Transcript/Ch 10: What's the commutator and the uncertainty principle?

Hi, everyone. Continuing our study of operators, in this video, I want to introduce arguably the most important mathematical tool when studying observables the commutator. I also want to show you how studying commutation tells us why uncertainty relations necessarily exist in quantum mechanics, and we'll build some intuition into where it comes from.

So with that, let's go ahead and dive right in. To begin, let's define a few things. If we have two operators, A and B, we say they commute.

If the action of operator AB is equal to the action of operator BA, in other words, you can flip the two operators without any issues. In general, operators do not commute. So you can get a totally different transformation when you switch the order of the operators.

However, there are times in quantum calculations that we would really like to switch their order. For example, say we are looking at the product of an energy and momentum operator and we wish to act this on an energy Eigenvector. If we could just flip their order, then we could use the eigenvector property and hopefully reduce this expression to a simpler one.

This motivates the definition of the commutator. Given two operators A and B, their commutator is designated with the following bracket and equal to the following difference. Simple, right? So if two operators commute, then the two terms on the right hand side are equal, and we get that their commutator is zero.

Likewise, if the commutator of two operators is zero, then it implies they commute. So how does this solve our problem? Well, in general, the commutator of two operators is equal to some third operator C. So moving one of the terms to the right hand side, we see that we can flip the product as long as we also add the commutator C.

Oftentimes, this operator C is simple. For example, it might be proportional to the identity, so it's worth flipping A and B in that case. Now, this is great in all, but so far this is just a mathematical sleight of hand.

I claimed this was the most important tool in studying observables. So what's the big deal? To begin seeing why this is so important in quantum physics, let's take some time to derive an amazing property of commuting observables and their Eigen bases. Let's say A and B are two physical observables, which means they both have orthonormal Eigen bases, and they both happen to commute.

We want to analyze how an Eigenvector of A relates to B. So let's take alpha to be an Eigenvector of A. To see how alpha relates to observable B, we'll analyze two transformations.

On the left, we'll first analyze how BA acts on alpha, where alpha is shown on the plane. First, since alpha is an Eigenvector of A, a simply scales it by its Eigenvalue lambda. We then act with B, which takes it someplace in our vector space.

Now, on the right, we'll analyze how AB acts on alpha acting b. First, it takes alpha to the same direction that it took lambda alpha. Since b is a linear operator, now we want to know how A acts on this vector.

Remember that the operators commute. So these two transformations must take alpha to the exact same place. This means we have just one choice for what A does.

It has to scale the vector by lambda in order for it to match the left hand side. Since A only scales b alpha, b alpha must be an eigenvector of A with the same eigenvalue lambda. Now, we have two cases here nondegenerate and degenerate eigenvalues.

If we have a nondegenerate eigenvalue, then we only have one eigenvector per eigenvalue. In that case, b alpha has to be the same eigenvector as alpha, since they both have eigenvalue lambda. So b alpha can only be a scaled alpha.

Therefore, alpha is an eigenvector of both A and b. So this is our first fact for commuting observables, nondegenerate eigenvectors of A must also be eigenvectors of b. We now want to show that we can make the same conclusion in the degenerate case, which means we now have more than one eigenvector for an eigenvalue to do.

So, let's work in R three and assume that A has a twofold degeneracy, meaning there are two eigenvectors with the same eigenvalue. These eigenvectors form a degenerate eigenspace, in this case, a 2D plane. When A acts on any vector in this plane, it simply scales it by lambda.

Now, what does b do to vectors in this eigenspace? We actually already figured this out, so let's review what we know. Let's take alpha to be some vector in this eigenspace. Then let's act a b on this alpha.

We previously showed that since A and b commute, this has to equal lambda b alpha, or in other words, b alpha is an eigenvector of A with eigenvalue lambda. Thus, b alpha must lie somewhere within this lambda eigenspace and therefore must still lie on this plane. So we found an important fact vectors inside the eigenspace stay inside when acted upon with b.

