Let's solve the one-dimensional quantum tunneling problem step by step. ### **Problem Setup** We have a potential barrier of height \( V_0 \) and width \( L \), as shown in the figure (standard text figure): \[ V(x) = \begin{cases} 0 & x < 0 \\ V_0 & 0 \leq x \leq L \\ 0 & x > L \end{cases} \] A particle of energy \( E \) (with \( E < V_0 \)) approaches from the left. --- ## **(a) Schrödinger Equations and Wavefunctions** ### **Regions** #### **Region I: \( x < 0 \), \( V(x) = 0 \)** \[ -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E\psi \] \[ \frac{d^2\psi}{dx^2} + k^2\psi = 0, \quad k = \frac{\sqrt{2mE}}{\hbar} \] **General solution:** \[ \psi_I(x) = e^{ikx} + R e^{-ikx} \] Where: - \( e^{ikx} \): incident wave - \( R e^{-ikx} \): reflected wave - \( R \): reflection amplitude --- #### **Region II: \( 0 \leq x \leq L, \; V(x) = V_0 \)** \[ -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V_0\psi = E\psi \] \[ \frac{d^2\psi}{dx^2} = \frac{2m}{\hbar^2}(V_0 - E)\psi = \kappa^2\psi \] \[ \kappa = \frac{\sqrt{2m(V_0-E)}}{\hbar} \] **General solution:** \[ \psi_{II}(x) = A e^{\kappa x} + B e^{-\kappa x} \] --- #### **Region III: \( x > L, \; V(x) = 0 \)** \[ \psi_{III}(x) = T e^{ikx} \] Where \( T \) is the transmission amplitude. --- ## **(b) Boundary Conditions** The wavefunction \(\psi(x)\) and its first derivative \(\psi'(x)\) must be continuous at \(x=0\) and \(x=L\): ### **At \( x = 0 \):** \[ \psi_I(0) = \psi_{II}(0) \] \[ \psi'_I(0) = \psi'_{II}(0) \] Substitute: \[ \psi_I(0) = 1 + R, \quad \psi_{II}(0) = A + B \] \[ \psi'_I(0) = ik(1 - R), \quad \psi'_{II}(0) = \kappa(A - B) \] **So:** \[ 1 + R = A + B \tag{1} \] \[ ik(1 - R) = \kappa(A - B) \tag{2} \] --- ### **At \( x = L \):** \[ \psi_{II}(L) = \psi_{III}(L) \] \[ \psi'_{II}(L) = \psi'_{III}(L) \] Substitute: \[ \psi_{II}(L) = A e^{\kappa L} + B e^{-\kappa L} \] \[ \psi_{III}(L) = T e^{ikL} \] \[ \psi'_{II}(L) = \kappa [A e^{\kappa L} - B e^{-\kappa L}] \] \[ \psi'_{III}(L) = ik T e^{ikL} \] **So:** \[ A e^{\kappa L} + B e^{-\kappa L} = T e^{ikL} \tag{3} \] \[ \kappa [A e^{\kappa L} - B e^{-\kappa L}] = ik T e^{ikL} \tag{4} \] --- ## **Summary of Equations** The four equations are: 1. \( 1 + R = A + B \) 2. \( ik(1 - R) = \kappa(A - B) \) 3. \( A e^{\kappa L} + B e^{-\kappa L} = T e^{ikL} \) 4. \( \kappa(A e^{\kappa L} - B e^{-\kappa L}) = ik T e^{ikL} \) --- ## **Final Answer (Summary and Interpretation)** ### **(a) Schrödinger's Equations (and solutions):** - **For \( x < 0 \):** \(\psi_I(x) = e^{ikx} + R e^{-ikx}\) - **For \( 0 \leq x \leq L \):** \(\psi_{II}(x) = A e^{\kappa x} + B e^{-\kappa x}\) - **For \( x > L \):** \(\psi_{III}(x) = T e^{ikx}\) Where: - \( k = \frac{\sqrt{2mE}}{\hbar} \) - \( \kappa = \frac{\sqrt{2m(V_0-E)}}{\hbar} \) --- ### **(b) Boundary Conditions:** At \( x = 0 \): \[ \begin{cases} 1 + R = A + B \\ ik(1 - R) = \kappa(A - B) \end{cases} \] At \( x = L \): \[ \begin{cases} A e^{\kappa L} + B e^{-\kappa L} = T e^{ikL} \\ \kappa(A e^{\kappa L} - B e^{-\kappa L}) = ik T e^{ikL} \end{cases} \] --- **These are the required equations and boundary conditions.** If you want to go further and solve for \( T \) (transmission coefficient), let me know!