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beebopbellopu
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Homework Statement
Consider reflection from a step potential of height V0 with E > V0, but now with an infinitely high wall added at a distance a from the step (x=0 at V=V0)
Solve the Schrodinger equation to find [tex]\varphi[/tex](x) for x<0 and 0 [tex]\leq[/tex] x [tex]\leq[/tex] a. Your soultion should contain only one unknown constant.
Homework Equations
Time independent Schrodinger equations for x<0 and 0 [tex]\leq[/tex] x [tex]\leq[/tex] a regions.
For x<0: [tex]\varphi_{1}[/tex] = A1eik1x + B1e-ik1x where k1 = [tex]\frac{\sqrt{2mE}}{\hbar}[/tex]
For 0 [tex]\leq[/tex] x [tex]\leq[/tex] a: [tex]\varphi_{2}[/tex] = A2eik2x + B2e-ik2x where k2 = [tex]\frac{\sqrt{2m(E-V0)}}{\hbar}[/tex]
The Attempt at a Solution
Sorry if my formatting is a little off, this is my first time posting. Anyways.
So I tried enforcing my boundary conditions:
1) [tex]\varphi_{1}[/tex](0) = [tex]\varphi_{2}[/tex](0)
2) [tex]\frac{d\varphi_{1}}{dx}[/tex] evaluated at x=0 is equal to [tex]\frac{d\varphi_{2}}{dx}[/tex] evaluated at x=0
3) [tex]\varphi_{2}[/tex](a) = 0
which resulted in the following (respectively to each boundary condition):
1) A1 + B1 = A2 + B2
2) k1(A1 - B1) = k2(A2 - B2)
3) A2eik2a = -B2eik2a
Then using the 3rd condition, you get A2 = -B2e-2ik2a
plugging that into condition 1 yields A1 + B1 = B2(1-e-2ik2a)
Then from the 2nd condition I get A1 = B1 - [tex]\frac{-B_{2}k_{2}}{k_{1}}[/tex](e-2ik2a+1) which I then plugged back into the equation from condition 1.
So along with the equation for A2, I get the following constants:
B1 = [tex]\frac{B_{2}}{2}[/tex][(1-e-2ik2a)+[tex]\frac{k_{2}}{k_{1}}[/tex](1+e-2ik2a)]
and A1 = B2[[tex]\frac{1}{2}[/tex] - [tex]\frac{k_{2}}{k_{1}}[/tex] - e-2ik2a([tex]\frac{1}{2}[/tex] + [tex]\frac{k_{2}}{k_{1}}[/tex])]
Do I then just plug these directly back into the schrodinger equations? My main question is that I feel like there is a way to simplify these constants in a way I'm just not seeing, and hoping if there is a way someone could point me in the right direction. Thanks.