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What happens to the continuity of wave function

by amalmirando
Tags: continuity, function, wave
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amalmirando
#1
Nov11-12, 11:14 PM
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In the presence of a delta potential, how does the continuity of the wave function gets violated?
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tom.stoer
#2
Nov12-12, 12:28 AM
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It's not violated
FedEx
#3
Nov12-12, 04:19 AM
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Quote Quote by amalmirando View Post
In the presence of a delta potential, how does the continuity of the wave function gets violated?
The wave function continuity is preserved. However, the derivative is not. A simple exercise with the Schrödinger equation should show that.

DrDu
#4
Nov12-12, 06:14 AM
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What happens to the continuity of wave function

Quote Quote by amalmirando View Post
In the presence of a delta potential, how does the continuity of the wave function gets violated?
Why should a wavefunction be continuous?
mfb
#5
Nov12-12, 07:56 AM
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Quote Quote by DrDu View Post
Why should a wavefunction be continuous?
To have a derivative.

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 well-defined derivatives. The limit of those wave functions exists, too, and can be considered as wave function of a delta distribution potential.
DrDu
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Nov12-12, 02:12 PM
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Quote Quote by mfb View Post
To have a derivative.

The delta function potential is "unphysical" - you won't see it in nature.
That's not quite true. There is no law that forces a wavefunction to have a derivative. Rather the operator of derivation is not too "good" an operator in being unbounded whence it is not defined on all possible vectors in Hilbert space.
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.
Jano L.
#7
Nov12-12, 04:17 PM
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I share some uneasiness about this delta-potential. 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 non-equivalent ways to describe reflecting wall?
Fredrik
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Nov12-12, 05:14 PM
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Quote Quote by DrDu View Post
There is no law that forces a wavefunction to have a derivative.
It has to be differentiable to satisfy the Schrödinger equation. (Twice with respect to the spatial variables).
mfb
#9
Nov12-12, 05:26 PM
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Quote Quote by DrDu View Post
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 coulomb potential is not better anyway, and the real coulomb problem for atoms has a finite charge distribution in the nucleus and no unbounded potential.
Try to evaluate the Schroedinger equation at r=0 in a perfect 1/r-potential ;).
DrDu
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Nov13-12, 01:47 AM
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Quote Quote by Fredrik View Post
It has to be differentiable to satisfy the Schrödinger equation. (Twice with respect to the spatial variables).
This clearly depends on the Schroedinger equation. Take for example H=(p-A)^2 with [itex] A=\pi \delta(x)[/itex] which can easily be obtained from H=p^2 applying a gauge transform [itex] \psi \rightarrow \exp (i\pi\theta(x)) \psi[/itex].
The original psi being continuous, the new psi is discontinuous.
DrDu
#11
Nov13-12, 01:57 AM
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Quote Quote by Jano L. View Post
I share some uneasiness about this delta-potential.
Right, that's why it is not defined mathematically the way you sketch. The proper way to define a Schroedinger equation with a delta function potential is via self-adjoint extensions as already described by von Neumann in his mathematical foundations of quantum mechanics.
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 so-called delta' interaction.
See Coutinho et al, Generalized point interactions in one-dimensional quantum mechanics, http://iopscience.iop.org/0305-4470/30/11/021
DrDu
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Nov13-12, 02:01 AM
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Quote Quote by mfb View Post
The coulomb potential is not better anyway, and the real coulomb problem for atoms has a finite charge distribution in the nucleus and no unbounded potential.
How about e.g. positronium?
mfb
#13
Nov13-12, 08:03 AM
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I would expect that the mechanism behind the Lamb-shift gives you something similar in positronium.
Fredrik
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Nov13-12, 01:17 PM
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Quote Quote by DrDu View Post
This clearly depends on the Schroedinger equation. Take for example H=(p-A)^2 with [itex] A=\pi \delta(x)[/itex] which can easily be obtained from H=p^2 applying a gauge transform [itex] \psi \rightarrow \exp (i\pi\theta(x)) \psi[/itex].
The original psi being continuous, the new psi is discontinuous.
Are you talking about the Dirac field? That's not even wavefunction.

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

Quote Quote by DrDu View Post
Right, that's why it is not defined mathematically the way you sketch. The proper way to define a Schroedinger equation with a delta function potential is via self-adjoint extensions as already described by von Neumann in his mathematical foundations of quantum mechanics.
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 so-called delta' interaction.
See Coutinho et al, Generalized point interactions in one-dimensional quantum mechanics, http://iopscience.iop.org/0305-4470/30/11/021
What I've always been thinking about infinite potentials is that they're just a way to specify what Hilbert space we're dealing with. For example, the box potential V(x)=0 for all x such that -L<x<L, and V(x)=+∞ for all other x, is just a way to specify that we are now working with the Hilbert space ##L^2((-L,L))## instead of ##L^2(\mathbb R)##.
DrDu
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Nov13-12, 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 S-operator, being an unbound operator, is not defined on all vectors of the Hilbert space.
The usual S-operators with smooth potentials are well defined on differentiable wavefunctions while the S-operators with delta functions are defined on functions with discontinuous derivatives.
You are quite mathematically interested. Are you familiar with the theory of self-adjoint extensions of operators?
See,
http://en.wikipedia.org/wiki/Extensi...tric_operators
Fredrik
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Nov14-12, 08:11 AM
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Quote Quote by DrDu View Post
No Fredrik, I was not talking about Dirac equation, just about some straight forward unitary transformation.
I guess I got confused by the appearance of ##\psi##, ##A## and "gauge transformation" all in one place.

Quote Quote by DrDu View Post
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].
This had me confused for a bit. For anyone who is still confused: θ is the step function, and we should write ##\tilde H=(U^\dagger p U)(U^\dagger p U)## and then figure out how to rewrite ##U^\dagger p U##.

Quote Quote by DrDu View Post
A Schroedinger operator with a delta potential also does not change the Hilbert space. Rather, the S-operator, being an unbound operator, is not defined on all vectors of the Hilbert space.
The usual S-operators with smooth potentials are well defined on differentiable wavefunctions while the S-operators with delta functions are defined on functions with discontinuous derivatives.
You are quite mathematically interested. Are you familiar with the theory of self-adjoint extensions of operators?
See,
http://en.wikipedia.org/wiki/Extensi...tric_operators
No, I'm not familiar with that. I will take a look at it when I have more time. Thanks for the tip.
DrDu
#17
Nov14-12, 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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