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Bose condensation in a harmonic potential

  1. Feb 9, 2015 #1

    CAF123

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    1. The problem statement, all variables and given/known data

    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.
    2. Relevant equations

    ##V = \int_0^{\epsilon_{max}} g(\epsilon) d \epsilon##

    3. The attempt at a solution

    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!
     
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  3. Feb 9, 2015 #2

    TSny

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    nx, ny, and nz are variables. What are the values of these variables in your expression for V?
     
  4. Feb 10, 2015 #3

    CAF123

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    Hi TSny,
    I could reexpress any one 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.
     
  5. Feb 10, 2015 #4

    TSny

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    The volume of the tetrahedron in n-space should depend only on C. Can you express V in terms of C?
     
  6. Feb 10, 2015 #5

    CAF123

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    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))##.
     
  7. Feb 10, 2015 #6

    TSny

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    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)##.
     
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