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What happens to the continuity of wave function 
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#1
Nov1112, 11:14 PM

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In the presence of a delta potential, how does the continuity of the wave function gets violated?



#2
Nov1212, 12:28 AM

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P: 5,366

It's not violated



#3
Nov1212, 04:19 AM

P: 336




#4
Nov1212, 06:14 AM

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P: 3,560

What happens to the continuity of wave function



#5
Nov1212, 07:56 AM

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P: 11,589

The delta function potential is "unphysical"  you won't see it in nature. It is the limit of a very deep, very small potential well. Every very deep, very small potential well gives a continuous wave function with welldefined derivatives. The limit of those wave functions exists, too, and can be considered as wave function of a delta distribution potential. 


#6
Nov1212, 02:12 PM

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Also the Hamiltonian for the motion in a Coulomb potential has undefined derivatives at r=0, so that is not a peculiarity of the delta potential. The delta potential becomes very important in relativistic qm as it is the only interaction potential compatible with Lorentz transformations, i.e. in relativistic qm particles can only interact via purely local interactions. 


#7
Nov1212, 04:17 PM

PF Gold
P: 1,148

I share some uneasiness about this deltapotential. The original use of potential energy was that its value [itex]U(x)[/itex] gives potential energy when the particle is at [itex]x[/itex]. But the object
[tex] U_A(x) = A\delta(x) [/tex] does not have any value at [itex]x[/itex], because it is not a function. Sometimes it is said that [itex]A\delta(x) = 0[/itex] on R\{0} and [itex]A\delta(0) = +\infty[/itex], thus suggesting that this describes infinite thin potential wall centred at 0, but this doesn't make sense to me. There is mathematically only one such potential wall (reflecting particles of any kinetic energy); but there is infinity of distributions [itex]U_A(x)[/itex], each producing different jump in the derivative of [itex]\psi[/itex]. How can we have infinity of nonequivalent ways to describe reflecting wall? 


#8
Nov1212, 05:14 PM

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PF Gold
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#9
Nov1212, 05:26 PM

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Try to evaluate the Schroedinger equation at r=0 in a perfect 1/rpotential ;). 


#10
Nov1312, 01:47 AM

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The original psi being continuous, the new psi is discontinuous. 


#11
Nov1312, 01:57 AM

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Basically you just specify the boundary conditions of the wavefunction. In the case of one dimensional Schroedinger equations, not all possibilities can be realized as limits of smooth potentials, especially the socalled delta' interaction. See Coutinho et al, Generalized point interactions in onedimensional quantum mechanics, http://iopscience.iop.org/03054470/30/11/021 


#12
Nov1312, 02:01 AM

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#13
Nov1312, 08:03 AM

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I would expect that the mechanism behind the Lambshift gives you something similar in positronium.



#14
Nov1312, 01:17 PM

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PF Gold
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What I'm saying is just that a function must be differentiable to satisfy the Schrödinger equation, and a differentiable function is continuous. 


#15
Nov1312, 03:30 PM

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No Fredrik, I was not talking about Dirac equation, just about some straight forward unitary transformation.
Take [itex]H=p^2[/itex] with p=id/dx and [itex] U=\exp(i\pi \theta(x)) [/itex], then [itex]\tilde{H}=U^\dagger H U=(p\pi\delta(x))^2 [/itex] has discontinuous eigenfunctions [itex] \tilde{\psi}=U^\dagger \psi=\mathrm{sgn}(x)\psi[/itex]. A Schroedinger operator with a delta potential also does not change the Hilbert space. Rather, the Soperator, being an unbound operator, is not defined on all vectors of the Hilbert space. The usual Soperators with smooth potentials are well defined on differentiable wavefunctions while the Soperators with delta functions are defined on functions with discontinuous derivatives. You are quite mathematically interested. Are you familiar with the theory of selfadjoint extensions of operators? See, http://en.wikipedia.org/wiki/Extensi...tric_operators 


#16
Nov1412, 08:11 AM

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PF Gold
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#17
Nov1412, 08:31 AM

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It is interesting that the Aharonov Bohm effect may discussed along these lines.
Just consider a particle moving on a circle and replace p by L_z. The magnetic vector potential due to an infinitely thin flux line at the center of the cycle may be gauged to be as [itex] A=\Phi/2\pi \theta(\phi)[/itex]. The interesting thing is that here you cannot trivialize the hamiltonian via a unitary transformation although you can switch to a smoother gauge. 


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