# Behaviour of 3-fermions in a Magnetic Field

1. Jun 10, 2014

### ranytawfik

When solving Pauil-Schrodinger Equation for 3-fermions in a potential well and a magnetic field, we basically solve the vector,

$\Psi(r_{1},r_{2},r_{3})=\left( \begin{array}{cc} \psi\uparrow(r_{1},r_{2},r_{3}) \\ \psi\downarrow(r_{1},r_{2},r_{3}) \end{array} \right)$

The $\psi\uparrow$ represents the wavefunction for all-spin up case, and $\psi\downarrow$ represents the wavefunction for all-spin down case. And for these cases both $\psi\uparrow$ and $\psi\downarrow$ has to be anti-symmetric to include the effect of Pauli Exclusion Principle since all the particles have the same spin.

But how about the wavefunction for the case when you have mixed spins? And how should you reflect this in the symmetric or anti-symmetric properties of $\Psi$?

2. Jun 12, 2014

### rigetFrog

This is a fun question.

I'm gonna focus on the ground states when magnetic field and confining potential try to counter act each other.

If the magnetic field is very strong, all the spins will be parallel, and the n=1,2,3 states will be occupied by one electron.

If the magnetic field is weak, n=1 will be occupied by two electrons one of each spin, and n=2 will be occupied with one electron

This is a very elegant way to understand the concept of Landau levels, and how the Fermi level shifts with increasing field.

3. Jun 13, 2014

### ranytawfik

Hi rigetFrog,

If the magnetic field is very strong, then (as you stated) it’s going to be straight forward (simply one fermion per 1, 2, and 3).

If the magnetic field is weak, you don’t have to have an opposing confining potential to have mixed spins. The magnetic interaction between the particles’ spin can lead to that scenario as well.

The question is: how by solving Pauli-Schrodinger Equation I can get the spin configuration that results in minimum ground state? I hope I don’t have to solve for all the combinations of spin ordering, and pick the case with the minimum energy.

4. Jun 13, 2014

### rigetFrog

The purpose of opposing confining potentials is to reduce the complexity of the system from a continuum of allowed energies (which is more challenging to solve), to several allowed energies that are multiples of n^2.

Don't worry about magnetic interaction between spins. Focus on how they (your Fermions) cannot have the same quantum numbers.

Unfortunately, you will probably have to write the full wave function and use the Slater determinant to make it anti-symmetric.

PSI =
+ psi_1(x1,s1)*psi_2(x2,s2)*psi_3(x3,s3)
- psi_1(x1,s1)*psi_2(x3,s3)*psi_3(x2,s2)
+ psi_2(x1,s1)*psi_1(x2,s2)*psi_3(x3,s3)
- psi_2(x1,s1)*psi_1(x3,s3)*psi_3(x2,s2)
+ .....

for each psi, you already know you can't get above n = 3 for the ground state, so you only need to use those three terms.

I imagine doing this will take up several pages of paper.

5. Jun 16, 2014

### ranytawfik

Now my problem with your formulation of $\Psi$ is the following: In Pauli-Schrodinger equation, there is coupling between $\psi\uparrow$ and $\psi\downarrow$ that occurs in the differential equation due to the Pauli's matrices. And this is captured through using $\Psi$ on a form of vector.

In the way you showed $\Psi$, how would that coupling be captured? Or how Pauli's matrices should be stated?

6. Jun 16, 2014

### rigetFrog

Syntactically, I absorbed the spinor (spin vector) term into psi.

I want to say the Pauli matrices will be diagonal, corresponding to spin states that are parallel with the field.

Explicitly

U = -B.S.spinor

U is energy, B is the magnetic field, and 'spinor' is the spin vector and S is a vector of pauli matrices

S = Sx x_hat + Sy y_hat + Sz z_hat

where _hat denotes the unit vector and Sx,Sy,Sz are the paul spin matrices. If you assume B = Bz z_hat, then Sx and Sy can be ignored.

Your question about coupling between spinors will become relevant if you try to measure the total spin of the system. Then you'll have to use the Clebsch Gordon coefficients to transform the basis.