# Quantum Step Potential Problem ## (a) **Solving the Schrödinger Equation** ### **Potential Profile** - For \( x < \): \( V(x) = \) - For \( x \geq \): \( V(x) = V_ \) - Particle energy: \( E > V_ \) ### **Schrödinger Equation** \[ -\frac{\hbar^2}{2m} \frac{d^2\psi}{dx^2} + V(x)\psi = E\psi \] ### **Region I (\(x < \))** \[ -\frac{\hbar^2}{2m} \frac{d^2\psi}{dx^2} = E\psi \implies \frac{d^2\psi}{dx^2} + k_1^2 \psi = \] where \( k_1 = \frac{\sqrt{2mE}}{\hbar} \). General solution: \[ \psi_1(x) = A e^{ik_1 x} + B e^{-ik_1 x} \] - \(A\): amplitude of wave traveling to the right (toward the step) - \(B\): amplitude of reflected wave traveling to the left ### **Region II (\(x \geq \))** \[ -\frac{\hbar^2}{2m} \frac{d^2\psi}{dx^2} + V_\psi = E\psi \implies \frac{d^2\psi}{dx^2} + k_2^2 \psi = \] where \( k_2 = \frac{\sqrt{2m(E-V_)}}{\hbar} \). General solution (for particle incident from right): \[ \psi_2(x) = C e^{-ik_2 x} \] Here, we take only the wave traveling to the left (\(e^{-ik_2 x}\)), as the particle is incident from the right and there is no wave coming from \(x = +\infty\). ### **Summary** Expressing in terms of a single unknown (let's use \(C\)): - For \(x < \): \(\psi_1(x) = A e^{ik_1 x} + B e^{-ik_1 x}\) - For \(x \geq \): \(\psi_2(x) = C e^{-ik_2 x}\) **But since the particle is incident from right, we set \(A = \):** - For \(x < \): \(\psi_1(x) = B e^{-ik_1 x}\) - For \(x \geq \): \(\psi_2(x) = C e^{-ik_2 x}\) Or, if you prefer to use \(C\) as the only unknown: - For \(x < \): \(\psi_1(x) = r C e^{ik_1 x} + C e^{-ik_1 x}\) - For \(x \geq \): \(\psi_2(x) = t C e^{-ik_2 x}\) where \(r\) and \(t\) are reflection and transmission amplitudes. --- ## (b) **Reflection Coefficient for \(E = 1.25V_\)** ### **Boundary Conditions at \(x = \):** 1. \(\psi_1() = \psi_2()\) 2. \(\psi_1'() = \psi_2'()\) From above: \[ \psi_1(x) = B e^{-ik_1 x} \] \[ \psi_2(x) = C e^{-ik_2 x} \] At \(x=\): \[ B = C \] \[ -ik_1 B = -ik_2 C \implies k_1 B = k_2 C \] But this leads to \(k_1 = k_2\), which is not generally true unless \(V_ = \). Let's include the general form: Let the general solution for \(x < \): \[ \psi_1(x) = A e^{ik_1 x} + B e^{-ik_1 x} \] For \(x \geq \): \[ \psi_2(x) = C e^{-ik_2 x} \] The incident wave is from the right (\(e^{-ik_2 x}\)), so in \(x \geq \) we have only the left-moving wave. Boundary conditions at \(x=\): 1. \(A + B = C\) 2. \(ik_1 A - ik_1 B = -ik_2 C\) Solving: \[ \begin{cases} A + B = C \\ k_1 (A - B) = -k_2 C \end{cases} \] Express \(A\) and \(B\) in terms of \(C\): From (1): \(A = C - B\) Plug into (2): \[ k_1 [(C - B) - B] = -k_2 C \implies k_1 (C - 2B) = -k_2 C \implies k_1 C - 2k_1 B = -k_2 C \implies k_1 C + k_2 C = 2k_1 B \implies B = \frac{(k_1 + k_2)}{2k_1} C \] Similarly, \[ A = C - B = C - \frac{(k_1 + k_2)}{2k_1} C = \frac{(2k_1 - k_1 - k_2)}{2k_1} C = \frac{(k_1 - k_2)}{2k_1} C \] ### **Reflection Coefficient \(R\):** \[ R = \left| \frac{A}{B} \right|^2 \] Plug in values: \[ \frac{A}{B} = \frac{\frac{k_1 - k_2}{2k_1}}{\frac{k_1 + k_2}{2k_1}} = \frac{k_1 - k_2}{k_1 + k_2} \] So: \[ R = \left| \frac{k_1 - k_2}{k_1 + k_2} \right|^2 \] Where: \[ k_1 = \frac{\sqrt{2mE}}{\hbar} \] \[ k_2 = \frac{\sqrt{2m(E - V_)}}{\hbar} \] For \(E = 1.25 V_\): \[ k_1 = \frac{\sqrt{2m \cdot 1.25 V_}}{\hbar} \] \[ k_2 = \frac{\sqrt{2m (1.25 V_ - V_)}}{\hbar} = \frac{\sqrt{2m \cdot .25 V_}}{\hbar} \] So, \[ R = \left| \frac{\sqrt{1.25} - \sqrt{.25}}{\sqrt{1.25} + \sqrt{.25}} \right|^2 \] Calculate: \[ \sqrt{1.25} \approx 1.118, \quad \sqrt{.25} = .5 \] \[ R = \left| \frac{1.118 - .5}{1.118 + .5} \right|^2 = \left| \frac{.618}{1.618} \right|^2 \approx (.382)^2 \approx .146 \] --- ## **Summary** **(a) Wavefunctions:** - \(x < : \psi_1(x) = \frac{k_1 - k_2}{2k_1} C e^{ik_1 x} + \frac{k_1 + k_2}{2k_1} C e^{-ik_1 x}\) - \(x \geq : \psi_2(x) = C e^{-ik_2 x}\) **(b) Reflection coefficient for \(E = 1.25V_\):** \[ R \approx .146 \] --- **Let me know if you want a more detailed algebraic derivation or further explanation!**