Let's break down each part of the problem: --- ### **Step Potential:** \[ V(x) = \begin{cases} 0, & x \leq 0 \\ V_0, & x > 0 \end{cases} \] A particle of energy \(E\) approaches the step from the left. --- ## **Part (a): Reflection Coefficient for \(E < V_0\)** ### **Step 1: Write the Schrödinger Equation Solutions** For \(x \leq 0\) (region I, \(V=0\)): \[ \psi_I(x) = Ae^{ik_1x} + Be^{-ik_1x} \] where \(k_1 = \sqrt{2mE}/\hbar\). For \(x > 0\) (region II, \(V=V_0\)): \[ \psi_{II}(x) = Ce^{-\kappa x} \] where \(\kappa = \sqrt{2m(V_0 - E)}/\hbar\) (since \(E < V_0\), the wave decays exponentially). ### **Step 2: Boundary Conditions at \(x = 0\)** - Continuity of wavefunction: \(\psi_I(0) = \psi_{II}(0)\) - Continuity of derivative: \(\psi_I'(0) = \psi_{II}'(0)\) So: 1. \(A + B = C\) 2. \(ik_1A - ik_1B = -\kappa C\) ### **Step 3: Solve for \(B/A\) (Reflection Amplitude)** From (1): \(C = A + B\) Plug into (2): \[ ik_1A - ik_1B = -\kappa (A + B) \] \[ ik_1A - ik_1B + \kappa A + \kappa B = 0 \] \[ (ik_1 + \kappa)A + (-ik_1 + \kappa)B = 0 \] \[ (ik_1 + \kappa)A = (ik_1 - \kappa)B \] \[ \frac{B}{A} = \frac{ik_1 + \kappa}{ik_1 - \kappa} \] ### **Step 4: Reflection Coefficient** The reflection coefficient is \(R = |B/A|^2\): \[ R = \left|\frac{ik_1 + \kappa}{ik_1 - \kappa}\right|^2 \] But, for \(E < V_0\), \(k_1\) and \(\kappa\) are both real, so: \[ |ik_1 + \kappa|^2 = k_1^2 + \kappa^2 \] \[ |ik_1 - \kappa|^2 = k_1^2 + \kappa^2 \] So, \[ R = 1 \] **Comment:** For \(E < V_0\), the reflection coefficient is 1. This means the particle is totally reflected and cannot penetrate the step, which matches classical expectations. --- ## **Part (b): Reflection Coefficient for \(E > V_0\)** ### **Step 1: Solutions** For \(x > 0\): \(\psi_{II}(x) = Fe^{ik_2x}\), where \(k_2 = \sqrt{2m(E-V_0)}/\hbar\). ### **Step 2: Boundary Conditions** 1. \(A + B = F\) 2. \(ik_1A - ik_1B = ik_2F\) ### **Step 3: Solve for \(B/A\)** From (1): \(F = A + B\) Plug into (2): \[ ik_1A - ik_1B = ik_2(A + B) \] \[ ik_1A - ik_1B - ik_2A - ik_2B = 0 \] \[ (i k_1 - i k_2)A + (-i k_1 - i k_2)B = 0 \] \[ (k_1 - k_2)A = (k_1 + k_2)B \] \[ \frac{B}{A} = \frac{k_1 - k_2}{k_1 + k_2} \] ### **Step 4: Reflection Coefficient** \[ R = \left| \frac{k_1 - k_2}{k_1 + k_2} \right|^2 \] --- ## **Part (c): Transmission Coefficient \(T\) for \(E > V_0\)** Given: \[ T = \frac{k_2}{k_1} \left| \frac{F}{A} \right|^2 \] From boundary conditions: \[ F = A + B \] Substitute \(B = \frac{k_1 - k_2}{k_1 + k_2}A\): \[ F = A + \frac{k_1 - k_2}{k_1 + k_2}A = \frac{2k_1}{k_1 + k_2}A \] So: \[ \left| \frac{F}{A} \right|^2 = \left| \frac{2k_1}{k_1 + k_2} \right|^2 \] \[ T = \frac{k_2}{k_1} \left| \frac{2k_1}{k_1 + k_2} \right|^2 = \frac{k_2}{k_1} \cdot \frac{4k_1^2}{(k_1 + k_2)^2} = \frac{4k_1 k_2}{(k_1 + k_2)^2} \] --- ## **Part (d): Check \(T + R = 1\) for \(E > V_0\)** \[ T = \frac{4k_1 k_2}{(k_1 + k_2)^2} \] \[ R = \left| \frac{k_1 - k_2}{k_1 + k_2} \right|^2 = \frac{(k_1 - k_2)^2}{(k_1 + k_2)^2} \] \[ T + R = \frac{4k_1 k_2 + (k_1 - k_2)^2}{(k_1 + k_2)^2} = \frac{4k_1 k_2 + (k_1^2 - 2k_1k_2 + k_2^2)}{(k_1 + k_2)^2} \] \[ = \frac{k_1^2 + 2k_1k_2 + k_2^2}{(k_1 + k_2)^2} = 1 \] --- ## **Summary of Answers** ### **(a) \(E < V_0\):** \[ R = 1 \qquad (\text{total reflection, classically expected}) \] ### **(b) \(E > V_0\):** \[ R = \left| \frac{k_1 - k_2}{k_1 + k_2} \right|^2 \] ### **(c) \(E > V_0\):** \[ T = \frac{4k_1 k_2}{(k_1 + k_2)^2} \] or as in the problem statement, \[ T = \sqrt{\frac{E - V_0}{E}} \cdot \left| \frac{F}{A} \right|^2 \] ### **(d) \(T + R = 1\)** is confirmed. --- #### **Where:** - \(k_1 = \sqrt{2mE}/\hbar\) - \(k_2 = \sqrt{2m(E-V_0)}/\hbar\) - \(\kappa = \sqrt{2m(V_0-E)}/\hbar\) (for \(E < V_0\)) If you need any step explained further, let me know!

Certainly! Here's a concise, one-line theoretical explanation for each part: --- ### **Part (a):** When \(E < V_0\), the wave cannot propagate into the barrier and is totally reflected, so the reflection coefficient \(R = 1\). ### **Part (b):** When \(E > V_0\), the wave partially transmits and reflects, with the reflection coefficient given by \(\left| \frac{k_1 - k_2}{k_1 + k_2} \right|^2\), reflecting the impedance mismatch. ### **Part (c):** The transmission coefficient for \(E > V_0\) is proportional to the ratio of transmitted to incident wave amplitudes, scaled by the ratio of their wave numbers, resulting in \(T = \frac{4k_1 k_2}{(k_1 + k_2)^2}\). ### **Part (d):** Because probability current is conserved, the sum of transmission and reflection coefficients satisfies \(T + R = 1\) for \(E > V_0\). --- Let me know if you'd like any further clarification!