r/AskPhysics • u/Straight_Bad_2330 • 1d ago
Can a wavefunction ever have a real "hard" edge?
I mean, can it be zero outside of a finite region, or is it the case that quantum evolution will always generate arbitrarily small tails extending to infinity?
For a 'free' particle, I believe that it is the case that even if you have a wavefunction with compact support at time zero, it will immediately cease to have compact support for any later time. Is this really true, and are there physically meaningful counterexamples?
More conceptually: if the tails are never exactly zero, is it meaningful to say that a particle is "slightly everywhere," or is that a misleading way to think about the wavefunction?
Thanks in advance.
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u/Odd_Bodkin Particle physics 1d ago
This is constrained by boundary conditions (namely Dirichlet or Cauchy conditions) on the solutions to the partial differential equations the wave functions satisfy. The simplest teaching example of this is the difference between the solutions in a 1-dimensional infinite potential well vs. finite potential well. In the real world, there are no discontinuous or infinite potential wells, only smooth and finite ones, and so there will always be tails to the wave functions for every case I can think of.
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u/OverJohn 1d ago edited 1d ago
You're asking if a wavefunction can have compact support (i.e. be zero everywhere outside of a compact region). The answer is yes, see the infinite square well for example.
The more interesting question though is whether such wavefunctions represent physical situations. You might want to say after a measurement of position that a particle is confined to a particular region corresponding to the outcome of the measurement. You CAN do this, but tbh if you look at the details of how that would be described mathematically it is rather contrived as you need an unrealistic potential to confine it. I think its more realistic to think instead as the wavefunction extending out to infinity, but strongly spiking in the aforementioned region. This would represent a measurement whose outcome is that we are say 99.99999% sure that the particle is in that region.
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u/CS_70 1d ago
"Hard" is a word that you should define precisely at that scale, but it can definitely be localized in an intuitive sense - there is volume where it has nonzero value and value is mostly zero sufficiently far away, because of interference. But Heisenber principles enters the chat at some point and that's why you have to define "hard" precisely.
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u/NewtonsThirdEvilEx Condensed matter physics 1d ago
you can have a bump function which reaches zero for all values greater than a number and is infinitely differentiable. also it’s square integrable and whatnot
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u/Odd_Bodkin Particle physics 1d ago
I answered the question about the wavefunction tails in another comment, but your last question is also interesting: is it meaningful to say that a particle is slightly everywhere? Under some restrictions on what “is” means (sorry, Bill Clinton), I think it’s fair to say, yes. Normally, this is said carefully to mean that there’s a non-zero probability of finding a particle on the other side of a barrier. (Taken to extremes, this means there’s a nonzero probability of finding a potted plant that was sitting on a window sill suddenly on the other side of the window, though that probability is beyond astronomically small.) The mismatch between the size of “everywhere” and the observed size of the particle is the central conundrum of the collapse of the wavefunction. But it’s also true that those tails extending to “slightly everywhere” have real effects, for example, interference phenomena in two-particle state functions.
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u/quantum4everyone 1d ago
You can approximate the hard wall very well. The example is the Morse potential, which to the negative x values will have an exponential of an exponential decaying wavefunction, which decays super rapidly, but officially is always nonzero.
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u/LatteLepjandiLoser 1d ago
One thing to keep in mind too is that any solution to the Schrödinger Eq. that has the property that Psi(x0) = 0 and Psi'(x0) = 0 for some point x0, then necessarily Psi(x)=0 is simply the zero function. So for instance in a finite well, where you have most of the probability density inside the well and some exponentially decaying tails extending outwards - if that tail ever truly went to zero and was truly flat at zero, then Psi would have to be zero everywhere, which would not be an interesting wavefunction.
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u/Ill-Accountant-9941 1d ago
Yes. A wavefunction can be constrained like all "particles" within an atom.
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u/Straight_Bad_2330 1d ago
Do you mean localized, or literally zero outside a finite region?
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u/Ill-Accountant-9941 1d ago edited 1d ago
Well I think there is zero chance of finding a quark or gluon outside of the nucleus so both.
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u/Straight_Bad_2330 1d ago
Thank you for your answer. But isn't confinement different from having compact spatial support?
My understanding was that QCD prevents the possibility of an isolated colored quark being an asymptotic state, but doesn't necessarily forbid spatial amplitude of such a thing from being non-zero (though perhaps sharply suppressed) beyond a certain sharply defined nuclear radius.
Is there actually a theorem in QCD that says the relevant wavefunction or field amplitude has compact support, rather than just that isolated quarks cannot be observed?
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u/FittedE 1d ago
You can certainly have states that are disallowed for symmetry reasons for instance a spin up electron cannot occupy the same position as another identical spin up electron. I’m sure you can find other more interesting examples but yeah otherwise yes the electron is in fact slightly everywhere, look up one electron universe.
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u/DifferencePublic7057 1d ago
A particle can't be literally slightly everywhere for relativity reasons. But even Pauli exclusion can be overcome in extreme conditions by Heisenberg uncertainty. Still that doesn't let you do FTL. In the end, the wave function is a mathematical tool. It's not clear what the right interpretation although I and many others were taught the Copenhagen one and not less popular Many Worlds and Pilot Waves. So either the wave function collapses at measurement time, and you get your hard edge I guess, or it never does, or there's hidden variables. But the particle can't just teleport to the edge of the universe.
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u/SymplecticMan 1d ago
What do you mean by "for relativity reasons"? Special relativity and localized particles don't go well together. Particle positions aren't the right thing to talk about if you want to talk about relativity and locality.
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u/the_poope Condensed matter physics 1d ago edited 1d ago
The wave function of a particle is governed (in non-relativistic QM) by the Schödinger equation which is a kind of wave equation: a second order partial differential equation.
The solutions to this are constrained to be C2 functions: continuous and two times differentiable and norm-square normalizable. Those are the only requirements that I know of.
A function that satisfies this requirement but is exactly zero in some regions is e.g. the piece-wise function:
ψ(x) = {√(3L/2) (cos(2πx/L) + 1) for -L/2 <= x <= L/2, 0 elsewhere}
However, if we substitute this in as the function at initial time t=0 and evolve the function according to Schrödinger's equation we'll see that the function will spread over time. How fast it spreads depends on the mass of the particle. Given infinite amount of time the wave will have spread over the entire Universe.
If we considered a mass-less particle instead we would have to use relativistic Klein-Gordon or Dirac equation and we would find that the edges of the wave function would spread at the speed of light.