Responsible party: vinc0053, Green Suit

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Main class wiki page: home

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.
  

notation: |1/2 -1/2>
What is the big picture? Why do we do the math?
<math>\chi_+ = \begin{pmatrix} 1
0\end{pmatrix}</math>
<math>\chi_- = \begin{pmatrix} 0
1\end{pmatrix}</math>
<math>|“m”| =< l</math>
S_e,p,n,q,nu = 1/2 is an intrinsic property that doesn't change. You can just say S_z,m_s,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).
S² = <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 even 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 S², S_z, S_x, S_y (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) hydrogen in H2O, then microwaves flip the spin over.
S² 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.

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