Transmission and Reflection Coefficients

[stunner5000pt]

Consider the step potential defined by
V(x) = 0 if x < 0
Vb > 0 if x=> 0

a) Compute te relfection and tranmission coefficients for a particle of energy E > Vb approaching from th left

For x < 0
Schrödinger equaion since V = 0
$$-\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} = E \psi(x)$$

$$\frac{\partial^2 \psi}{\partial x^2} = -\frac{2mE}{\hbar^2} \psi(x)$$

define $$k_{1}^2 = \frac{2mE}{\hbar^2}$$

$$\frac{\partial^2 \psi}{\partial x^2} = -k^2 \psi(x)$$

$$\psi(x) = A \cos(k_{1}x) + B \sin(k_{1} x)$$

for x=> 0
$$-\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} = (E-V_{B}) \psi(x)$$
$$k_{2}^2 = \frac{2m}{\hbar^2} (E - V_{B})$$
$$\frac{\partial^2 \psi}{\partial x^2} = k_{2}^2 \psi(x)$$
$$\psi_(x) = C \exp(ik_{2} x) + D\exp(-ik_{2}x)$$
but we only care about the one with the negative exponential
hence C = 0
$$\psi_(x) = D\exp(-ik_{2}x)$$

are these solutions correct? I am just afraid i may have made a stupid mistake by switching the negative sign somewhere :(

ok to proceed i apply the boundary conditions s.t.

$$\psi_{I}(0) = \psi_{T}(0)$$
$$\psi_{I}'(0) = \psi_{T}'(0)$$

we calculate probability current density j

$$j_{trans} = \frac{\hbar k_{2}}{m} |D|^2$$
$$j _{inc} = \frac{\hbar k_{1}}{m} |A|^2$$
$$j_{refl} = \frac{\hbar k_{1}}{m} |B|^2$$

and transmission coefficient is calculated like this

$$T = \left| \frac{j_{trans}}{j_{inc}} \right|$$
$$R = \left| \frac{j_{refl}}{j_{inc}} \right|$$

thank you for your input

 

[nrqed]

Consider the step potential defined by
V(x) = 0 if x < 0
Vb > 0 if x=> 0

a) Compute te relfection and tranmission coefficients for a particle of energy E > Vb approaching from th left

For x < 0
Schrödinger equaion since V = 0
$$-\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} = E \psi(x)$$

$$\frac{\partial^2 \psi}{\partial x^2} = -\frac{2mE}{\hbar^2} \psi(x)$$

define $$k_{1}^2 = \frac{2mE}{\hbar^2}$$

$$\frac{\partial^2 \psi}{\partial x^2} = -k^2 \psi(x)$$

$$\psi(x) = A \cos(k_{1}x) + B \sin(k_{1} x)$$

for x=> 0
$$-\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} = (E-V_{B}) \psi(x)$$
$$k_{2}^2 = \frac{2m}{\hbar^2} (E - V_{B})$$
$$\frac{\partial^2 \psi}{\partial x^2} = k_{2}^2 \psi(x)$$
$$\psi_(x) = C \exp(ik_{2} x) + D\exp(-ik_{2}x)$$
but we only care about the one with the negative exponential
hence C = 0
$$\psi_(x) = D\exp(-ik_{2}x)$$

are these solutions correct? I am just afraid i may have made a stupid mistake by switching the negative sign somewhere :(

You want a wave traveling to the right so you must keep the C wavefunction and set D =0.

Patrick

 

[FunkyDwarf]

Sorry to resurrect a long dead thread, but if i wanted to impose the condition D = 0, ie incoming waves only from one direction on standing wave computed numerically, how would i go about doing that? I know Fourier analysis can give me the momentum components, but that's about it. Would i have to do it in the time dependent case?

 

[kuruman]

Say you have a wave incident from the left. The wavefunction would be

ψ_I = A*Exp(ik₁x)+B*Exp(-ik₁x) in region I (x ≤ 0)
ψ_II = C*Exp(ik₂x) in region II (x ≥ 0)

You have an incident and reflected wave in region I (constants A and B respectively) and a transmitted wave in region II. Of course, you have to impose the proper boundary conditions at x=0.

For a lovely simulation of this and potential barriers in general, you may wish to go to
http://phet.colorado.edu/en/simulation/quantum-tunneling

 

[FunkyDwarf]

I understand that perfectly from an analytic point of view, but what I've got is the equation (actually want to do the KG equation but good to start with Schrod) with a set of boundary conditions, mass of the particle and potential and that's it. I then solve numerically and get a standing wave. It is in this process that i am unsure as to how to impose the correct condition to kill the other incoming wave. In fact, I'm really lost because essentially mathematica will spit out what are essentially shifted cos waves in either regime, which when you take a Fourier transform will have momentum components in both directions, both with equal magnitude (as you expect from a cosine/sine wave). I thought perhaps a Fourier decomposotion would pull out the momentum components and i could kill them off accordingly but it didnt work really :(

thanks!
-G