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Potential Well

  • #1
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.

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

This is a time independent Schrodinger wave equation.


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

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Answers and Replies

  • #2
malawi_glenn
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Show work done, attempt to solution!!!
 
  • #3
From the Schrodinger equation, if V(x)=0, we can find the following:
[tex] \frac{d^2\psi(x)}{dx^2} =-\frac{2mE}{\hbar^2}\psi=-k^2\psi[/tex]

I do not know how I am suppose to solve it in terms of wave numbers.
 
  • #4
malawi_glenn
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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?
 
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  • #5
The question is only asking for energies less than V0.

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

[tex] k=\frac{\sqrt(2m(V0-E))}{\hbar}[/tex]

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

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?
 
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  • #6
and for first region [tex]\psi=Ae^{ikx}+Be^{-ikx}[/tex]
and for 2nd region [tex]\psi=Ce^{Kx}+De^{-Kx}[/tex]
and C=0
 
  • #7
malawi_glenn
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The teacher must mean "for a particle with energy less then V0"
 
  • #8
yes, I know that, but how can I get a single equation from two regions?
 
  • #9
Gokul43201
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The region of interest, I imagine, is the one with V(x)=0.
 
  • #10
I didn't understand the question, should I find the transmission probability, tunneling?
if yes, how?
 
  • #11
malawi_glenn
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yes, I know that, but how can I get a single equation from two regions?
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
thanks for reply, could you please help e to solve this, I already write what I know
 
  • #13
Gokul43201
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I believe you need to find the solutions [itex] \psi_k(x) [/itex] 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
malawi_glenn
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lets take it from the begining.

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

Now what is the soloution for that differential equation?

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

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

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

Second region:
[tex]\psi(x)_2 =D e^{- \gamma x}[/tex]

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

[tex]\frac{ d\psi (x)_1 }{dx}= \frac{ d\psi(x)_2}{dx} [/tex] 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)
 
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