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wood
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I am working through a past exam paper and this one has me stumped
1. Homework Statement
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 Schrödinger ¨
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
Schrödinger 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?
TISE and ##R=|\frac{B}{A}|^2##
[/B]
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
1. Homework Statement
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 Schrödinger ¨
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
Schrödinger 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?
Homework Equations
TISE and ##R=|\frac{B}{A}|^2##
The Attempt at a Solution
[/B]
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