# Ground state splitting due to spin interaction

1. Apr 20, 2014

### CAF123

1. The problem statement, all variables and given/known data
The interaction between the spins of the two particles in a hydrogenic atom can be described by the interaction Hamiltonian $$\hat{H_I} = A \hat{S_1} \cdot \hat{S_2}.$$ Compute the splitting of the ground state due to $\hat{H_I}$. Both particles have spin 1/2.

2. Relevant equations

$\hat{S}^2 = \hat{S_1}^2 + \hat{S_2}^2 + 2 \hat{S_1} \cdot \hat{S_2}$

3. The attempt at a solution
Using the relevant equation, I can rewrite the Hamiltonian in terms of operators that are diagonal in the coupled basis.

So $$\hat{H_I} |S,S_z \rangle = \frac{1}{2} A (\hat{S}^2 - \hat{S_1}^2 - \hat{S_2}^2)|S,S_z \rangle,$$ where $|S,S_z \rangle$ is the spin part of the system wavefunction expressed as a linear combination of the coupled basis vectors.

In the ground state, the orbital part of the wavefunction will be symmetric and since we are dealing with a 2 spin 1/2 particle system, the spin part must be antisymmetric. Therefore $|S,S_z \rangle = |0,0 \rangle.$
So, total spin is 0 and the z components of the individual electrons are oriented in different direction, with the same magnitude. So the result is, by plugging in numbers into the eqn above, $H_I |0,0\rangle = -3/4 A \hbar^2 |0,0\rangle$ So does the eigenvalue there give the interaction energy of the two electrons? Why does this correspond to the splitting?

Many thanks

Last edited: Apr 20, 2014
2. Apr 20, 2014

### TSny

What are the two spin 1/2 particles that you are concerned with here? Are they identical particles? Why must the spin part be anitsymmetric?

3. Apr 21, 2014

### CAF123

Hi TSny,
We are considering a hydrogen-like atom, and the preamble to the problem talks about a bound state of a muon (or positron) and an electron. I suppose they are identical in every way except in their masses (mass of positron = mass of electron ≠ mass of muon.) But if we are only concerned with their spins, then for this part of the problem perhaps we can treat them as identical.
My reasoning was in the ground state, the orbital component of the wavefunction is of the form $\Phi = u_{100}(\underline{x_1})u_{100}(\underline{x_2})$, which is symmetric. Since we are dealing with a fermionic system (both particles spin 1/2) the total wave function must be antisymmetric. From this, I thought then that the spin component must be antisymmetric. The only candidate for the spin state would then be $|0,0\rangle$. Is that right?

4. Apr 21, 2014

### TSny

If one particle is an electron, then the other particle would also need to be an electron if you are dealing with a system of identical particles. An electron and a muon are distinguishable particles. The wavefunction of the system does not need to be antisymmetric under the exchange of the particles.

So besides the $|0,0\rangle$ spin state, what other spin state(s) could you have?

5. Apr 21, 2014

### CAF123

I see, so then there is no constraint on the total wavefunction being antisymmetric. Given that both particles are spin 1/2, the total spin of the coupled system is 1 or 0. Label a coupled state like $|S_1, S_2, S, S_z\rangle \rightarrow |S, S_z \rangle$ where $S_1, S_2$ are the total spins of particles 1 and 2 (which are fixed = 1/2, so I will shorten the notation), $S$ is the total spin of the composite system and $S_z$ is the total z component.

We have that $S_{1,z} + S_{2,z} = S_z$, so possible values of $S_z$ are 1,0,-1 if total spin is 1 and 0 if total spin is 0. So the other spin states are $|1,0\rangle, |1,-1\rangle, |1,1\rangle$.

6. Apr 21, 2014

### TSny

What do you get if you operate on these triplet states with $\hat{H}_I$?

7. Apr 21, 2014

### CAF123

I get $$\hat{H_I}|1,i\rangle = \frac{A}{4}\hbar^2|1,i\rangle$$ $i \in \left\{-1,0,1\right\}$

8. Apr 21, 2014

### TSny

OK, so what are the energy levels for the singlet and triplet states? What is the "splitting"?

