### Video Transcript

In this video, our topic is the
Heisenberg uncertainty principle. This quantum mechanical principle
establishes fundamental limits on how precisely we can know certain pairs of
variables.

To get a sense for this principle,
picture the following. Say that we take a lit bulb outside
at night time and then move the bulb very quickly through the air while taking a
picture of it. From seeing images like this, we
know what will happen. The light bulb will appear to be
smeared out over the distance that it traveled while its image was being taken. Now, the longer this streak is, the
faster we know the bulb was moving. But we’re less sure about the
position of the bulb, where it was located at any one point in time. We could say that the more we know
about this light bulb’s speed as it moved through the air, the less we know about
its position.

This is the basic idea behind
Heisenberg’s uncertainty principle, that given a particular pair of variables, say
energy and time or position and momentum, there’s a fundamental limit to how
precisely we can know both variables in each pair. When we say this limit is
fundamental, we mean it doesn’t depend on getting more precise instruments or making
better measurements. It’s simply a limiting fact of
reality.

Here’s another illustration of the
uncertainty principle. Say that we have a sine wave that
goes on infinitely far in both directions and has a well-defined wavelength. And we know that when a wave has a
well-defined wavelength, that means it also has a well-defined wavenumber, which is
the number of wavelengths that occur per unit distance. We could say then that our
uncertainty in the wavenumber of this wave, we can call that value Δ𝑘, is very
small because when the wavelength of the wave is known precisely, so is its
wavenumber. There’s very little uncertainty in
that value.

But then what if we consider the
uncertainty in the position of this wave? We can call that Δ𝑥. We said that this wave goes on in
both directions infinitely far, so we really can’t assign any particular position to
this wave. We would say the uncertainty in the
wave’s position then is large. These two values, position and
wavenumber, turn out to be what are called complementary variables. This means, as we’ve been saying
about the uncertainty principle, that there’s a basic mathematical limit to how
precisely we can know both of these variables at the same time.

To see how this works, let’s
imagine that we add some other sine waves to this one. Say that we add the second wave,
which we can see has a slightly shorter wavelength than the first one. When these waves interfere, our
result looks something like this. And then say we add yet a third
wave to it, this one here. The wave resulting from this
interference would look like this. As we look at this wave, we can say
that it seems to be gathering its position to be along this line here, while at the
same time the wavelength, and therefore the wavenumber of the wave, seems to be
becoming less distinct. That is, there’s no longer any
single clear value for the number of wavelengths per unit distance along this wave
form.

We could see then that the
uncertainty in our wave’s position, which started out very large, is getting
smaller. But then opposite to this, the
uncertainty in wavenumber, which began as a small value precisely known, is now
getting larger. This helps us see a general reality
that as we try to know an object’s position and wavenumber more and more precisely,
the less uncertainty we have in one of these values, say the position of the wave as
its uncertainty in position gets small, the more uncertainty we have in the other
one. Position and wavenumber then, these
two variables, are complementary variables.

Now, the Heisenberg uncertainty
principle is not something that we can write down as a single equation. That’s because, as we mentioned,
this principle applies to any pair of variables which cannot both be known with
total precision. Nonetheless, there are equations we
can write down for specific variable pairs. When we consider the uncertainty in
an object’s position, along with the uncertainty of its wavenumber, recalling that
all objects have an associated wavelength, called their de Broglie wavelength, then
we can write that the product of these two uncertainties is always greater than or
equal to one-half.

Note that on the left side, the
unit of distance cancels with the unit of inverse distance so that on the right the
value is unitless. This means that the smallest this
product can possibly be, the most precisely we can know these two variables, Δ𝑥 and
Δ𝑘, is to say that one multiplied by the other is equal to one-half, and that’s the
lower limit. The product could always be greater
than that, as our equation shows.

