Quantum Mechanics Problem

  • #1
I am asked to determine the expressions for coefficients of the solutions [tex]\psi_1[/tex] and [tex] \psi_2 [/tex] to the Schrodinger's Equation in a system where a particle travelling to the right encounters a potential step when [tex] E < V_0 [/tex], where E is the total energy.

What I was able to come up with is that
[tex]
\psi_1 = Ae^{kx} + Be^{-kx}, k = \frac{\sqrt{2mE}}{\hbar}
[/tex]

[tex]
\psi_2 = Ce^{qx} + De^{-qx}, q = \frac{\sqrt{2m(E-V_0)}}{\hbar}
[/tex]

What I also know is that [tex]Ce^{qx}[/tex] is unacceptable because the wave must decay exponentially when it hits the barrier.

My first question is whether or not this statement is true:

[tex]
\int_{-\infty}^{0}|\psi_1|^2dx + \int_{0}^{+\infty}|\psi_2|^2dx = 1
[/tex]

My second question is if [tex]Be^{-kx}[/tex] is an acceptable solution to [tex]\psi_1[/tex]. In my opinion i believe it is not acceptable because it diverges as x negative approaches infinity.

Thanks for any help.
 

Answers and Replies

  • #2
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relinquished™ said:
[tex]
\psi_1 = Ae^{kx} + Be^{-kx}, k = \frac{\sqrt{2mE}}{\hbar}
[/tex]
For [itex] \psi_1\ E>V[/itex] right? So you should have oscillating solutions on the left side of the barrier. As you have written the equations you have oscillating solutions on the right side of the barrier.

How is the [itex]k[/itex] defined?
 
  • #3
dextercioby
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You have his definition there...[tex]k=:\frac{\sqrt{2mE}}{\hbar} [/tex]

If the barrier in in the origin,then yes,you must make sure that:
1.Incoming wave is sinusoidal type (yours is not,so check it again).
2.Reflected wave is sinusoidal type (yours is not,so check it again).
3.Transmitted wave is exponentially decaying in amplitude...(i think you somehow got this part right).

Check again the signs & the substitutions you make in the initial SE for each domain.
Impose continuity conditions fo the wavefunction & for the its first space derivative.
Your wave functions are NOT normalizable!

Daniel.
 
  • #4
I made a mistake in my [tex]\psi_1[/tex], i think the solution should be
[tex]
\psi_1 = Ae^{ikx} + B^{-ikx}
[/tex]
[tex]
\psi_2 = Ce^{-iqx}
[/tex]

Hmmm... If I remember correctly, in Eisberg's book he used [tex]\psi_2 = Ce^{-qx}[/tex] (a real function) and then equated it to [tex]\psi_1[/tex] (a complex function) at x=0, then applied the boundary conditions for continuity... is [tex]\psi_2 = Ce^{-qx}[/tex] really not normalizable?

Thanks again for all the help
 
Last edited:
  • #5
dextercioby
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That one is...The sinusoidal is not...Therefore,the coefficients can be found only imposing the continuity conditions...

Daniel.

P.S.Eisberg or whatever is right.
 
  • #6
Is my new [tex]\psi_1[/tex] normalizable now? i.e.,

[tex]
\psi_1 = Ae^{ipx} + Be^{-ipx}
[/tex]

Normalization is quite confusing to me sometimes...

Thanks again for the help
 
Last edited:
  • #7
dextercioby
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Nope.Try to see.Look,my maple says: [tex]\left| Ae^{ipx}+Be^{-ipx}\right| ^2=\left( A\cos px+B\cos px\right) ^2+\left( A\sin px-B\sin px\right) ^2 [/tex]

Now,open the brackets and then integrate the simplified function between (-oO,0)...You'll see that it diverges.

Daniel.
 
  • #8
Um... is it necessary to convert [tex]Ae^{ipx} + Be^{-ipx}[/tex] to its ciskx and coskx - isinkx form before normalizing it? I trashed the idea of thinking that the wave function at the left had the form of either sine or cosine because they would definitely diverge when I integrate them from (-o0, 0)when they are normalized...
 
  • #9
Galileo
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relinquished™ said:
I made a mistake in my [tex]\psi_1[/tex], i think the solution should be
[tex]
\psi_1 = Ae^{ikx} + B^{-ikx}
[/tex]
[tex]
\psi_2 = Ce^{-iqx}
[/tex]
If you rewrite the Schrodinger equation for x>0 correctly:

[tex]-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi(x)+V_0\psi(x)=E\psi(x)[/tex]
[tex]\Rightarrow \frac{d^2}{dx^2}\psi(x)=\frac{2m(V_0-E)}{\hbar^2}\psi(x)=q^2\psi(x)[/tex]
You'll find [itex]q = \frac{\sqrt{2m(V_0-E)}}{\hbar}[/itex] is real and positive. So you get exponentially decaying functions.
 
  • #10
dextercioby
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He did that,Galileo,we were discussing the sollution for negative "x"...For positive "x" there was nothing more to discuss...

Daniel.
 

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