Eigenstates of composite spin hamiltoninan

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Homework Help Overview

The discussion revolves around the eigenstates and energies of a Hamiltonian for a system comprising a spin-3/2 particle and a spin-1/2 particle. The Hamiltonian includes terms involving the interaction of the spins and their respective z-components, raising questions about the appropriate formulation and interpretation of the spin operators involved.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the addition of angular momenta and the use of Clebsch-Gordan coefficients for constructing composite spin states. Questions arise regarding the definition of the scalar product between the spin operators and the implications of expressing the Hamiltonian in terms of total spin operators.

Discussion Status

Participants are actively questioning the definitions and relationships between the spin operators, with some suggesting potential reformulations of the Hamiltonian. There is a recognition of the complexity involved in expressing the Hamiltonian purely in terms of total spin and z-components, and some guidance has been offered regarding the use of Clebsch-Gordan coefficients in constructing the states.

Contextual Notes

There is an ongoing discussion about the nature of the spin operators and the assumptions being made about their representation in the Hamiltonian. The original poster expresses uncertainty about the formulation of the Hamiltonian and the role of the individual spin operators in the context of the composite system.

center o bass
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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?
 
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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.
 
S1 should be a vector having 4 entries, S2 a vector having 2 entries.

Why is that? Is'nt the hamiltonian originally defined in a classical sense 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?
 
center o bass said:
S_{tot}^2 = S_1^2 + S_2^2 + 2 \vec S_1 \cdot \vec S_2
Use this relation and solve for \vec S_1 \cdot \vec S_2.
 
susskind_leon said:
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.
You're confusing the states with the operators.
 
Yes, you're right, sorry!
 
vela said:
Use this relation and solve for \vec S_1 \cdot \vec S_2.

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?
 
center o bass said:
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.
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.
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?
Yes. (So far you've referred to them as CB and GB coefficients. I assume you mean the Clebsch-Gordan coefficients. :smile:)
 
[QOUTE]Yes. (So far you've referred to them as CB and GB coefficients. I assume you mean the Clebsch-Gordan coefficients. :smile:)[/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
Yup.
 

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