Spin inversion under dipole-dipole interaction of fermions

In summary: I don't really know where to start from. This exercise is included in the Identical Particles homework series of my Quantum Mechanics II course.
  • #1
Siberion
33
1

Homework Statement



Consider two spin 1/2 particles interacting through a dipole-dipole potential

[tex]\hat{V} = A\frac{(\hat{\sigma_1} \cdot \hat{\sigma_2})r^2 - (\sigma_1 \cdot \vec{r})(\sigma_2 \cdot \vec{r})}{r^5} [/tex]

If both spins are fixed at a distance d between each other, and at t = 0 one of them is parallel to the radius vector r, whereas the other one is anti-parallel to r, find the first moment when the orientations of the spins have inverted.

(Please note that the wording of the problem doesn't mention what does A nor sigma mean, but I assume A must be a normalisation constant and sigma/2 the corresponding spin operators)

Homework Equations


[/B]
Time-dependent Schrödinger equation:

[tex]-\frac{\hbar^2}{2m}\nabla^2\Psi + V(\mathbf{r})\Psi
= i\hbar \frac{\partial\Psi}{\partial t}[/tex]

The Attempt at a Solution


[/B]
I don't really know where to start from. This exercise is included in the Identical Particles homework series of my Quantum Mechanics II course. I'm not sure what does "find the first moment" really mean. How can there be "a first moment" when in general all we tan talk about is probabilities of measuring certain observables?

As I have to deal with the time variable, I was guessing it might have something to do with the time-dependent Schödinger equation. I guess the wave function needs to evolve with time, and the orientation of the spins would have to change either continuously or spontaneously. Plus the wave function must be anti-symmetric because we are dealing with fermions... Uhm... I am kind of lost. Any kind of help or reference would be greatly appreciated.
 
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  • #2
Siberion said:
(Please note that the wording of the problem doesn't mention what does A nor sigma mean, but I assume A must be a normalisation constant and sigma/2 the corresponding spin operators)
##A## is not a normalization constant, but rather parametrizes the strength of the interaction.

Siberion said:

Homework Equations


[/B]
Time-dependent Schrödinger equation:

[tex]-\frac{\hbar^2}{2m}\nabla^2\Psi + V(\mathbf{r})\Psi
= i\hbar \frac{\partial\Psi}{\partial t}[/tex]
The particles being fixed, I don't know why you have a momentum operator there. The first part of such a problem is determining what the Hamiltonian is, and then you can write the proper Schrödinger equation.

Siberion said:
I'm not sure what does "find the first moment" really mean. How can there be "a first moment" when in general all we tan talk about is probabilities of measuring certain observables?
But what if the probabilities are time dependent?

Siberion said:
As I have to deal with the time variable, I was guessing it might have something to do with the time-dependent Schödinger equation. I guess the wave function needs to evolve with time, and the orientation of the spins would have to change either continuously or spontaneously. Plus the wave function must be anti-symmetric because we are dealing with fermions...
Exactly. You have to start by writing the initial state of the system, then find its time evolution (which will be continuous, I don't know why it would be "spontaneous"). As for the parity of teh wave function, just make sure that the initial wave function is anti-symmetric, and the time evolution will preserve the parity.
 
  • #3
Thank you very much for your reply. I am getting some problems with the algebra:

As the particles are fixed, the hamiltonian is going to be equal to the potential energy:

[itex]\hat{H} = \hat{V}[/itex]

Generally speaking,

[itex]\hat{\sigma_1} = \sigma_{1x} \hat{i} + \sigma_{1y} \hat{j} + \sigma_{1z} \hat{k}[/itex]
[itex]\hat{\sigma_2} = \sigma_{2x} \hat{i} + \sigma_{2y} \hat{j} + \sigma_{2z} \hat{k}[/itex]

Let [itex]\vec{r}[/itex] point in the [itex]\hat{k}[/itex] direction, i.e. [itex]\vec{r}=z\hat{k}[/itex]

At t = 0, we have:

[itex]\hat{\sigma_1} = \sigma_{1z} \hat{k}[/itex]
[itex]\hat{\sigma_2} =-\sigma_{2z} \hat{k}[/itex]

[itex] (\hat{\sigma_1} \cdot z\hat{k})(\hat{\sigma_2} \cdot z\hat{k}) = -z^2 \sigma_{1z} \sigma_{2z}[/itex]

whereas

[itex] \hat{\sigma_1} \cdot \hat{\sigma_2} = - \sigma_{1z} \sigma_{2z}[/itex]

thus the potential vanishes at t = 0. I might be messing up at some very basic mistake, right?

