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Why do particles have non-zero position amplitude across all space?

  1. Sep 8, 2013 #1
    I keep coming across the claim that it is a consequence of the Schrodinger dynamics, that even if one postulates wave-function-collapse, so that the wave-function for a particle is entirely confined within some small region at the moment of collapse, an instant later the wave-function will have acquired "small tails" spreading to infinity in all directions.

    Can someone please tell me what the mechanism behind this is? Why does the Schrodinger dynamics make the amplitude nonzero everywhere?
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  3. Sep 8, 2013 #2
    One way that might help you to understand this phenomenon is to look at the Heisenberg uncertainty principle. The Heisenberg uncertainty principle sets an upper bound on how much one can in principle know about the motion of a particle, and roughly stated it says that: "If you know the position of a particle very exactly, then you cannot know its momentum exactly." Mathematically it is written ΔxΔp≥h/4π, where Δx is the uncertainty in the particle's position, Δp is the uncertainty in the particle's momentum, and h is Planck's constant.

    This principle implies that if one measures a particle's position very precisely (thereby reducing Δx to a very small number), then the particle's momentum, and thus its velocity, are not precisely known (Δp must increase), meaning that the particle could be moving very slow or very fast in any direction. When we create a model to predict what the outcomes of future position measurements will be, the model reflects the fact that the particle could have been moving very fast since the last measurement was performed. The result is that the spatial probability distribution spreads out very quickly after a very precise position measurement is performed.
    Last edited: Sep 8, 2013
  4. Sep 8, 2013 #3


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    If you drop a pebble in a pond causing a localized distubance, there will be ripples spreading out after that. Schroedinger's equation is a sort of wave equation, so in the same way a wave function that is localized at one time will spread out at later times.
  5. Sep 8, 2013 #4
    Thanks, this is a useful analogy. But there is seems to be a difference in the two cases: in the pond case the force of the pebble clearly causes the waves to ripple outward to infinity. But in the case of wave function collapse, there appears to be no force that pushes the wave function outwards, to spread it out. There is only a force (or perhaps better: a postulated fundamental process) that causes the wave function to "spread in", that is, to localise.

    So then the question is, what, for quantum waves, is the analog for the force of the pebble that pushes the pond ripples outward? What causes quantum waves to spread out so vastly and so rapidly?
  6. Sep 8, 2013 #5
    Thanks. I understand that if one localises a particle's position wave-function, then it must be that one spreads out its momentum wave-function. I take this to mean that relatively equal probabilities are then assigned to a large range of possible outcomes for momentum measurements. Now if I've understood, you're saying that if were to find the particle to have a very high momentum, we had better account for that in terms of change of position over time, and so we had better account for that by spreading out the position wave function directly after collapse.

    Well that is a very interesting and intuitive explanation! The one thing I'm still unclear on though is that the authors I read claim that the "wave-function tails" actually smear out in all directions in space infinity far, directly after collapse. I'm confused as to why that is necessary. It still seems consistent with your explanation that the to account for the possible states of momentum that gain non-zero amplitude given the position-collapse, one only then needs to ascribe non-zero amplitude to a finite region of position states.

    Do you know of a derivation that actually entails your result?
  7. Sep 8, 2013 #6
    Well, I can sort of show you the mathematical version of what I'm saying but it doesn't qualify as any particular derivation--it's more just a fundamental part of quantum mechanics that has to do with the mathematics of Fourier transforms.

