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Double Delta Function Potential Well

  • Thread starter doublemint
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  • #1
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Hello Again!

My question:
Find the bound energy spectrum of the potential that contains two delta-function
wells: [tex]V(x) = -V_{0}\delta(x-\frac{a}{2}) -V_{0}\delta(x+\frac{a}{2})[/tex] under the assumption that the wells are located very far away from each other. Find and plot the associated stationary states. Verify that your result is consistent with that for a single well in the limit of infinite a.
I pretty much followed Griffith's method in solving for the energy spectrum. However, would the assumption that the wells are really far from each other affect the calculations?
Also, how would you do verify the result using a single well?
Any help would be appreciated!
Thanks
DoubleMint
 

Answers and Replies

  • #2
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Well look at the probability of transmission (once you've found the transmission coefficient), and try to plot it for various distances of the potentials.

You should get some interesting behavior ;)

I think Flugge has something simmilar in it...

HINT: resonant scattering :)
 
  • #3
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Apparently, for large a, the energy looks similar for a single well! I have one more question, when I solve the transcendental equation, griffiths sets c=1.
The equation is e^-z = cz - 1 why do they set it to 1?
 
  • #4
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Normalization maybe...
 
  • #5
vela
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It would help if you showed the work that led to the equation. It's hard to say why he set c=1 without seeing the context.
 
  • #6
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This is from Griffiths solutions manual. At the top of the second page, he talks about the equation and then chooses c=1.
 

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  • #7
vela
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It looks like part of the question in Griffiths had to do with various values of α. There's no other reason to set c=1 otherwise.
 
  • #8
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Oh I think he was just going for the dimensionless units...

Oh and I totally didn't see that there is - sign XD

Still try to look at positive V_0, and plot the probability for transmission, the graph is pretty epic :)
 
  • #9
Still try to look at positive V_0, and plot the probability for transmission, the graph is pretty epic :)
(This is my first post; so, please let me know if I'm going about this wrong.) Griffiths actually asks the reader to calculate the transmission coefficient in the next problem for a fixed [itex]\alpha[/itex]. How does the transmission coefficient vary from the single-well case? Does anyone have any pointers on where to begin?
 
  • #10
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For a single well you mean infinite square well or?

Because this is the case with delta peaks at certain places. As you change the height and distance of the potentials you get resonant scattering.

Say you have potential
[tex]V(x)=V_0\delta (x)+V_0\delta (x-R)[/tex], and after you make the necessary assumptions for the continuity of the wave functions and it's derivatives, you put [tex]u=u_0 R=\frac{2mV_0}{\hbar^2}R[/tex], and after calculating the transmission coefficient you'll get these pictures:

nPnXe.png

PAlex.png
 
  • #11
For a single well you mean infinite square well or?
Sorry, I should have been more specific! I meant a single delta function potential at some point x=a. However, it seems that you grasped my real interest---understanding how to arrive at the transmission coefficient when we have two symmetric delta potential wells:
[tex]V(x)=V_0\delta (x)+V_0\delta (x-R)[/tex].

If we translate this so that R/2 ---> 0, it's an even potential about the origin, and we can assume the WF is even or odd.

What's really baffled me is how to use the assumption of continuity (with regards to the WF) and what we know about the discontinuity of its derivative to calculate [tex] T = ? [/tex].

I'm really curious to see how the transmission coefficient differs in this case. If I'm not totally misunderstanding your point, it seems that it becomes a function of "a", half the distance separating the wells.

I'll have at the derivation of T once more tomorrow and post what I come up with.

PS: Do the graphs show "T vs. a", where T is the transmission coefficient and a is the distance from the origin of either well?
 
  • #12
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Hmm I'm not sure about the fact that it's even about the origin :\

First of all, delta function isn't really a function so I dk if it's good to even talk about it being odd or even :\

But say that you look at point x=R/2, then it would be symmetrical around origin, no?


First you divide the problem in 3 areas: 1st where you have incoming wave, what happens to it when it comes to a barrier (even delta one)?
2nd you have a space between two barriers, and the last one is where you have only transmitted wave.

you write Schrodinger equation and you can integrate it (around the potentials, and for some small value). Ofc you'll look at each potential separately (the first and then the second delta).

When you're dealing with barriers there are several general conditions that must be satisfied, one of those are that the wave function and it's derivative have to be continuous at the place of barrier.

From that you'll get 4 different equation and you can fiddle with them to get the transmission coeff.

The graph shows the probability (transmission coeff. absolute squared) vs the kR, R is the distance and k is wave number (it's assumed that your incident wave is plane wave).
 
  • #13
First you divide the problem in 3 areas: 1st where you have incoming wave, what happens to it when it comes to a barrier (even delta one)?
2nd you have a space between two barriers, and the last one is where you have only transmitted wave.
Allow me to attempt :)!!!

