# Thread: Electron in a Box

1. ## Electron in a Box

Given an electron is contained in a box which has length $\displaystyle 20nm$ (from $\displaystyle x=0$ to $\displaystyle x=20nm$):

1.) Write what $\displaystyle \hat{H}$ is for the electron.

2.) Find $\displaystyle \hat{H}$ 's eigenfunctions and eigenvalues. Then, determine what the general solution $\displaystyle \psi(x,t)$ is for the system. (Sol'n to time-dep. Schrodinger Eq.).

3.) What do the first five energy levels look like. Suppose an electron is excited to the 4th energy level. Determine the possible wavelengths of the light emitted when it deexcites.

4.) If this were a golf ball, would you be able to see energy quantization?

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So 4.) is clearly no, since quantum mechanics deals with particles!

2. Originally Posted by DiscreteW
Given an electron is contained in a box which has length $\displaystyle 20nm$ (from $\displaystyle x=0$ to $\displaystyle x=20nm$):

1.) Write what $\displaystyle \hat{H}$ is for the electron.

2.) Find $\displaystyle \hat{H}$ 's eigenfunctions and eigenvalues. Then, determine what the general solution $\displaystyle \psi(x,t)$ is for the system. (Sol'n to time-dep. Schrodinger Eq.).

3.) What do the first five energy levels look like. Suppose an electron is excited to the 4th energy level. Determine the possible wavelengths of the light emitted when it deexcites.

4.) If this were a golf ball, would you be able to see energy quantization?

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So 4.) is clearly no, since quantum mechanics deals with particles!
First of all, ignore the electron spin. In non-relativistic QM you have to put that in by hand.

1. So what is the Hamiltonian for a particle in a box? All particles in a box have the same form for the Hamiltonian, in terms of the mass and width of the potential well. Then use the mass of the electron and your width. Basically this is just plug'n'chug.

2. Start by solving the time-independent S-equation: $\displaystyle H \psi = E \psi$. Once you have that, then use the time evolution operator to find solutions for the time-dependent S-equation. (We can get away with this because the potential is not a function of time.)

3. The energies will depend on a integer, usually chosen to be n. So n = 1, 2, 3, 4, 5 in your energy eigenvalue equation. Then if the electron is in n = 4, it can fall to 3, 2, and 1. So find the difference in energy between the levels and then $\displaystyle E = \frac{hc}{\lambda}$.

4. Yeah, golf balls don't do well in QM, but you have to prove you can't see the energy quantization. Think of it this way: What is the approximate mass of the golf ball? The golf ball has to sit in a potential well that is at least as wide as the golf ball. So what is the diameter of a golf ball? (Order of magnitude numbers should work just fine here. You don't need to go out and buy a golf ball.) The largest difference in energy levels will be between the 1st level (n = 1) and the "infinith" level (n = infinity.) So find this difference to demonstrate that (realistically) the energy difference is too small to measure. (The energy spectrum has an upper bound, so what is $\displaystyle \lim_{n \to \infty}E_n$?)

-Dan

3. Originally Posted by topsquark
First of all, ignore the electron spin. In non-relativistic QM you have to put that in by hand.

1. So what is the Hamiltonian for a particle in a box? All particles in a box have the same form for the Hamiltonian, in terms of the mass and width of the potential well. Then use the mass of the electron and your width. Basically this is just plug'n'chug.

2. Start by solving the time-independent S-equation: $\displaystyle H \psi = E \psi$. Once you have that, then use the time evolution operator to find solutions for the time-dependent S-equation. (We can get away with this because the potential is not a function of time.)

3. The energies will depend on a integer, usually chosen to be n. So n = 1, 2, 3, 4, 5 in your energy eigenvalue equation. Then if the electron is in n = 4, it can fall to 3, 2, and 1. So find the difference in energy between the levels and then $\displaystyle E = \frac{hc}{\lambda}$.

4. Yeah, golf balls don't do well in QM, but you have to prove you can't see the energy quantization. Think of it this way: What is the approximate mass of the golf ball? The golf ball has to sit in a potential well that is at least as wide as the golf ball. So what is the diameter of a golf ball? (Order of magnitude numbers should work just fine here. You don't need to go out and buy a golf ball.) The largest difference in energy levels will be between the 1st level (n = 1) and the "infinith" level (n = infinity.) So find this difference to demonstrate that (realistically) the energy difference is too small to measure. (The energy spectrum has an upper bound, so what is $\displaystyle \lim_{n \to \infty}E_n$?)

-Dan
Okay, lets take a look at this:

So, we know that the total energy operator, or the hamiltonian is

$\displaystyle \hat{H} = \frac{\hat{p}^2}{2m} + V(x)$.

So, then it'd be $\displaystyle \hat{H} = \frac{\hat{p}^2}{2(20)} + V(x)$. Where does the width come into play, unless it's in the $\displaystyle \hat{p}$. Another way to write the Hamiltonian is:

$\displaystyle \frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x)$

Partial of what, however. We'd need something to distribute this with to make use of it.

For 2 and 3, I'm assuming we're using:

$\displaystyle \left[-\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + V(x)\right]u(x) = Eu(x)$ (what you have).

Since, this is just $\displaystyle \hat{H}u(x) = Eu(x)$.

But I thought the hamitonian was the total energy operator. So we can use $\displaystyle \hat{H}$ for E?

Looking at my notes, we have:

$\displaystyle E_n = \frac{n^2\pi^2\hbar^2}{2ma^2}, n=1,2,3,\ldots$.

Or, since $\displaystyle E_1 = \frac{\pi^2\hbar^2}{2ma^2}$ we have:

$\displaystyle E_n = n^2E_1$ and this is our energy quantization. I feel as though I need to use this for 3.

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So I don't have to use the time-indep schrod. eq.?

$\displaystyle \psi_n(x,t) = \sqrt{\frac{2}{a}}\sin{\frac{n\pi x}{a}}\, e^{\frac{-iEt}{\hbar}}, 0 < x < a$ or:

$\displaystyle i\hbar\frac{\partial}{\partial t}\psi = \left(\frac{\hat{p}^2}{2m} + V(x)\right)\psi$

This all gets confusing pretty quickly, especially when used to classical physics.