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GR9677 #77
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Alternate Solutions |
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Comments |
astro
2008-10-15 09:19:34 | If you multiply out all terms of (D), doesn't it just reduce to (B)?
medellin
2008-11-01 13:02:31 |
I agree with you!!
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99percent
2008-11-06 08:08:27 |
Yes..!! and I marked B while solving this problem only to see that it is wrong..!!
FUCK the ETS..!!
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Herminso
2009-08-21 15:41:36 |
That's true only if you assume that S1 and S2 commute, but it's not true in general. Then the best answer choice is (D) since the angular momentum operators S1 and S2 not necessarily commute!!
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Herminso
2009-08-21 15:51:38 |
Now, I can see that you are really right since they use S1S2=\frac{1}{2}\left((S_1+S_2)^2 - S_1^2 - S_2^2\right), that's mean that S1 and S2 commute truly. Thus the choices are really confuse since (D) can be reduced to (B).
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munster
2009-10-06 17:34:51 |
I made the same mistake initially, but after thinking about it a little bit, the difference between B and D is actually quite important. Particularly, we are not looking at the spin in a particular direction which has an operator  , and is well-defined. However, the total spin is only well-defined for  , i.e. \Psi) . For this reason, the energy of the ground state should not strictly be "defined" as b, but as d. ETS IS trying to trick us, but we should be wary of the fact that the Hamiltonian itself should not be defined by total spin operators that are not squared (i.e. they are asking which of the correct answers is "more correct").
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echo
2009-10-26 23:33:48 |
Scalars always commute, and 1 + 1 is equally correct as 2 as far as arithmetic is concerned. They must have messed up.
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|  | scorinaldi
2007-10-18 21:03:50 | hi yosun, great site.
what is the "proper" ( i.e. traditional ) way to solve this problem?
| | Richard
2007-09-28 13:31:11 | I believe there is a typo.
You have on the last line, \psi=(A+B)(A+B+1)\psi) .
I believe it should be, ^2\psi=(A+B)(A+B+1)\psi) by generalization.
|  | pablojm
2006-10-28 16:13:15 | Is it really ok to generalize A^2psi=A(A+1)psi and B^2psi=B(B+1)psi to (A+B)^2psi=(A+B)(A+B+1)psi?
Maybe I'm just confused, but it doesn't seem very obvious why that works.
tonyhong
2008-10-04 08:32:00 |
this generalization validates only if we want the ground state energy(minimum), at this time A vector + B vector = (A+B) vector, and (A+B)vector^2=(A+B)(A+B+1)
caution: this is only ground state, for other cases A vector + B vector ?= (A+B) vector
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