# Quantum - determine reflection coefficient

1. Jun 10, 2015

### wood

I am working through a past exam paper and this one has me stumped

1. The problem statement, all variables and given/known data

Consider a particle of mass m with kinetic energy E incident from the left upon a
step-up potential:

$$U(x)=\begin{cases} 0 & \quad \text{for } x <0\\ V & \quad \text{for } x>0\\ \end{cases}$$

assuming V > 0. Assume that E > V .

(a) For the two distinct regions, write down the relevant time-independent Schrodinger ¨
equation in the form

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

where k is real. Clearly specify the values of k^2 in each region, using kL to
represent the value of k to the left of the step, and kR to represent the value of k to
the right of the step. For each case, write down the general solution of the
Schrodinger equation in terms of real or complex exponentials.

(b) Impose the relevant boundary conditions for this system.

(c) Solve your boundary conditions to show that the reflection coefficient is given by
$$R=\left(\frac{k_R-k_L}{k_R+k_L}\right)^2$$

[3]
(d) Obtain a formula for the transmission coefficient. How does it differ from that
obtained in the classical case?

2. Relevant equations

TISE and $R=|\frac{B}{A}|^2$

3. The attempt at a solution

So I can get my equations for part a
$$\psi_L(x)=Ae^{ik_Lx}+Be^{-ik_Lx}$$
$$\psi_L(x)=Ce^{ik_Lx}+De^{-ik_Lx}$$
D=0 as there is no reflected wave on x>0 so
$$\psi_L(x)=Ce^ik_Lx$$
and impose the boundary conditions at x=0 for part be and get
$$\psi_L(0)=\psi_R(0)$$
$$A+B=C$$
and
$$\psi'_L(0)=\psi'_R(0)$$
$$ik_L(A-B)=ik_RC$$
$$C=\frac{k_L(A-B)}{k_R}=A+B$$
And this is where it all goes wrong. I cannot get from there to $$\frac{B}{A}=\left(\frac{k_R-k_L}{k_R+k_L}\right)$$
I end up with $\left(\frac{k_L-k_R}{k_R+k_L}\right)$ when I try to compute B/A. Is there another way to solve the boundary conditions for the reflection coefficient that gets the correct answer?

Thanks

2. Jun 10, 2015

### Orodruin

Staff Emeritus
You did get the correct answer ...

3. Jun 10, 2015

### wood

No, I keep getting kL-kR when I try to solve for A and B algebraically.

4. Jun 10, 2015

### Orodruin

Staff Emeritus
What happens when you square a negative number?

5. Jun 10, 2015

### wood

I get a positive number so I thnk I might be an idiot and have had the correct answer all along...ie
$$\left(\frac{k_R-k_L}{k_R+k_L}\right)^2=\left(\frac{k_L-k_R}{k_R+k_L}\right)^2$$