Let's break the problem into two parts: **bosons** and **fermions**. --- ## **1. Two Free Bosons (Spin 0) in 1D Box of Length L** ### **Single Particle Solutions** A particle in a 1D box of length \(L\) has normalized wavefunctions: \[ \psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right),\quad n=1,2,3,\ldots \] with energies: \[ E_n = \frac{n^2 \pi^2 \hbar^2}{2mL^2} \] ### **Two Bosons (Symmetric Wavefunction)** For two indistinguishable bosons (spin 0), the total spatial wavefunction must be **symmetric**: \[ \Psi_B(x_1,x_2) = \frac{1}{\sqrt{2(1+\delta_{n_1 n_2})}} \left[\psi_{n_1}(x_1)\psi_{n_2}(x_2) + \psi_{n_2}(x_1)\psi_{n_1}(x_2)\right] \] #### **Ground State** Both in the lowest state (\(n_1 = n_2 = 1\)): \[ \Psi_{B,0}(x_1,x_2) = \psi_1(x_1)\psi_1(x_2) \] Energy: \(E_0 = 2E_1 = 2\frac{\pi^2\hbar^2}{2mL^2} = \frac{\pi^2\hbar^2}{mL^2}\) #### **First Excited State** Both cannot occupy the same state (but for bosons they can!), so next possible energy state is one in \(n=1\), one in \(n=2\): \[ \Psi_{B,1}(x_1,x_2) = \frac{1}{\sqrt{2}}\left[\psi_1(x_1)\psi_2(x_2) + \psi_2(x_1)\psi_1(x_2)\right] \] Energy: \(E_1 = E_1 + E_2 = \frac{\pi^2\hbar^2}{2mL^2} + \frac{4\pi^2\hbar^2}{2mL^2} = \frac{5\pi^2\hbar^2}{2mL^2}\) --- ## **2. Two Fermions (Spin 1/2) in 1D Box** Fermions require the **total wavefunction (space × spin) to be antisymmetric**. ### **Spatial Part (Antisymmetric)** \[ \Psi_F(x_1, x_2) = \frac{1}{\sqrt{2}} \left[\psi_{n_1}(x_1)\psi_{n_2}(x_2) - \psi_{n_2}(x_1)\psi_{n_1}(x_2)\right] \] But **if \(n_1 = n_2\), this vanishes** (Pauli exclusion). ### **Spin Part** - **Singlet (antisymmetric):** \(\chi_A = \frac{1}{\sqrt{2}}[\uparrow\downarrow - \downarrow\uparrow]\) - **Triplet (symmetric):** \(\chi_S\) #### **Ground State** - Lowest two different states: \(n_1=1\), \(n_2=2\) - Spatial: antisymmetric, so spin must be **symmetric** (triplet) - **Wavefunction:** \[ \Psi_{F,0}(x_1,x_2) = \frac{1}{\sqrt{2}} [\psi_1(x_1)\psi_2(x_2) - \psi_2(x_1)\psi_1(x_2)] \chi_S \] - **Energy:** \(E_0 = E_1 + E_2 = \frac{5\pi^2\hbar^2}{2mL^2}\) #### **First Excited State** Next, \(n_1=1\), \(n_2=3\) or \(n_1=2\), \(n_2=3\), etc. - For \(n_1=1\), \(n_2=3\): \[ \Psi_{F,1}(x_1,x_2) = \frac{1}{\sqrt{2}}[\psi_1(x_1)\psi_3(x_2) - \psi_3(x_1)\psi_1(x_2)] \chi_S \] - **Energy:** \(E_1 = E_1 + E_3 = \frac{1\pi^2\hbar^2}{2mL^2} + \frac{9\pi^2\hbar^2}{2mL^2} = \frac{10\pi^2\hbar^2}{2mL^2} = \frac{5\pi^2\hbar^2}{mL^2}\) --- ## **Summary Table** | | Bosons (Spin 0) | Fermions (Spin 1/2) | |---------|------------------------------------------------|---------------------------------------------------------| | Ground | \(\psi_1(x_1)\psi_1(x_2)\) | \(\frac{1}{\sqrt{2}}[\psi_1(x_1)\psi_2(x_2) - \psi_2(x_1)\psi_1(x_2)]\chi_S\) | | | \(E_0 = \frac{\pi^2\hbar^2}{mL^2}\) | \(E_0 = \frac{5\pi^2\hbar^2}{2mL^2}\) | | 1st Exc | \(\frac{1}{\sqrt{2}}[\psi_1(x_1)\psi_2(x_2) + \psi_2(x_1)\psi_1(x_2)]\) | \(\frac{1}{\sqrt{2}}[\psi_1(x_1)\psi_3(x_2) - \psi_3(x_1)\psi_1(x_2)]\chi_S\) | | | \(E_1 = \frac{5\pi^2\hbar^2}{2mL^2}\) | \(E_1 = \frac{5\pi^2\hbar^2}{mL^2}\) | --- ### **Final Answers (Explicit Forms)** **Bosons:** - Ground: \(\Psi_{B,0}(x_1,x_2) = \psi_1(x_1)\psi_1(x_2)\) - 1st Excited: \(\Psi_{B,1}(x_1,x_2) = \frac{1}{\sqrt{2}}[\psi_1(x_1)\psi_2(x_2) + \psi_2(x_1)\psi_1(x_2)]\) **Fermions:** - Ground: \(\Psi_{F,0}(x_1,x_2) = \frac{1}{\sqrt{2}}[\psi_1(x_1)\psi_2(x_2) - \psi_2(x_1)\psi_1(x_2)]\chi_S\) - 1st Excited: \(\Psi_{F,1}(x_1,x_2) = \frac{1}{\sqrt{2}}[\psi_1(x_1)\psi_3(x_2) - \psi_3(x_1)\psi_1(x_2)]\chi_S\) Where \(\psi_n(x) = \sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})\), and \(\chi_S\) is any symmetric spin triplet state. --- **Let me know if you want the explicit spin functions or further details!**