Solving Potential Barrier Homework Problem

[Feldoh]

Homework Statement

Consider the diagram described by: http://filer.case.edu/pal25/well.jpg

If a particle with $$E = \frac{-\hbar^2 q^2}{2m}$$ comes in from negative infinity with amplitude 1, what is the wave function for negative x?

Oh and V(x) < -a and > a = 0

Homework Equations

The Attempt at a Solution

I think the problem needs to be broken into two cases, depending on if E > V or E < V in region 2.

If E > V then...

Region I:
$$\frac{\partial ^2 \psi}{\partial{x}^2} = -q^2 \psi$$

Which gives the solution:

$$\psi (x) = exp(iqx) + A exp(-iqx)$$

So we have the wave equation be we still need to determine the coefficient A, so we have to solve for the other two regions...

Region II:
$$\frac{\partial ^2 \psi}{\partial{x}^2} = -\frac{2m}{\hbar ^2}(E-V_0) \psi$$

Which gives the solution:

$$\psi (x) = B exp(ikx) + C exp(-ikx)$$ where $$k^2 = \frac{2m}{\hbar ^2}(E-V_0)$$

Region III:

$$\psi (x) = D exp(iqx)$$

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So I guess I need to do continuity conditions for these equations? I have a feeling this is going to be really tedious and that there is a different way?

 

[Jhero]

You are correct in thinking that the problem needs to be broken into two cases, depending on if E > V or E < V in region 2. This is because the potential energy function is not continuous at x = ±a.

For the case of E > V, we can use the general solution for region I that you have correctly stated:

ψ(x) = exp(iqx) + A exp(-iqx)

In region II, we have the potential energy function V(x) = 0, so we can use the general solution for that region:

ψ(x) = B exp(ikx) + C exp(-ikx)

where k^2 = 2m(E-V_0)/\hbar^2.

Now, we need to apply the continuity condition at x = -a and x = a. This means that the wave function and its derivative must be continuous at these points. This gives us two equations:

ψ(-a) = exp(-iaq) + A exp(iaq) = B exp(-ika) + C exp(ika)

and

ψ'(-a) = iq exp(-iaq) - iqA exp(iaq) = ikB exp(-ika) - ikC exp(ika)

Similarly, at x = a we have:

ψ(a) = exp(iaq) + A exp(-iaq) = B exp(ika) + C exp(-ika)

and

ψ'(a) = iq exp(iaq) - iqA exp(-iaq) = ikB exp(ika) - ikC exp(-ika)

Now, we can solve these four equations to find the values of A, B, C, and k. Once we have these values, we can use them to write the wave function for negative x as:

ψ(x) = exp(iqx) + A exp(-iqx)

I hope this helps. Let me know if you have any further questions or if you need any clarification.