9. Apr 21, 2014

### CAF123

So would the energies for each of the levels be the energy of the ground state + the energy due to the interaction between the spins?

Energy of ground state is $E_n$~$-\frac{1}{n^2}$ and the energy of interaction is either $-3/4 A\hbar^2$ or $A/4 \hbar^2$ depending on the spin state. So the splitting is $A\hbar^2$?

10. Apr 21, 2014

### TSny

Yes. That sounds right to me. Good.

11. Apr 21, 2014

### CAF123

Thanks. In the energy level derivation for the hydrogen atom to produce the 1/n2 dependency, the spin interaction of the proton and the electron was not factored. Why was this? If we were to include such effects, would we just be increasing the accuracy of the model?

12. Apr 21, 2014

### TSny

Yes, that's right. The spin-spin interaction in hydrogen is a very small correction ( see hyperfine ) . The splitting of the ground state in hydrogen is only about 6 x 10-6 eV compared to the ground state energy (-13.6 eV).

13. Apr 21, 2014

### CAF123

Lorentz violating interactions may be described by new terms in the Hamiltonian of the form $$\delta V = \frac{g^2}{2M}(\underline{E}^2 - \underline{E_z}^2),$$ where $\underline{E}$ is the electric field due to the Coulomb interaction. The wavefunction for the ground state of a hydrogen-like atom is of the form $u_o = Ca_o^{-3/2} \exp(-r/a_o)$. Ignoring the spin of the particles, the shift of the ground state energy can be computed at first order in perturbation theory under the assumption that the electric field vanishes for r < ao. Evaluate the dependence of the energy shift on ao.

I think this just comes down to evaluating an integral of the form $\langle u_o | \delta V | u_o \rangle = \int_V u_o^2 \delta V \,\text{d}V$. I think I can write the electric field like $\underline{E} = e/kr \underline{e}_r \Rightarrow E_z = (e/kr) \cos \theta \underline{e}_z$, with $\theta$ the angle from the z axis (usual coordinate in spherical polars.) Then I would evaluate the integral, with the r bounds from ao to ∞. Does that setup look okay?

14. Apr 21, 2014

### TSny

I've never seen such a "Lorentz violating interaction". I don't understand the motivation for the expression δV nor the reason for the assumption that E vanishes for r < ao.

But your procedure seems sound to me!

15. Apr 21, 2014

### CAF123

Indeed, those were two of my follow up questions. g as in Lande g factor? Although I did not read much about that yet. My other question was what would happen if we did consider the spin of the particles? Would the perturbation theory not be applicable?

Thanks, I made a small typo above - the electric field should have been proportional to 1/r2. With that, my r integral is like $$\int_{a_o}^{\infty} \frac{e^{-2r/a_o}}{r^2}\,\text{d}r$$ but I don't think that exists.

I was trying to make my integrands dimensionless so that I did not have to evaluate them, but I could not.

16. Apr 21, 2014

### TSny

I don't know enough to answer those questions.

That integral would exist (but maybe not expressible in terms of elementary functions). But I don't think that's the integral you need to evaluate. Did you use the correct expression for the volume element in spherical coordinates?

17. Apr 22, 2014

### CAF123

Hi TSny,
No problem
Indeed it would exist. I dismissed the form of the limits.

I think so. My integral should be $$\frac{g^2}{2M} \int_V u_o^2 (E^2 - E_z^2) \,\text{d}V$$ where $E^2 - E_z^2 = \frac{e^2}{k^2r^4}\sin^2 \theta$ using my expressions previously. $dV = r^2 \sin \theta dr d\theta d\phi$ and if I sub this in, the $1/r^4$ on the denominator coming from the electric field term reduces to $1/r^2$ from the volume element.

18. Apr 22, 2014

### TSny

Yes, you're right. My mistake. So, the integral is not elementary. But you can still find how the shift in energy depends on $a_0$.

19. Apr 22, 2014

### CAF123

Yes, introduce the dimensionless variable $u = r/a_0$. I am getting then that the shift in energy is proportional to 1/ao4.

Last edited: Apr 22, 2014
20. Apr 22, 2014

### TSny

That looks right.