    In the position basis, you can write a position eigenstate (a state in which the particle has a precisely known position) as [itex]\psi(x)=\delta(x-x_0)[/itex] where I'm assuming the particle is at the position x0 of our coordinate frame [and I'll do this derivation in 1D, with easy generalizations to 3D]. Mathematically, we can represent this wavefunction equally well in momentum space using a fourier transform:
    [tex]\tilde{\psi}(p) = \frac{1}{\sqrt{2\pi}}\int e^\frac{-ipx}{\hbar} \delta(x-x_0) dx=\frac{e^\frac{-ipx_0}{\hbar}}{\sqrt{2\pi}}=\int \frac{e^\frac{-ip_0x_0}{\hbar}}{\sqrt{2\pi}}\delta(p-p_0) dp_0[/tex]
    where [itex]\delta(p-p_0)[/itex] is a momentum eigenstate [i.e. a state with a precisely known momentum p0.]
    Thus no matter where the particle is located, as long as it is precisely located, its state is a superposition of momentum eigenstates, with all the probabilities for each momentum having norm (2π)-1/2. In other words, the exactly-known position state is composed of all momentum states from p0=-∞ to p0=∞, where the weightings of each momentum state are complex numbers of the same magnitude.

    Now each of these momentum eigenstates can roughly be thought of as a wave which "moves" along at a rate proportional to its momentum p0. Since the Schrodinger equation is a linear equation, we can think of the composite superposition of all these momentum eigenstates [which sum to give the precisely known position state] as being built of waves moving at all the velocities from -∞ to +∞. [So I guess with regards to your book saying the position eigenstate immediately gets a contribution at infinity, this mathematical argument shows that's not unreasonable since infinite velocity is what gets you to spatial infinity instantly.] The less intuitive but more mathematically satisfying way to see this is to plug [itex]\psi(x)[/itex] into the schrodinger equation and caclulate what happens.
    Last edited: Sep 8, 2013
  8. Sep 8, 2013 #7


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    Well, force is a made up concept to explain what we see. The question is can we make up a force that predicts what we see very precisely? In the case of the pond we can. We also know the forces that govern masses (gravity) and classical charged particles (electromagentism).

    In the case of the spreading out of the wave function, the made up concept we use is the energy (or more precisely, the Hamiltonian). Then the question is what is the precise mathematical expression of such an energy for different particles and their interactions? High energy physics when new particles were being discovered consisted of guessing such mathematical expressions, and testing if those experssions made correct predictions.
  9. Sep 9, 2013 #8


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    The state of a particle can be expressed using the position basis or the momentum basis. When you work in the position basis, you just have the usual wave function ##\psi(x,t)##, but you can also work in the momentum basis, where you have an analogous function ##\phi(p,t)##. The two functions are related by the Fourier transform
    $$\psi(x,t) = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^\infty \phi(p,t)e^{ipx/\hbar}\,dp.$$ You can think of this as saying the state ##\psi(x,t)## results from adding up waves of every wavelength. The function ##\phi## specifies the relative amplitude and phase of each component.

    For the wave function ##\psi## to vanish in a certain region of space at t=0, all of the momentum components have to be arranged just so so that the components interfere destructively in that region and sum to 0. But a moment later, each components will have propagated a bit, each moving at its own velocity. The result is that they won't interfere perfectly anymore, and the wave function won't vanish in that region anymore. The wave function spreads out.
  10. Sep 9, 2013 #9
    Hi James MC........

    I doubt anyone can.....I know I cannot. I doubt this has a specific widely agreed upon answer...... but good question.....QM has lots of them!! Perhaps you are coming from the perspective we know what a 'particle' actually is??

    The 'moment of collapse' to which you refer might be a measurement, a detection, or quantum confinement within an expanding cosmological particle horizon.

    Another way to think about the return to wave function tails is as 'a return to normalcy'.

    Three posts on this subject I liked from others in these forums:

    The last quote might support your perspective that the localized particle is 'real', the wave not so much?? I've come to the opposite conclusion,myself, but remain open minded, after many discussions in these forums [see one link below].

    The line of demarcation between mathematical representations and the physical, observable, world is often not so clear as we might prefer:

    why...as atyy posted, it's a mathematical form of expression that has proved useful.

    Carlo Rovelli provides this perspective:

    [This relates to vela's description].

    In some views, 'collapse of the wavefunction' is not even required.....there is no universal agreement that it even happens. [more in the link below]

    I personally find that a bit fanciful. I'd suggest alternatively the wavefunction has remained confined, bounded, at the cosmological particle horizon', just beyond the Hubble sphere. Nobody can prove local detection 'removes' the wavefunction. [An analogy might be....detecting the voltage of a lightning strike does not eliminate the bolt of lightning itself...all we may 'see' is a local phenomena.

    yes....OR you can think instead of those energy states always being present in spacetime, and what we detect is a local phenomena ala Rovelli's description.

