One of the simplest problems solved by applying the time-independent Schrödinger equation is the one-dimensional “particle in a box”, where the potential *V*(*x*) is zero for 0<*x*<*L*, and infinite everywhere else. Then we have *ψ*(*x*)=0 for *x*≤0 and *x*≥*L*, and for 0<*x*<*L*, Schrödinger equation becomes

with boundary conditions *ψ*(0)=*ψ*(*L*)=0. The general solution to the differential equation, which can be rewritten as

, is

, where , or solving for energy, . The boundary condition *ψ*(0)=0 tells us that *B*=0, and the condition *ψ*(*L*)=0 thus tells us that , and thus , *n* a positive integer. Thus, in terms of that *n*,

, and , and the energy is quantized. Normalizing via , we get normalized wavefunction .

Now, let us consider the momentum. The one-dimensional quantum momentum operator is , and we see ; our energy eigenstates are obviously not momentum eigenstates. However, note that the eigenfunctions of are of the form (with eigenvalues ). Noting that , we see that our energy eigenstate is the linear combination of a state of momentum and a state of momentum , with equal probabilities for these two possible measurements. Compare this to the classical particle in a one dimensional box, where it has kinetic energy , half the time travelling with momentum of magnitude *p* in one direction, the other half moving in the opposite direction with momentum of magnitude *p*.

Now, let us consider a three-dimensional cubic box:

*V*(*x*, *y*, *z*)=0 when 0<*x*<*L*, 0<*y*<*L*, and 0<*z*<*L*, and is infinite at all other points.

Then we have for the interior of our cube:

,

with boundary conditions .

using the definition of nabla for Cartesian coordinates:

,

and using separation of variables via ,

we get

with . These are just copies of the one-dimensional problem, so we obtain normalized 3d wavefunction

, where *n _{x}*,

*n*, and

_{y}*n*are positive integers. The energy is . A given state is defined entirely by the triplet of positive integers

_{z}*n*,

_{x}*n*,

_{y}*n*. Note that except for the ground state

_{z}*n*=

_{x}*n*=

_{y}*n*=1 (), all of the energy levels are degenerate; all states for which have the same value will have the same energy.

_{z}[Note also that the proper linear combination of the plane wave with momentum and the seven other plane waves formed by reflections in

*x*,

*y*,

*z*, and combinations thereof, will produce our energy eigenfunction.

Again compare to the possible momentum values of a classical particle in a box making perfectly elastic collisions with the walls.]

Tags: Degeneracy, Friday Physics, Momentum, Particle-in-a-Box, physics, Quantum Mechanics, Schrödinger Equation

September 29, 2009 at 3:06 pm |

[…] confined to a cubical box of volume V=L3; I demonstrated the quantum particle in a cubical box here; we have the allowed orbital states given by , , where nx, ny, and nz are positive integers. Thus, […]

October 16, 2009 at 12:15 am |

[…] neglecting the ground state. First, let us examine the spacing between the energy levels. For our quantum particle in a box, we have energy levels , where nx, ny, and nz are positive integers. Using , we rewrite as . The […]