2 — Ehrenfest’s Theorem: Explained

An intuitive bridge between classical and quantum mechanics

Ehrenfest and students at the Institute of Theoretical Physics at Leiden University. (Noord Hollands Archief, Haarlem, Archive T. van Aardenne/P. Ehrenfest, Jr., box 5.)

This article will explore one of the most intuitive theorems of quantum mechanics. The previous article delved into the role of quantum operators and commutators. If you aren’t familiar with already, I’d suggest you read up:

Part One: Ehrenfest’s Theorem

The (generalized) Ehrenfest’s Theorem

Anyway, presupposing you know what quantum operators and commutators are, the last concept we really need to understand this theorem is expectation values. Since we’ll be working with wave functions, here’s a primer on that too:

The Language of Quantum Physics

Having established that the wave function is a mathematical description of everything we know about our quantum system and that its square is directly linked to the probability distribution of our system, we can go diving in the deep end.

One quantity that we can mathematically calculate for a given quantum system is an expectation value. An expectation value is essentially an average. Or rather a weighted average of a probability distribution. Simply put, if the probability distribution for a particle over some region of space looks something like this:

Then the expectation value of the position of the electron is an average of the positions. Which, in this case, would be the largest value on the probability axis, i.e, the purple arrow 3rd from the left. Now, this is a fairly simplified explanation. The distribution is symmetric about the purple arrow. This is rather never the case.

It is also worth noting that the expectation value of our particle is not necessarily the position at which it is most likely to be found. The expectation value is not the modal value. It’s the mean value. Another example that would make this clearer is if we considered our function to look like this:

The r axis is some arbitrary axis I've used to denote position in space.

It’s evident that since the values are largest at 2 and 6, we’re most likely to find our particle at those positions in space. Since the values are small at 0, 4, and 8, we’re least likely to find our particle there. Also, the values at 0, 4, and 8 are zero, so we won’t find our particle there at all; 0² is still 0. What we’re particularly interested in here, is the particle’s expectation value. Well, this probability distribution just happens to be symmetric again. I certainly did not tweak it for mathematical convenience. And so taking the average of all values, its expectation value is right down the middle, at r = 4. But as we’ve already established, there’s actually no chance that we’ll find our electron at this position.

A quick note about expectation values: when we find the expectation value of a particular operator — say, the position operator — we write it like this:

The angled brackets tell us that we’re finding an expectation value. Whatever operator is inside those brackets tells us what we’re finding an expectation value of.

However, this is a shortened notation for convenience’s sake. In quantum mechanics, it’s technically written like this:

But why’s any of this even remotely useful? What do expectation values really tell us? How should it matter?

More often than not, they can be used as a bridge between quantum and classical mechanics. Sometimes, certain expectation values behave like classical quantities. This is particularly useful to understand when exploring what Ehrenfest’s theorem postulates.

With the understanding of quantum operators, commutators, and expectation values, Ehrenfest’s theorem shouldn’t be difficult to understand.

Now, a couple things might look particularly familiar. For starters, we see the operators:

We’ve also got a couple of square brackets that tell us we’re dealing with commutators. And engulfing angled brackets that tell us that we’re exploring expectation values. It’s also worth pointing out here that we’ve got the Hamiltonian operator.

Now the Hamiltonian is the big guy of quantum operators. You’ll see this written everywhere, even in the Schrodinger equation. This operator is the total energy of the quantum system. For a single particle system, it accounts for kinetic and potential energy. The expression for a single-particle system’s Hamiltonian would look something like this:

Where P hat is the momentum operator and V hat represents the potential energy

Ĥ = kinetic energy+ potential energy+ any other energy in the system

However, for larger systems, we need to start considering electrostatic attractions and repulsions. For example, the generalized Hamiltonian for the Helium atom looks pretty daunting

It contains the following terms (in order of appearance): the kinetic energy of the first electron, the kinetic energy of the second electron, the electrostatic attraction between the nucleus and the first electron, the same thing with the second electron, and the electrostatic repulsion between the two electrons.