What about a vector orthogonal to the eigenspace? Let p be some vector orthogonal to the lambda eigenspace. We want to know what happens when b acts upon p. Well, let's take the inner product with alpha, which is an arbitrary vector in the eigenspace.

B is an observable, which we proved means that b is her mission. Therefore, we can move it to the other side of the inner product. We just showed that b keeps everything inside the eigenspace.

So b alpha is still within this plane and therefore orthogonal to p. Hence, the inner product between b alpha and p is zero. Since alpha is an arbitrary vector in the eigenspace, this now implies that after acting b on p, it must remain orthogonal to the entire eigen space.

Therefore, we have another fact vectors orthogonal to the eigenspace stay orthogonal when acted upon with b. Now, how do these two facts help us? Well, let's take some general eigenvector of b called beta, which without loss of generality, is neither in the eigenspace nor perpendicular to it. We can decompose beta into a unique sum of a component along the eigenspace and a component orthogonal to it, which we label as beta parallel and beta perpendicular respectively.

Since beta is an eigenvector of b, acting b on it will simply scale it by an eigenvalue mu. Now, b is a linear operator. So we can also act b on beta by acting b on each of the components individually and then take their sum.

We just showed that the component inside the eigenspace must stay inside and the component orthogonal to it must stay orthogonal when acted upon with b. Therefore, when we act b on beta and scale it by mu, the only thing b can do to each component is also scale them by mu. Neither of them can move any other way.

Do you see that? If b didn't have these properties, then b could move the two components elsewhere. But we proved that they must either stay in the eigen space or stay orthogonal to it. But this means that the eigenvector beta is actually composed of two independent eigenvectors one inside the eigen space and one orthogonal to it.

Since b also scaled them by mu. This is a nuanced argument, so feel free to take some time to digest it. So we have our final very important fact.

We can take the eigenvectors of b to lie either inside the eigenspace of A or orthogonal to it. There is no in between. So why is this final fact useful? Well, this means that in order for the eigenbases of b to span the eigenspace of A, the b eigenvectors that span this space must lie within the eigenspace.

Since the only other choice is to be orthogonal to this space and then it be impossible to span it. Since these b eigenvectors lie inside the eigen space of A, they must also be eigenvectors of A. Hence, we have found eigenvectors in the degenerate eigen space that are eigenvectors of both A and b.

So A and b share eigenvectors even in the degenerate case. I remember finding this argument confusing when I first learned it. So let's do a quick recap of how we made this conclusion.

Let's say A has some degenerate eigenspace. We proved that vectors in the eigenspace stay inside when acted upon with b, and vectors orthogonal to the eigenspace stay orthogonal. We then use this to show that the eigenvectors of b must lie either inside the eigenspace or perpendicular to it.

Therefore, in order for the eigenvectors of b to span the eigenspace of A, b must have spanning eigenvectors in this same eigenspace. Since everything in this eigen space is an eigenvector of A, we have found eigenvectors in this eigenspace that are eigenvectors of both A and B. Although we looked at a twofold degeneracy, this argument easily scales to any fold degeneracy.

So what does this imply about commuting observables? Well, let's look at both operators grab any eigenvector from a's eigen basis. If that eigenvector is nondegenerate, then we showed that this must also be an eigenvector of B. Let's add this to a set of shared eigenvectors.

If A has degenerate eigenvectors, then we showed that there exist eigenvectors of B that also span this eigen space and hence our eigenvectors of both. So we can add them to the shared set and we can continue this for all eigenvectors. And in the end we are left with something incredible a set of eigenvectors that forms an eigen basis of both A and B.

A simultaneous eigen basis. This is the truly incredible fact about commuting observables. If two observables A and B commute, then they share a simultaneous eigen basis.

