Lesson Video: The Heisenberg Uncertainty Principle | Nagwa Lesson Video: The Heisenberg Uncertainty Principle | Nagwa

Lesson Video: The Heisenberg Uncertainty Principle Physics

In this video, we will learn how to calculate the minimum uncertainty in the momentum of a particle given the minimum uncertainty in its position, and vice versa.

13:33

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.

Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

  • Interactive Sessions
  • Chat & Messaging
  • Realistic Exam Questions

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy