Quantum Mechanics Problem. Help!!

scorinaldi

Hey gang,

This is a physicsGRE problem ( GR9677 question #77 ) that I have yet to see anyone answer well.

Please let me know if I'm not being clear. Here goes:

Question:

"Two ions, 1 and 2, at fixed separation, with spin angular momentum operators S_1 and S_2, have
the interaction Hamiltonian H = -JS_1 (dot) S_2, where J > 0.

The values of (S_1)^2 and (S_2)^2 are fixed at S_1(S_1 + 1) and S_2(S_2 + 1), respectively.
Which of the following is the energy of the ground state of the system?"

Ok, that's the question. The answer is:

-(J/2)[( S_1 + S_2)( S_1 + S_2 + 1) - S_1(S_1 + 1) - S_2(S_2 + 1)]

How did we get this answer? Can someone fill in the steps?

javierR

...which is, by the way, simply E= -J*s_1*s_2. You can write this at the start if you see that the most negative energy will be when the spins align with each other and have the largest individual magnitudes in the z-direction (taking the states to be eigenstates of Sz), as in classical dipole problems.

Gokul43201

A reminder of the rules, to all above, in case you haven't had a chance to read them yet:
http://www.physicsforums.com/showthread.php?t=5374

Thanks, and welcome to PF.

scorinaldi

hi gokul,

this problem was not assigned to me ( or anyone else ) for homework. it is a problem from a practice physics GRE from the 90's where the answers were distributed by ETS.

the distributed packet is here: ( http://phys.columbia.edu/~hbar/Physics-GRE.pdf )

answers and discussion of this particular problem, and all old physics GRE problems are freely available to anyone here: http://grephysics.net/ans/9677/77

i apologize if i posted this in the wrong forum, in that it is not homework. I just wanted to get another insight into ways to arrive at the problem besides the one given on the Physics GRE bulletin boards.

thanks

ZapperZ

It doesn't really matter where the question comes from. When it is the TYPE that involves working out a solution, the rules apply and it belongs on this forum.

Zz.

JoAuSc

I'm having problems with this question too, even though I've checked the grephysics.net page and this thread at physicsgre.com, not to mention this thread. The issue is that if [tex]S_1[/tex] and [tex]S_2[/tex] are scalars, then when you carry out the multiplication this answer (D) is identical to answer B, which is [tex]-J S_1 S_2[/tex], and we can't have two identical answers.

Here's some work, borrowing heavily from the grephysics.net site:

We start with:
[tex]H = -J \bold{S_1} \cdot \bold{S_2}[/tex]
Using the identity
[tex]\bold{a} \cdot \bold{b} = \frac{1}{2} ( (\bold{a} + \bold{b})^2 - a^2 - b^2 ) [/tex]
we get
[tex]H = - \frac{J}{2} ( (\bold{S_1} + \bold{S_2})^2 - (S_1)^2 - (S_2)^2 ) [/tex]

Now, [tex](S_1)^2 \psi = s_1 (s_1 + 1) \psi[/tex], and the same goes for [tex]S_2[/tex], where I use lowercase s for eigenvalues, a non-bold uppercase S for non-vector operators, and a bold uppercase [tex]\bold{S}[/tex] for vector operators. We now have

[tex]H = - \frac{J}{2} ( (\bold{S_1} + \bold{S_2})^2 - s_1 (s_1 + 1) - s_2 (s_2 + 1)) [/tex]

Finally, a comment on the physicsgre.net site says that
[tex](\bold{S_1} + \bold{S_2})^2[/tex]
becomes
[tex](s_1 + s_2) (s_1 + s_2 + 1)[/tex]
and thus we get answer D. (I don't understand this step too well.) But the problem is, as I've stated above, if the S_1 and S_2 values are scalars, then answer B and answer D are identical. The answer is supposed to be an energy, rather than a Hamiltonian. Is the problem wrong or are answers B and D supposed to be different?

borgwal

I don't have access to the full question, but here's what I gather:

first, the problem is clearly about how to add two angular momenta. The answer D is indeed correct: the total angular momentum of the two systems together lies between
two values (look them up in your textbook!), and the maximum possible value gives rise to the lowest energy.

second, if answer B is in fact what it is claimed to be here (as I said, I haven't seen the full question) then that is the same answer. So, someone screwed up the question in that case.