# Quantum Mechanics particle in a box question

1. May 11, 2009

### umagongdi

1. The problem statement, all variables and given/known data

Consider an electron that is constrained to be in a one dimensional box of size L, but is otherwise free to move inside the box.

i.) Write down the (time independent) Schrodinger equation for this particle, the boundary conditions for the wavefunction Ψ and find an expression for the energy levels.

ii.) Consider the process where the electron decays from the nth energy level to the ground state by emitting a photon. Find the wavelength of the emitted photon as function of L, n and m.

iii.) Consider now an electron that can freely move in a two dimensional square box. What are the energy levels in this case. Please motivate your answer.

2. Relevant equations

I let, h' = h/2π

EΨ(x) = h'/2m * d²Ψ(x)/dx² + U(x)Ψ(x)

p = h/λ

k = nπ/L = 2π/λ

3. The attempt at a solution

HI thanks for taking the time to help me. I have completed part i and ii and need them to be checked. As for part iii i dont have a clue any help is greatly appreciated. These are not h/w questions but past paper questions.

i.) EΨ(x) = -h'/2m * d²Ψ(x)/dx² + U(x)Ψ(x)

0<=x<=L

For the particle in the box U(x) = 0

EΨ(x) = -h'/2m * d²Ψ(x)/dx²

Ψ(x) = A1e^(ikx)+ A2e^(-ikx)
= (A1+A2)cos(kx) + i(A1-A2)sin(kx)
Ψ(0) = (A1+A2) = 0, therefore
A1 = -A2

Ψ(x) = i2A1sin(kx)
d²Ψ(x)/dx² = -i2A1k²sin(kx)

E[i2A1sin(kx)] = (-h'/2m)(-i2A1k²sin(kx))

E = (h')²k²/2m

ii.)

En = p²/2m
= h²/2mλ²
= h²k²/2m(2π)²
= (h')²n²π²/2mL²
= n²π²(h')²/2mL²

E = En - E1
= n²π²(h')²/2mL² - π²(h')²/2mL²
= π²(h')²/2mL²(n-1)

λ = hc/E
= hc/[π²(h')²/2mL²(n-1)]
= hc2mL²(n-1)/π²(h')²
= 8cmL²/h(n-1)

iii.) ???

Last edited: May 11, 2009
2. May 11, 2009

### diazona

For part (i), you should probably have an expression in terms of the quantum number n, not in terms of k.

In part (ii), I think your n2 turned into an n somewhere.

For part (iii), conceptually you can say that the electron is free to move in two dimensions independently. So there will be an energy in the x direction and an energy in the y direction. If you want to prove it to yourself mathematically, use separation of variables on the Schrödinger equation (i.e. assume a spatial wavefunction of the form $$\psi(x, y) = \psi_x(x)\psi_y(y)$$, plug in, and separate the equation into two equations, one in each dimension)

3. May 12, 2009