# 2 particles in an infinite quantum well

1. Nov 12, 2009

### Joqe

If we have 2 particles in ground state in an infinite quantum well of lenght L (one-dimetion). Suppose then that the unperturbed energy is two times the ground energy of only one particle in an infinite qunatum well of lenght L.

Then I want to use first order perturbation theory to find the perturbed energy if the perturbation is A*d(x2-x1) (diracs-delta function), interaction between the particles or something, where A is a constant.

How do I construct the wave equation for this system en how do I calculate the perturbation?

I have tried with various linear combinations with no convincing success.

2. Nov 13, 2009

### criz.corral

Hi Joqe!

First of all you have to find the wave function of the two-particles system.
Do you know if there are identical or not? This condition and the spin of
each particle give it to you if the wave function will be symmetrical or
anti-symmetrical. Then you can use perturbation theory to found the value

$$<\Psi_{12}|A\delta(x_{1} - x_{2})|\Psi_{12}>$$

3. Nov 13, 2009

### Joqe

Yes the particles are indeed identical and have spin 1/2.

As for the perturbation, a friend of mine suggested that I should ues degenerat perturbation theory, which I'm not familliar with and I don't see why I cannot use that you have suggested.

Later I'm supposed to answer why the anti-symmetrical wave functions does'nt give any shift of the energy. And what the total spin quantum numer S is.

4. Nov 13, 2009

### criz.corral

Hi again Joqe:

Roughly speeking, first order degenerate perturbation theory diagonalizes
function and the eigenvalues to the energy. In this case you may use it,
because you have two different states with the same energy. Particle
1 on $$n_{1}=1$$ and spin "up" and particle 2 on $$n_{2}=1$$ and
spin "down". There are different states but with the same energy. But, if you
see, you doesn't have the problem, because you interaction is diagonal.
I think that you doesn't have any problem to calculate:

$$<\Psi_{12}|A\delta(x_{2}-x_{1})|\Psi_{12}>$$

Now, we still have the wave function problem. How we found it?

If there are Fermions (half-integer spin) you know that the total wave function
have to be Anti-symetric. But, this gave us two system configuration.

$$\Psi_{n_{1},n_{2}}^{AS}(x_{1},x_{2}) = \Phi^{AS}(x_{1},x_{2})\chi^{S}(S_{1},S_{2})$$

or

$$\Psi_{n_{1},n_{2}}^{AS}(x_{1},x_{2}) = \Phi^{S}(x_{1},x_{2})\chi^{AS}(S_{1},S_{2})$$

Where $$\chi^{AS}$$ is the triplet state for the two 1/2 spin particles
and $$\chi^{S}$$ is the singlet state. And:

$$\Phi^{AS}\equiv\frac{1}{\sqrt{2}}(\psi_{n_{1}}(x_{1})\psi_{n_{2}}(x_{2}) - \psi_{n_{1}}(x_{2})\psi_{n_{2}}(x_{1}))$$

$$\Phi^{S}\equiv\frac{1}{\sqrt{2}}(\psi_{n_{1}}(x_{1})\psi_{n_{2}}(x_{2}) + \psi_{n_{1}}(x_{2})\psi_{n_{2}}(x_{1}))$$

Are the spatial part of your total wave function

The interaction pontential $$A\delta(x_{2} - x_{1})$$ doesn't depends of the
spin part of the wave function. When you calculate:

$$<\Psi^{AS/S}|A\delta(x_{2} - x_{1})|\Psi^{AS/S}> = <\psi^{AS/S}|A\delta(x_{2} - x_{1})|\psi^{AS/S}><\chi^{AS/S}(S_{1},S_{2})|\chi^{AS/S}(S_{1},S_{2})> = <\psi^{AS/S}|A\delta(x_{2} - x_{1})|\psi^{AS/S}>$$

Because the spin parts are orthogonal.

Note what happens when $$n_{1}=n_{2}=1$$ you only have $$\Phi^{S}$$
because the anti-symmetrical part annihilates. Then:

$$\Phi^{S}(x_{1},x_{2})=\frac{2}{L}Sin(\frac{\pi x_{1}}{L})Sin(\frac{\pi x_{2}}{L})$$

Sorry about my english, I'm still learning.

Cordially

Cristóbal.

Last edited: Nov 13, 2009
5. Nov 13, 2009

### Joqe

Hi again criz.corral!

Thank you so much, I think you had a terrific answer and everything is sorted out in a sens. Still somethings to concider but I'm very pleased with your answer.

Thank you again.
Joqe