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Homework Help: Simple (Constant) Wavefunction -- Find Uncertainty In p^2

  1. Mar 22, 2016 #1
    1. The problem statement, all variables and given/known data
    Given the following wave function valid over [itex]-a \le x \le a[/itex] and which is 0 elsewhere,
    [tex]\psi(x) = 1/\sqrt{2a}[/tex]
    Find the uncertainty in [itex]\left<\left(\Delta p\right)^2\right>[/itex] momentum, and the uncertainty product [itex]\left<\left(\Delta x\right)^2\right>\left<\left(\Delta p\right)^2\right>[/itex]

    2. Relevant equations
    Expectation value of an observable [itex]\underline{\alpha}[/itex]
    [tex]\left<\alpha\right> = \int_{-\infty}^{\infty} \psi^*\alpha\psi\; dx[/tex]

    Heaviside / unit step function
    [tex]H(x) = \begin{cases}1, x\ge 0\\0, \mathrm{otherwise}\end{cases}[/tex]

    Relationship between Heaviside and Dirac delta function
    [tex]\delta(x) = \frac{dH(x)}{dx}[/tex]

    Uncertainty of an observable [itex]\underline{\alpha}[/itex]
    \left<\left(\Delta\alpha\right)^2\right> &= \left<\left(\alpha -
    &= \left<\alpha^2\right> - \left<\alpha\right>^2
    3. The attempt at a solution
    To find the uncertainty in momentum, which is just the difference between the expectation value of the square of momentum and the expectation value of the momentum, I need to find [itex]\left<p\right>[/itex] and [itex]\left<p^2\right>[/itex]

    First, the wave function and its conjugate can be written in terms of Heaviside functions.
    [tex]\begin{align*}\psi(x) &= \frac{1}{\sqrt{2a}}\left[H(x+a)-H(x-a)\right]\\\psi^*(x) &= \frac{1}{\sqrt{2a}}\left[H(x+a)-H(x-a)\right]\end{align*}[/tex]

    Next, [itex]\left<p\right>[/itex]

    [tex]\left<p\right> = \int_{-\infty}^{\infty} \psi^*(x)\left[\frac{\hbar}{i}\frac{d}{dx}\right]\psi(x)\; dx[/tex]

    The derivative of [itex]\psi(x)[/itex] is
    [tex]\begin{align*}\frac{d\psi(x)}{dx} &= \frac{d}{dx}\left(\frac{1}{\sqrt{2a}}\left[H(x+a)-H(x-a)\right]\right)\\
    &= \frac{1}{\sqrt{2a}}\left[\delta(x+a)-\delta(x-a)\right]\end{align*}[/tex]

    So, [itex]\left<p\right>[/itex] is
    [tex]\begin{align*}\left<p\right> &= \int_{-\infty}^{\infty} \psi^*(x)\left[\frac{\hbar}{i}\frac{d}{dx}\right]\psi(x)\; dx\\
    &= \int_{-\infty}^{\infty} \frac{1}{\sqrt{2a}}\left[H(x+a)-H(x-a)\right]\frac{\hbar}{i} \frac{1}{\sqrt{2a}}\left[\delta(x+a)-\delta(x-a)\right]\; dx\\
    &= \frac{\hbar}{i2a}\int_{-\infty}^{\infty} \left[H(x+a)-H(x-a)\right]\left[\delta(x+a)-\delta(x-a)\right]\; dx\end{align*}[/tex]

    Distributing out results (FOILing), in all but the [itex]H(x-a)\delta(x+a)[/itex] integrand component producing 1. The [itex]H(x-a)\delta(x+a)[/itex] part produces 0 in the course of integration. In other words,

    [tex]\begin{align*}\left<p\right> &= \frac{\hbar}{i2a}(1 - 1 + 0 + 1)\end{align*} = \frac{\hbar}{i2a}[/tex]

    First of all, is this reasonable for the momentum to have an expectation value that is imaginary?

    Second, how would I compute [itex]\left<p^2\right>[/itex] given that the second derivative of the wave function, which is summation of Heavisides, would result in a derivative of a Dirac delta function?

    Thanks very much everyone! :)
  2. jcsd
  3. Mar 22, 2016 #2

    Charles Link

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    Suggestion: Write the wavefunction in momentum (k) space by taking the F.T. I think you might then be able to compute the mean and ## \sigma^2 ## for the k-space wavefunction. Not sure why your above calculation isn't working, but it just doesn't seem to be getting good results. Perhaps writing the wavefunction as a linear combination of two step functions doesn't accurately describe its momentum content, but I'm not quite sure why.
    Last edited: Mar 22, 2016
  4. Mar 23, 2016 #3


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    One problem is your claim that
    $$\int_{-\infty}^\infty H(x)\delta(x)\,dx = 1.$$ Instead, you should have
    $$\int_{-\infty}^\infty H(x)\delta(x)\,dx = \frac 12.$$ Using this result, you find that ##\langle p \rangle = 0##. You have to be careful when using the delta function and the Heaviside step function. Take a look at http://mathworld.wolfram.com/HeavisideStepFunction.html.

