Share
Explore BrainMass

# Bose-Einstein condensate of atoms in a potential well

This exercise is for a Bose-Einstein condensate of indistinguishable atoms which do not interact with each other and are in a 3-dimensional harmonic well. The system is described by the following Hamiltonian (see attached file).

#### Solution Preview

I've attached the solution (PDF, LaTeX and PS format).

P_{i}^2 = P_{i,x}^2 + P_{i,y}^2 + P_{i,z}^2

and, of course:

X_{i}^2 = X_{i,x}^2 + X_{i,y}^2 + X_{i,z}^2

Note that in your attachment there is a typo: x_{i} should be x_{i}^2

The one dimensional Hamiltonian

H = P^2/(2m) + ½m omega^2 X^2

has eigenvalues E_{n} = (n + 1/2)h-bar omega

Since the three dimensional harmonic oscillator is the sum of three one dimensional harmonic oscillator terms that each commute with each other, the eigenvalues are given by the sum of three terms of the above form with three quantum numbers:

E_{nx,ny,nz} = E_{nx} + E_{ny} + E_{nz} = (nx + ny + nz + 3/2)h-bar omega (0)

The energy eigenvalues for the one dimensional oscillator can be derived as follows:

You can rewrite the Hamiltonian for the one dimensional harmonic oscillator as:

H = h-bar omega [a^(+) a^(-) + ½] (1)

Here:

a^(-) = Squareroot[m omega/(2 hbar)] [X + i/(m omega) P] (2)

and

a^(+) = Squareroot[m omega/(2 hbar)] [X - i/(m omega) P] (3)

The operators a^(-) and a^(+) are each other Hermitian conjugates. Using (2) and (3) and the fact that [X,P] = i h-bar you find that the commutator is:

[a^(-) ,a^(+)] = 1 (4)

You can now easily derive that the eigenvalues of the operator N = a^(+)a^(-) are integers greater than or equal to zero. Using (4) you easily derive that:

[N, a^(-)] = -a^(-) (5)

[N, a^(+)] = a^(+) (6)

Let's denote by |y> the eigenstate of N with eigenvalue x. Then using (5) let's find out how the operator N acts on the state a^(-)|y>:

Na^(-) |y> - a^(-) N|y> = -a^(-)|y> --->

Na^(-) |y> - xa^(-) |y> = -a^(-)|y> --->

Na^(-) |y> = (y - 1)a^(-)|y>

So, if there is an eigenstate with eigenvalue y, denoted as |y>, then a^(-)|y> is an eigenstate with eigenvalue y - 1. This is provided we can properly normalize this new state. Similarly you find using (6) that a^(+)|y> is an eigenstate with eigenvalue y + 1. We must properly normalize these states, so let's find out what the norm is. The inner product of a^(-)|y> with itself is:

<y|a^(+) a^(-)|y> = <y|N|y> = y

So, the square of the norm of a^(-)|y> is y. Since this can never be negative, the allowed eigenvalues y must be larger or equal to zero. This seems to be contradictory because if you start with some arbitrary state you can lower the eigenvalue by one by applying the operator
a^(-), so you should be able to make y negative by applying this operator often enough. The only thing wch prevents you from going to a state with a lower eigenvalue ...

#### Solution Summary

A detailed solution is given.

\$2.19