Standing waves in a funny Potential distribution

In summary, the conversation is discussing the solution to the Schrodinger equation in regions x<0 and x between 0 and a, with a potential distribution given in an attached image. The question centers around determining the coefficients A and B, and whether they are equal due to the condition of complete reflection at x=a. The expert advises against assuming this and instead suggests using boundary conditions to find the coefficients.
  • #1
frankcastle
10
0

Homework Statement



The description of the potential distribution is given in the attached image.
The particle arrives from the left with E>V0.

write the solutions to the S.E in regions x<o and x between o and a


Homework Equations


I believe psi(x)= e^ikx+Re^-ikx in x<0
and psi(x)=Ae^iqx+Be^-iqx for x b/w o and a.


The Attempt at a Solution


My question is, since there is complete reflection occurring at x=a, can A=B in region x b/w 0 and a? If so, there will be destructive interference in the region, giving R=1, which is what we are asked to prove in the question. Is this approach of equating coefficients of wave traveling in +-x directions in this region applicable?
 

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  • #2
The potential:
V=0 for x<0
V=V0 for 0<x<a
V->infinity for x>a
frankcastle said:
write the solutions to the S.E in regions x<o and x between o and a

I believe psi(x)= e^ikx+Re^-ikx in x<0
and psi(x)=Ae^iqx+Be^-iqx for x b/w o and a.
You have one boundary condition at x=a that relates A and B. You have two boundary conditions at x=0 that relate A, B and R. Once you have determined k and q (which I'm assuming you know how to do), then I believe you simply apply these boundary conditions.
 
  • #3
thanks turin, I understand the problem well. My question is regarding the relation of the coefficients, A and B; with the respective intensities.

Since R=1 at x=a, I would immediately assume that B=A
instead of having to use Boundary conditions to find coefficients.
Would this be correct logic?
 
  • #4
frankcastle said:
Since R=1 at x=a, I would immediately assume that B=A
instead of having to use Boundary conditions to find coefficients.
Would this be correct logic?
No. You're contradicting yourself. You specified R for x<0, and now you want to talk about R at x=a, which doesn't even make sense, unless this is somehow a different R than the coefficient of the exponential that you originally gave. It's been a while since I solved one of these problems, so I can't remember if it should turn out that A=B is, in fact true; however, your logic to arrive at this conclusion is flawed.
 

Related to Standing waves in a funny Potential distribution

1. What are standing waves?

Standing waves are a type of wave that occurs when two waves with the same frequency and amplitude travel in opposite directions and interfere with each other. This creates a pattern of nodes and antinodes where the amplitude of the wave remains constant.

2. What is a funny potential distribution?

A funny potential distribution refers to a potential energy graph that has a unique or unusual shape, such as a sinusoidal curve or a sawtooth pattern. This type of potential distribution can lead to interesting and unexpected standing wave patterns.

3. How are standing waves formed in a funny potential distribution?

In a funny potential distribution, standing waves are formed when the potential energy of the system oscillates between high and low values, creating regions of maximum and minimum potential. This leads to the formation of nodes and antinodes, which correspond to points of maximum and minimum amplitude in the standing wave.

4. What factors affect the formation of standing waves in a funny potential distribution?

The formation of standing waves in a funny potential distribution is affected by the frequency and amplitude of the waves, as well as the shape and properties of the potential energy graph. The boundary conditions and properties of the medium in which the waves are traveling can also have an impact on the standing wave pattern.

5. What are the applications of studying standing waves in a funny potential distribution?

Studying standing waves in a funny potential distribution can have practical applications in various fields such as acoustics, optics, and electronics. It can also be used to understand the behavior of particles in quantum mechanics and the properties of matter on a microscopic level. Additionally, understanding standing waves in a funny potential distribution can aid in the design and optimization of devices and systems that utilize wave phenomena.

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