Physics Friday 85

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 xL, 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 nx, ny, and nz are positive integers. The energy is . A given state is defined entirely by the triplet of positive integers nx, ny, nz. Note that except for the ground state nx=ny=nz=1 (), all of the energy levels are degenerate; all states for which have the same value will have the same energy.

[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.]


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2 Responses to “Physics Friday 85”

  1. Physics Friday 88 « Twisted One 151’s Weblog Says:

    […] 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, […]

  2. Physics Friday 93 « Twisted One 151’s Weblog Says:

    […] 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 […]

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