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Am I crazy, or is there a sign error in the eigenvalue in eq 4.136? As written it is <math> S_\pm \, |s\, m> = \hbar \sqrt{s(s+1) - m(m\pm1)} \, |s \,(m\pm1)> </math>

But when I do

<math>S_{+} \chi_-</math> with this, I get <math>\frac{\hbar}{\sqrt{2}}</math> as the eigenvalue. Shouldn't I get an eigenvalue of <math>\hbar</math> ??
What am I doing wrong?

Oh. Embarrassed. Arithmetic error.

One more picky point. This factor is not an eigenvalue strictly speaking since the vectors on the LHS and RHS are not the same.

Why is it that energy is discrete but linear momentum is not? Aren't they related? I'm having a hard time understanding this.

Energy and momentum are certainly related: <math>E=\frac{p^2}{2m}=\frac{\hbar^2 k^2}{2m}</math>. Note that energy can be discrete as well as continuous (i.e. the free particle). Also, depending on the situation momentum p can have discrete eigenvalues in concordance with the energy, for example the infinite square well has <math>p=\pm \frac{n\pi}{L}\hbar </math>.

If that is true than what does it mean at the end of last weeks discussion problem? Because what that said and what you said seem to be contradicting. I don't get it either.

I've got a question I've been meaning to ask for a while but keep forgetting to post…A while back I was reading through the Angular Momentum chapter again, and came across the part where they're talking about how the Lx, Ly, and Lz commutators go. In equation 4.97, they jump to equation 4.98 while dropping two terms and simplifying the remaining….Now it looks like all they did was pull a y px out of the first term, and an x py out of the second term, even though each is only associated with one of the two terms. Is there some identity that lets you do this, or does it just work out this way by expanding the commutators and doing algebra?

This is simply a matter of expanding it and noting that y commutes with px, and x with py, as in equation 4.10.

In class today, Yuichi expressed the relations from Clebsch-Gordan table as being a sum of the two possible states (with respective probabilities) as being equal to a third state (of probability one). My question is how did he get the expressions for the components inside of the kets of the third relation. I believe it was |5/2> (maybe 3/2) and

Love,

Esquire

In class, he was specifically asking what combinations of the lz and sz states would give you that combined state of L + S = J, so he just picked one of the possible combinations given l = 2, s = 1/2.

On page 186, I understood <math> S^{(1)}*S^{(2)}(\downarrow\uparrow) = \hbar^2/4 *(2\uparrow\downarrow -\downarrow\uparrow ) and S^{(1)}*S^{(2)}(\uparrow\downarrow) = \hbar^2/4 *(2\downarrow\uparrow -\uparrow\downarrow) </math>. But, on the Eq. 4. 180, <math> S^{(1)}*S^{(2)} | 1 , 0 > = \hbar^2/4*1/sqrt{2}*(2\downarrow\uparrow - \uparrow\downarrow +2 \uparrow\downarrow-\downarrow\uparrow) </math>, I couldn't understand how to get those. Could somebody explain how these equations turn out?

I also didn't get that.

We are trying to prove that the triplet states and the singlet state are eigenvectors of the <math>S^2</math> operator with eigenvalues of <math>2\hbar^2</math> and zero, where<math>S^2</math> operator is given by 4.179. <math>S^2 |1,0>=(S^{(1)}^2+S^{(2)}^2+2S^{(1)}*S^{(2)})|1,0></math>. the fisrt two terms are straight forward with eigenvalues of <math>3/4*\hbar^2</math> so we only have to worry about the last term. <math>2S^{(1)}*S^{(2)}|1,0>=2S^{(1)}*S^{(2)}[1/sqrt2(\uparrow\downarrow+\downarrow\uparrow)]</math> from 4.177. to evaluate this last term Griffiths first finds <math> S^{(1)}*S^{(2)}(\downarrow\uparrow) = \hbar^2/4 *(2\uparrow\downarrow -\downarrow\uparrow ) and S^{(1)}*S^{(2)}(\uparrow\downarrow) = \hbar^2/4 *(2\downarrow\uparrow -\uparrow\downarrow) </math>. then add the together and multiply it by <math>1/sqrt2</math> to get equation 1.80:<math> S^{(1)}*S^{(2)} | 1 , 0 > = \hbar^2/4*1/sqrt{2}*(2\downarrow\uparrow - \uparrow\downarrow +2 \uparrow\downarrow-\downarrow\uparrow)=\hbar^2/4|1,0> </math>. eventually he puts them all together to get <math>S^2 |1,0>=(3/4\hbar^2+3/4\hbar^2+(\hbar^2)/2)|1,0>=2\hbar^2|1,0></math> proving that |1,0> state is indeed an eigenvector of the <math>S^2</math> operator with eigenvalue of <math>2\hbar^2</math>. the procedure is similar for the other 3 states. hope this helped.

Thank you for answering. But i cannot understand still why we need to multiply <math>1/sqrt2</math>.

I thought the <math>\frac{1}{sqrt{2}}</math> came from normalization of the “spinor” state.

It mainly comes from Eq. 4.177 in part that we are dealing with antisymmetric state for this situation.

For example, let's look at the first term, (Sx(1)*Up-spin)=(h_bar/2*Down-spin). As you remember this, Sx can be written by (S+ + S-)/2. If you substitue this for Sx(1), then you would get the first term and the rest of terms can be solved in similar way. I hope this helped.

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