I'm currently considering a rectangular window $\psi$ function: $$ \psi(x) = \begin{cases}\left(2a\right)^{-1/2}&\text{for } |x|<a \\ 0&\text{otherwise.} \end{cases} $$ I am interested in the momentum uncertainty of this function. I expect it to be a function of a, the 'width' of $\psi$ in the x-space.

I claim that $\langle{p}\rangle = 0$ because this is a stationary state and so $m\frac{d\langle x\rangle}{dt} = 0 $. This spares me a trial to take the derivative of a step function, which I'm about to have to do though for $\langle{p^2}\rangle$.

For $\langle{p^2}\rangle$ I should calculate:

$$ {-\hbar^2\int_{-a}^a\psi^*\frac{d^2}{dx^2}\psi} \,dx . $$

This requires taking the second derivative of a square window function, which I imagine will lead to infinite values. So instead, I will work in the momentum space after calculating the Fourier transform of $\psi$ for which I got the sinc function: $$ \phi(p) = \sqrt{\frac{\hbar}{a\pi}}\frac{\sin(\frac{a\pi p}{\hbar})}{p}. $$ And then I tried to calculate $\langle p^2 \rangle$ $$ \langle p^2\rangle = \frac{\hbar}{a\pi}\int_{-\infty}^{\infty}\frac{\sin^2(\frac{a\pi p}{\hbar})}{p^2}p^2dp = \frac{\hbar}{a\pi}\int_{-\infty}^{\infty}{\sin^2(\frac{a\pi p}{\hbar})} dp \rightarrow \infty $$

Am I mistaken? This would say that the uncertainty in momentum, $\sqrt{\langle p^2\rangle-\langle p\rangle^2}$ , is infinite independently of the width $a$ of the $x$-localization.

  • $\begingroup$ Comment to the question (v3): Yes, this wave function packs infinite kinetic energy $\langle\frac{p^2}{2m}\rangle=\infty$, and is hence unphysical. Related: physics.stackexchange.com/q/38181/2451 $\endgroup$ – Qmechanic Jan 27 '15 at 21:49
  • $\begingroup$ Whoops, thanks! @Qmechanic That's an interesting read. To show that this state has infinite kinetic energy by showing that $\frac{\langle p^2\rangle}{2m}$ is proportional to the integral of (dirac-delta)^2. I'll have to work that out explicitly. $\endgroup$ – jeau_von_shrau Jan 27 '15 at 22:38
  • $\begingroup$ The discontinuous derivative of the square function makes the momentum infinite. Remember that the momentum is the second derivative. $\endgroup$ – DanielSank Jan 27 '15 at 22:43

Just for your self verification, you could though do the calculus in the $x$ representation,

$$ {(1) \ \langle p^2 \rangle = -\hbar^2\int_{-a}^a\psi^*\frac{d^2}{dx^2}\psi} \,dx . $$

where I understand that you assume units in which $2m = 1$.

Now, the derivative of the step ascending function is which goes up at $x=-a$ is $\delta(x + a)$ and since the descending step function that goes down at $x=a$ can be written as 1 - the step function, its derivative is $-\delta(x - a)$. Therefore

$$ (2) \frac {d W(x;a)}{dx} = \frac {\delta(x+a) - \delta(x-a)}{\sqrt {2a}} .$$

Now, let's use (2) in (1).

$$ \ \langle p^2 \rangle = - \frac {\hbar^2}{2a} \int_{-a}^a \frac {d}{dx} \left[ \delta(x+a) - \delta(x-a) \right] dx .$$

$$ \ \ \ \ \ \ = - \frac {\hbar^2}{2a} \left[ \delta(x+a) - \delta(x-a) \right] \Bigg|_{-a}^{a} = \frac {\hbar^2}{a} \cdot \infty \ \ \ . $$

It is exactly the same as what you get with your calculus in the $p$ space, if you correct the mistake in $\phi (p)$, where should appear $\hbar $, not $\sqrt {\hbar}$ .

  • $\begingroup$ This is really great, thank you so much. So I never had to switch over to momentum space at all. $\endgroup$ – jeau_von_shrau Jan 28 '15 at 18:34
  • $\begingroup$ I'm still trying to find where I went wrong with that $\sqrt{\hbar}$ because my understanding is that there is a $\frac{1}{\sqrt{\hbar}}$ in the Fourier transform definition from $\psi$ to $\phi$. I also just checked out my $\phi$ and it's normalized. $\endgroup$ – jeau_von_shrau Jan 28 '15 at 18:43
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    $\begingroup$ @hopsital I apologize, I saw you comment only now. The Fourier transform of a function $f(x)$ is $\frac {1}{\sqrt {2 \pi}} \int_{- \infty} ^{\infty} f(x) e^{ikx} dx$, where $k = p/ \hbar$. Introducing your window function you get, $\frac {1}{2 \sqrt {\pi a}} \int_{-a}^a e^{ikx} dx$, whose result is $\frac {1}{2ik \sqrt {\pi a}} (e^{ika} - e^{-ika}) dx$. Introducing here the expression of $k$ you get $\frac {\hbar}{p \sqrt {\pi a}} sin(pa/ \hbar)$. $\endgroup$ – Sofia Jan 28 '15 at 23:07
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    $\begingroup$ @hospital I refer to your remark I never had to switch over to momentum space at all. Why not switch? Calculating in two ways and getting the same result would give you confidence that the result is good. Now, one more thing: in your sine function $sin(a \pi p/ \hbar)$ the $\pi$ is futile. The Fourier transform, as I said in my previous comment, is obtained by multiplying your function with $e^{ikx}$ and integrating over $x$. Just for rigor, over $k$ should be a bar indicating that $k$ is defined as $2\pi /\lambda$. But usually we omit the bar over $k$ as I did in my previous comment. $\endgroup$ – Sofia Jan 29 '15 at 2:02
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    $\begingroup$ @jeau_von_shrau : Aaaa ! I believe that I understand his motivation. When you transform back, i.e. $\psi (x) = \frac {1}{\sqrt {2 \pi}} \int _{-\infty}^{\infty} dk \phi(k) e^{ikx} dk$, he probably wishes to write the inverse transform as $\psi (x) = \frac {1}{\sqrt {2 \pi}} \int _{-\infty}^{\infty} dk \phi(p) e^{ipx/\hbar} dp$. The product of the two elements of integration, dx \ dp, has the dimension of $\hbar$. But this sophistication is not generally used. People work in general as I say. $\endgroup$ – Sofia Jan 29 '15 at 22:27

You can also look at this problem from a practical point of view. If you even take a smoothed (allowed) version of a rectangular window $\psi$ function, you will end up with very steep "edges" of your function. And now from mathematical point of view you will have very big values of derivative around $-a$ and $a$ (which means big values of momentum in QM), and consequently a huge $\langle p^2 \rangle$, while $\langle p \rangle$ will remain zero due to the opposite signs of derivative values.
In your Fourie transform, this smoothed wave function means integration over large but finite region, which will lead you again to a huge uncertainty.


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