Two particle in a square potential well?

[H Psi equal E Psi]

Hi guys!

I'm struggling with the following problem:

Consider two distinguishable (not interacting) particles in a quadratic 2 dimensional potential well. So

$V(x,y)=\left\{\begin{matrix} 0,\quad\quad-\frac { L }{ 2 } \le \quad x\quad \le \quad \frac { L }{ 2 } \quad and\quad -\frac { L }{ 2 } \le \quad y\quad \le \quad \frac { L }{ 2 } \\ \infty ,\quad \quad \quad rest \end{matrix}\right.$

I am now asked to find the normalized wave function in the ground state for two particles within the given potential. I tried to solve the schroedinger equation by means of the method of separation of variables:

$\psi ({ x }_{ 1 },{ x }_{ 2 },{ y }_{ 1 }{ y }_{ 2 })=\alpha ({ x }_{ 1 })\beta ({ x }_{ 2 })\delta ({ y }_{ 1 })\varepsilon ({ y }_{ 2 })$

This was harder then i thought so i didn't quiet got an sensible answer...

The second part of the exercise is to replace the two distinguishable particle with two spin ½ (Not interacting) electrons. Now pauli's principle has to be taken into account. Since i didn't manged to find radial function I am not able to construct the anti symmetric wave function for the electrons:

$\psi ({ x }_{ 1,2 },{ y }_{ 1,2 })=\phi ({ x }_{ 1,2 },{ y }_{ 1,2 })\cdot \frac { 1 }{ \sqrt { 2 } } ((\left| \left \uparrow \downarrow \right> \right) -(\left| \left \downarrow \uparrow \right> \right))$

While $\phi (x,y)$ is the searched function (has to be symmetric).

Sorry for my English ( not my mother tongue )

Thanks and Cheers!

 

[mfb]

I moved the thread to our homework section.

If the particles are distinguishable and not interacting, you can solve for their wave functions individually and then combine them.

Since i didn't manged to find radial function

Why do you want to find a radial function (a function of r?)?

Finding an antisymmetric wave function works the same as for every problem, you just have to find the ground state for a single electron.