# Allowable energies for a Particle in a box

1. Nov 15, 2015

### Nick.

Hi,

I am wondering about additional solutions to the particle in a box problem. In the one dimension the wave functions allowed all make sense with nodes pinned to an infinite potential barrier at either end and then divided into node increments related to the length of the box. However, in 2 or more dimensions there seam to be other solutions which I can't find a discussion or clarity on.

For example;

In a 2d equal box we could now also have wave paths at angles to the box faces. So there would be one path at 45deg to the faces with a wavelength different to those running perpendicular - but I don't see this solution discussed? There are also similar paths at 45deg where we get say a half wavelength reflected into a full wavelength (or more) - now we could assemble a wide variety of different allowable wavelengths.

Extend this principle to 3d and we would get numerous solutions.

What am I missing?

2. Nov 15, 2015

### blue_leaf77

Putting the origin in one corner of the 2D box, one can write the eigenfunction to be $\psi_{mn}(x,y) = A \sin (n k_x x) \sin(mk_y y)$. Writing each of the sine term in the complex exponential form, you will get
$$\psi_{mn}(x,y) = B \left( e^{i(n k_x x+mk_y y)} -e^{i(n k_x x-mk_y y)} - e^{i(-n k_x x+mk_y y)} + e^{i(-n k_x x-mk_y y)} \right)$$.
In other words, the solution of 2D infinite box is a superposition of 2D plane waves. You also see, for a given $m$ and $n$, there can be no reflected wave whose wavelength differs from the incoming one.

3. Nov 15, 2015

### Nick.

Ok let's look at the wave number kx=nπ/L for the typical wave in 1D and this 2D solution. But we could also produce a standing wave, bound at nodes, at 45deg located in the middle of the side L equal to n45π/√L/22+L/22 - where n45is a not a whole number increment from n. This solution requires the wave to follow a path - rather than fill the space - but why is that not a suitable solution ?

4. Nov 15, 2015

### blue_leaf77

By following a path, do you mean similar to a vibrating rope instead of a vibrating surface?

5. Nov 15, 2015

### Nick.

Yes - that's a great analogy. Why aren't the rope solutions suitable as well as the surface?

6. Nov 15, 2015

### blue_leaf77

Wavefunction with a functional form like a vibrating rope might exist as an arbitrary wavefunction in a 2D infinite box, but it cannot serve as one of the eigenfunctions of the Hamiltonian because it doesn't satisfy the eigenvalue equation $H\psi_{mn} = E_{mn}\psi_{mn}$.