Go to the U of M home page
School of Physics & Astronomy
School of Physics and Astronomy Wiki

User Tools


classes:2009:fall:phys4101.001:lec_notes_1120

Differences

This shows you the differences between two versions of the page.

Link to this comparison view

Both sides previous revisionPrevious revision
Next revision
Previous revision
classes:2009:fall:phys4101.001:lec_notes_1120 [2009/11/22 17:10] x500_vinc0053classes:2009:fall:phys4101.001:lec_notes_1120 [2009/11/23 21:52] (current) yk
Line 1: Line 1:
-===== Nov 20 (Fri)  =====+===== Nov 20 (Fri) Spin operators, expectation values =====
 ** Responsible party: vinc0053, Green Suit **  ** Responsible party: vinc0053, Green Suit ** 
  
Line 20: Line 20:
 notation: |1/2 -1/2>\\ notation: |1/2 -1/2>\\
 What is the big picture?  Why do we do the math?\\ What is the big picture?  Why do we do the math?\\
-chi<sub>+</sub><sup>T</sup> = (1 0)\\ +<math>\chi_= \begin{pmatrix} 1\\ 0\end{pmatrix}</math>\\ 
-chi<sub>-</sub><sup>T</sup> = (0 1)\\ +<math>\chi_- = \begin{pmatrix} 0\\ 1\end{pmatrix}</math>\\ 
-<math>|"m"| <l</math>\\+<math>|"m"| =< l</math>\\ 
 +S<sub>e,p,n,q,nu</sub> = 1/2 is an intrinsic property that doesn't change.  You can just say S<sub>z</sub>,m<sub>s</sub>,m = +/- 1/2 <math>\hbar</math> <-(the <math>\hbar</math> is really supposed to be there, but it is usually left out of the notation).\\ 
 +S<sup>2</sup><math>\frac{1}{2}(\frac{1}{2}+1)\hbar^2</math> Compare to the shallow well with only 2 states.\\ 
 +<math>\psi = c_1\psi_1 + c_2\psi_2</math> => <math>\begin{pmatrix} c_1\\ c_2\end{pmatrix}</math>, then <math>\begin{pmatrix} 1\\ 0\end{pmatrix}</math> is <math>\psi_1</math> the ground state and <math>\begin{pmatrix} 0\\ 1\end{pmatrix}</math> is the first excited state <math>\psi_2</math>.\\ 
 +When you have spin momentum it creates a magnetic dipole, <math>\psi_1</math> and <math>\psi_2</math> are degenerate states when without a magnetic field.  Apply magnetic field and <math>\psi_1</math>, <math>\psi_2</math> take different energies.\\ 
 +\\ 
 +Q: How do we know they are degenerate without a magnetic field?\\ 
 +A: <math>\psi = \psi_{nlm} s_{z(=\pm\frac{1}{2}}</math> When it decays there's emission of photon with energy equal to the difference.  The spin decay <math>\pm\frac{1}{2}\rightarrow\pm\frac{1}{2}</math> makes no difference of photon energy without a magnetic field, but with it, experiments started showing a difference.  Two peaks on a plot of measurements (which collapse to one peak without a magnetic field).\\ 
 +We could use convention <math>\chi_+^x</math> = <math>\begin{pmatrix} 1\\ 0\end{pmatrix}</math> <math>\chi_-^x</math> = <math>\begin{pmatrix} 0\\ 1\end{pmatrix}</math> for spin up and spin down along the //x// direction, but it's confusing.   
 + 
 +Also as experimental evidence for existence of spins, (in the S-G experiment,) if the z-component is discrete, not continuous, then it snaps into allowed spots (up or down).  There can only be an odd <del>even</del> number of allowed spots, why is that?\\ 
 + 
 +Now that we know how to represent quantum states using vectors, what else would we wisk to be able to reprent so that various calculations can be done? -> operator in terms of matrix such as S<sup>2</sup>, S<sub>z</sub>, S<sub>x</sub>, S<sub>y</sub> (sometimes the x and y components play an important role) like a microwave in MRI has a magnetic field that lines up perpendicular to the large DC B field (which is usually defined to be the //z// direction) <del>hydrogen in H2O</del>, then microwaves flip the spin over.\\ 
 +S<sup>2</sup> has eigenvalues, <math>\frac{3}{4}\hbar^2</math> for both spin up and down states.  If it is represented by a matrix, <math>\frac{3}{4}\hbar^2\begin{pmatrix} 1 & 0\\ 0 & 1\end{pmatrix}</math> then <math>\frac{3}{4}\hbar^2\begin{pmatrix} 1 & 0\\ 0 & 1\end{pmatrix}\begin{pmatrix} 1\\ 0\end{pmatrix} \frac{3}{4}\hbar^2\begin{pmatrix} 1\\ 0\end{pmatrix}</math>, which satisfy our desire (eigenvalue for at least <math>\begin{pmatrix} 1\\ 0\end{pmatrix}</math> vector is what we want.  The same can be shown for <math>\begin{pmatrix} 0\\ 1\end{pmatrix}</math> vector, too.\
 + 
 +Similarly, <math>S_z=\frac{1}{2}\hbar\begin{pmatrix} 1 & 0\\ 0 & -1\end{pmatrix}</math> will satisfy our desire that it is eigenvalues <math>s_z=\pm\frac{1}{2}\hbar</math> for vectors representing up/down spin states.
 ------------------------------------------ ------------------------------------------
 **To go back to the lecture note list, click [[lec_notes]]**\\ **To go back to the lecture note list, click [[lec_notes]]**\\
classes/2009/fall/phys4101.001/lec_notes_1120.1258931426.txt.gz · Last modified: 2009/11/22 17:10 by x500_vinc0053