|Dec18-08, 09:22 AM||#1|
Infinite potential barrier
1. The problem statement, all variables and given/known data
write the solutions to the S.E in regions x<o and x between o and a
2. Relevant 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.
3. The attempt at a solution
My question is, since there is complete reflection occuring 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?
|Dec18-08, 12:43 PM||#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.
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