Let's address the problem step by step as requested. --- ## Step 1: **Sketch and General Solution for \( < E < V_ \) (Parts a & b)** ### a) **Sketch the Situation** - **Potential Step:** - For \( x < \) (Region I): \( V(x) = \) - For \( x \geq \) (Region II): \( V(x) = V_ \) - Draw an axis: - \( x \) on the horizontal. - \( V(x) \) on the vertical. - At \( x = \), a step up in potential from \( \) to \( V_ \). **Explanation:** This is a classic quantum step potential, with the particle approaching from left (\( x < \)), encountering a sudden jump in potential at \( x = \). --- ### b)General Solution for the Wave Function (\( < E < V_))** #### **Region I (\( x < ), \( V = \)):** The time-independentödinger equation: \[ -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E\psi \] \[ \frac{d^2\psi}{dx^2} + k_1^2\psi = ,\quad k_1 = \frac{\sqrt{2mE}}{\hbar} \] **General solution:** \[ \psi_I(x) = Ae^{ik_1 x} + Be^{-ik_1 x} \] Where: - \( Ae^{ik_1 x} \): incident wave (moving right) - \( Be^{-ik_1 x} \): reflected wave (moving left) --- #### **Region II (\( x \geq \), \( V = V_ \)):** \[ -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V_\psi = E\psi \] \[ \frac{d^2\psi}{dx^2} = \frac{2m}{\hbar^2}(V_ - E)\psi \] Let \( \kappa = \frac{\sqrt{2m(V_ - E)}}{\hbar} \) (since \( V_ > E \)), so solution is exponential: \[ \psi_{II}(x) = Ce^{-\kappa x} + De^{\kappa x} \] But for \( x \to +\infty \), \( e^{\kappa x} \) blows up, so set \( D = \): \[ \psi_{II}(x) = Ce^{-\kappa x} \] **Explanation:** For \( E < V_ \), the wavefunction in Region II decays exponentially (classically forbidden region). In Region I, there are incident and reflected waves. --- ## Final Answers for Step 1 - **Sketch:** Potential step at \( x = \); Region I (\( V= \)), Region II (\( V=V_ \)). - **General solutions:** - Region I: \( \psi_I(x) = Ae^{ik_1 x} + Be^{-ik_1 x} \) - Region II: \( \psi_{II}(x) = Ce^{-\kappa x} \) --- **Let me know if you'd like to proceed to Step 2!**

## Step 2: **Boundary Conditions and Final Wave Function** ### a) **Apply Boundary Conditions** To find the coefficients \( A \), \( B \), and \( C \), we need to apply boundary conditions at \( x = 0 \) (the potential step). 1. **Continuity of Wave Function:** \[ \psi_I(0) = \psi_{II}(0) \] This gives: \[ A + B = C \] 2. **Continuity of Derivative:** \[ \left.\frac{d\psi_I}{dx}\right|_{x=0} = \left.\frac{d\psi_{II}}{dx}\right|_{x=0} \] Evaluating the derivatives: \[ ik_1(A - B) = -\kappa C \] **Explanation:** The continuity of the wave function and its derivative ensures that there are no discontinuities at the boundary, which is a fundamental requirement in quantum mechanics. --- ### b) **Solve for Coefficients** From the equations obtained: 1. **From the first equation:** \[ C = A + B \] 2. **Substituting \( C \) into the second equation:** \[ ik_1(A - B) = -\kappa (A + B) \] Rearranging gives: \[ ik_1 A - ik_1 B + \kappa A + \kappa B = 0 \] \[ (ik_1 + \kappa)A + (-ik_1 + \kappa)B = 0 \] This is a system of linear equations in \( A \) and \( B \). The coefficients can be related to the transmission and reflection coefficients, \( T \) and \( R \), defined as: - \( R = \left| \frac{B}{A} \right|^2 \) - \( T = \left| \frac{C}{A} \right|^2 \) **Explanation:** The relationship between the coefficients helps us to determine the probability of reflecting or transmitting the wave at the potential step. --- ## Final Answers for Step 2 - **Boundary Conditions:** - \( A + B = C \) - \( (ik_1 + \kappa)A + (-ik_1 + \kappa)B = 0 \) - **Coefficients:** - \( R \) and \( T \) can be derived from the relationships above. --- This concludes the solution for the quantum step potential problem. If there are further questions or clarifications needed, please let me know!