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On the last page, Esquire asked whether a non-square matrix could represent a hermitian operator. A hermitian matrix must be square and, more-over, must be equal to its “transpose conjugate” (which is the same as a hermitian conjugate). The transpose conjugate is attained by transposing the matrix (flipping it along the diagonal) and taking the complex conjugate of each element. See pages 443-444. To get some insight into why this is so, try finding <math><\alpha |\hat{Q} \beta></math> and <math><\hat{Q} \alpha | \beta ></math> where <math>\displaystyle \hat{Q}= \left(\begin{array}{ccc} 1 & 0 & 1
1 & 0 & 1
1 & 0 & 1 \end{array} \right)</math>, <math>\displaystyle \vec{\alpha} = \left(\begin{array}{c} \alpha_1
\alpha_2
\alpha_3 \end{array} \right)</math>, and <math>\displaystyle \vec{\beta} = \left(\begin{array}{c} \beta_1
\beta_2
\beta_3 \end{array} \right)</math>.

I get <math>(\alpha_1 + \alpha_2 + \alpha_3)^*(\beta_1 + \beta_3)</math> and <math>(\alpha_1 + \alpha_3)^* (\beta_1 + \beta_2 + \beta_3)</math>. These are clearly not equal. If you don't find that experiment convincing, try using more general matrix elements (a la <math>q_{1,2}</math>); you should come to the conclusion that <math>q_{m,n}=(q_{n,m})^*</math> for all m and n for any hermitian matrix. If the matrix isn't square, there are m and n such that one of those terms exists and the other doesn't. In fact, as Spherical Chicken said on the last page, you'll find that you can't even calculate both of those quantities with a non-square matrix. Hopefully some of this was helpful (as opposed to redundant).

I guess might as well ask what people thought of the test. I didn't know what to expect going in seeing as how the material we covered was mostly just the solution to the hydrogen atom, which doesn't lend itself well to a test format. The first problem seemed a little out of place to me; the practice test is the only place I recall seeing anything like it so far in the course. (I knew the units of <math>\eps_0</math>, so I didn't have any problems, but I'm curious if that was the kind of thing the proctors would answer; did anyone ask during the test?) All of the problems were straight-forward, though, and I felt like I had enough time. I don't know what to think of problem 3; using the spherical gradient made the problem trivial, so I tried to use the other method as well and made some errors. As a result, though, I now have a somewhat better understanding of partial vs. regular derivatives. Does anyone else have any comments? Agreement/disagreement? Something completely different?

If we look at the probability density of the spherical harmonic <math>\|Y({\theta},{\phi})\| ^2</math>, we can see that it is always expressed only in terms of a polynomial of degree 2l in <math>cos\theta</math>. How can or has this relationship been exploited?

Could you explain how to derive from 4.97 to 4.98?

is the bonus point problem the only HW problem?

Yeah, thats the only one.

And that's due a week from tomorrow, not tomorrow. It's my Thanksgiving break project to read your solutions.

On page 164, Griffiths solves for the eigenvalue of <math>L^2</math>, <math>\lambda={\hbar^2}l(l+1)</math> but how did he get the eigenvalue for <math>L_z</math>, <math>\mu={\hbar}m</math> ? On equation [4.115] Griffiths just sets it equal to <math>\hbar{l}</math> how did he know to do this?

From what I can tell <math>L_zm</math> is proportional to <math>{\hbar}m</math> as shown in equation [4.118]. Here m is a quantum value that goes from -l, -l+1,…,l-1, l. In equation [4.115] he set m = -l, since l_bar = -1, to find the angular momentum for the bottom rung. In equation [4.111] Griffiths does this again, except this time it is for the top rung (m = l) of the angular momentum states.

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