Question

A gas of non-interacting \(^4\text{He}\) atoms (of mass \(m\)) is in a three-dimensional trap whose energy levels can be approximated by those of a harmonic oscillator potential \(V(x,y,z) = \dfrac{1}{2} m \omega^2 (x^2+y^2+z^2)\). The chemical potential of the gas at \(T=0\) K is

• A. 0
• B. \(\dfrac{1}{2}\hbar\omega\)
• C. \(\dfrac{3}{2}\hbar\omega\)
• D. \(3\hbar\omega\)

Hint:

\(^4\)He atoms are bosons, so at \(T=0\) K all of them sit in the trap's ground state.
For an isotropic 3D harmonic oscillator, the ground-state (zero-point) energy is \(\frac32\hbar\omega\), and \(\mu(T=0)\) equals that.

Answer 1

The Correct Option is C

Step 1: Start from the Bose-Einstein occupation formula instead of the energy levels.
$$n(\varepsilon) = \frac{1}{\exp\left(\dfrac{\varepsilon-\mu}{k_BT}\right)-1}$$
For this occupation number to stay positive and finite for every level, the chemical potential $\mu$ must always stay below the lowest available energy level $\varepsilon_0$, otherwise the ground state occupation would turn negative or blow up in an unphysical way.

Step 2: Use the condensation condition at $T=0$.
As $T\to 0$, essentially the entire population of $N$ atoms has to pile into the ground state alone, since all excited levels get frozen out once $k_BT \ll \hbar\omega$. The only way the ground-state occupation number $n(\varepsilon_0)$ can hold a macroscopic number $N$ of atoms as $T\to 0$ is for the gap $\varepsilon_0 - \mu$ to shrink to exactly zero, that is $\mu \to \varepsilon_0$ from below. So at $T=0$, $\mu$ locks onto the ground-state energy itself: $\mu(T=0) = \varepsilon_0$.

Step 3: Now just need $\varepsilon_0$ for this particular trap.
Separate the 3D harmonic oscillator Hamiltonian into $x$, $y$ and $z$ pieces, each an independent 1D oscillator with its own zero-point energy $\frac12\hbar\omega$. The lowest total energy adds all three:
\[ \varepsilon_0 = \frac{1}{2}\hbar\omega + \frac{1}{2}\hbar\omega + \frac{1}{2}\hbar\omega = \frac{3}{2}\hbar\omega \]

Step 4: Final Answer.
Combining the condensation argument, $\mu(T=0)=\varepsilon_0$, with the trap's actual ground-state energy, $\varepsilon_0 = \frac32\hbar\omega$, gives the same result as the direct method.
$$\mu(T=0) = \frac{3}{2}\hbar\omega$$ \[ \boxed{\text{Option (C)}} \]