*Michael Fowler, University
of Virginia*

As we have said, the *wave
nature* of particles implies that we cannot know *both* position *and* momentum
of a particle to an arbitrary degree of accuracy—if $\Delta x$ represents the fuzziness in our knowledge of
position, and $\Delta p$ that of momentum, then

$\Delta x\Delta p\sim h,$

where $h$ is Planck’s constant. We formulated this by constructing a wave packet moving along a line. In the real world, particles are three dimensional and we should say

$\Delta x\Delta {p}_{x}\sim h,$

with corresponding equations for the other two spatial
directions. The fuzziness about position
is related to that of momentum *in the
same direction*.

Let’s see how this works by trying to measure $y$ position and $y$ momentum very accurately. Suppose we have a source of electrons, say, an electron gun in a CRT (cathode ray tube, such as an old-fashioned monitor). The beam spreads out a bit, but if we interpose a sheet of metal with a slit of width $w,$ then for particles that make it through the slit, we know $y$ with an uncertainty $\Delta y=w.$ Now, if the slit is a long way downstream from the electron gun source, we also know ${p}_{y}$ very accurately as the electron reaches the slit, because to make it to the slit the electron’s velocity would have to be aimed just right.

But does the measurement of the electron’s $y$ position—in other
words, having it go through the slit—affect its
$y$ momentum?
The answer is yes. If it didn’t,
then sending a stream of particles through the slit they would all hit very
close to the same point on a screen placed further downstream. But we know from
experiment that this is not what happens—a single
slit diffraction pattern builds up, of angular width $\theta \sim \lambda /w,$ where the electron’s de Broglie wavelength *lambda* is given by ${p}_{x}\cong h/\lambda $ (there is a negligible contribution to *lambda* from the $y$ -momentum).
The consequent uncertainty in ${p}_{y}$ is

$\Delta {p}_{y}/{p}_{x}\sim \theta \sim \lambda /w.$

Putting in ${p}_{x}=h/\lambda ,$ we find immediately that

$\Delta {p}_{y}\sim h/w$

so the act of measuring the electron’s $y$ position has fuzzed out its $y$ momentum by precisely the amount required by the uncertainty principle.

In order to understand the Uncertainty Principle better, let’s try to see what goes wrong when we actually try to measure position and momentum more accurately than allowed.

For example, suppose we look at an electron through a
microscope. What could we expect to
see? Of course, you know that if we try
to look at something *really* small
through a microscope it gets blurry—a small
sharp object gets diffraction patterns around its edges, indicating that we are
looking at something of size comparable to the wavelength of the light being
used. If we look at something much
smaller than the wavelength of light—like the
electron—we would
expect a diffraction pattern of concentric rings with a circular blob in the
middle. The size of the pattern is of
order the wavelength of the light, in fact from optics it can be shown to be $\sim \lambda f/d$ where $d$ is the diameter of the object lens of the
microscope, $f$ the focal distance (the distance from the lens
to the object). We shall take $f/d\sim 1,$ as it usually is. So looking at an object the size of an
electron should give a diffraction pattern centered on the location of the
object. That would seem to pin down its
position fairly precisely.

What about the *momentum
*of the electron? Here a problem
arises that doesn’t matter for larger objects—the light
we see has, of course, bounced off the electron, and so the electron has some
recoil momentum. That is, by bouncing
light off the electron we have given it some momentum. Can we say how much? To make it simple, suppose we have good eyes
and only need to bounce one photon off the electron to see it. We know the initial momentum of the photon
(because we know the direction of the light beam we’re using to illuminate the
electron) and we know that after bouncing off, the photon hits the object lens
and goes through the microscope, but we don’t know *where* the photon hit the object lens. The whole point of a microscope is that all
the light from a point, light that hits the object lens *in different places*, is all focused back to one spot, forming the
image (apart from the blurriness mentioned above). So if the light has wavelength $\lambda ,$ its constituent photons have momentum$\sim h/\lambda ,$ and from our ignorance of where the
photon entered the microscope we are uncertain of its $x$ -direction momentum by an amount $\sim h/\lambda .$ Necessarily, then, we have the same
uncertainty about the electron’s $x$ -direction momentum, since this was imparted
by the photon bouncing off.

But now we have a problem.
In our attempts to minimize the uncertainty in the electron’s momentum,
by only using one photon to detect it, we are not going to see much of the
diffraction pattern discussed above—such
diffraction patterns are generated by *many
*photons hitting the film, retina or whatever detecting equipment is being
used. A single photon generates a single
point (at best!). This point will most
likely be within of order $\lambda $ of the center of the pattern, but this leaves
us with an uncertainty in position of order $\lambda $.

