\gamma S</math> where <math>\gamma</math> is the gyromagnetic ratio, the torque changes the angular momentum of the particle in a perpendicular fashion causing <math>\mu</math> to precess about the direction of the B-field. Note that under normal circumstances the torque would tend to line up with the magnetic moment with the B-field, which corresponds to the minimum energy of the particle. Moreover, there are many useful applications of this phenomena in a wide array of fields (read about it!).

The Hamiltonian for this situation is:

<math>H=\left(\frac{p^2}{2m}+V\right)-\mu B=-\gamma SB=-\gamma S_zB=-\frac{\gamma B_o\hbar}{2}\begin{pmatrix} 1&0\\0&-1 \end{pmatrix}</math>

the first chunk is not so important so we exclude it since one assumes that the particle under examination is stationary, except for the angular momentum. One can write the initial state of the particle as follows:

<math>\chi(0)=\begin{pmatrix} a_o\\b_o \end{pmatrix}=\begin{pmatrix} \cos\alpha\\e^{i\delta}\sin\alpha \end{pmatrix} </math>.

The coefficients were replaced with a cosine and a sine due to the idea (pointed out by Zeno) that the angular momentum must remain constant i.e. the rotational and magnetic torques balance in such a way that a constant angle <math>2\alpha</math> can be sustained. In addition, a phase angle <math>\delta</math> was included so as not to lose generality. Just for fun, here are the expectation values that define the point in the xyz plane where the <math><S></math> lies:

  * <math><S_x>=\sin\alpha\cos\delta</math>
  * <math><S_y>=\sin\alpha\sin\delta</math>
  * <math><S_z>=\cos\alpha</math>

To examine the time dependence of the particle, <math>\chi(0)</math> can be used, noting that the coefficients carry the t-dependence:

<math>\chi(t)=\begin{pmatrix} a(t)\\b(t) \end{pmatrix} \Rightarrow\ \left\{ {\text{a(t)=a_o e^{\frac{-iE_+t}{\hbar}\chi_+}\atop
          \text{b(t)=b_o e^{\frac{-iE_-t}{\hbar}\chi_-}} \right.</math>

Using the Hamiltonian above to find <math>E_\pm=\mp\gamma\frac{\hbar}{2}B</math>, the a(t) equation can be re-written:

<math>a(t)=a_o exp\left[-i\frac{-\gamma<del>\hbar</del> B}{2<del>\hbar</del>}t\right]</math>\\
The expectation values for the spin operators can be found in terms of <math>\chi(t)</math>:

  * <math><S_x>=\chi(t)^{\dag}S_x\chi(t)=\frac{\hbar}{2}\sin(2\alpha)\cos(\gamma Bt-\delta)</math>
  * <math><S_y>=-\frac{\hbar}{2}\sin(2\alpha)\sin(\gamma Bt-\delta)</math>
  * <math><S_z>=\frac{\hbar}{2}\cos(2\alpha)</math>

Note the slight difference in this answer with respect to the textbook-this is because of the assignment of an angle <math>\alpha</math> in the re-definitions of <math>a_o</math> and <math>b_o</math>. These expectation values confirm that <math><S></math> sits at an angle of <math>2\alpha</math> from the direction of the B-field. Also, the Larmor frequency, which is the frequency at which the particle precesses, can be established as <math>\omega=\gamma Bt</math>, dropping the delta term.

==== III. Stern-Gerlach ====
This experiment involves an inhomogeneous magnetic field <math>B_z=B_o+\alpha z</math>; the x-component averages to zero. Performing the same sort of work as was done for Larmor precession (this is already done in the textbook), the conclusion is that <math>\chi(t)</math> ends up having terms that closely resemble that of the plane wave, in the z-direction.

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