Sunday, March 13, 2011

Mastering Physics: Particle in a 3-D Box: Counting States

Problems that require solving the three-dimensional Schrödinger equation can often be reduced to related one-dimensional problems. An example of this would be the particle in a cubical box.
Consider a cubical box with rigid walls (i.e., U(x,y,z)=\infty outside of the cube) and edges of length L. The general solution for this problem is
\psi(x,y,z) = A \sin \left(\frac{n_x\pi x}{L} \right)\sin \left(\frac{n_y\pi y}{L} \right)\sin \left(\frac{n_z\pi z}{L} \right),
where n_x, n_y, and n_z are all positive integers. Note that this solution is just the product of three solutions to the one-dimensional particle in a box. The energy corresponding to the three-dimensional solution is just the sum of the energies for each of the three one-dimensional solutions:

E=\frac{\pi^2 \hbar^2}{2mL^2}(n_x^2+n_y^2+n_z^2).

Part A
What is the smallest allowed energy E_0 for a particle in a cubical box?
Express your answer in terms of hbar, m, and L.

  E_0  = \frac{3}{2}\frac{{\pi}^{2}{\hbar}^{2}}{mL^{2}}

When using quantum mechanics to describe large collections of particles, such as the electrons in a piece of metal, it is important to know how many quantum states exist below a certain energy. For the three-dimensional box potential, there is a convenient graphical way to do this. Since there are three different quantum numbers ( n_x, n_y, and n_z), picture three-dimensional space with each number measured down one of the axes. The energy is proportional to the sum of the squares of the three numbers, which corresponds to the distance from the origin squared. To ask how many states have energy less than or equal to some number n_{\rm rs}^2 \pi^2\hbar^2/(2mL^2) is equivalent to asking how many states are inside of the sphere (sketched in the figure) defined by n_x^2+n_y^2+n_z^2 = n_{\rm rs}^2. This sphere has radius n_rs. (The subscript "rs" just stands for "radius of sphere".) Because all of the integers must be positive, you are really only working with one octant (one-eighth) of the sphere.

Part B
If you are dealing with a very large n_rs, you can assume that each state (point with integer coordinates) corresponds roughly to one unit of volume inside of the sphere. So, the number of states is approximately equal to the volume of the octant of the sphere. Use this idea to find the number N of states with energy less than or equal to n_{\rm rs}^2 \pi^2\hbar^2/(2mL^2) for a large n_rs.
Express your answer in terms of n_rs.

  N  = \frac{{\pi}}{6}{\cdot}n_{rs}{^{3}}

Part C
Suppose that you have N particles, each of which has to go into its own state, in the cubical box. Assume that no state is occupied if a lower energy state remains unoccupied. What is the highest energy E_max that one of these particles has? Assume that N is large so that the analysis from the previous part applies.
Express your answer in terms of N, hbar, m, and L.

  E_max  = \frac{\left({\pi}^{2}{\hbar}^{2}\right)}{2mL^{2}}{\cdot}\left(\frac{6}{{\pi}}N\right)^{\frac{2}{3}}

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