| Both sides previous revisionPrevious revision | |
| classes:2009:fall:phys4101.001:lec_notes_1123 [2009/12/02 04:18] – x500_santi026 | classes:2009:fall:phys4101.001:lec_notes_1123 [2009/12/02 05:57] (current) – x500_santi026 |
|---|
| 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: | 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\\b \end{pmatrix}=\begin{pmatrix} \cos\alpha\\e^{i\delta}\sin\alpha \end{pmatrix} </math>. | <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 fact that | 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. |
| |
| ------------------------------------------ | ------------------------------------------ |