Let's break down and solve the problem step by step: --- ## **Given:** - **System:** Two noninteracting identical particles in a common external harmonic oscillator potential. - **Tasks:** Find energy levels and ground state wave functions for: - (a) Two spin-1 particles, and - (b) Two spin-½ particles. --- ## **Step 1: Single Particle in a 1D Harmonic Oscillator** The energy levels for a single particle in a 1D harmonic oscillator are: \[ E_n = \left( n + \frac{1}{2} \right) \hbar\omega \quad \text{with} \quad n = 0,1,2,... \] The corresponding normalized wavefunctions are denoted as \( \psi_n(x) \). --- ## **Step 2: Two Particles (Non-interacting)** **Total Hamiltonian:** \[ H = H_1 + H_2 \] So the total energy is: \[ E = E_{n_1} + E_{n_2} \] --- ## **Step 3: Symmetrization (Identical Particles)** - **Bosons (integer spin, e.g., spin-1):** Total wavefunction **must be symmetric** under particle exchange. - **Fermions (half-integer spin, e.g., spin-½):** Total wavefunction **must be antisymmetric**. **Total wavefunction** = (Spatial part) × (Spin part) --- ## **Step 4: Ground State (Lowest Energy)** ### **(a) Two Spin-1 Particles (Bosons)** #### **Spatial Part:** - Both particles in lowest state: \( n_1 = n_2 = 0 \) - \[ E_\text{ground} = \hbar\omega \] - **Spatial wavefunction:** \[ \Psi_\text{space}(x_1, x_2) = \psi_0(x_1)\psi_0(x_2) \] This is symmetric under exchange. #### **Spin Part:** - Each particle has spin-1, so possible combined spin states: \( S = 2, 1, 0 \) - The spin wavefunction can be symmetric (triplet, quintet, singlet, etc.) #### **Total Wavefunction:** - For bosons, both spatial and spin parts can be symmetric. - **Wavefunction:** Symmetric in both spin and space. #### **Ground State:** - \[ \Psi_\text{total}(x_1, x_2) = \psi_0(x_1)\psi_0(x_2) \otimes \chi_\text{sym} \] where \( \chi_\text{sym} \) is any symmetric spin state. - **Degeneracy:** The spin-1 combination allows for 9 states (3 × 3). --- ### **(b) Two Spin-½ Particles (Fermions)** #### **Spatial Part:** - Both particles cannot occupy the exact same quantum state due to Pauli exclusion principle. - So, one particle in \( n = 0 \), other in \( n = 1 \): \[ E_\text{ground} = \left(\frac{1}{2} + \frac{1}{2} + 1 \right)\hbar\omega = 2\hbar\omega \] - **Spatial wavefunction:** Must be **antisymmetric** for fermions: \[ \Psi_\text{space}(x_1, x_2) = \frac{1}{\sqrt{2}} [\psi_0(x_1)\psi_1(x_2) - \psi_1(x_1)\psi_0(x_2)] \] #### **Spin Part:** - **Spin states:** For two spin-½, combined spin can be **singlet (antisymmetric)** or **triplet (symmetric)**. - To maintain total antisymmetry, **spatial antisymmetric** ⟺ **spin symmetric** (triplet). #### **Total Wavefunction:** - \[ \Psi_\text{total}(x_1, x_2) = \Psi_\text{space}^\text{antisym}(x_1, x_2) \otimes \chi_\text{sym} \] - **Degeneracy:** The spin triplet has 3 symmetric states. --- ## **Step 5: Final Answers** ### **(a) Two Spin-1 Particles (Bosons):** - **Ground state energy:** \(\boxed{E = \hbar\omega}\) - **Wavefunction:** \(\boxed{\Psi_\text{total}(x_1, x_2) = \psi_0(x_1)\psi_0(x_2) \otimes \chi_\text{sym}}\) - **Degeneracy:** 9 (from spin states) --- ### **(b) Two Spin-½ Particles (Fermions):** - **Ground state energy:** \(\boxed{E = 2\hbar\omega}\) - **Wavefunction:** \(\boxed{\Psi_\text{total}(x_1, x_2) = \frac{1}{\sqrt{2}}\left[\psi_0(x_1)\psi_1(x_2) - \psi_1(x_1)\psi_0(x_2)\right] \otimes \chi_\text{sym}}\) - **Degeneracy:** 3 (from spin triplet) --- If you need the explicit forms of the spin or spatial wavefunctions, let me know!