Thanks in advance!
 
  • #4
Sorry, I think my reasoning was erroneous (it still might be)

I think it is better to say that the initial state of the system can be described by the singlet state (because they point in opposite directions at t = 0, and the wave function needs to be antisymmetric):

[itex] \Psi(0) = 1/ \sqrt{2} ( |+ \rangle |- \rangle - |-\rangle |+ \rangle)[/itex]

Now, to construct the time evolution operator, the hamiltonian of the system is given by the dipole interaction. As I said previously, in general we have

[itex] \hat{\sigma_1} \cdot \hat{\sigma_2} = \sigma_{1x}\sigma_{2x} + \sigma_{1y}\sigma_{2y} + \sigma_{1z}\sigma_{2z}[/itex]

and (assuming that [itex] \vec{r} [/itex] points in the k direction

[itex] \hat{\sigma_i} \cdot \vec{r} = d \sigma_{iz}[/itex]

so the second term in the potential would be [itex]d^2 \sigma_{1z}\sigma_{2z}[/itex], while the first term corresponds to [itex] d^2(\sigma_{1x}\sigma_{2x} + \sigma_{1y}\sigma_{2y} + \sigma_{1z}\sigma_{2z})[/itex]

This way, the hamiltonian would read
[itex] \hat{H} = \frac{A}{d^3}(\sigma_{1x}\sigma_{2x} + \sigma_{1y}\sigma_{2y} + \sigma_{1z}\sigma_{2z} - \sigma_{1z}\sigma_{2z})[/itex]
[itex] \hat{H} = \frac{A}{d^3}(\sigma_{1x}\sigma_{2x} + \sigma_{1y}\sigma_{2y})[/itex]

I am not sure how to proceed from here. I was hoping to get a single-matrix expression for the hamiltonian so I could diagonalise it and construct the evolution operator. Does this procedure make some sense at all?
 
  • #5
The standard trick is to use ##\vec{S}^2 = (\vec{S}_1+\vec{S}_2)^2 = \vec{S}_1^2+\vec{S}_2^2+2\vec{S}_1\cdot\vec{S}_2##.
 
  • #6
vela said:
The standard trick is to use ##\vec{S}^2 = (\vec{S}_1+\vec{S}_2)^2 = \vec{S}_1^2+\vec{S}_2^2+2\vec{S}_1\cdot\vec{S}_2##.

do you suggest using that relation on the last part, or did I make a wrong choice from the beginning?
 

1. What is spin inversion under dipole-dipole interaction of fermions?

Spin inversion refers to the phenomenon where the spin state of a fermion (a particle with half-integer spin) is flipped or reversed due to its interaction with other particles. This can occur when fermions interact through their dipole moments, which are created by their electric charge and spin. Under certain conditions, the spin of a fermion can be inverted or reversed, resulting in a change in its magnetic moment.

2. How does dipole-dipole interaction affect the spin state of fermions?

When two fermions with dipole moments interact, they can exchange energy and momentum through their electric and magnetic fields. This exchange of energy and momentum can cause the spin of one fermion to flip, resulting in spin inversion. The strength of the dipole-dipole interaction depends on the distance between the fermions and their respective dipole moments.

3. What are the applications of spin inversion under dipole-dipole interaction?

Spin inversion under dipole-dipole interaction has many applications in various fields of physics, such as quantum computing, nuclear magnetic resonance, and solid-state physics. In quantum computing, spin inversion can be used to manipulate the spin state of qubits, the basic units of quantum information. In nuclear magnetic resonance, it is used to study the structure and dynamics of molecules. In solid-state physics, it is used to understand the behavior of magnetic materials.

4. Are there any experimental evidence of spin inversion under dipole-dipole interaction?

Yes, there have been several experiments that have observed spin inversion under dipole-dipole interaction. One example is the study of spin-polarized hydrogen gas, where the spin inversion of hydrogen atoms was observed due to their interaction with each other. Another example is the observation of spin inversion in ultracold atomic gases, where the dipole-dipole interaction between fermionic atoms was controlled and studied.

5. How does spin inversion under dipole-dipole interaction affect the behavior of fermions?

Spin inversion can have a significant impact on the behavior of fermions, especially in systems where the dipole-dipole interaction is strong. It can lead to changes in the magnetic properties of fermions, such as their magnetic moment and susceptibility. It can also affect the dynamics of fermionic systems, leading to phenomena such as spin-flipping and spin relaxation. Understanding spin inversion under dipole-dipole interaction is crucial for studying the behavior of fermions in various physical systems.

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