# 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).

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

H = âˆ‘ (piÂ²/2m + mÏ‰Â²xi/2) with [xj,Î±,pk,Î²] = Î´jkÎ´Î±Î² iÄ§ Î±, Î² = 1,2,3 and the sum runs from 1 to N

a) Show that the grandcanonical partition function is given by

Zgr = Î Î Î [1 âˆ’ exp(âˆ’Î²(Ä§Ï‰(nx + ny + nz + 3/2) âˆ’ Î¼))]^(-1), where the products are taken over nx ,ny, nz respectively and run from 0 to âˆž

The number of particles in the state with quantum numbers (nx,ny,nz) is given as

N(nx,ny,nz) = [exp(Î²(Ä§Ï‰(nx + ny + nz + 3/2) âˆ’ Î¼)) âˆ’ 1]^( âˆ’1)

and the total number of particles is found to be

N = N0 + âˆ‘ âˆ«dx âˆ«dy âˆ«dz exp(âˆ’lÎ²(Ä§Ï‰(x + y + z + 3/2) âˆ’ Î¼)), where N0 is the number of atoms in the Bose-Einstein condensate, the summation runs from l = 1 to âˆž and the integrals are from 0 to âˆž

b) Investigate under which circumstances the ground state is occupied with macroscopically many atoms (which means N0 is very large and is proportional to the total number of particles N). Explain why Î¼ = 3Ä§Ï‰/2 is a suitable choice

c) Express the number of atoms in excited states for Î¼= 3Ä§Ï‰/2 in terms of the Riemann-Zeta function

Î¶(3) = âˆ‘ (1/l)Â³.

Argue that the critical temperature of the phase transition to the Bose condensed state is given by kTc = Ä§Ï‰ Â³âˆš(N/Î¶(3))

https://brainmass.com/physics/energy/bose-einstein-condensate-atoms-potential-well-112236

#### 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) + Â½m 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^(-) + Â½] (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. The expert examines Bose-Einstein condensate of atoms in a potential well. A 3-dimensional harmonic well is analyzed.