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: Understanding the Concept:
\(^4\text{He}\) has 2 protons, 2 neutrons and 2 electrons, an even total number of these spin-1/2 particles, so a \(^4\text{He}\) atom carries integer total spin and behaves as a boson. A gas of non-interacting bosons can undergo Bose-Einstein condensation, and at exactly \(T=0\) K every single particle must sit in the lowest available single-particle energy level, the ground state of the trap. For an ideal Bose gas at \(T=0\), the chemical potential is pinned to equal that ground-state energy, \(\mu(T=0) = \varepsilon_0\).

Step 2: Key Formula or Approach:
The trap is an isotropic 3D harmonic oscillator, \(V(x,y,z) = \frac{1}{2}m\omega^2(x^2+y^2+z^2)\), which separates into three independent 1D oscillators along \(x\), \(y\) and \(z\). The energy levels of a single 1D quantum harmonic oscillator are \(\left(n+\frac12\right)\hbar\omega\), so the full 3D levels are:
\[ \varepsilon(n_x,n_y,n_z) = \left(n_x + n_y + n_z + \frac{3}{2}\right)\hbar\omega, \qquad n_x,n_y,n_z = 0,1,2,\ldots \]

Step 3: Detailed Explanation:
The ground state has the lowest possible quantum numbers in all three directions, \(n_x=n_y=n_z=0\). Plugging these in:
\[ \varepsilon_0 = \left(0+0+0+\frac{3}{2}\right)\hbar\omega = \frac{3}{2}\hbar\omega \]
This \(\frac{3}{2}\hbar\omega\) is simply the sum of three independent zero-point energies, one \(\frac12\hbar\omega\) from each of the \(x\), \(y\) and \(z\) oscillators, which never vanish even in the ground state because of the uncertainty principle. Since \(\mu(T=0) = \varepsilon_0\), we get \(\mu(T=0) = \frac{3}{2}\hbar\omega\).

Step 4: Why the other options are wrong.
Option (A), \(\mu=0\), would only be correct for a classical ideal gas or a trap with no zero-point energy, it ignores the quantum mechanical fact that a trapped particle can never have exactly zero energy. Option (B), \(\frac12\hbar\omega\), is the zero-point energy of just one of the three 1D oscillators, not the sum over all three directions of the actual 3D trap. Option (D), \(3\hbar\omega\), would require \(n_x+n_y+n_z+\frac32 = 3\), that is \(n_x+n_y+n_z = \frac32\), which is not an integer and so is not an allowed combination of quantum numbers.

Final Answer:
At \(T=0\) K, every \(^4\text{He}\) atom condenses into the trap's ground state, whose energy is \(\frac32\hbar\omega\). \[ \boxed{\mu(T=0) = \frac{3}{2}\hbar\omega} \]