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| classes:2009:fall:phys4101.001:q_a_1204 [2009/12/06 21:38] – ely | classes:2009:fall:phys4101.001:q_a_1204 [2009/12/13 21:09] (current) – youmans |
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| 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? | 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? |
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| | === 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. |
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| ==== Esquire 12/4 1611 (Age of Military Time)==== | ==== Esquire 12/4 1611 (Age of Military Time)==== |
| ===Andromeda 12/4 8:53=== | ===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. | 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. |
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| | == Asprin 12/6 == |
| | Thank you for answering. But i cannot understand still why we need to multiply <math>1/sqrt2</math>. |
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| | == ice IX 12/7 01:50 == |
| | I thought the <math>\frac{1}{sqrt{2}}</math> came from normalization of the "spinor" state. |
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| | ==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. |
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| | ===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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