But besides the Hamiltonian operator, we’ve also got these A hats in place. Although unconventional, A is just a generic placeholder for any quantum operator. In other words, A in itself is not an operator but merely represents that an operator can take that spot. The idea is that we can take another operator, say the position operator, and substitute it into Ehrenfest’s theorem.

Or we could do this with the momentum operator. Even the Hamiltonian. So  by itself doesn’t mean anything. Now one more thing that we need to take a deeper look at is the d/dt bit. The things the theorem starts and ends with.

For the uninitiated: d/dt basically tells us that we’re looking at the rate of change of something. To draw a parallel to the classical word, when we measure the speed of a ball, we’re measuring the rate of change of its position with respect to time. The only difference in the quantum mechanical world is that we’re now measuring the rate of change of an operator’s expectation value with respect to time.

Now you can measure the rate of change of something with respect to a lot of variables. There is a world of applied situations. Here, however, the denominator holds the “dt” part. This tells us that it’s with respect to time. t, of course, standing for time. Put this together and we understand that we’re measuring the change of a quantum operator’s expectation value with respect to time. This may sound awfully confusing but it’ll all make sense in a minute. But before we delve into that, the right-hand side:

One of the things we’ve got (highlighted in red) is the expectation value of the rate of change of an arbitrary operator, A with respect to time, t. Now, these are not the same ds we explored earlier but for the purposes of this article, we won’t delve too deep into that. But if you do want to know, look up partial derivatives. Now in the case of the position and momentum operators, neither of them change with time. If you recall, an operator — for classical purposes — is equivalent to making a measurement. The operator itself does not change with time but the end result of what we apply the operator to may change.

What I mean by that is we can get different results if we make measurements and different points in time. But the process of measuring itself does not change with time. And so the operator itself does not change with time. This means that the rate of change of our operator is 0. And the expectation value of 0 is 0.

Now coming to the other terms:

i, here, is the imaginary unit which is equal to the square root of negative 1. hbar, on the other hand, is the modified Planck’s constant. For our purposes, we don’t necessarily need to worry about what these mean but instead just that they are constants to our equation and we treat them as such. But what we’re more interested in, is the rest of the term. It reads as the expectation value of the commutator between A and H. This is what it “reads” as. But we’re particularly interested in what it means. To do that, we’ll substitute A with any operator — say the position operator, and follow through with what that means.

We won’t delve too deep into the mathematics behind this, but we start by substituting the position operator into Ehrenfest’s Theorem.

As we’ve already established, the last term is equal to zero. Since the change of a measurement with respect to time remains zero, we can write this:

The expectation value of the rate of change of the position operator with respect to time is zero

Then we can follow this through with the commutator. There’s some jargon-y math involved but we don’t need to worry about that yet. For the purposes of this article, we’re not interested in how we solve it, we’re interested in whether we solve it or not. The answer is of relevance, the method? Not so much. But for those of you who are interested here’s how that works out:

As I said, the important thing is what we get after all the mathematics. When the dust settles, what we find is:

In other words, this means that the rate of change of the expectation value of the position of our particle, multiplied by its mass is equal to the expectation value of its momentum.

Heading back to classical physics for a second. If you recall that the rate of change of position with respect to time is the same thing as velocity. And when we multiply the velocity by mass, we get momentum.

This is where Ehrenfest’s Theorem links classical physics with quantum mechanics. Here’s the difference:

The equations look similar but with quantum mechanical systems, we deal with the expectation value and not the actual measurement itself. At this part, I’d like to say that although quantum mechanics is counterintuitive in many ways, Ehrenfest’s Theorem successfully draws a bridge between the two.

Now I would stop here but for a more intuitive understanding, we could substitute the momentum operator and see where that gets us.

Which, to a great degree, looks behaves like Newton’s second law. But we won’t go into that too much here because it’s a very subtle parallel. The key takeaway here is that although quantum mechanics employs probabilities if we study expectation values, they will behave like their classical counterparts.

With all that being said, I’m going to end the article here. If there’s something I haven’t explained entirely or if you find incomplete, please let me know in the comments. Thank you so much for reading.