Michael Fowler,
As we have shown for wavepackets, the wave nature of particles implies that we cannot know both position and momentum of a particle to an arbitrary degree of accuracy—if represents the
uncertainty in our knowledge of position, and
that of momentum, then
where h is Planck’s constant. In the real world, particles are three-dimensional and we should say
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 . Now, if the slit is
a long way downstream from the electron gun source, we also know py 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 , where the electron’s de Broglie wavelength
is given by
(there is a negligible
contribution to
from the y-momentum). The consequent uncertainty in py is
Putting in , we find immediately that
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 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 ~
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 , its constituent photons have momentum
, and from our ignorance of where the photon entered the
microscope we are uncertain of its x-direction
momentum by an amount
. 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
of the center of the pattern, but this leaves us with an
uncertainty in position of order
.
Therefore, in attempting to observe the position and
momentum of a single electron using a single photon, we find an uncertainty in
position , and in momentum
. These results are in
accordance with Heisenberg’s Uncertainty Principle
.
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. 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
. (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 . Although this is
correct, recall from above that we stated the size of the diffraction pattern
was
, 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
, gets bigger by just the amount that
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
.
If the light used to see which slit the electron goes
through generates an uncertainty in the electron’s y momentum , in order not to destroy the diffraction pattern we must
have
(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, . Now, the uncertainty
in the electron’s y momentum, looking
for it with a microscope, is
.
Substituting these values in the inequality above we find the condition for the diffraction pattern to survive is
,
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/2p.
Then one wavelength of the electron’s de Broglie wave just fits around the
circle, . Clearly, as we
shrink the circle’s radius r,
goes down proportionately, and the
electrons momentum
increases. Adding the electron’s electrostatic potential energy we find the total energy for a circular orbit of radius r is:
.
Notice that for very large r, the potential energy dominates, the kinetic energy is
negligible, and shrinking the atom lowers the total energy. However, for small enough r, the (always positive) kinetic energy
term wins, and the total energy grows
as the atom shrinks. Evidently, then, there must be a value of r for which the total energy is a minimum.
Visualizing a graph of the total energy given by the equation above as a
function of r, at the minimum point
the slope of is zero,
.
That is,
giving
.
The total energy for this radius is the exact right answer, which is reassuring (but we don’t deserve it, because we have used a 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 trade off between potential energy minimization and kinetic energy minimization. Thus the physical sizes of atoms, molecules and ultimately ourselves are determined by Planck’s constant.