Responsible party: Ekrpat, chap0326

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* Making sense of the Del Operator

* Beginning spin

When shown that

<math>\mathbf{\vec{ \nabla}} = \mathbf{\hat r} \partial r + \mathbf{\hat \theta} {\partial \theta \over r} + \mathbf{\hat \phi}{\partial \phi \over r\sin(\theta)} </math>

we were asked “does this make sense?” (DTMS)

We then recalled that <math>\mathbf{\vec{ \nabla}}</math> comes from

<math>\mathbf{\vec{ \nabla}} = \mathbf{\hat x}\partial x + \mathbf{\hat y}\partial y + \mathbf{\hat z}\partial z</math>

and that you could use a brute force type method to get the solution we are looking for from that. Also explained today was another method that involved introducing another Cartesian system: x',y',z', where

<math>\mathbf{\vec{ \nabla}} = \mathbf{\hat x'}\partial x' + \mathbf{\hat y'}\partial y' + \mathbf{\hat z'}\partial z'</math>,

and rotating the coordinates such that it lined up with the r-vector, <math>\theta</math> and <math>\phi</math>. Then it is possible to make a direct transformation from the [x',y',z'] coordinates to the [<math>\mathbf{\hat r}, \theta, \phi</math>] coordinates.

We find that <math>\partial x' = \partial r</math>, <math>\partial y' = {\partial \theta \over r}</math>, and <math>\partial z' = {\partial \phi \over r\sin(\theta)}</math>

Thus in a more elegant, though perhaps not as readily apparent way, we arrive at the desired expression:

<math>\mathbf{\vec{ \nabla}} = \mathbf{\hat r} \partial r + \mathbf{\hat \theta} {\partial \theta \over r} + \mathbf{\hat \phi}{\partial \phi \over r\sin(\theta)} </math>

We first made the note that we started with a quantum state represented by <math>\psi_n</math> which implies that it is a function of x, <math>\psi_n(x)</math>. Now we use a more abstract representation |<math>\psi_n</math», which doesn't necessarily imply that <math>\psi_n</math> is a function of x.

Then from this notation, |<math>\psi_n</math», we can go to a matrix vector notation:

<math>\psi_n</math> ~ |<math>\psi_n</math» ~ <math>\Sigma c_n \psi_n</math> ~ <math>$\begin{pmatrix} c_1\\ c_2\\ c_3\\ etc. \end{pmatrix}$</math>

In spin we have something analogous:

~|<math>\chi</math» ~ <math>c_+\chi_+ + c_-\chi_-</math> which goes to <math>$\begin{pmatrix} c_+\\ c_-\end{pmatrix}$</math>

We paused here to understand what characterizes <math>\chi_+</math>.

When you operate with the z-component of the angular momontum, <math>L_z</math>, you get:

<math>L_z \chi_+ = \frac{\hbar}{2} \chi_+ </math>

This says to us that <math>\chi_+</math> is an eigenvector of the <math>L_z</math> operator with an eigenvalue of <math>\frac{\hbar}{2}</math>

Then we said that

<math>\chi_+</math> ~ <math>|s, s_z></math> = <math>| \frac{1}{2}, \frac{1}{2}></math>

We also made note that the book also calls |<math>s, s_z</math»:

<math>|s, s_z></math> = <math>|s, m></math> = <math>|s, m_s></math>

and wondered why we use all these different notations. To explain we recalled from angular momentum that

<math>{Y_l}^m </math> ~ <math>{F_l}^m</math> ~<math> |l, m></math>

If you apply <math>L^2</math> on it

<math>L^2 |l, m> = l(l+1){\hbar}^2 |l,m></math>

<math>L_z |l, m> = m{\hbar} |l,m></math>

<math>L^2 \chi_+ = s(s+1){\hbar}^2 \chi_+</math>

we can begin to see how it makes sense. The 'm' is related to the magnetic quantum number, <math>m_s</math> is to help distinguish between the spin part of the magnetic. <math>s_z</math> indicates we are talking about the z-component.

Yuichi says to take a good look at section 4.4 to find what does not make sense to us, so that we know what questions to ask next lecture.

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