I've the following application of Heisenberg's uncertainty principle.

If a beam of particles in localised in the $x$-direction by a long slit, what is the uncertainty in position?

Firstly, I believe that uncertainty is equivalent to the standard deviation in this case, as I have seen the equation is written $\sigma_x \sigma_p \ge \frac{h}{4 \pi}$ and $\Delta x \Delta p \ge \frac{h}{4 \pi}$ (why is this swap allowable? Surely the uncertainty is the range of possible values it can take, whereas the SD is simply the average deviation from the mean?).

Now, the question:

What is $\sigma_x$? Is it the width of the slit (because that is the range of (classically) allowed values of $x$*), or half-width of the slit (as the middle of the slit is assumed to be the average position, and the standard deviation measures the deviation from the average)?

*I have a problem with those possible answers if one of them is correct: particles really have a nonzero probability of existing in classically forbidden regions (i.e. inside the metal or wood that surrounds the slit), so why do we base the values of a quantum mechanical number ($\sigma_x$) on a value (the slit width, which is the classically allowed region) that is classical?

Additionally, does the particle have a nonzero probability of existing everywhere? If so, and uncertainty takes the meaning I thought it did in the first paragraph, isn't the uncertainty infinite?

I hope that badly written popularisations haven't misguided me too much.


1 Answer 1


The precise, mathematical statement of the uncertainty principle is $\sigma^2_x \sigma^2_k \geq 1/4$. The use of deltas is just an informal way of talking about it. Nevertheless, it's pretty common to say, for instance, that the width of a peak is either the standard deviation or some quantity proportional to it--see, for example, full width at half maximum, which ends up being about $2.35\sigma$.

I'm not really sure what a slit would look like in 1 dimension. It's easier for me to consider a particle in a 1d infinite square well. Note that the infinite well absolutely forbids any leakage of the particle into the forbidden region, just like the classical case. In this case, the variances depend on the energy of the particle. For a particle in one of the $n$th energy eigenstate of an infinite well with width $L$, the variances are (per wikipedia)

$$\sigma^2_x = \frac{L^2}{12} \Bigg ( 1 - \frac{6}{n^2 \pi^2} \Bigg), \quad \sigma_k^2 = \frac{n^2 \pi^2}{L^2}$$

The product of the variances is then $\sigma_x^2 \sigma_k^2 = (n^2 \pi^2/12 - 1/2)$. For $n=1$, this is about $.322 \geq .25$, as required.

You can't really see what the uncertainties will be by inspection, by the geometry of the problem. These are the uncertainties for energy eigenstates, and there's no reason to expect that a particle will be in an eigenstate (which would then make the computation more complicated).

Really, one just calculates the variances of the wavefunction with respect to $x$ and $k$. You might be able to get a rough idea from the quantities in the problem (for instance, the standard deviation with respect to $x$ is indeed proportional to $L$, but only proportional, not exactly $L$), but that's all.

You ask if a particle has nonzero probability of existing everywhere. To be pedantic, a particle has zero probability of existing at any specific point, but it typically has a nonzero probability of existing in a region of any finite size. This infinite square well is an exception, as the infinite potential around the box absolutely forbids particles.

Uncertainty really is just a loose, loose word to use. It almost always really means standard deviation of the wavefunction.

  • $\begingroup$ Thank you, I'll stop using that useless term. Why do you use the wavenumber $k(=\frac{2 \pi}{\lambda})$, rather than momentum, which I have seen written more? I assume it's something to do with $\lambda = \frac{h}{p}$, but I can't see it. $\endgroup$
    – Meow
    Commented Jan 5, 2013 at 18:51
  • $\begingroup$ Unless the variance of $k$ is linear wrt $k$, then it just seems to be missing a factor of $2$. $\endgroup$
    – Meow
    Commented Jan 5, 2013 at 19:00
  • 1
    $\begingroup$ Yes, $p=\hbar k$. I find using $k$ easier than $p$ because it eliminates a lot of pesky factors of $\hbar$, but that's merely a personal preference. $\endgroup$
    – Muphrid
    Commented Jan 5, 2013 at 19:23
  • $\begingroup$ $k$ is varying with respect to $n$, right? And isn't $k=\frac{\pi n}{L}$, so why is its variance equal to itself? $\endgroup$
    – Meow
    Commented Jan 5, 2013 at 19:25
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    $\begingroup$ $k_n = \pi n/L$ are the eigenvalues of the wavenumber operator $\hat k$ on the energy eigenstates. Note that $\langle k \rangle = 0$ for any of the energy eigenstates, so $\sigma^2_k = \langle k^2 \rangle$. As a result, this will be $k_n^2$ for any particular energy eigenstate. $\endgroup$
    – Muphrid
    Commented Jan 5, 2013 at 19:44

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