QM: Potential, Finite Step Function

  • #1
RJLiberator
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Homework Statement



A beam of particles of mass m and energy E is incident from the right unto a square well potential given by ##V(x)=-V_0## for ##-a<x<0##, and ##V(x)=0## otherwise.

Solve the Schrodinger equation to determine the wave function which describes this situation. Determine the transmission and reflection coefficients, and show R+T=1.

Homework Equations



The Schrodinger Equation
[tex] \frac{-ħ}{2m} \frac{ \partial^2 \psi}{\partial x^2} + V(x) \psi(x) = E \psi(x)[/tex]

The Attempt at a Solution



This is a pretty standard question apparently as there are a plethora of examples available from my notes/textbook/INTERNET.

My problem with this question is that the instructor has switched it such that the energy is incident FROM the right.

I'm working with the example from my book, so I will skip some of the calculations, but present you this:

I split the regions into 3 obvious regions. Region 1 will be the beginging region from the right. Region 2 will be the finite well, and region 3 will be the ending region.
Now, solving the Schrodinger equation for each region has left me with the following:

[tex] \psi (x) = Ae^{ikx}+Be^{-ikx}, Region 1 [/tex]
[tex] \psi (x) = I \sin{- \alpha x}+J \cos{ \alpha x}, Region 2 [/tex]
[tex] \psi (x) = Ce^{ikx}+De^{-ikx}, Region 3 [/tex]

Now my questions that I need help on and the reason why I am posting here is:

1) I need to understand which of these parts I can use.
For example, in Region 1, in class we had the incident wave being the Ae part. However, in this question, I feel like A = 0 since it is incident from the right so that means we must use the Be part.
Similarly, in region 3, we would then use the De part as C = 0.

Is my thinking correct here?

2) For Region 2, my book and notes differ. My book uses the sin and cosine representation as it is a 'even potential problem' so we can then just use the cosine part.
However, my notes uses the exponentials for a similiar problem.

Am I correct in using the cosine/sine representation? Does it really matter since e^(ix) can be represented as a sin cosine wave anyway?

Thank you for helping me with these 2 questions.
 

Answers and Replies

  • #2
Dr Transport
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three questions you need to answer so that you can set up the problem correctly
  1. what is the form of the incoming wave function in region 1
  2. what is the form of the wave function in region 3
  3. the boundary conditions
I'd not use the sine/cosine form in region 2. this problem is best done via exponential functions all the way through out.
 
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  • #3
RJLiberator
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I'd not use the sine/cosine form in region 2. this problem is best done via exponential functions all the way through out.
Ok. Thanks for the tip.

what is the form of the incoming wave function in region 1
Form? Hm. I'm not sure I understand what you are asking me.

It is a wave heading from the right to the left. It reaches x = 0 and then enters a potential well of -V_0 until x = -a in which case the potential then becomes 0 again.

Region 1 = Region 3

the boundary conditions
I see/know that the boundary conditions exist at x = 0 and x = -a. I then set the regions equal to each other at these points, once I understand what my wave function actually is.

For example, In the first region, which is the right most region, we see that calculating the Schrodinger equation gives us:
[tex]
\psi (x) = Ae^{ikx}+Be^{-ikx}, Region 1[/tex]

However, in all the examples of this problem, the incident wave is going from the left, and so they use this to say B = 0.
Since this problem has the incident wave coming from the right, my sense is to set A = 0 instead and let the B part be the incident wave.

Can I do this?
 
  • #4
Dr Transport
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Region 1 has the incident and a reflected wave, correct?? for expediency sake, set the incident wave amplitude to 1.

How many waves are transmitted??

Draw a diagram of what is happening and label all the waves......it becomes readily apparent.
 
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  • #5
RJLiberator
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Ah, this is where the book (Griffiths page 79 if you are interested) and my notes differ.

Griffiths sets the reflected wave to 0, whereas my notes (and your advice) does not.

So then, in the first region we now have the full
[tex] \psi (x) = Ae^{ikx}+Be^{-ikx}, Region 1 [/tex]
As A represents the refracted part and B represents the incident part.
Then, we have
[tex]
\psi (x) = Ie^{ikx}+Je^{-ikx}, Region 2 [/tex]
For the middle part, where we have J as the transmitted part and I as the refracted part.

Finally, we have
[tex]\psi (x) = Ce^{ikx}+De^{-ikx}, Region 3 [/tex]

As the final part, however we can set C = 0 as there is nothing reflected in region 3.

Note: In all cases, the incident/transmitted part of the wave is the one with the NEGATIVE ikx in the exponential. This is signaling the DIRECTION of the wave.

now do I have the set up of the problem correct? :)
 
  • #6
Dr Transport
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so far so good..... match waves and their derivatives at the boundaries and you are done....
 
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  • #7
RJLiberator
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Thank you for your help with my 2 concerns. I think I can do the rest. I will post my solution once I've reached it (might take some time tonight).

Cheers!
 

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