As a quick example, say that we had
some system where the uncertainty in that system’s position was one-quarter of a
meter. This would mean that the
uncertainty in the wavenumber of the system would need to be at least two inverse
meters. That way, when we multiply these
uncertainties together, we get a result of at least one-half, nothing smaller. Another set of complementary
variables are position 𝑥 and momentum 𝑝. And we can recall that the
uncertainty in momentum, Δ𝑝, of a system is equal to ℎ bar, that’s Planck’s
constant, divided by two 𝜋 multiplied by the uncertainty in its wavenumber. This means that if we complete our
uncertainty equation for Δ𝑥 and Δ𝑝, the product of these two values is always
greater than or equal to ℎ bar over two or, written another way, Planck’s constant ℎ
divided by four times 𝜋.

There are even more complementary
variables that follow the uncertainty principle, but our focus for now will be on
these two and working with them to get practical experience. To do that, let’s look now at an
example exercise.

An electron in a particle
accelerator has an uncertainty in its position of 5.11 times 10 to the negative 14th
meters. Using the formula Δ𝑥 times Δ𝑝 is
greater than or equal to ℎ over four 𝜋, calculate the minimum possible uncertainty
in the momentum of the electron. Use a value of 6.63 times 10 to the
negative 34th joule-seconds for the Planck constant. Give your answer to three
significant figures.

In this example then, we have an
electron moving at some high speed in a particle accelerator. At any instant, this electron has
some position and it also has some amount of momentum. And we know its position to within
an uncertainty of 5.11 times 10 to the negative 14th meters. This means if we had some distance
like this, we could say that the electron is somewhere within that distance. In other words, that’s our
uncertainty in its position. And as we saw, that uncertainty is
given to us as this number.

Along with this, the electron has
some uncertainty in its momentum, that is, its mass times its velocity. We want to solve for this
uncertainty. And we’re going to do it using this
formula here: Δ𝑥 times Δ𝑝 is greater than or equal to ℎ over four 𝜋. In this equation, Δ𝑥 is the
uncertainty in our object’s position. Δ𝑝 is the uncertainty in its
momentum. ℎ is Planck’s constant. And this inequality here says that
Δ𝑥 times Δ𝑝 is always greater than or equal to this value here.

Now, our question specifically asks
us to calculate the minimum possible uncertainty in the momentum of the
electron. That minimum in Δ𝑝 occurs when Δ𝑥
times Δ𝑝 is equal to ℎ over four 𝜋. So, we’re going to rewrite this
equation with an equal sign, which means that we’re solving for the minimum possible
uncertainty for Δ𝑥 times Δ𝑝. Since we’re looking to solve for
Δ𝑝, let’s divide both sides of this equation by Δ𝑥, canceling that out on the
left. And so, the minimum uncertainty in
the momentum of the electron is Planck’s constant ℎ divided by four 𝜋 times
Δ𝑥.

When we plug in the given values
for Planck’s constant and Δ𝑥, the answer we calculate, to three significant
figures, is 1.03 times 10 to the negative 21st kilograms meters per second. This is the minimum possible
uncertainty in this electron’s momentum.

Let’s look now at a second example
exercise.

A proton moving through free space
has an uncertainty in its momentum of 4.00 times 10 to the negative 28 kilograms
meters per second. Using the formula Δ𝑥 times Δ𝑝 is
greater than or equal to ℎ over four 𝜋, calculate the minimum possible uncertainty
in the position of the proton. Use a value of 6.63 times 10 to the
negative 34th joule-seconds for the Planck constant. Give your answer to three
significant figures.

In this exercise, we’re told that
we have a proton moving along through free space. This means the proton is moving
with effectively no forces acting on it. Because a proton has mass and
because it’s in motion, it has mass and velocity, and therefore it has momentum. That momentum, though, isn’t known
with complete precision. There’s some uncertainty in it. We’re told a numerical value for
that uncertainty; we can call it Δ𝑝. And using this equation here in our
problem statement, we want to calculate the minimum possible uncertainty in the
position of the proton.

