# Bose condensation in a harmonic potential

• CAF123
In summary: This simplifies to $$\frac{1}{6} \left(\frac{\epsilon}{\hbar w} - \frac{3}{2}\right)^3$$ In summary, the density of states is given by the volume of a tetrahedron, the number of states with energy less than ##\epsilon## is given by the volume of a tetrahedron, and the attempted solution states that the volume of the tetrahedron is proportional to C.
CAF123
Gold Member

## Homework Statement

[/B]
Consider a gas of N weakly interacting bosons trapped in a 3d harmonic potential ##V = \frac{1}{2}mw^2 (x^2 + y^2 + z^2)##. The single particle quantum states have energies ##\epsilon = \hbar w (n_x + n_y + n_z + 3/2)##.

Calculate the total number of quantum states with energies less than ##\epsilon## and from this deduce that the density of states ##g(\epsilon)## is $$g(\epsilon) \approx \frac{\epsilon^2}{2 (\hbar w)^3} \,\,\,\text{for large}\,\,\,\,\epsilon$$
Hint: ##\epsilon## is a function of ##\mathbf n## rather than just the magnitude ##n##. A surface of constant energy is a plane in ##n## space and the number of states with energy less than ##\epsilon## is given by the volume of a tetrahedron.

## Homework Equations

[/B]
##V = \int_0^{\epsilon_{max}} g(\epsilon) d \epsilon##

## The Attempt at a Solution

[/B]
A plane intersecting the ##n_x, n_y ## and ##n_z## axes form a tetrahedron with the sides the ##n_x - n_y## plane, ##n_y - n_z## plane and so on. The volume of such a tetrahedron is ##V = n_x n_y n_z/6##. Therefore per the hint, $$\frac{n_x n_y n_z}{6} = \int_0^{\epsilon_{max}} g(\epsilon) d \epsilon = G(\epsilon_{max}) - G(0)$$ I am just a bit unsure of how to progress. Many thanks!

CAF123 said:
The volume of such a tetrahedron is ##V = n_x n_y n_z/6##.

nx, ny, and nz are variables. What are the values of these variables in your expression for V?

Hi TSny,
TSny said:
nx, ny, and nz are variables. What are the values of these variables in your expression for V?
I could reexpress anyone of ##n_x, n_y## or ##n_z## using the equation for the energy ##\epsilon = \hbar w(n_x + n_y + n_z + 3/2)## to reduce the number of variables by one. Is it what you meant? The plane in n-space can be written as ##n_x + n_y + n_z = C## for C a constant given by rearranging the above equation. Thanks.

The volume of the tetrahedron in n-space should depend only on C. Can you express V in terms of C?

TSny said:
The volume of the tetrahedron in n-space should depend only on C. Can you express V in terms of C?
Yes sorry, I was confusing my own notation. The volume is ##C^3/6## and so we have the integral $$\frac{C^3}{6} = \int_0^{\epsilon_{max}} g(\epsilon) d \epsilon = G(\epsilon_{\max}) - G(0) = \frac{1}{6} \left(\frac{\epsilon_{max}}{\hbar w} - \frac{3}{2}\right)^3$$ Then $$\frac{d}{d\epsilon} \left(G(\epsilon) -G(0)\right) = \frac{d}{d\epsilon} \frac{1}{6} \left(\frac{\epsilon}{\hbar w} - \frac{3}{2}\right)^3 \approx \frac{\epsilon^2}{2 (\hbar w)^3}$$ Is this along the right lines? I feel like I have sort of fudged the result in getting ##g(\epsilon)##, in particular how I can neglect further terms in the expansion of the square at the end and in going from ##G(\epsilon_{max}) - G(0)## to ##d/d\epsilon (G(\epsilon) - G(0))##.

That looks good.

##\epsilon## is assumed "large". Presumably, that means relative to ##\hbar \omega##. So, you can neglect the 3/2 relative to ##\epsilon/(\hbar \omega)##.

CAF123

## What is Bose condensation in a harmonic potential?

Bose condensation in a harmonic potential refers to the phenomenon where a large number of bosons (particles with integer spin) in a confined space are cooled to a low temperature, causing them to "condense" into the lowest energy state of the system. This occurs when the bosons are subjected to a harmonic potential, which is a type of potential energy that increases quadratically with distance from the center.

## What is the significance of Bose condensation in a harmonic potential?

Bose condensation in a harmonic potential has been observed in various systems, such as ultracold atomic gases, superconductors, and superfluids. It is a unique state of matter that exhibits macroscopic quantum phenomena, such as superfluidity and coherence. Understanding Bose condensation in a harmonic potential can provide insights into fundamental principles of quantum mechanics and has potential applications in technologies such as quantum computing and precision measurement.

## How does Bose condensation in a harmonic potential occur?

Bose condensation in a harmonic potential occurs when the bosons are cooled to a temperature close to absolute zero, also known as the Bose-Einstein condensation temperature. At this temperature, the bosons occupy the lowest energy state of the system, forming a macroscopic quantum state. This occurs due to the Pauli exclusion principle, which states that no two identical fermions (particles with half-integer spin) can occupy the same quantum state, but bosons do not follow this rule and can occupy the same state.

## What is the Bose-Einstein distribution in a harmonic potential?

The Bose-Einstein distribution is a statistical distribution that describes the distribution of bosons in a system at thermal equilibrium. In the case of Bose condensation in a harmonic potential, the majority of bosons occupy the lowest energy state, while a small fraction follows the Bose-Einstein distribution at higher energy states. This distribution is dependent on temperature and the energy levels of the harmonic potential.

## What factors affect Bose condensation in a harmonic potential?

The main factors that affect Bose condensation in a harmonic potential are temperature, the number of bosons in the system, and the strength of the harmonic potential. Lower temperatures and a larger number of bosons increase the likelihood of Bose condensation occurring, while a stronger harmonic potential can make it more difficult for the bosons to occupy the lowest energy state. Other factors such as interactions between bosons and external perturbations can also play a role in the formation of Bose condensates.

### Similar threads

• Advanced Physics Homework Help
Replies
8
Views
1K
• Advanced Physics Homework Help
Replies
3
Views
2K
• Quantum Physics
Replies
16
Views
1K
• Advanced Physics Homework Help
Replies
2
Views
1K
• Advanced Physics Homework Help
Replies
4
Views
2K
• Quantum Physics
Replies
3
Views
791
• Advanced Physics Homework Help
Replies
6
Views
1K
• Advanced Physics Homework Help
Replies
1
Views
1K
• Advanced Physics Homework Help
Replies
2
Views
208
• Introductory Physics Homework Help
Replies
6
Views
1K