First, I consider the case of
[tex] \begin{array}{lr}
-\frac{h^2}{2m} \frac{d^2 \psi}{d x^2} - \alpha [\delta (x+a) + \delta (x-a)] \psi = E \psi & [1]
\end{array} [/tex]
, where [tex] E > 0 [/tex] (scattering states) and [tex] x \neq |a| [/tex]

There are three such regions, and for each the equation simplifies to
[tex] -\frac{h^2}{2m} \frac{d^2 \psi}{d x^2} = E \psi [/tex]

Thus, the regions present solutions (with presumably unique coefficients) of the form
[tex] \psi (x) = Ae^{-ikx}+Be^{ikx} [/tex]
, with
[tex] k = Im( \sqrt{\frac{-2mE}{h}} )[/tex]

Now the ugly part! Suppose the three regions are described by the three equations
[tex] \begin{array}{llll}
\psi (x) & = & Ae^{-ikx}+Be^{ikx}, & x < -a \\
\psi (x) & = & Ce^{-ikx}+De^{ikx}, & -a < x < a \\
\psi (x) & = & Fe^{-ikx}+Ge^{ikx}, & x > a
\end{array}
[/tex]

Assume the wave comes in "from the left" (G = 0). Then continuity at x = |a| tells us that
[tex] \begin{array}{llll}
Ae^{ika}+Be^{-ika} & = & Ce^{ika}+De^{-ika} \\
Ce^{-ika}+De^{ika} & = & Fe^{-ika}
\end{array}
[/tex]

In order to use the discontinuity of the derivative at x = a (extended WLOG to x= -a), we evaluate
[tex] \lim_{\epsilon \rightarrow 0} \int_{a- \epsilon}^{a+ \epsilon} [1] [/tex]
, which simplifies to
[tex] \Delta (\frac{d \psi}{d x}) = -\frac{2m \alpha}{h^2} \psi(a) [/tex]
, or in other words, the difference of the derivative from the left deducted from the derivative from the right is a constant times the WF at x = a.

Now applying this we can deduce two more functions. For simplicity, lets call
[tex] \frac{2m \alpha}{h^2} = j [/tex]
Then the third and fourth equations of the system are
[tex] \begin{array}{llll}
[ik][Ae^{ika}-Be^{-ika} - Ce^{ika}+ De^{-ika}] & = & [-j][Ae^{ika}+Be^{-ika}] \\
[ik][Ce^{-ika}-De^{ika} - Fe^{-ika}] & = & [-j][Fe^{-ika}]
\end{array}
[/tex]

Simplifying these last two equations, we're left with a system of five coefficients and 4 equations
[tex] \begin{array}{llll}
- Ce^{ika}+ De^{-ika} & = & -[1-ij/k][Ae^{ika}]+[1+ij/k][Be^{-ika}] \\
Ce^{-ika}-De^{ika} & = & [1+ij/k][Fe^{-ika}] \\
Ae^{ika}+Be^{-ika} & = & Ce^{ika}+De^{-ika} \\
Ce^{-ika}+De^{ika} & = & Fe^{-ika}
\end{array}
[/tex]

Let me stop here before I try to solve for T. Have I set up the system correctly? If so, I have had a lot of difficulty finding any "pretty" answer to the system that relates A, B, and F, and I would appreciate some insight into what I might be overlooking. If not, any thoughts on where I went wrong? Thanks so much!

Viele Gruesse,
Jeff
 
  • #14
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If the incoming wave is [tex]e^{ikx}[/tex], then the F=0, not G, and I don't think that you'll loose on generality if you put A=1.

Oh and I would recommend that you look at delta potentials situated at 0 and some distance (a, 2a...), that way you'll look at the continuity on x=0 and x=a (for instance). A lot of exponential functions will disappear (e^0=1), an it's going to be a lot easier for you to solve it.

But besides your putting the incoming wave as [tex]e^{ikx}[/tex] (I guess you could do that, depending on how you define k, no?), the system looks ok. Try using Mathematica to solve it, and by hand...
 
  • #15
Thanks dingo. I did notice that I had the wrong sign in the exponent shortly after posting.

As for solving it:
Unfortunately, I have to give some thought on how to make Mathematica obey me (since solving for A,B,F is an overdetermined system in its opinion, right?). By hand I'm having a little luck, but the solution is uuuuuuugly. I believe the trick is using the boundary conditions at a to rewrite C and D in terms of F, and then using the boundary conditions at -a to write F or B in terms of A.

I'm working on it right now and will try to post in an hour or so.
 
  • #16
211
0
Try solving it by writing it in matrix form and using Cramer's rule...
 
  • #17
Closure! I'm pretty sure I solved it a little while back. It turns out the answer was not pretty as I had imagined it would be. The real problem was that I had failed to simplify it into something comparable. Thanks so much!
 

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