# Transmission and Reflection Coefficients

• stunner5000pt
In summary, the conversation discusses the computation of reflection and transmission coefficients for a particle with energy greater than the potential barrier, approaching from the left. The Schrodinger equation is used to find the wavefunctions in the two regions, with the condition that the wavefunction in region II only contains the negative exponential term. The boundary conditions are then applied to calculate the probability current densities and the transmission coefficient is calculated using the ratio of the transmitted current density to the incident current density.
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
Schrodinger 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

stunner5000pt said:
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
Schrodinger 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

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?

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

ψI = A*Exp(ik1x)+B*Exp(-ik1x) in region I (x ≤ 0)
ψII = 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

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

## 1. What are transmission and reflection coefficients?

Transmission and reflection coefficients are mathematical values that represent the percentage of energy that is transmitted or reflected when an electromagnetic wave encounters a boundary between two different materials.

## 2. How are transmission and reflection coefficients calculated?

Transmission and reflection coefficients are calculated using the Fresnel equations, which take into account the angle of incidence, the indices of refraction of the two materials, and the polarization of the wave.

## 3. What factors affect the transmission and reflection coefficients?

The transmission and reflection coefficients are affected by the angle of incidence, the type of material the wave is passing through, the wavelength of the wave, and the polarization of the wave.

## 4. What is the difference between transmission and reflection coefficients?

The transmission coefficient represents the percentage of energy that passes through the boundary into the second material, while the reflection coefficient represents the percentage of energy that is reflected back into the first material.

## 5. How can transmission and reflection coefficients be used in practical applications?

Transmission and reflection coefficients are important in understanding and designing a variety of devices, such as optical coatings, lenses, and filters. They also have applications in fields such as optics, telecommunications, and materials science.

• Advanced Physics Homework Help
Replies
30
Views
1K
• Advanced Physics Homework Help
Replies
10
Views
559
• Advanced Physics Homework Help
Replies
1
Views
1K
• Advanced Physics Homework Help
Replies
14
Views
1K
• Advanced Physics Homework Help
Replies
10
Views
325
• Advanced Physics Homework Help
Replies
24
Views
790
• Advanced Physics Homework Help
Replies
29
Views
116
• Advanced Physics Homework Help
Replies
1
Views
868
• Advanced Physics Homework Help
Replies
1
Views
581
• Advanced Physics Homework Help
Replies
4
Views
934