Why is this important in physics? Remember that when we make a measurement, for example, of momentum, our quantum state will collapse into an eigen state and we measure the corresponding eigenvalue. Now, if, say, momentum and energy commuted, then after making the momentum measurement and collapsing into its eigen state, we can also make a measurement of energy and our quantum state would collapse into a definite eigenvector for both momentum and energy since they share an eigen basis. In other words, our particle can be in a definite state of both at the same time and we can simultaneously know the value for momentum and energy.

What about observables that don't commute? Well, in that case you can prove that they cannot have a simultaneous eigen basis. I encourage you to try this proof out. As a hint, assume they did have a simultaneous eigen basis, which we'll label with C and write both observables in the form we derived last episode.

Now show that AB equals BA and hence you have a contradiction. So if two operators don't commute, then there exist eigenstates of one observable that are always in a superposition of the other's eigenstates. So for example, if energy and momentum don't commute, then when you measure the momentum and collapse into a momentum eigenstate, you can still be in an energy superposition and be uncertain about the energy until you make a dedicated energy measurement.

But when you measure energy and collapse into an energy eigenstate, you can now be in a momentum superposition and be uncertain about the momentum. You may not be able to measure both at the same time. Collapsing into the eigenstate of one can put you into a superposition of the other.

You might be starting to notice the fundamental connection between the commutator and a sort of uncertainty principle. And just to get to the punchline, although we haven't derived the exact forms of the position and momentum operators, you might have guessed that their commutator is nonzero. It actually equals a constant times the identity.

This means that they do not share an eigen basis. In fact, they don't share any eigenvectors. So whenever you are in a position definite state, you are always in a momentum superposition.

And whenever you're in a momentum definite state, you're always in a position superposition. Because of their commutator, you can never have a quantum state that has a definite value for both position and momentum at the same time. They simply do not share an eigen basis.

I really cannot overstate how significant this commutator is. This one equation is responsible for so much of quantum physics's. Weirdness, some authors even build the framework of quantum mechanics based off this one commutation relation.

And as you hopefully see, this commutator is the root of the ########## uncertainty principle. This principle says that the standard deviation of possible positions times the standard deviation of the possible momentum must always be greater than zero. There does not exist a quantum state with a definite value of both position and momentum.

Therefore, we can never have a certain measurement of both at the same time. If we are certain about one, we are uncertain about the other and vice versa. This inequality seems to mystify many people, and it's one of the quirks of a quantum theory.

But hopefully you see that this is a natural outcome of using linear algebra to describe our world. If we want to use operators to describe the universe, some of them won't commute. And therefore sometimes we can't be in an eigenstate of both at the same time, and hence we have a limit on how certain we can be about both.

Now, the ########## uncertainty principle is actually derived from what some people call the generalized uncertainty principle. I won't derive this here since there are many online sources and textbooks that show the full derivation. But notice the focus of this relation is the commutator, which now hopefully makes intuitive sense since we show that non commutation is the root of uncertainty in quantum mechanics.

Although this is sometimes called the generalized uncertainty principle, I want to let you know that there actually exists more than one uncertainty relation. For example, here is the stronger Macone Paddy uncertainty relation for the sum of variances. Although it looks more complicated, note that the commutator is still a central object here.

It shows up in the first term on the left, and if you expand the second term, you'll get some modified commutation expressions between A and B. So hopefully you better understand why the commutator is so important in our quantum theory. Not only is it a mathematical tool, but it tells us whether or not observables have simultaneous eigenbases and therefore can be simultaneously measured.

It may seem strange, but you now understand that because AB doesn't always equal BA in linear algebra, we cannot measure position and momentum at the same time. That, to me, is absolutely amazing. In the next episode, we're going to start our journey towards deriving the Schrodinger equation and our first stop will be discussing unitary operators in quantum mechanics and what they intuitively are with that.

Thank you so much for watching, as always. Let me know if you have any questions. Hope to see you in the rest of the series.