Can A and B be equal in the region between x=0 and x=a?

In summary, the question is about writing the solutions to the Schrodinger equation in regions x<0 and x between 0 and a. The equations for these regions are psi(x)= e^ikx+Re^-ikx and psi(x)=Ae^iqx+Be^-iqx respectively. The question is whether A and B can be equal in the region between 0 and a, resulting in destructive interference and a value of R=1, which is the desired outcome. This approach involves equating coefficients of the wave traveling in opposite directions in this region and applying boundary conditions to the general solution for the wave function.
  • #1
frankcastle
10
0

Homework Statement



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

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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
You haven't defined your potential over the domain of x. My guess based on the info given is that the potential V(x) = infinity for regions less than or equal to x = 0 and greater than or equal to x = a.

To solve for the equations, you must impart the boundary conditions on the general solution for the wave function. So, psi(x) must vanish at x = 0 and x = a. For example, suppose that psi(x) = Asin(kx) + Bcos(kx) is a general solution to the time-independent schrodinger equation. Now, for the potential I expressed in the first paragraph, we must have psi(x) = 0 at x = 0 and x = a. When x = 0, psi(x = 0) = B; therefore choose B = 0, and now psi(x) = Asin(kx). Now fit the wavefunction to x = a: psi(x=a) = 0 = Asin(ka). Under what conditions for k (the angular wavenumber) will the sine term vanish?

You can apply this idea to your solutions. As a check, verify that your solution satisfies the time-independent schrodinger equation.
 

1. What is an infinite potential barrier?

An infinite potential barrier is a concept in physics that describes a situation where a particle is confined within a certain region by an infinitely high potential energy. This means that the particle cannot escape from the region, as it would require an infinite amount of energy to do so.

2. How does an infinite potential barrier affect the behavior of particles?

An infinite potential barrier forces particles to exist within a confined region, leading to unique behaviors. For example, particles can only exist within specific energy levels and cannot move freely like they would in an unconfined space. This confinement also affects the probability of particles being found in certain regions.

3. What are some real-world examples of an infinite potential barrier?

One example of an infinite potential barrier is a particle trapped in a nucleus, as the strong nuclear force acts as a barrier that prevents the particle from escaping. Another example is an electron confined within a semiconductor, where the potential barrier created by the material's band gap keeps the electron within the material.

4. Can an infinite potential barrier be penetrated by particles?

No, an infinite potential barrier cannot be penetrated by particles. As mentioned before, it would require an infinite amount of energy for a particle to escape the barrier, which is impossible. However, particles can tunnel through the barrier, which is a quantum mechanical phenomenon that allows particles to pass through barriers that they normally couldn't overcome.

5. How does the width of an infinite potential barrier affect particle behavior?

The width of an infinite potential barrier has a direct impact on the energy level and probability of particles within the barrier. As the width decreases, the energy levels become more confined and the probability of finding particles in certain regions increases. Conversely, as the width increases, the energy levels become more spread out and the probability of finding particles decreases.

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