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--- classes:2009:fall:phys4101.001:q_a_1204 [2009/12/03 20:01] – yk
+++ classes:2009:fall:phys4101.001:q_a_1204 [2009/12/13 21:09] (current) – youmans
@@ Line 8: / Line 8: @@
 **Main class wiki page: ** [[home]]
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 ====Daniel Faraday 12/3 11am====
 Am I crazy, or is there a sign error in the eigenvalue in eq 4.136? As written it is
@@ Line 24: / Line 22: @@
 One more picky point.  This factor is not an eigenvalue strictly speaking since the vectors on the LHS and RHS are not the same.
+==== Ralph 12/4 10:50 am====
+Why is it that energy is discrete but linear momentum is not? Aren't they related? I'm having a hard time understanding this.
+=== ice IX 12/6 18:37 ===
+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>.
+===Dark Hemet 12/6 21:37===
+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.
+==== joh04684 12/4 1607 ====
+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?
+=== Jake22 12/13 1707 ===
+This is simply a matter of expanding it and noting that y commutes with px, and x with py, as in equation 4.10.
+==== Esquire 12/4 1611 (Age of Military Time)====
+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
+|1/2>
+Love,
+//Esquire//
+=== joh04684 12/4 1615 ===
+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.
+==== Asprin 12/4 6:00pm ====
+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?
+==Devlin12/5==
+I also didn't get that.
+===Andromeda 12/4 8:53===
+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.
+== Asprin 12/6 ==
+Thank you for answering. But i cannot understand still why we need to multiply <math>1/sqrt2</math>.
+== ice IX 12/7 01:50 ==
+I thought the <math>\frac{1}{sqrt{2}}</math> came from normalization of the "spinor" state.
+==Pluto 4ever 12/10 7:21PM==
+It mainly comes from Eq. 4.177 in part that we are dealing with antisymmetric state for this situation.
+===Blackbox 12/10===
+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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