Solving schrodinger, reflection coefficient

In summary, the conversation discusses the potential function V(x) for a particle with energy E coming in from the left, and the solutions to the Schrodinger equation for the three regions defined by the potential function. The solution for regions I and III have two terms, while the solution for region II only has one term due to the exponential decay of the probability in that region. The second term in regions I and III represents a particle coming from the right. The conversation also touches on the complex nature of the solution for region II due to the potential function being greater than the particle's energy.
  • #1
SoggyBottoms
59
0
Consider the potential

[tex]
V(x) =
\begin{cases}
0, & x < -a & (I) \\
+W, & -a < x < a & (II) \\
0, & x > a & (III)
\end{cases}
[/tex]

for a particle coming in from the left ([itex]-\infty[/itex]) with energy E (0 < E < W). Give the solution to the Schrodinger equation for I, II and III and use these to calculate the reflection coefficient.

I have the answer to this problem in front of me, but I don't understand. First they calculate the solution to the Schrodinger equation for I, II and III:

[itex]\psi_I(x) = Ae^{ikx} + Be^{-ikx}, \ \mbox{with} \ k = \frac{\sqrt{2mE}}{\hbar}[/itex]

[itex]\psi_{II}(x) = Ce^{\kappa x} + De^{-\kappa x}, \ \mbox{with} \ \kappa = \frac{\sqrt{2m(E - W)}}{\hbar}[/itex]

[itex]\psi_{III}(x) = Fe^{i k x}, \ \mbox{with} \ k = \frac{\sqrt{2mE}}{\hbar}[/itex]

I understand [itex]\psi_I[/itex], but not [itex]\psi_{II}[/itex] and [itex]\psi_{III}[/itex]. Why is there no i in [itex]\psi_{II}[/itex]? And why is [itex]\psi_{III}[/itex] only a single term? I imagine it has something to do with the particle coming from the left?
 
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  • #2
SoggyBottoms said:
Why is there no i in [itex]\psi_{II}[/itex]?
Because these are classically forbidden solutions, and the probability of the particle being found in such a region exponentially decays.

And why is [itex]\psi_{III}[/itex] only a single term? I imagine it has something to do with the particle coming from the left?

Correct, the other term would represent some particle coming in from the right (which is usually not assumed).
 
  • #3
Nabeshin said:
Because these are classically forbidden solutions, and the probability of the particle being found in such a region exponentially decays.

If I were to solve the equation for II, then [itex]Ce^{i \kappa x} + De^{-i \kappa x}[/itex] is mathematically still a valid solution right? I don't understand why it is not allowed.

Nabeshin said:
Correct, the other term would represent some particle coming in from the right (which is usually not assumed).

But why do I and II have two terms if the second term represents a particle coming from the right? Shouldn't they have only one term too then?
 
  • #4
SoggyBottoms said:
If I were to solve the equation for II, then [itex]Ce^{i \kappa x} + De^{-i \kappa x}[/itex] is mathematically still a valid solution right? I don't understand why it is not allowed.

Nope. You should go through and solve why these are the solutions here, but essentially you have something like (don't quote me exactly on this) [itex]\kappa \sim \sqrt{E-V}[/itex] So in the regions I and III, V=0 and this is real and fine. But in the region where V=V0>E, this is imaginary, which is why the thing you would normally write, e^(i*k*x), goes to e^(-kx) (Where now k in this expression is understood to be the real part). [I think there is a typo in what you wrote somewhere, and the sign inside your definition of kappa should be switched so that it is by definition real]


But why do I and II have two terms if the second term represents a particle coming from the right? Shouldn't they have only one term too then?

In region I, the term moving to the left represents the reflected wave off of the barrier between regions I/II. In region II, similarly, the term moving to the left represents the reflected wave off the barrier between II/III.
 
  • #5
Thanks!
 

1. What is Schrödinger's equation and how is it used to solve problems?

Schrödinger's equation is a mathematical equation used in quantum mechanics to describe the behavior of a quantum system over time. It is used to calculate the wave function of a particle, which represents the probability of finding the particle in a specific location at a specific time.

2. How is the reflection coefficient related to Schrödinger's equation?

The reflection coefficient is a measure of how much of an incident wave is reflected by a potential barrier or potential well in a quantum system. It is related to Schrödinger's equation because the wave function calculated by the equation can be used to determine the reflection and transmission coefficients for a given potential.

3. What factors influence the value of the reflection coefficient?

The value of the reflection coefficient is influenced by the energy of the incident particle, the width and height of the potential barrier or well, and the shape of the potential. It is also affected by the mass and spin of the particle.

4. How can the reflection coefficient be calculated using Schrödinger's equation?

The reflection coefficient can be calculated by solving Schrödinger's equation for the wave function on both sides of the potential barrier or well. The ratio of the reflected wave function to the incident wave function at the boundary gives the value of the reflection coefficient.

5. What are some applications of using Schrödinger's equation to calculate the reflection coefficient?

Schrödinger's equation and the reflection coefficient have many applications in quantum mechanics, such as in studying electron behavior in semiconductors, calculating the transmission of particles through potential barriers, and understanding the properties of quantum tunneling. They are also used in the development of quantum computing and in understanding the behavior of atoms and molecules in chemical reactions.

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