Consider an attractive square-well potential, \(U(x)=0\) for \(x<-\alpha, U(x)=-U_{0}\) for \(-\alpha \leq x \leq \alpha\) where \(U_{0}\) is a positive constant, and \(U(x)=0\) for \(x>\alpha .\) For \(E>0,\) the solution of the Schrödinger equation in the 3 regions will be the following: For \(x<-\alpha, \psi(x)=e^{i \kappa x}+R e^{-i \kappa x}\) where \(\kappa^{2}=2 m E / \hbar^{2}\) and \(R\) is the amplitude of a reflected wave. For \(-\alpha \leq x \leq \alpha, \psi(x)=A e^{i \kappa^{\prime} x}+B e^{-i \kappa^{\prime} x}\) and \(\left(\kappa^{\prime}\right)^{2}=2 m\left(E+U_{0}\right) / \hbar^{2}\). For \(x>\alpha, \psi(x)=T e^{i \kappa x}\) where \(T\) is the amplitude of the transmitted wave. Match \(\psi(x)\) and \(d \psi(x) / d x\) at \(-\alpha\) and \(\alpha\) and find an expression for \(R\). What is the condition for which \(R=0\) (that is, there is no reflected wave)?

Step by step solution

01

Write down the boundary conditions for \(\psi(x)\)

At \(x=-\alpha\) and \(x=\alpha\), the wave functions and their first-derivatives must be continuous. Write down these boundary conditions: 1. \(\psi(x=-\alpha)\): \(e^{i \kappa (-\alpha)}+R e^{-i \kappa (-\alpha)} = A e^{i \kappa' (-\alpha)}+B e^{-i \kappa' (-\alpha)}\) 2. \(\psi(x=\alpha)\): \(T e^{i \kappa \alpha} = A e^{i \kappa' \alpha}+B e^{-i \kappa' \alpha}\) 3. \(\frac{d\psi}{dx}(x=-\alpha)\): \(i \kappa (e^{i \kappa (-\alpha)}-R e^{-i \kappa (-\alpha)}) = i \kappa' (A e^{i \kappa' (-\alpha)}-B e^{-i \kappa' (-\alpha)})\) 4. \(\frac{d\psi}{dx}(x=\alpha)\): \(i \kappa T e^{i \kappa \alpha} = i \kappa' (A e^{i \kappa' \alpha}-B e^{-i \kappa' \alpha})\)

02

Solve the system of equations

We have four equations with four unknowns (R, T, A, and B). Solve this system of equations in order to find the amplitude of the reflected wave R. Divide equation 3 by equation 1 and equation 4 by equation 2, to eliminate A and B. 5. \(\frac{i \kappa (e^{i \kappa (-\alpha)}-R e^{-i \kappa (-\alpha)})}{e^{i \kappa (-\alpha)}+R e^{-i \kappa (-\alpha)}} = \frac{i \kappa' (A e^{i \kappa' (-\alpha)}-B e^{-i \kappa' (-\alpha)})}{A e^{i \kappa' (-\alpha)}+B e^{-i \kappa' (-\alpha)}}\) 6. \(\frac{i \kappa T e^{i \kappa \alpha}}{A e^{i \kappa' \alpha}+B e^{-i \kappa' \alpha}} = \frac{i \kappa' (A e^{i \kappa' \alpha}-B e^{-i \kappa' \alpha})}{T e^{i \kappa \alpha}}\) Now, we can find R by solving equations 5 and 6.

03

Find the expression for R

Divide equation 6 by equation 5 in order to eliminate all the terms without R. Then, solve for R: \(R = \frac{\kappa - \kappa'}{\kappa + \kappa'}\)

04

Find the condition for R = 0

In order to find the condition for which there is no reflected wave (\(R=0\)), we will set R equal to zero and solve for the relationship between \(\kappa\) and \(\kappa'\): \(\frac{\kappa - \kappa'}{\kappa + \kappa'}=0\) This condition is satisfied when \(\kappa=\kappa'\), which means the wave has no reflection, and all of the wave energy is transmitted through the potential.

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