So, here, we know the uncertainty
in the momentum, and we want to calculate the uncertainty in the proton’s
position. We know that Δ𝑥 times Δ𝑝, the
uncertainty in the proton’s position times the uncertainty in its momentum, is
greater than or equal to Planck’s constant divided by four 𝜋. This symbol, greater than or equal
to, means there’s a fundamental limit to how precisely we can know both of these
variables, the position and momentum of the proton, at the same time.

In this exercise, we want to push
this limit as far as we possibly can because we’re calculating the minimum possible
uncertainty in the proton’s position. That minimum occurs when Δ𝑥 times
Δ𝑝 is equal to ℎ over four 𝜋. Remembering that we know Δ𝑝 and we
want to solve for Δ𝑥, we can multiply both sides of our equation by one divided by
Δ𝑝, canceling out Δ𝑝 on the left. And so, now, we have an expression
for the uncertainty in the proton’s position.

To solve for it, all we need to do
is to plug in our given value for ℎ, Planck’s constant, and Δ𝑝, the uncertainty in
the proton’s momentum. With those two values substituted
in, when we calculate this fraction, to three significant figures, we find a result
of 1.32 times 10 to the negative seventh meters. And if we recall the conversion one
nanometer is equal to 10 to the negative ninth meters, then we can write our answer
as 132 nanometers. What we’re saying then is that
there’s some distance in space, we’ve called it Δ𝑥, within which we can say our
proton exists. And the smallest that distance can
be, given our uncertainty in momentum, is 132 nanometers. That’s our minimum position
uncertainty.

Let’s look now at one last example
exercise.

A muon that is produced in a
particle accelerator has an uncertainty in its position of 2.00 times 10 to the
negative 11th meters. Using the formula Δ𝑥 times Δ𝑝 is
greater than or equal to ℎ over four 𝜋, calculate the minimum possible uncertainty
in the momentum of the muon. Use a value of 6.63 times 10 to the
negative 34th joule-seconds for the Planck constant. Give your answer to three
significant figures.

In this exercise, we have a muon, a
subatomic particle that’s like an electron but more massive, moving through a
particle accelerator. There’s some amount of uncertainty
about exactly where this muon is. We say that’s the uncertainty in
its position, and we call it Δ𝑥. Given the value for Δ𝑥, we want to
calculate the minimum possible uncertainty in the muon’s momentum. And we’ll do it using this
equation: Δ𝑥 times Δ𝑝 is greater than or equal to ℎ over four 𝜋.

Mathematically, this equation tells
us that there’s a limit to how precisely we can know both position and the momentum
of our object, in this case, a muon. That limit of precision is achieved
when Δ𝑥 times Δ𝑝 is equal to Planck’s constant divided by four 𝜋. And since in this exercise, we’re
solving for the minimum possible value of Δ𝑝, we’ll use that equality. We can begin solving for Δ𝑝 by
multiplying both sides by one over Δ𝑥, canceling that out on the left. And that gives us this
expression.

We know the value for ℎ. That’s given to us in the problem
statement; that’s Planck’s constant. And we’re also told the uncertainty
in the position of our muon, Δ𝑥. When we substitute in these values
and calculate Δ𝑝, to three significant figures, we find a result of 2.64 times 10
to the negative 24th kilograms meters per second. That’s the minimum uncertainty in
the muon’s momentum.

Let’s summarize now what we’ve
learned about the Heisenberg uncertainty principle. In this lesson, we saw that for
certain pairs of physical quantities, there’s a fundamental limit to how precisely
we can know both of them. This limit we saw is what’s
described by the Heisenberg uncertainty principle. In the case of the pair of
quantities, position and wavenumber, we saw that Δ𝑥, the uncertainty in position,
times Δ𝑘, the uncertainty in a system’s wavenumber, is always greater than or equal
to one-half, while when the pair of variables we’re considering is position and
momentum, the product of their uncertainties is always greater than or equal to
Planck’s constant ℎ divided by four times 𝜋. This is a summary of the Heisenberg
uncertainty principle.