Let's solve each part step by step: --- ### (i) Identical Particles and Symmetry #### **What does this mean for \(\phi(\mathbf{r}_1, \mathbf{r}_2)\)?** - The total wave function for two identical fermions must be **anti-symmetric** under particle exchange due to the Pauli exclusion principle. - The total state = spin part × spatial part. - Given: Spin wave function is **anti-symmetric** (i.e., singlet state). - Therefore, the **spatial wave function must also be anti-symmetric**: \[ \phi(\mathbf{r}_1, \mathbf{r}_2) = -\phi(\mathbf{r}_2, \mathbf{r}_1) \] #### **What can be said about the spin of the identical particles?** - The spin wave function is anti-symmetric (singlet state). - For two spin-1/2 particles, this means the total spin \( S = 0 \). - **Thus, the system is in a singlet state** (anti-parallel spins, total spin 0). --- ## (ii) One-dimensional Potential ### **a) Sketch the potential** \[ V(x) = \begin{cases} 0 & (x \leq 0) \\ -2V_a & (0 < x \leq a) \\ V_a & (a < x \leq 3a) \\ 0 & (x > 3a) \end{cases} \] - At \(x = 0\): jumps from 0 to \(-2V_a\) - At \(x = a\): jumps from \(-2V_a\) to \(V_a\) - At \(x = 3a\): jumps from \(V_a\) back to 0 *Sketch:* - Draw x-axis. - From \(x \leq 0\), \(V(x) = 0\). - From \(0 < x \leq a\), \(V(x) = -2V_a\) (below x-axis). - From \(a < x \leq 3a\), \(V(x) = V_a\) (above x-axis). - From \(x > 3a\), \(V(x) = 0\). --- ### **b) For which energies do you expect bound state and scattering solutions?** - **Bound states:** The energy \(E\) must be less than the value of the potential at infinity (here, 0). So, **bound states exist for \(E < 0\)**. - **Scattering states:** The energy is above the asymptotic potential, so **scattering states occur for \(E > 0\)**. --- ### **c) Solve the Schrödinger equation for \(0 < E < V_a\), particle coming from the right** #### **Regions:** 1. \(x \leq 0\): \(V=0\) 2. \(0 < x \leq a\): \(V=-2V_a\) 3. \(a < x \leq 3a\): \(V=V_a\) 4. \(x > 3a\): \(V=0\) #### **Wave numbers:** Let \(k = \sqrt{2mE}/\hbar\), \(k_1 = \sqrt{2m(E+2V_a)}/\hbar\), \(\kappa = \sqrt{2m(V_a-E)}/\hbar\). #### **Wavefunctions:** 1. \(x \leq 0\): \(V=0\) \[ \psi_1(x) = Ae^{-ikx} + Be^{ikx} \] 2. \(0 < x \leq a\): \(V = -2V_a\) \[ \psi_2(x) = C e^{-ik_1 x} + D e^{ik_1 x} \] 3. \(a < x \leq 3a\): \(V = V_a\) \[ \psi_3(x) = F e^{\kappa x} + G e^{-\kappa x} \] 4. \(x > 3a\): \(V=0\) \[ \psi_4(x) = H e^{ikx} \] (Assume a beam comes from the right, so only transmitted wave to the right.) #### **Integration constants:** There are 8 constants: \(A, B, C, D, F, G, H\). #### **Boundary conditions:** At each interface (\(x = 0, a, 3a\)), both \(\psi\) and \(d\psi/dx\) must be continuous. - 3 boundaries × 2 conditions each = 6 equations. Thus, **2 integration constants will remain undetermined** (typically related to the normalization of the incoming wave). --- ### **d) Reflection and transmission probabilities** - **Reflection probability:** \(R = \left| \frac{A}{H} \right|^2\) - **Transmission probability:** \(T = \left| \frac{H}{A} \right|^2\) (Assuming \(A\) is the amplitude of the incident wave, and \(H\) is the amplitude of the transmitted wave.) --- ## (iii) Particle in one-dimensional finite potential well ### **a) Solve the stationary Schrödinger equation for bound states** \[ V(x) = \begin{cases} \infty & (x \leq 0) \\ 0 & (0 < x \leq L) \\ V_c & (x > L) \end{cases} \] #### **Regions:** 1. \(x \leq 0\): \(\psi = 0\) 2. \(0 < x < L\): \(V = 0\) \[ \psi_1(x) = A \sin(kx) \] where \(k = \sqrt{2mE}/\hbar\) 3. \(x > L\): \(V = V_c\) \[ \psi_2(x) = B e^{-\kappa(x-L)} \] where \(\kappa = \sqrt{2m(V_c - E)}/\hbar\) #### **Boundary conditions:** - At \(x = 0\): \(\psi(0) = 0 \implies\) already satisfied. - At \(x = L\): Continuity of \(\psi\) and \(d\psi/dx\): \[ A \sin(kL) = B \] \[ A k \cos(kL) = -B \kappa \] Divide the second equation by the first: \[ k \cot(kL) = -\kappa \] This is the **transcendental equation for the bound state energies**. --- ### **b) Requirement on \(V_c\) to have at least one bound state** - At least one bound state exists if the well is deep enough. - The lowest energy bound state occurs when \(kL \to 0\), so the requirement is: \[ V_c > \frac{\pi^2 \hbar^2}{2m L^2} \] (This is the minimum potential required for at least one solution to the transcendental equation.) --- ## **Summary of Final Answers** --- **(i)** - \(\phi(\mathbf{r}_1, \mathbf{r}_2) = -\phi(\mathbf{r}_2, \mathbf{r}_1)\) - The particles are in a singlet state, total spin \(S=0\). **(ii)** - (a) Sketch as described above. - (b) Bound states: \(E < 0\), Scattering states: \(E > 0\). - (c) Write general solutions in each region, match at boundaries; 2 constants remain undetermined. - (d) \(R = |A/H|^2\), \(T = |H/A|^2\). **(iii)** - (a) Solutions: \(\psi(x) = A\sin(kx)\) for \(0<x<L\), \(\psi(x) = B e^{-\kappa(x-L)}\) for \(x>L\); energy given by \(k\cot(kL) = -\kappa\). - (b) Condition: \(V_c > \frac{\pi^2 \hbar^2}{2m L^2}\).

Certainly! Let's go through each part carefully, step by step, with detailed explanations and calculations. --- ## **(i) Symmetry of the two-particle wavefunction** ### Given: - Two identical particles in a 3D spherical potential. - The spin wave function is **antisymmetric**. - The position space wave function \(\phi(\mathbf{r}_1, \mathbf{r}_2)\) is **antisymmetric**. --- ### **What does this imply for \(\phi(\mathbf{r}_1, \mathbf{r}_2)\)?** **Step 1:** Recall the total wavefunction for two identical particles is a product of spin and spatial parts: \[ \Psi_{\text{total}} = \text{(spin part)} \times \text{(spatial part)}. \] **Step 2:** For **fermions** (like electrons), the total wavefunction must be **antisymmetric** under exchange: \[ \Psi_{\text{total}}(\mathbf{r}_1, \mathbf{r}_2) = - \Psi_{\text{total}}(\mathbf{r}_2, \mathbf{r}_1). \] **Step 3:** Given: - The **spin wave function is antisymmetric** (singlet state), so: \[ \text{spin part} = \chi_{\text{antisym}}. \] - The **spatial wave function** \(\phi(\mathbf{r}_1, \mathbf{r}_2)\) is **also antisymmetric**. **Step 4:** Since the total wavefunction is a product: \[ \Psi_{\text{total}} = \phi(\mathbf{r}_1, \mathbf{r}_2) \times \chi_{\text{antisym}}, \] and both are antisymmetric, their product is **symmetric**: \[ \text{antisymmetric} \times \text{antisymmetric} = \text{symmetric}. \] But **this contradicts** the requirement that the total wavefunction for fermions must be antisymmetric. **Step 5:** **Therefore**, the initial assumption that the spatial wavefunction is antisymmetric **must be** for the **spin-symmetric case**. Since the spin wavefunction is antisymmetric, the spatial part must be **symmetric** to ensure the total wavefunction is antisymmetric: \[ \phi(\mathbf{r}_1, \mathbf{r}_2) = + \phi(\mathbf{r}_2, \mathbf{r}_1). \] --- ### **Summary:** - Because the spin wavefunction is **antisymmetric**, the **spatial wavefunction must be symmetric** to produce an overall antisymmetric total wavefunction. --- ### **What can be said about the spin of the particles?