This problem asks us to find the configuration of 4 spin-\(\frac{1}{2}\) fermions with different masses (m and 4m) in a harmonic potential that results in the lowest ground state energy. The potential is given by \(V(x) = \frac{1}{2} kx^2\).

For a one-dimensional simple harmonic oscillator (SHO), the energy levels are quantized and given by the formula:

\(E_n = (n + \frac{1}{2})\hbar\omega\)

where \(n = 0, 1, 2, \dots\) is the principal quantum number, \(\hbar\) is the reduced Planck constant, and \(\omega\) is the angular frequency of the oscillator, determined by the stiffness of the potential (k) and the mass (μ) of the particle:

\(\omega = \sqrt{\frac{k}{\mu}}\)

The ground state energy for a single particle in this potential is for \(n=0\), which is \(E_0 = \frac{1}{2}\hbar\omega\).

Energy Levels Based on Mass

Let's calculate the angular frequency and energy levels for the two types of particles:

• For particles of mass m: \(\omega_m = \sqrt{\frac{k}{m}}\). The energy levels are \(E_n^{(m)} = (n + \frac{1}{2})\hbar\sqrt{\frac{k}{m}}\). Let's define \(\mathcal{E}_0 = \hbar\sqrt{\frac{k}{m}}\) for convenience. The levels are \(\frac{1}{2}\mathcal{E}_0, \frac{3}{2}\mathcal{E}_0, \frac{5}{2}\mathcal{E}_0, \dots\)
• For particles of mass 4m: \(\omega_{4m} = \sqrt{\frac{k}{4m}} = \frac{1}{2}\sqrt{\frac{k}{m}} = \frac{1}{2}\omega_m\). The energy levels are \(E_n^{(4m)} = (n + \frac{1}{2})\hbar(\frac{1}{2}\omega_m) = \frac{1}{2}(n + \frac{1}{2})\hbar\omega_m = \frac{1}{2}E_n^{(m)}\). The levels are \(\frac{1}{2}(\frac{1}{2}\mathcal{E}_0), \frac{3}{2}(\frac{1}{2}\mathcal{E}_0), \frac{5}{2}(\frac{1}{2}\mathcal{E}_0), \dots\), which are \(\frac{1}{4}\mathcal{E}_0, \frac{3}{4}\mathcal{E}_0, \frac{5}{4}\mathcal{E}_0, \dots\)

Note that the energy levels for particles of mass 4m are lower than the corresponding levels for particles of mass m.

Ground State Configuration for Fermions

Fermions are subject to the Pauli Exclusion Principle. For spin-\(\frac{1}{2}\) fermions in a 1D SHO, each energy level (defined by the quantum number n) can be occupied by a maximum of two particles: one with spin up and one with spin down.

To find the lowest ground state energy for a given configuration of 4 particles, we must fill the lowest available energy levels for each type of particle, respecting the Pauli principle.

Calculating Total Ground State Energy for Each Option

Let's evaluate the total energy for each proposed configuration:

Option 1: 4 particles of mass m

We have 4 mass m particles. The lowest energy levels for mass m are \(\frac{1}{2}\mathcal{E}_0\) (\(n=0\)) and \(\frac{3}{2}\mathcal{E}_0\) (\(n=1\)).

• 2 particles go into the \(n=0\) level (\(\frac{1}{2}\mathcal{E}_0\)). Energy contribution: \(2 \times \frac{1}{2}\mathcal{E}_0 = \mathcal{E}_0\).
• The remaining 2 particles go into the \(n=1\) level (\(\frac{3}{2}\mathcal{E}_0\)). Energy contribution: \(2 \times \frac{3}{2}\mathcal{E}_0 = 3\mathcal{E}_0\).

Total energy for Option 1: \(\mathcal{E}_0 + 3\mathcal{E}_0 = 4\mathcal{E}_0\).

