Discussion
A gas of non-interacting 4He atoms (of mass \(m\)) is in a three-dimensional trap whose energy levels can be approximated by those of a harmonic oscillator potential \(V(x, y, z) = \frac{1}{2} m\omega^{2}(x^{2} + y^{2} + z^{2})\). The chemical potential of the gas at \(T\)=0 K is
Answer: C
The chemical potential at T = 0 is \(\frac{3}{2}\hbar\omega\).
4He atoms are bosons. For an ideal Bose gas the chemical potential must stay below the lowest single-particle energy, and as T falls to zero all the atoms condense into that ground state and \(\mu\) rises to meet it. The ground state of the 3D isotropic oscillator has zero-point energy \(\frac{1}{2}\hbar\omega\) in each of the three directions:
\[ \varepsilon_{0} = \tfrac{3}{2}\hbar\omega \]
So \(\mu \to \frac{3}{2}\hbar\omega\). Zero would be the answer only if energies were measured from the ground state, \(\frac{1}{2}\hbar\omega\) counts one direction only, and \(3\hbar\omega\) doubles the zero-point energy.
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