How Does Quantum Tunneling Affect Particle Reflection?

[roam]

Homework Statement

A beam of particles of energy E is incident from the left on a rectangular step of infinite width and height U₀, situated at the origin, with E > U₀, as shown in the diagram.

http://img39.imageshack.us/img39/5918/problemda.jpg

(a) Write down the Schrodinger equation for a particle in Region 1, and give a general expression for its wave function in terms of the wave number k₁ in Region 1. Give the expression for k₁ also. Repeat the same process for Region 2 and write an expression for k₂.(b) By applying the boundary conditions at the origin, show that the reflection coefficient R is given by

$R=\frac{(k_1-k_2)^2}{(k_1+k_2)^2}$

Homework Equations

Relevant boundary conditions:

$A+B=C+D$

$ikA-ikB=-\alpha C + \alpha D$

The Attempt at a Solution

(a) In region 1 I believe the potential is 0 so the Schrodinger equation in this region is:

$-\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} = E \psi$

or

$\frac{d^2 \psi}{dx^2} =k^2 \psi$

$k_1= \frac{\sqrt{-2mE}}{\hbar}$

A general solution to this would be:

$\psi (x) = Ae^{ik_1x} + Be^{-ik_1x}$

Now for Region 2 the potential is U(x)=U₀ so the equation is

$-\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2}-U_0 \psi = E \psi$

$\implies \frac{d^2 \psi}{dx^2} = -k^2 \psi$

$k_2= \frac{\sqrt{2m(E+U_0)}}{\hbar}$

And wave function is $\psi(x) = Ce^{-\alpha x} + De^{\alpha x}$

where $\alpha= \frac{\sqrt{2m(U-E)}}{\hbar}$

I hope everything is correct so far. I would appreciate it if anyone could correct me if I'm wrong.

(b) So I'm not sure if my approach is correct, but since in my notes the reflection coefficient is defined as

$R= \frac{|B|^2}{|A|^2}$​

I have substituted k₁ and k₂ into (k₁-k₂)²/(k¹-k₁)² to see if we can simplify it to yield the same expression:

$\frac{\left(\frac{\sqrt{-2mE}}{\hbar} - \frac{\sqrt{2m(E+U_0)}}{\hbar} \right)^2}{\left(\frac{\sqrt{-2mE}}{\hbar} + \frac{\sqrt{2m(E+U_0)}}{\hbar} \right)^2}$

$= \frac{-4mE+2mU_0}{2mU_0}$

So how can I show that this is equal to the reflection coefficient R?
Any guidance is appreciated.

 

[mark726]

Thank you for your post. Your approach to solving this problem is correct so far. However, in order to show that the expression you have derived is equal to the reflection coefficient R, you need to apply the boundary conditions at the origin.

From the Schrödinger equation, we know that the wave function and its first derivative must be continuous at the boundary. This means that at x=0, we have:

\psi_1(0) = \psi_2(0)

and

\frac{d\psi_1}{dx}(0) = \frac{d\psi_2}{dx}(0)

where \psi_1 is the wave function in Region 1 and \psi_2 is the wave function in Region 2.

Using the general expressions for the wave functions that you have derived, we can write:

A + B = C + D

and

ik_1A - ik_1B = -\alpha C + \alpha D

Solving these equations for B and D, we get:

B = \frac{ik_1-\alpha}{ik_1 + \alpha} A

and

D = \frac{ik_1 + \alpha}{ik_1 + \alpha} C

Substituting these values into the reflection coefficient R, we get:

R = \frac{|B|^2}{|A|^2} = \frac{\left(\frac{ik_1-\alpha}{ik_1 + \alpha} \right)^2}{\left(\frac{ik_1 + \alpha}{ik_1 + \alpha} \right)^2} = \frac{(k_1-\alpha)^2}{(k_1+\alpha)^2}

Now, we can substitute the expressions for k_1 and \alpha that you have derived in part (a) into this expression to get:

R = \frac{\left(\frac{\sqrt{-2mE}}{\hbar} - \frac{\sqrt{2m(E+U_0)}}{\hbar} \right)^2}{\left(\frac{\sqrt{-2mE}}{\hbar} + \frac{\sqrt{2m(E+U_0)}}{\hbar} \right)^2} = \frac{-4mE+2mU_0}{2mU_0}

which is the same expression