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Particle's Probability Density Momentum Distribution

  1. Dec 28, 2007 #1
    1. The problem statement, all variables and given/known data

    Assuming that the probability density of a particle is A^2 sin^2(kx), is the particle localised in space? Using the uncertainty principle determine the degree to which the momentum of the particle is specified.

    A is just a constant, k is the wavenumber, x is position.

    2. Relevant equations

    Heisenberg's Uncertainty Principle dpdx ≥ hbar i think that's the only one...

    3. The attempt at a solution

    I thought about what the graph would look like, like a sin(x) graph but with all the dips going below y = 0 inverted because its squared, the kx bit would shrink the graph a bit i think or something like that, but anyway the point is i thought since it looks the same, hilly, over all of x then maybe it can't be localised?

    I mean the fact there are maximums means it's more likely to find the particle in certain regions looking over infinity doesn't it?

    But the particle is equally likely to be at any of the maximums over infinity. So i thought dx = infinity therefore the uncertainty in momentum dp = 0 therefore the momentum can be precisely specified just like a free particle? But i'm confused because i might be wrong and this isn't a free particle is it?

    I thought a free particle was one which had a constant probability density just like a straight line, equal probability of being found everywhere?

    Any insight would be appreciated thanks
    Last edited: Dec 29, 2007
  2. jcsd
  3. Dec 31, 2007 #2
    I'm starting to get the feeling i didn't do a very good job at explaining myself XD

    I have a particle which i worked out the probability density for as [tex]A^2\sin{(kx)}^2[/tex]

    I think the graph would look something like this:


    Using Heisenberg's Uncertainty Principle:

    [tex]\triangle p\triangle x \geq \hbar[/tex]

    We are meant to determine the degree to which the momentum of the particle is specified.

    I'm confused because the particle's probability density has a dependency on its position, wouldn't that imply the particle is localised?

    On the other hand the maximums (all of which are equal height) continue to infinity along the x axis so wouldn't this mean the particle isn't localised because you can still find it at an infinite number of locations?

    Would i be right in saying [tex]\triangle x = \infty[/tex] and therefore [tex]\triangle p = \frac{\hbar}{\infty} = 0[/tex] ?

    Also sorry for double posting the edit button for my initial post is gone it seems :S
    Last edited: Jan 1, 2008
  4. Dec 31, 2007 #3


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    Your wavefunction can be expressed as a linear combination of plane waves. e^(ikx) and e^(-ikx). So it's a superposition of a left going plane wave (free particle) and a right going plane wave (a 'standing wave'). Both of those have perfectly definite momenta (basically +k and -k) and are NOT localised. Where this stands via the uncertainty principle? Hmm. The momentum is a bit uncertain (there are two possible values) and the position is a bit localized (it's lumpy). But <x> is certainly not well defined, <p>=0. But is it really 'localized'? Guess I'd say not.
  5. Jan 1, 2008 #4
    Thanks for the response :)

    Would their momenta be dependent on +k and -k because:

    [tex]\lambda = \frac{h}{p}[/tex]

    [tex]k = \frac{2\pi}{\lambda}[/tex]

    [tex]k = \frac{2\pi p}{h}[/tex] ?

    Also this might be a stupid question but why do you write <x> instead of x? That must be special notation i'm guessing maybe to mean the range in or the uncertainty in? Or maybe just to make it easier to read? :P

    So you're saying it's neither but since the question forces us to choose you'd say it's not localized and the uncertainty in momentum is small enough to say [tex]\triangle p = 0[/tex]?
  6. Jan 1, 2008 #5


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    <x> and <p> are expectation values of x and p. The expectation value of p is zero just because the +k and -k parts cancel out (the average). But [tex]\triangle p[/tex] is NOT zero. It's basically a standard deviation and is finite. The only reason to waver is just that I don't know your exact definition of localized. If it is that [tex]\triangle x[/tex] is finite, it's certainly not. The wavefunction isn't even normalizable.
  7. Jan 1, 2008 #6
    I understand perhaps it would help if i gave the original wavefunction in the question it's:

    [tex]\psi (x,t) = A\sin (kx) e^{-i(kx-\omega t)}[/tex]

    We're supposed to sketch the probability density for a particle as a function of position x at a fixed time t described by that wavefunction. (A constant)

    The exact wording for the next part is:

    "Determine whether the probability density depends on time."


    "Is the particle localised in space?"