    So there may be an infinite number of underlying constituent waves.....

    For lots more on particles and discussion of the above descriptions, check out this discussion:

    What is a particle:
  11. Sep 24, 2013 #10
    Thanks that's helpful! But why does the exactly known position state have to be composed of all momentum states from p0=-∞ to p0=∞? Firstly, how can the momentum have a negative value? Secondly, aren't positive momentum states (e.g. where v > c) unphysical?

    Again, negative velocities and infinite positive velocities? Perhaps the response to say that the unphysicality of infinite velocities corresponds to the unphysicality of quantum delta functions for position, that is, all amplitude at one point. But the wave function in position basis supposedly spreads everywhere instantaneously not matter the initial state, so still a bit confused here...
  12. Sep 24, 2013 #11
    That's really useful. Why ALL of the momentum components? Why not only use momentum components that are greater than 0 and less than one? I'm sure we have no empirical confirmation of the claim that the wave function in position basis has non-zero amplitude at every point in an infintiely large space no matter the inital wave function. So this doesn't seem like a datum that the mathemetics absolutely must recover?
  13. Sep 24, 2013 #12
    When a wave function it's located somewhere as a wave packet, it has a gaussian form and therefore its amplitude decays very fast as move away from the center of the gaussian.

    Since amplitude decays so quickly, it would take too long to make so many tests to confirm that you cannot find the particle more far than certain point but with no other evidence I think it's better to follow the Cophenhagen interpretation.
  14. Sep 24, 2013 #13
    Why? Why not a function with compact support? You appear to agree that there is no empirical evidence that wave functions are Gaussians, so is Gaussian form somehow supported by the formalism? As I said to the others above, if there's no need to think of position as a sum of all conceivable momenta, but only a small subset (e.g. 0 to 300,000) then there appears to be no reason to think of position wave functions as being everywhere.

    Or is there some further reason I'm missing?
  15. Sep 24, 2013 #14
    [tex]<x|p> = \frac{1}{{(2\pi)}^{1/2}}\exp ( ipx )[/tex] shows that a localized particle has evenly distributed support over all momenta.
  16. Sep 24, 2013 #15


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    If you form a linear combination of only a small subset of the momentum eigenstates, you'll get a perfectly good solution to the Schrodinger equation - but it won't be one with a tightly localized position. The solution to the Schrodinger equation in which the position has a definite value at a particular moment is an equal superposition of all the momentum eigenstates.

    So this isn't a case of us asking the mathematics to recover a particular result; instead the result comes from the math and we're left to scratch our heads and decide what it means.
  17. Sep 24, 2013 #16

    This doesn't seems so puzzling - every measurement has the potential to reveal a very definite location but since no measurement apparatus is perfect, a perfectly defined position is unobtainable in practice(but such a position surely exists with every measurement being made). So it all boils down to what information can be obtained about the measured system, not whether the particle is perfectly localized(it is always perfectly localized upon measurement, but such accuracy isn't achievable).
  18. Sep 24, 2013 #17

    There isn't anything more basic behind the uncertainty relationship at this point that can cause this. Maybe someone can help you with how the equation of the HUP was arrived at.
  19. Sep 25, 2013 #18
    The probability is so low at the tails, that is can be considered nil.

    Thus the tails in reality is not spreading to infinity.

    Take the case of weight (or height) distribution of humans.

    Let's say the mean is 50 kgs (or height 5.5 feet), say.

    What is the probability per the probability distribution graph of a human having 1 gram weight? Its not zero but its so low that .....da da da ...;)

    The status of liberty could wave its hand if all the atoms happened to be on one side/direction etc......the probability of that happening could be something like once in a trillion, raised to power billion/trillion, years.
    Last edited: Sep 25, 2013
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