** - The **spin wavefunction is antisymmetric**, indicating the particles are in the **singlet state**. - **Total spin \(S=0\)**. - **Particles are spin-1/2 fermions (electrons, protons, etc.)**. --- ## **(ii) One-dimensional potential** --- ### **a) Sketch of the potential \(V(x)\)** \[ V(x) = \begin{cases} 0 & x \leq 0 \\ -2V_a & 0 < x \leq a \\ V_a & a < x \leq 3a \\ 0 & x > 3a \end{cases} \] **Description:** - For \(x \leq 0\): potential is zero. - Between \(0\) and \(a\): potential dips to \(-2V_a\) (a well). - Between \(a\) and \(3a\): potential rises to \(V_a\). - Beyond \(3a\): potential returns to zero. *Graph:* Starting at zero, downward step to \(-2V_a\) in \([0,a]\), then upward to \(V_a\) in \([a, 3a]\), then back to zero. --- ### **b) Bound and scattering states** - **Bound states:** occur when the particle is confined in the potential well, i.e., energies **less than** the asymptotic potential at infinity (which is 0). - **Thus,** for \(E < 0\), the particle can be trapped, forming bound states. - **Scattering states:** occur when \(E > 0\), where the particle can escape to infinity. **Answer:** | Energy \(E\) | Type of solutions | |--------------|-------------------------| | \(E < 0\) | Bound states | | \(E > 0\) | Scattering states | --- ### **c) Solving the stationary Schrödinger equation \(E < V_a\)** Given: - The incoming beam comes from the right, with energy \(E\), where \(0 < E < V_a\). --- ### **Step 1:** Write the Schrödinger equation in each region: \[ -\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} + V(x)\psi = E \psi. \] --- ### **Step 2:** Define wave numbers: \[ k = \frac{\sqrt{2mE}}{\hbar}, \quad k_1 = \frac{\sqrt{2m(E + 2V_a)}}{\hbar}, \quad \kappa = \frac{\sqrt{2m(V_a - E)}}{\hbar}. \] Note: - For \(V = 0\), the general solution is oscillatory with wave number \(k\). - For \(V = -2V_a\), the potential is negative, so the wave function oscillates with wave number \(k_1\). - For \(V = V_a\), potential is positive, leading to exponential decay or growth with \(\kappa\). --- ### **Step 3:** Write solutions in each region: 1. **Region I (\(x \leq 0\), \(V=0\)):** \[ \psi_1(x) = A e^{i k x} + B e^{-i k x}. \] (Incident wave moving from right to left, so for incoming from the right, the wave coming from the right is \(e^{-i k x}\).) But since the beam comes from the right, better to write: \[ \boxed{ \psi_1(x) = e^{-i k x} + R e^{i k x} } \] - Here, the incident wave is normalized to 1, \(e^{-i k x}\), traveling to the left. - \(R\) is the reflection amplitude. 2. **Region II (\(0 < x \leq a\), \(V=-2V_a\)):** \[ \psi_2(x) = C e^{i k_1 x} + D e^{-i k_1 x}. \] 3. **Region III (\(a < x \leq 3a\), \(V=V_a\)):** \[ \psi_3(x) = F e^{\kappa x} + G e^{-\kappa x}. \] Since \(E < V_a\), the solutions are exponential (decaying or growing). 4. **Region IV (\(x > 3a\), \(V=0\)):** \[ \psi_4(x) = T e^{i k x}. \] - \(T\) is the transmission amplitude to the right. --- ### **Summary of solutions:** \[ \boxed{ \begin{aligned} & \textbf{Region 1 (x \leq 0):} \quad \psi_1(x) = e^{-i k x} + R e^{i k x} \\ & \textbf{Region 2 (0 < x \leq a):} \quad \psi_2(x) = C e^{i k_1 x} + D e^{-i k_1 x} \\ & \textbf{Region 3 (a < x \leq 3a):} \quad \psi_3(x) = F e^{\kappa x} + G e^{-\kappa x} \\ & \textbf{Region 4 (x > 3a):} \quad \psi_4(x) = T e^{i k x} \end{aligned} } \] --- ### **Number of undetermined constants:** - Constants: \(R, C, D, F, G, T\)—a total of **6**. - Boundary conditions at \(x=0, a, 3a\) (continuity of wavefunction and derivative) give **6 equations**. - One of these constants (say, the incident amplitude normalized to 1) is fixed, so **5 constants** are determined. - **Two constants** (say, \(R\) and \(T\)) correspond to reflection and transmission coefficients, while the others are internal. **Answer:** - **Number of undetermined constants:** **2** (the incident amplitude set to 1, and the transmitted amplitude \(T\), which is related to transmission probability). --- ### **d) Reflection and transmission probabilities** - Reflection probability: \[ R_P = |R|^2, \] - Transmission probability: \[ T_P = \frac{\text{probability flux in transmitted wave}}{\text{incident flux}}. \] Since the incident wave is normalized to amplitude 1, and the wavefunctions are oscillatory in the free regions, the **probability flux**: \[ J = \frac{\hbar k}{m} |\text{amplitude}|^2. \] Thus, \[ \boxed{ \begin{aligned} R &= |R|^2, \\ T &= \frac{k}{k} |T|^2 = |T|^2, \end{aligned} } \] assuming equal fluxes in the incident and transmitted waves (since they have same wave number \(k\)). --- ## **(iii) Bound states in a finite potential well** Given: \[ V(x) = \begin{cases} \infty & x \leq 0, \\ 0 & 0 < x < L, \\ V_c & x \geq L, \end{cases} \] with \(V_c > 0\). --- ### **a) Solve for bound states** **Step 1:** The wavefunction must vanish at \(x=0\): \[ \psi(0) = 0. \] **Step 2:** In the well \((0 < x < L)\): \[ -\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} = E \psi, \] with \(E < V_c\) (bound state), so: \[ k = \sqrt{\frac{2mE}{\hbar^2}}. \] Solution: \[ \psi_1(x) = A \sin(k x), \] since \(\psi(0)=0\) implies no cosine term. **Step 3:** Outside the well (\(x \geq L\)), the wavefunction decays exponentially: \[ \psi_2(x) = B e^{-\kappa (x - L)}, \] where \[ \kappa = \sqrt{\frac{2m(V_c - E)}{\hbar^2}}. \] **Step 4:** Boundary conditions at \(x=L\): - Continuity of \(\psi\): \[ A \sin(k L) = B, \] - Continuity of derivative: \[ A k \cos(k L) = - B \kappa, \] which yields: \[ A k \cos(k L) = - (A \sin(k L)) \kappa, \] divide both sides by \(A\): \[ k \cos(k L) = - \sin(k L) \kappa, \] or \[ k \cot(k L) = - \kappa. \] This is the **transcendental equation** for the bound state energies: \[ \boxed{ k \cot(k L) = - \sqrt{\frac{2m(V_c - E)}{\hbar^2}}. } \] Expressed explicitly in terms of \(E\): \[ k = \sqrt{\frac{2m E}{\hbar^2}}, \quad \kappa = \sqrt{\frac{2m(V_c - E)}{\hbar^2}}. \] --- ### **b) Condition on \(V_c\) for at least one bound state** - The lowest bound state corresponds approximately to the ground state of an infinite well: \[ E_1 \approx \frac{\pi^2 \hbar^2}{2m L^2}. \] - For this energy to be less than \(V_c\): \[ V_c > E_1 \implies V_c > \frac{\pi^2 \hbar^2}{2m L^2}. \] **Thus**, **the critical potential depth**: \[ \boxed{ V_c > \frac{\pi^2 \hbar^2}{2m L^2} } \] **must be satisfied for at least one bound state to exist.** --- # **Final Summary** | Part | Key Results | |---------|-------------------------| | (i) | Spatial wavefunction symmetric; particles are in spin singlet (\(S=0\)). | | (ii)a | Sketch as described: step potentials at specified points. | | (ii)b | Bound states for \(E<0\); scattering states for \(E>0\). | | (ii)c | Wavefunctions in each region with constants \(A,B,C,D,F,G,T\); 2 undetermined constants due to boundary conditions. | | (ii)d | Reflection probability \(R=|R|^2\), transmission