Quantum Mechanics particle in a box question

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The discussion focuses on solving a quantum mechanics problem involving an electron in a one-dimensional box and later in a two-dimensional box. The Schrödinger equation for the one-dimensional box is derived, along with boundary conditions and expressions for energy levels. The participant successfully calculates the energy levels and the wavelength of emitted photons when the electron decays from a higher energy state to the ground state. For the two-dimensional box, the need for separation of variables is emphasized to derive energy levels, with a suggestion to apply boundary conditions in both dimensions. The conversation highlights the importance of expressing results in terms of quantum numbers and correcting any miscalculations in the process.
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Homework Statement



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.

Homework 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

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 don't 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.) ?
 
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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)
 
sin(2pi+kx) = sin(kx)...you need to add n in somewhere for your answer of part i.
about the problem of part III i think the method is in many textbook.
1.Separation of variable.
2.Apply boundary conition for both x and y
3.Get energy eigen value
 

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