Therefore, in attempting to observe the position and momentum of a single electron using a single photon, we find an uncertainty in position $\Delta x\sim \lambda ,$ and in momentum $\Delta {p}_{x}\sim h/\lambda .$ These results are in accordance with Heisenberg’s Uncertainty Principle $\Delta x\Delta {p}_{x}\sim h.$

Of course, we could pin down the position much better if we used $N$ photons instead of a single one. From statistical theory, it is known that the remaining uncertainty $\sim \lambda /\sqrt{N}.$ But then $N$ photons have bounced off the electron, so, since each is equally likely to have gone through any part of the object lens, the uncertainly in momentum of the electron as a consequence of these collisions goes up as $\sqrt{N}.$ (The same as the average imbalance between heads and tails in a sequence of $N$ coin flips.)

Noting that the uncertainty in the momentum of the electron arises because we don’t know where the bounced-off photon passes through the object lens, it is tempting to think we could just use a smaller object lens, that would reduce $\Delta {p}_{x}.$ Although this is correct, recall from above that we stated the size of the diffraction pattern was $\sim \lambda f/d,$ where $d$ is the diameter of the object lens and $f$ its focal length. It is easy to see that the diffraction pattern, and consequently $\Delta x,$ gets bigger by just the amount that $\Delta {p}_{x}$ gets smaller!

Suppose now that in the double slit experiment, we set out
to detect which slit each electron goes through by shining a light just behind
the screen and watching for reflected light from the electron immediately after
it had passed through a slit. Following
the discussion in Feynman’s *Lectures in
Physics, Volume III*, we shall now establish that if we can detect the
electrons, we ruin the diffraction pattern!

Taking the distance between the two slits to be $d,$ the *dark*
lines in the diffraction pattern are at angles

$\left(n+{\scriptscriptstyle \frac{1}{2}}\right){\lambda}_{\text{elec}}=d\mathrm{sin}\theta .$

If the light used to see which slit the electron goes through generates an uncertainty in the electron’s $y$ momentum $\Delta {p}_{y},$ in order not to destroy the diffraction pattern we must have

$\Delta {p}_{y}/p<{\lambda}_{\text{elec}}/d,$

(the angular uncertainty in the electron’s direction must not be enough to spread it from the diffraction pattern maxima into the minima). Here $p$ is the electron’s full momentum, $p=h/{\lambda}_{elec}.$ Now, the uncertainty in the electron’s $y$ momentum, looking for it with a microscope, is $\Delta {p}_{y}\sim h/{\lambda}_{\text{light}}.$

Substituting these values in the inequality above we find the condition for the diffraction pattern to survive is

${\lambda}_{\text{light}}>d,$

the wavelength of the light used to detect which slit the
electron went through must be greater than the distance between the slits. Unfortunately,
the light scattered from the electron then gives one point in a diffraction
pattern of size the wavelength of the light used, so *even if we see the flash this does not pin down the electron
sufficiently to say which slit it went through*. Heisenberg wins again.

It is interesting to see how the actual physical size of the
hydrogen atom is determined by the wave nature of the electron—in effect,
by the Uncertainty Principle. In the
ground state of the hydrogen atom, the electron minimizes its total energy. For
a classical atom, the energy would be minus infinity, assuming the nucleus is a
point (and very large in any case) because the electron would sit right on top
of the nucleus. However, this cannot
happen in quantum mechanics. Such a very
localized electron would have a very large uncertainty in momentum—in other
words, the kinetic energy would be large. This is most clearly seen by
imagining that the electron is going in a circular orbit of radius* r* with angular momentum $h/2\pi .$ Then
one wavelength of the electron’s de Broglie wave just fits around the circle, $\lambda =2\pi r.$ Clearly, as we shrink the circle’s radius $r,\text{\hspace{0.17em}}\lambda $ goes down proportionately, and the electrons
momentum $p=h/\lambda $ increases. Adding the electron’s electrostatic
potential energy we find the total energy for a circular orbit of radius $r$ is:

$$K.E.+P.E.=\frac{{h}^{2}}{8m{\pi}^{2}{r}^{2}}-\frac{{e}^{2}}{4\pi {\epsilon}_{0}r}.$$

Notice that for small enough $r,$ the kinetic energy term dominates, and the total energy grows as the atom shrinks. Evidently, there is a value of $r$ for which the total energy is a minimum. This is easy to find by differentiating the total energy with respect to $r$ and finding where it is zero. This gives

$$-\frac{{h}^{2}}{4m{\pi}^{2}r}+\frac{{e}^{2}}{4\pi {\epsilon}_{0}}=0,$$

from which we find

$${r}_{\text{min}}=\frac{{\epsilon}_{0}{h}^{2}}{\pi m{e}^{2}}.$$

The total energy for this radius is, incredibly, the exact right answer for the ground state of the hydrogen atom. Very nice, but we don’t deserve it, because we have used a very naïve picture, as will become clear later.

The point of this exercise is to see that in quantum mechanics, unlike classical mechanics, a particle cannot position itself at the exact minimum of potential energy, because that would require a very narrow wave packet and thus be expensive in kinetic energy. The ground state of a quantum particle in an attractive potential is a tradeoff between potential energy minimization and kinetic energy minimization. Thus the physical sizes of atoms, molecules and ultimately ourselves are determined by Planck’s constant.

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