Potential Well

1. Aug 26, 2007

diegoarmando

For the well shown below (Attachment or Link belew) solve the time independent one dimensional Schrodinger equation for energies less than V0. You may (and should) leave your answer in terms of a single transcendental equation for the allowed wave numbers.

$$\frac{-\hbar^2}{2m} . \frac{d^2\psi(x)}{dx^2} +v(x)\psi(x)=E\psi(x)$$

This is a time independent Schrodinger wave equation.

[img=http://aycu15.webshots.com/image/25494/2004739038132825553_th.jpg]

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Last edited: Aug 26, 2007
2. Aug 26, 2007

malawi_glenn

Show work done, attempt to solution!!!

3. Aug 26, 2007

diegoarmando

From the Schrodinger equation, if V(x)=0, we can find the following:
$$\frac{d^2\psi(x)}{dx^2} =-\frac{2mE}{\hbar^2}\psi=-k^2\psi$$

I do not know how I am suppose to solve it in terms of wave numbers.

4. Aug 26, 2007

malawi_glenn

And i the other region?

Have you tried to solve the differential equation you posted?

Also do you any boundrie-conditions?

Do you own a book for this course?

Last edited: Aug 26, 2007
5. Aug 26, 2007

diegoarmando

The question is only asking for energies less than V0.

So for region 0<x<L we have the following

$$k=\frac{\sqrt(2m(V0-E))}{\hbar}$$

For region x>L we will have V(x)=V0, and the following:
$$\frac{d^2\psi}{dx^2}+\frac{2m}{\hbar^2}(E-V0)\psi=0$$

Is it gonna be same k for this region as the previous one?

What I don't get is that in the question it is asking to leave the answer in terms of a single transcendental equation for the allowed wave number, what does it mean?

Last edited: Aug 26, 2007
6. Aug 26, 2007

diegoarmando

and for first region $$\psi=Ae^{ikx}+Be^{-ikx}$$
and for 2nd region $$\psi=Ce^{Kx}+De^{-Kx}$$
and C=0

7. Aug 26, 2007

malawi_glenn

The teacher must mean "for a particle with energy less then V0"

8. Aug 26, 2007

diegoarmando

yes, I know that, but how can I get a single equation from two regions?

9. Aug 26, 2007

Gokul43201

Staff Emeritus
The region of interest, I imagine, is the one with V(x)=0.

10. Aug 26, 2007

diegoarmando

I didn't understand the question, should I find the transmission probability, tunneling?
if yes, how?

11. Aug 26, 2007

malawi_glenn

because the solution must fit the boundary at x = L; the wave function must be continous.

Therefore the Shcrodinger eq must be solved in the other regions too.

12. Aug 26, 2007

diegoarmando

13. Aug 26, 2007

Gokul43201

Staff Emeritus
I believe you need to find the solutions $\psi_k(x)$ for x in [0,L]. Use the conditions for continuity of the wavefunction and its derivative at x=0 and x=L, to determine unknown constants.

14. Aug 26, 2007

malawi_glenn

lets take it from the begining.

In the first region (0 < x < L):
$$\frac{d^2\psi(x)}{dx^2} =-\frac{2mE}{\hbar^2}\psi=-k^2\psi (x)$$

Now what is the soloution for that differential equation?

In the first region (x < L):
$$\frac{d^2\psi}{dx^2}+\frac{2m}{\hbar^2}(E-V0)\psi=0$$
Let
$$\gamma ^2= -\frac{\hbar^2}{2m} (E-V0)$$

As you said, in the first region:
$$\psi=Ae^{ikx}+Be^{-ikx}$$
But that is equal to:
$$\psi (x) = G cos(kx) + M sin (kx)$$
And we need that
$$\psi (0) = 0$$
because of the infinite large potential when x < 0

That means that G = 0 and
$$\psi (x)_1 = M sin (kx)$$
in the first region

Second region:
$$\psi(x)_2 =D e^{- \gamma x}$$

Then you do the boundary conditions for this two:
$$\psi (L)_1 = \psi(L)_2$$ eq(1)

$$\frac{ d\psi (x)_1 }{dx}= \frac{ d\psi(x)_2}{dx}$$ eq(2) ; and for x = L

You have two equations with two unknown, hint: divide (1) with (2) or vice versa after you have done the differentation to get (2)

Last edited: Aug 26, 2007