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Uncertainty values for non-Gaussian functions

  1. Apr 6, 2012 #1
    Uncertainty values for non-Gaussian wave packet functions

    1. The problem statement, all variables and given/known data
    [itex]\phi(k)= \left\{ \begin{array}{cc} \sqrt{\frac{3}{2a^3}}(a-|k|), & |k| \leq a \\ 0, & |k|>a \end{array} \right.[/itex]
    [itex]\psi(x)= \frac{4}{x^2}sin^2 (\frac{ax}{2})[/itex]
    Calculate the uncertainties [itex]\Delta x[/itex] and [itex]\Delta p[/itex] and check whether they satisfy the uncertainty principle.

    2. Relevant equations
    [itex]\Delta x\Delta p \geq h/2[/itex]

    3. The attempt at a solution
    The solution is worked out in the book, which is [itex]\Delta k=a[/itex] and [itex]\Delta x=\pi /a[/itex]. I understand that for a Gaussian distribution, you can use the standard deviation as [itex]\Delta x[/itex] and [itex]\Delta k[/itex], and this leads to the lowest limit of the uncertainty relation, [itex]h/2[/itex]. I don't see how I'm supposed to come up with [itex]\Delta x[/itex] and [itex]\Delta k[/itex] for non-Gaussian functions, though. The book seems to just pick [itex]\Delta k=a[/itex] and [itex]\Delta x=\pi /a[/itex] somewhat arbitrarily without explaining why these values were chosen. Could someone tell me if there is a method for figuring out what the uncertainties should be for non-Gaussian functions like this one?
     
    Last edited: Apr 6, 2012
  2. jcsd
  3. Apr 6, 2012 #2

    fzero

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    If we are measuring the eigenvalue [itex]a[/itex] of an observable [itex]A[/itex], then we may characterize the uncertainty [itex]\Delta a[/itex] in [itex]a[/itex] using the statistical variance:

    [tex](\Delta a)^2 = \langle A^2 \rangle - \langle A\rangle^2.[/tex]

    This definition works for any state we use to compute the expectation values. In fact, this definition is the one used in the derivation of the generalized uncertainty principle.
     
  4. Apr 7, 2012 #3
    [itex]\langle A\rangle=0[/itex] for both functions, so [itex](\Delta a)^2= \int ^\infty _{-\infty}a^2 f(a) da[/itex]

    [itex](\Delta k)^2= \int ^\infty _{-\infty} k^2 |\phi(k)|^2 dk = \frac{3}{2a^3} [\int ^0 _{-a} k^2 (a+k)^2 dk + \int ^a _0 k^2 (a-k)^2 dk][/itex]
    [itex]=\frac{3}{2a^3} [(\frac{1}{3} a^2 k^3 + \frac{1}{2} a k^4 + \frac{1}{5} k^5)| ^0 _{-a} + (\frac{1}{3} a^2 k^3 - \frac{1}{2} a k^4 + \frac{1}{5} k^5)| ^a _0[/itex]
    [itex]=\frac{3}{2a^3} \frac{2}{30} a^5 = \frac{a^2}{10}[/itex]
    [itex]\Delta k = \frac{a}{\sqrt{10}}[/itex]
    [itex]\Delta p = \frac{a \hbar}{\sqrt{10}}[/itex]

    [itex](\Delta x)^2= \int ^\infty _{-\infty} x^2 |\psi(x)|^2 dx = 16 \int ^\infty _{-\infty} \sin ^4 (\frac{ax}{2}) dx / x^2[/itex]
    [itex]=16[\frac{1}{2} a Si(ax) - \frac{1}{4} a Si(2ax) - \sin ^4 (\frac{ax}{2}) / x]|^\infty _{-\infty} = 16(\frac{a \pi}{2} - \frac{a \pi}{4}) = 4a \pi[/itex]
    [itex]\Delta x = 2 \sqrt{a \pi}[/itex]

    [itex]\Delta p \Delta x = \frac{2a \hbar \sqrt{a \pi}}{\sqrt{10}}[/itex]
    So whether or not this is greater than [itex]\hbar / 2[/itex] depends on what a is. Did I do something wrong here?
     
    Last edited: Apr 7, 2012
  5. Apr 7, 2012 #4

    fzero

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    I agree with this, but we'd also want to calculate [itex]\Delta p[/itex] using [itex] \hat{x} = -i\hbar d/dp[/itex] in the momentum representation. Note that [itex]\psi(x)[/itex] is not the Fourier transform of [itex]\phi(k)[/itex], so we really have two problems to solve here.

    Again, this is a completely different wavefunction, so we can't use the uncertainty in momentum that we computed from the other one. Use [itex]\hat{p} = -i\hbar d/dx[/itex] to compute it.
     
  6. Apr 8, 2012 #5
    Actually, [itex]\psi (x)[/itex] is the Fourier transform of [itex]\phi (k)[/itex], or at least it is supposed to be...

    [itex]\psi (x)= \frac{1}{\sqrt{2 \pi}} \int ^\infty _{-\infty} \phi(k) e^{ikx} dk[/itex]
    [itex]=\frac{\sqrt{3}}{2 \sqrt{\pi a^3}} [\int ^0 _{-a} (a+k)e^{ikx} dk + \int ^a _0 (a-k)e^{ikx} dk][/itex]
    [itex]=\frac{\sqrt{3}}{2 \sqrt{\pi a^3}} [\int ^0 _{-a} ke^{ikx} dk - \int ^a _0 ke^{ikx} dk + a \int ^a _{-a} e^{ikx} dk][/itex]
    [itex]=\frac{\sqrt{3}}{2 \sqrt{\pi a^3}} [\frac{a}{ix} e^{-iax} + (1 - e^{-iax} )/ x^2 - \frac{a}{ix} e^{iax} - (e^{iax} - 1)/ x^2 + 2a \sin (ax) /x][/itex]

    At this point, the book just says that after some calculations, the answer is [itex]\psi (x) = 4 \sin ^2 (\frac{ax}{2}) / x^2[/itex]. I don't know how they came up with that; the answer I got is [itex]\frac{\sqrt{3}}{x^2 \sqrt{\pi a^3}} [1 - \frac{ax}{2} \sin (ax) - \cos (ax)][/itex].
     
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