    Charles has a good suggestion, but you'll still run into a problem calculating ##\langle p^2 \rangle##. The Fourier transform of ##\psi(x)## will give you a sinc function, but when you throw in the factor of ##p^2##, you end up with the integral of ##\sin^2 p##, which won't converge.
  5. Mar 23, 2016 #4
    That's an interesting perspective, though the next comment says that it might not be so helpful for the end result.

    That's useful. I'll give it a shot, but then after finding [itex]\left<p\right>[/itex], I am still left with the situation where I have to take a derivative of a Dirac delta in order to compute [itex]\left<p^2\right>[/itex]. How do I tackle that? Thanks to both of you for your suggestions!
  6. Mar 23, 2016 #5

    Charles Link

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    In college, we worked a similar calculation with a Gaussian and we had no trouble showing that the (delta p)(delta x)=h/(2 ## \pi ##) or thereabouts. Could it be that the sharp corners make this one mathematically difficult? The delta x is approximately "a" which makes for a quick estimate of delta p. I would hope it is possible to get an exact answer for this one...And a follow-on-I verified vela's calculation that p^2 gives ## \int \sin^2(ka)dk ## and diverges. What if we were to just look qualitatively at ## | \phi(k)|^2=A \sin^2(ka)/k^2 ## and assign the uncertainty in k by taking the width of the central lobe, i.e. ## \delta k=2 \pi /a ## so that ## \delta p=h/a ## ? Also we can say qualitatively that ## \delta x=a ## so that ## (\delta x)(\delta p)=h ##
    Last edited: Mar 23, 2016
  7. Mar 23, 2016 #6


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    If you have a regular old function ##f##, you can show that
    $$\int f(x)\delta'(x)\,dx = -f'(0)$$ and if you naively apply it to H(x), you find the integral diverges, which kinda makes sense because you also find ##\langle p^2 \rangle## diverges using the Fourier transform.
  8. Mar 29, 2016 #7
    I just don't think I know enough math to prove that, so, after trying for a bit, I gave up on trying to prove that equation you provided is true, and just took it for granted. I'm not even sure it makes intuitive sense to me, but taking the derivative of a distribution produces some weird results for me in my mind -- like the doublet.

    Anyway, I was able to use that to show that [itex]\left<p^2\right>[/itex] is infinite and therefore the uncertainty in momentum is infinite. However, I find that [itex]\left<x\right>[/itex] is 0 and [itex]\left<x^2\right>[/itex] is [itex]\frac{a^2}{3}[/itex]. This means that the uncertainty in position is [itex]\frac{a^2}{3}[/itex]. So, how would I find the uncertainty product if the uncertainty in momentum is infinite and the uncertainty product is just the product of the uncertainty in position and the uncertainty in momentum?

    I suspect I should find that the Gaussian satisfies the minimum uncertainty product relation -- meaning that it should give an uncertainty product of [itex]\frac{\hbar^2}{4}[/itex], but, given that my uncertainty in momentum is infinite, I'm not sure how to do that.

    Any further help would be very appreciated. I'm just stuck and it's the little tricks I'm blind to that help me crack this problem!
    Last edited: Mar 29, 2016
  9. Mar 29, 2016 #8

    Charles Link

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    Please read my last posting more carefully. It has two different parts. 1) The Gaussian is somewhat straightforward, but a separate calculation. 2) The ## sin^2(ka)/k^2 ## has infinite expectation for k^2 as vela first mentioned, but you can still assign an uncertainty to it in a different manner. If you look at the graph of the function, (it is incidentally the same form that shows up in the intensity pattern for the single slit diffraction calculation), it has a central lobe and setting ## ka=\pi ## gives you the first zero of the function. The central lobe goes from ## ka=\pi ## to ## ka=-\pi ## . This seems to be one of the better ways to assign a width of the spread in k or p (## p=hk/(2 \pi) ##) to this problem.
  10. Apr 14, 2016 #9
    I think I follow, now. Basically I can specify the height of the delta function at the origin as an idealization of a distribution that gets arbitrarily tall.

    Thank you! :)
  11. Apr 14, 2016 #10

    Charles Link

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    The integrated area under the delta function (spike) is equal to 1, thereby when integrated with another function, it will pick off the value of that function.
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