- #1
CAF123
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Homework Statement
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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
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##V = \int_0^{\epsilon_{max}} g(\epsilon) d \epsilon##
The Attempt at a Solution
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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!