GR9677 #77



Comments 
NeuralLotus 20140926 19:22:40  The problem I am facing with this question is that if S1=S2, then B gives, H = 2JS1(S1+1/2). But, in this case, you should have H= JS1(S1+1), since S1^2 = S1(S1+1). None of the answers fit this constraint, though. Well, besides the trivial answer, A, assuming which, you'd have to assume that S1 is always 0, in which case S2 is as well.   mpdude8 20120420 18:13:30  I looked at this problem, and said, there must be elimination. No way I have time to do any algebra.
A is gone. E is gone, there's no division anywhere here. C is gone, taking the dot product is way more complicated than that.
B was an interesting choice, but on the basis of where the ions were "fixed", I picked D.   KanHengLee 20111010 02:34:18  I am really confused with the official answer.
In the last step (A+B)^2 = (A+B)(A+B+1)
is used. However, is that valid?
is the eigenfunction of both S1^2, S2^2,
but is it so for (S1+S2)^2?
I don't know how to prove that [(S1+S2)^2, S1^2]=0.
If I think in a wrong way, please help me!!!!
wavicle 20111013 13:42:28 
The point of this problem is that we don't know the eigenvalue of . We do however know the . Using the identityrnrnrnrnrnThe leap I believe you're supposed to make is that the eigenvalue of is which is the same as the other eigenvalues replacing the with rnrnSee also reply by 99percent to astro regarding sameness of (B) and (D)rnrnI suppose the lesson here is, always chose the unsimplified version of the answer?

  keradeek 20110923 00:41:27  The lowest energy state will be when the two ions have their angular momentum vectors most appositely aligned. That happens when m_z of ion 1 is S_1, and m_z of ion 2 is S_2. Thus the answer is B, which is the same as D, so this question has two answers.
keradeek 20110923 01:54:31 
(forgot to mention) taking the dot product leads to the answer.

keradeek 20110923 17:50:54 
Actually, I made a mistake. The lowest energy is when angular momenta are aligned, not oppositely aligned. I should also mention that the x and y components average out to zero. Is there a way to 'modify' posts? I find it annoying that I can't modify my post

keradeek 20110923 18:24:03 
I need to point out that the operator (S_1 + S_2)^2 (with hats on S_1 and S_2 to signify that they are operators, not scalars) does not automatically yield the eigenvalue (S_1 + S_2)(S_1 + S_2 +1) as the solution suggests. The operator J = S_1 + S_2 (with hats on them) is the angular momenta addition operator, and J can take on a range of values from (S_1 + S_2) down to S_1  S_2  in integer steps. For the ground state, the eigenfunction with J = S_1 + S_2 (scalar) will yield the lowest energy, hence the answer is D, and B.

keradeek 20110923 18:47:08 
J = S_1 + S_2 means of course, that they are aligned (as much as the uncertainty principle will allow). Conclusion: the ETS screwed up!

  flyboy621 20101105 21:13:13  In hindsight, I think you can arrive at a reasonable guess here without doing all that algebra.
The ground state energy should be negative, which rules out A.
It's not going to be a simple product of S1 and S2, since quantum angular momentum is never that straightforward. So B is pretty much out.
C doesn't make sense because it isn't symmetric upon interchange of S1 and S2 (the overall sign changes).
I can't rule out D.
E looks unlikely because it doesn't seem likely that you would need to divide by anything.
Pretty rough, I know, but heyyou only get a couple of minutes.   astro 20081015 09:19:34  If you multiply out all terms of (D), doesn't it just reduce to (B)?
medellin 20081101 13:02:31 
I agree with you!!

99percent 20081106 08:08:27 
Yes..!! and I marked B while solving this problem only to see that it is wrong..!!
FUCK the ETS..!!

Herminso 20090821 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!!

Herminso 20090821 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).

munster 20091006 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 welldefined. However, the total spin is only welldefined for , i.e. . 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").

echo 20091026 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.

raymondtco 20101110 16:59:39 
I totally agree D is equivilant to B since the S's in the express are nothing but scalar quantities. Of course they commute.

keradeek 20110922 02:18:08 
so what is the final word on this? is D the same as B?rn

ewcikewqikd 20140714 15:17:23 
"This problem can be solved (without much knowledge of quantum mechanics) by noting the following general arithmetic trick: "
This is COMPLETELY WRONG!
, it equals only if commute. In the context of the problem, .
So in their solution, they immediately made the assumption that commute.
This means that if "the trick" is true, , then answer B is equivalent to answer D.
The solution is BS, PROVE TO ME THAT THE ANSWER IS D WITHOUT USING "THE TRICK" THEN I WILL BELIEVE IT.

cgrman1 20150325 10:12:35 
Instead of the "trick"
I thought of it as:
defined as (+ )
so
S^2=( + )dot ( + )
=S1^2+S2^2+2 *
Rearranging,
* = 1/2[  S1^2  S2^2]
Let S^2= S(S+1) with S= S1+S2, and plug in:
* =1/2[(S1+S2)(S1+S2+1)S1(S1+1)S2(S2+1)]
Then you just have to multiple by J to get your answer.
See The Heisenberg model for ground state in the ferromagnetic case for similar calculations of spin.

cgrman1 20150325 10:16:25 
But I did not mean to imply that it fixes the commute issue, just that instead of using the formula/trick, I did that

  scorinaldi 20071018 21:03:50  hi yosun, great site.
what is the "proper" ( i.e. traditional ) way to solve this problem?
  Richard 20070928 13:31:11  I believe there is a typo.
You have on the last line, .
I believe it should be, by generalization.
  pablojm 20061028 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 20081004 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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