Eigenstates of composite spin hamiltoninan

1. Oct 20, 2011

center o bass

A spin-3/2 particle (labeled 1) and a spin-1/2 particle (labeled 2) are interacting via
the Hamiltonian

$$\lambda \vec S_1 \cdot \vec S_2 + g(S_{1z} + S_{2z})^2$$

where S1 is the spin operator of particle 1 (the spin-3/2) and S2 is the spin operator
of particle 2 (the spin-1/2). λ and g are positive constants (λ ≫ g).

I am asked to find the eigenstates and the corresponding energies for this hamiltonian (neglecting any spatial degrees of freedom).

Since we have a two particle system I guess one must consider addition of angular momenta and the CB-coefficients, but I am really unsure about this hamiltonian. Should one reformulate in in terms of total spin operators? Do we know how the total spin operator S acts without squaring it?

How do I get started here?

2. Oct 20, 2011

susskind_leon

Just one question: How exactly is the scalar product between S1 (spin 3/2) and S2 (spin 1/2) defined? S1 should be a vector having 4 entries, S2 a vector having 2 entries.

3. Oct 20, 2011

center o bass

Why is that? Is'nt the hamiltonian originally defined in a classical sence so that we think of theese vectors as classical 3-dimensional spin vectors? I have the habit of doing so until they act on quantum states, then I think of them as QM operators.

I guess you could define that scalar product as the product of the direct products

$$(S \otimes I)(I \otimes S),$$

furthermore we have the relation

$$S_{tot}^2 = S_1^2 + S_2^2 + 2 \vec S_1 \cdot \vec S_2 = (S \otimes I)^2 + (I \otimes S)^2 + 2 (S \otimes I)(I\otimes S)$$

which maybe could be used to define the action of

$$(S \otimes I)(I \otimes S)$$

on a state?

4. Oct 20, 2011

vela

Staff Emeritus
Use this relation and solve for $\vec S_1 \cdot \vec S_2$.

5. Oct 20, 2011

vela

Staff Emeritus
You're confusing the states with the operators.

6. Oct 20, 2011

susskind_leon

Yes, you're right, sorry!

7. Oct 20, 2011

center o bass

Why would I want to do that? :) I'm a little bit bothered by the fact that I can not express the Hamiltonian purely in terms of the total spin and the total z spin. I could rewrite it as

$$H = \frac{\lambda}{2} ( S_{tot}^2 - S_1^2 - S_2^2 ) + g S_{ztot}^2$$

but then I have the pesky S1 and S2 operators in there. Btw is it correct to assume that any such composite spin state of particle 1 and two can be written as

$$|s m\rangle = \sum_{m1,m2} C |\frac{3}2,m_1\rangle |\frac{1}2, m_2\rangle$$

where C are the GB coeffs?

8. Oct 20, 2011

vela

Staff Emeritus
That's not a problem because S12, S22, S2, and Sz all commute with each other.

If you think about it, you can't really describe everything in terms of just S2 and Sz. You have two quantum numbers for each particle, so the combined state needs four quantum numbers. It's not surprising that the Hamiltonian will depend on S12 and S22 as well.
Yes. (So far you've referred to them as CB and GB coefficients. I assume you mean the Clebsch-Gordan coefficients. )

9. Oct 20, 2011

center o bass

[QOUTE]Yes. (So far you've referred to them as CB and GB coefficients. I assume you mean the Clebsch-Gordan coefficients. )[/QUOTE]

Hehe, the _CG_ coefficients ;P If i assume that this is the case I get

$$H|sm\rangle = \left[\frac{\lambda}{2}(S_{tot}^2 - S_1^2 -S_2^2) +gS_{ztot}^2\right] |sm\rangle$$

$$= \left[\frac{\lambda}2 (\hbar^2s(s+1) - S_1^2 - S_2^2) + g\hbar^2 m\right] \sum C |\frac{3}2, m_1\rangle|\frac{1}{2} m_2\rangle$$
$$= \left[ \lambda \hbar^2 \left( \frac{2 s(s+1) - 9}{4}\right) + gm\hbar \right] |sm\rangle.$$

Is the states and the allowed energies now dictated by the allowed s and m's given s_1 and s_2?

10. Oct 20, 2011

vela

Staff Emeritus
Yup.