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

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SUMMARY

The discussion centers on solving the time-independent Schrödinger equation (S.E) in the regions x < 0 and 0 < x < a, with specific boundary conditions. The proposed wave functions are psi(x) = eikx + Re-ikx for x < 0 and psi(x) = Aeiqx + Be-iqx for 0 < x < a. The key question raised is whether coefficients A and B can be equal in the region between 0 and a, leading to destructive interference and proving R = 1. The potential V(x) is assumed to be infinite at the boundaries, necessitating that psi(x) vanishes at x = 0 and x = a.

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  • Knowledge of angular wavenumber (k) and its implications
  • Concept of destructive interference in quantum mechanics
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frankcastle
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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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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 Schrödinger 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 Schrödinger equation.
 

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