Half Infinite Well Homework: Solve for E<0

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Homework Statement



4hw7il.jpg


Homework Equations



-h^2/2m d^2F(x)/dx^2 = EF(x)

The Attempt at a Solution



i just need to a part. for E<0 i can find for 0<x<L side F(x) = ACos(Lx) + BSin(Lx)

at the L<x side, F(x) = e^(Kx) where

L^2= 2m(E+V)/h^2

K^2= -2mE/h^2

but i do not know what will i do. can you help me for a part of question.
 
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This is an infinite barrier with a finite well in front of it.
1. Solve the schrodinger equation for the eigenvalues.
2. Find out what is needed for an eigenvalue to be negative.

[edit] Me and Mute crossed posts - I've decided to support Mute's approach.

Have you seen the finite square well? This is similar.
 
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You probably want to use a different symbol than L in the Cos(Lx) and Sin(Lx) terms.

For L<x (using the length L, here), your solution should be a decaying exponential - you have chosen the exponential that grows as x grows.

Now that you have the solutions in the two regions, you need to apply the boundary conditions. Using the boundary condition that your wave function should go to zero at infinity is what gives you the decaying exponential for L<x. There are two more boundaries: The boundary at x = 0, and the boundary at x = L. Do you know what the boundary conditions are at these boundaries? (Edit: Simon Bridge has given you the conditions, so get to it!)
 
aghh i get all A,B,C = 0 :confused::confused:

i attached the solution;

http://imgur.com/rtBpCl.jg

where am i doing wrong?
 
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You picked the "trivial" solutions - zero everywhere is, indeed, a solution.
But you only need the different sections to agree at one point.

Recap:
##\psi_I(0)=\psi_{II}(0)##

You need ##\psi_{II}(x)=A\cos(kx)+B\sin(kx)## to be 0 at x=0, which must mean that A=0, because ##\cos(0)=1##... so ##\psi_{II}(x)=B\sin(kx)## ... so far so good.

At the other end it is more complicated:
##A\sin(kL)=Ce^{-KL}##
##Ak\cos(kL)=-CKe^{-KL}##
... which appears to give you four variables in only two equations doesn't it?
But I think you'll find that k and K have to be related, so that's really only three variables.

In the end - the entire wavefunction has to be normalized ... so $$\int_0^L \psi_{II}^\star(x)\psi_{II}(x)dx + \int_L^\infty \psi_{III}^\star(x)\psi_{III}(x)dx = 1$$ which should give you the third equation.

You'll find that only specific values of k (hence K) will satisfy these conditions... (particularly for E<0) so providing discrete energy levels.
Hint: what are k and K both functions of?

Like I said before - the method of solving these last two is very similar to that for a finite square well ... which you can look up.
You should also be able to sketch the basic shape of the (amplitudes) first few bound-states (if they exist) ... eg. the first one starts at 0 when x=0, has a peak between 0 and L, is still >0 at x=L then decays exponentially from there.
 
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thank you very much. i think i can solve it after your help.
 
To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
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