Option 2: 4 particles of mass 4m

We have 4 mass 4m particles. The lowest energy levels for mass 4m are \(\frac{1}{4}\mathcal{E}_0\) (\(n=0\)) and \(\frac{3}{4}\mathcal{E}_0\) (\(n=1\)).

• 2 particles go into the \(n=0\) level (\(\frac{1}{4}\mathcal{E}_0\)). Energy contribution: \(2 \times \frac{1}{4}\mathcal{E}_0 = \frac{1}{2}\mathcal{E}_0\).
• The remaining 2 particles go into the \(n=1\) level (\(\frac{3}{4}\mathcal{E}_0\)). Energy contribution: \(2 \times \frac{3}{4}\mathcal{E}_0 = \frac{3}{2}\mathcal{E}_0\).

Total energy for Option 2: \(\frac{1}{2}\mathcal{E}_0 + \frac{3}{2}\mathcal{E}_0 = 2\mathcal{E}_0\).

Option 3: 1 particle of mass m and 3 particles of mass 4m

We have 1 mass m particle and 3 mass 4m particles. The lowest energy levels are \(\frac{1}{2}\mathcal{E}_0\) (mass m) and \(\frac{1}{4}\mathcal{E}_0\) (mass 4m). The lowest available level overall is \(\frac{1}{4}\mathcal{E}_0\) for the mass 4m particles.

• Place the 3 mass 4m particles:
  • 2 particles go into the mass 4m, \(n=0\) level (\(\frac{1}{4}\mathcal{E}_0\)). Energy contribution: \(2 \times \frac{1}{4}\mathcal{E}_0 = \frac{1}{2}\mathcal{E}_0\).
  • The remaining 1 particle goes into the mass 4m, \(n=1\) level (\(\frac{3}{4}\mathcal{E}_0\)). Energy contribution: \(1 \times \frac{3}{4}\mathcal{E}_0 = \frac{3}{4}\mathcal{E}_0\).
• Place the 1 mass m particle:
  • The 1 particle goes into the mass m, \(n=0\) level (\(\frac{1}{2}\mathcal{E}_0\)). Energy contribution: \(1 \times \frac{1}{2}\mathcal{E}_0 = \frac{1}{2}\mathcal{E}_0\).

Total energy for Option 3: \(\frac{1}{2}\mathcal{E}_0 + \frac{3}{4}\mathcal{E}_0 + \frac{1}{2}\mathcal{E}_0 = (1 + \frac{3}{4})\mathcal{E}_0 = \frac{7}{4}\mathcal{E}_0 = 1.75\mathcal{E}_0\).

Option 4: 2 particles of mass m and 2 particles of mass 4m

We have 2 mass m particles and 2 mass 4m particles. The lowest energy levels are \(\frac{1}{2}\mathcal{E}_0\) (mass m) and \(\frac{1}{4}\mathcal{E}_0\) (mass 4m).

• Place the 2 mass 4m particles:
  • 2 particles go into the mass 4m, \(n=0\) level (\(\frac{1}{4}\mathcal{E}_0\)). Energy contribution: \(2 \times \frac{1}{4}\mathcal{E}_0 = \frac{1}{2}\mathcal{E}_0\).
• Place the 2 mass m particles:
  • 2 particles go into the mass m, \(n=0\) level (\(\frac{1}{2}\mathcal{E}_0\)). Energy contribution: \(2 \times \frac{1}{2}\mathcal{E}_0 = \mathcal{E}_0\).

Total energy for Option 4: \(\frac{1}{2}\mathcal{E}_0 + \mathcal{E}_0 = \frac{3}{2}\mathcal{E}_0 = 1.5\mathcal{E}_0\).

Comparing Total Energies

Let's summarize the total ground state energy for each configuration:

Comparing the total energies, the configuration with the lowest ground state energy is 2 particles of mass m and 2 particles of mass 4m, with a total energy of \(1.5\mathcal{E}_0\).