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A QM problem on finding eigenvalues

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  • #26
CarlB
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If you just want to get your homework done, and you don't want to go through a lot more work than you have to (or have your homework look like it was solved for you by a string theorist), then take a look at my previous comment. That breaks the problem down into an inhomogenous differential equation. There is an obvious particular solution, [tex]e^{-\kappa |x|}[/tex]. To find the general solutions, solve the homogeneous DE to get [tex]e^{kx}[/tex] as the form of the general solutions of the homogeneous equation where [tex]k[/tex] is a constant. Then the arbitrary solutions of the general equation can be obtained by substituting the following form:
[tex]\psi(x) = e^{-\kappa|x|} + A e^{-Bx} + C e^{-Dx} [/tex]

where A, B, C and D are constants you will have to determine, and you will likely have to stitch together two different solutions at the origin using the continuity arguments you are already familiar with.

The author of this problem had no intention on turning it into a bad PhD thesis, LOL. For more information on solving inhomogeneous differential equations, see, for example:
http://hyperphysics.phy-astr.gsu.edu/hbase/math/deinhom.html

Carl
 
  • #27
dextercioby
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I don't know how u came up with a nonhomgenous diff.eq.It looked like an integro-diff.one to me.:confused:

Daniel.
 
  • #28
CarlB
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dextercioby said:
I don't know how u came up with a nonhomgenous diff.eq.It looked like an integro-diff.one to me.:confused:

Daniel.
The potential operator converts all wave functions to some multiple (depending on the particular wave function) of [tex]e^{-\kappa |x|}[/tex], so it's not a very complicated integral equation.

You can easily evaluate the action of the potential operator on all the possible functions of the form [tex]e^{kx}[/tex], which happens to cover the solutions to the homogeneous problem as well as the particular solution. Thus the "integro-diff" equation is converted to a standard inhomogeneous differential equation with very little work. Then you can apply the usual methods of solving that sort of DE.

Carl
 
  • #29
reilly
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I see things a bit differently. First, there's no need for any integral equation. A straightforward application of appropriate unitary transformations gives;

<x| H0 |y> = - GRAD squared delta(x-y)

And, by definition of the problem, <x|v>V<v|y> = V exp(-2kx) delta(x-y) for x>0

The Hamiltonian is diagonal in configuration space.

To get past the formal stuff, set y = exp(ax), and you will find that the ensuing Schrodinger Eq. is Bessel's Eq (in y), given the right a.

Regards,
Reilly Atkinson
 
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reilly said:
And, by definition of the problem, <x|v>V<v|y> = V exp(-2kx) delta(x-y) for x>0
To me, it looks like

<x|v>V<v|y> = kV exp(-k|x|)exp(-k|y|)

How did you get a delta function?
 
  • #31
reilly
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HackaB Think about a slighly different version, say a spin dependent problem, with

H0(m,n) = S*S + V (|m><n| where S is spin (S*S has eigenvalues s(s=1), and the |m> states are Sz eigenstates. The problem specifies the potential to be proportional to a "projector", a single state projection operator. The only choice of m & n that conforms to the properties of a single-state projection operator is m=n. Generalize this reasoning, and get the delta function.
Regards,
Reilly Atkinson
 
  • #32
reilly
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By the way, Seratend's solution is the correct one for the non-diagonal version -- the projection operator is replaced by a transition operator. -- which is the equation and solution that Cooper used in his original Cooper Pair paper.
Regards,
Reilly Atkinson
 
  • #33
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So you do not agree that <x|v>V<v|y> = kV exp(-k|x|)exp(-k|y|) ?

reilly said:
The only choice of m & n that conforms to the properties of a single-state projection operator is m=n. Generalize this reasoning, and get the delta function.
I'm sorry, but I don't follow this reasoning. |v> is not a "single state" in the x-representation. So I do not see how a delta function would arise. Could you show the mathematical steps to get it?

By the way, Seratend's solution is the correct one for the non-diagonal version -- the projection operator is replaced by a transition operator.
What is a transition operator? Is it similar to a projection operator?
 
  • #34
reilly
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HackaB -- I blew it with the Bessels -- too good to be true, I suppose. So with egg on my face, I retract the delta function, and remain,
Reilly Atkinson
 
  • #35
CarlB
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If the assistance this thread has given the OP, has not convinced him that quantum mechanics is impossibly difficult, I wonder what it would take.

Carl
 

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