    I wasn't sure what definition of localised the question is using, i was thinking it meant something like the particle is confined to a finite region along the axis by a potential... but then i thought of the gaussian and that does actually trail off to infinity and yet we say that's localised.

    My friend said it's a probability density not equal at all points in space which i guess would make this particle localised, i've searched through my notes and there doesn't seem to be a clear definition.

    edit: Almost forgot my manners thanks for the response again :)
    Last edited: Jan 1, 2008
  8. Jan 1, 2008 #7


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    Ok, so we get to make our own definition. Let's just say localized means there's a lot of it in some finite region of space and not much outside that region. Your wave function definitely fails to meet even a crappy definition like that. I'm not crazy about your friend definition because I would consider a gaussian (or an electron in a hydrogen atom to be 'localized').
  9. Jan 2, 2008 #8
    Oh i wasn't going to believe my friend anyway i never do it's all part of the fun to form an opposing view against him whether or not he's right XD

    Well even though you say it's crappy I agree with your definition more so than mine or my friends mainly because you're probably an expert and like you said the electron in the hydrogen atom is localized.

    So i'm convinced this particle is not localized in space.

    As for the uncertainty part it's worth 1 mark so it can't be expecting much. Well it's actually worth 3 marks but the other 2 are for different wavefunctions which are:

    [tex]\psi = Ae^{i(kx-\omega t)}[/tex]


    [tex]\psi = Ae^{\frac{-x^2}{2\sigma^2}}e^{i(kx-\omega t)}[/tex]

    I think i can do these ones fine...

    For the first one [tex]\triangle p \geq 0[/tex]

    and for the second [tex]\triangle p \geq \frac{\hbar}{2\sigma}[/tex]

    But for this one i'm not sure, i think it wants me to put something definite like the other 2 (which might be wrong) [tex]\triangle p \geq[/tex]...

    I was thinking along the lines of...

    [tex]\frac{2\pi n}{k} < x < \frac{2\pi (n+1)}{k}[/tex] for all integer values of n but i don't think that'll go anywhere

    Is the only thing i can say that it has 2 possible values +k and -k like you said?

    Thanks for the help i appreciate it.
    Last edited: Jan 2, 2008
  10. Jan 2, 2008 #9


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    You mean [tex]\triangle P[/tex] EQUALS those quantities, right? And I don't think you want to write [tex]e^{i(kx-\omega t)}[/tex] after all of your wavefunctions. What method did you use to compute the first two?
  11. Jan 2, 2008 #10
    Well those wavefunctions are given to us in the question i didn't mention them earlier because i thought i could handle them.

    Here's the questions scanned in:


    Earlier i was using the uncertainty principle:

    [tex]\triangle p \triangle x \geq \hbar[/tex]

    and i reliased last post i changed the [tex]\geq[/tex] into an = at some points which isn't right i'm guessing you have to leave it as [tex]\geq[/tex] because that's what the uncertainty principle says?

    The way i calculated [tex]\triangle p[/tex] for the first two is initially i found the probability density for both:

    For the first it'd be:

    [tex]A^2[/tex] which graphically is just a straight line since A is a constant.

    Therefore [tex]\triangle x = \infty[/tex]

    and for the second it'd be:

    [tex]A^2e^{\frac{-x^2}{\sigma^2}}[/tex] which is a gaussian where most of the bell shape is between [tex]- \sigma[/tex] and [tex]\sigma[/tex].

    Therefore [tex] - \sigma \leq x \leq \sigma[/tex] so [tex]\triangle x = 2 \sigma[/tex].

    When i had [tex]\triangle x[/tex] for both equations i just plugged the values into the uncertainty principle and rearranged for [tex]\triangle p[/tex].

    Thanks for the continued help, very much appreciated =)
    Last edited: Jan 2, 2008
  12. Jan 2, 2008 #11


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    Ok, so you aren't finding [tex]\triangle p[/tex] directly. You are just finding [tex]\triangle x[/tex] and trying to use it to set some bound on [tex]\triangle p[/tex]. In that case, [tex]\triangle x[/tex] is basically infinite for the first two which doesn't really give you much to say about [tex]\triangle p[/tex] except that it's greater than or equal to zero. If you were computing [tex]\triangle p[/tex] directly, you would see it's zero in the first case and finite in the second. But I'm still not sure how you are even computing [tex]\triangle x[/tex]. I don't see anyplace where you are actually finding the mean and expectation values of x.
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