Exploring Geometric Algebra via Quantum Mechanics: Introduction

Transcript

In this video, I'd like us to explore some of the abstract topics concerning vector spaces, which are often swept under the rug in any quantum mechanics course. In doing so, I will introduce to you a somewhat underrated and undervalued branch of mathematics called geometric algebra. Throughout this introduction, we will inadvertently encounter topics such as duality, the levy chivvita tensor, the cross productduct, and so on. And I will use these topics to highlight the elegance and simplicity of geometric algebra. My goal with this video is not to fully explain these topics, but rather highlight their connections.

And perhaps in the future, I can take more time explaining each topic individually. To start, I think everyone is already familiar with scalers and vectors. In geometric algebra, we can construct higher dimensional or higher grade objects, for example, by vectors. And to do that we use the outer product or often called the wedge product. The bi vector ued v is simply the oriented area given by the parallelogram formed by u andv.

The orientation will depend on which vector you start with. So for u wedge v you curl your fingers from u towards v. And you can easily see that this operation is anti-ymmetric. So u wedge v is equal to minus v wedge u and we can extend this idea to even higher dimensions. So say we are working in 3D space with basis vectors e1 e2 and e3 then we can construct the unit cubic volume by wedging all three together.

And I'm going to omit the orientation for now for simplicity because we don't really need it for this video. Now when we wedge all the basis vectors together like this, we call the resulting unit volume the pseudo scalar and it will come up time and time again in our discussion. And you can imagine extending this notion even further. So constructing what is called multiv vectors or hypervols using the outer product. And now whenever you see the outer product this is geometrically what it means.

Now at the core of geometric algebra is a multiplication operation called the geometric product. And it might look intimidating at first. you know adding scalers using the inner product to by vectors basically two different grade objects but this is no different from complex numbers which we don't often question say we have Z1 and Z2 we simply add them by adding the real and imaginary components respectively often times though this geometric product simplifies for example if we're working in 2D and we take the geometric product of the orthonormal basis vectors E1 and E2 then we know that their inner product vanishes. In this case the geometric product is simply the outer product of the two. If however we bring E2 towards E1, the area between them vanishes and the geometric product E1 E1 reduces to just the inner product which we know is one because the vectors are normalized.

Now this generalizes to any orthonormal bis vectors I and J where I is not equal to J. And this simplification is going to be our bread and butter in geometric algebra. And I'm going to be employing it quite frequently. Now, in order to see geometric algebra in action, I'd like to anchor us to something familiar. The crossroduct for vectors A and B, the product A cross B is a vector that lies in the plane perpendicular to both.

Now perhaps not so familiar is this definition of the cross productduct where I employ the levichevita tensor epsilonn j k. I'm also using summation notation where you would expand a and b in terms of our basis set. Now if you replace a and b with any two basis vectors you get this. Clearly, this leviga tensor is quite involved. And while it will often make an appearance, my goal here is not to explain exactly what that object represents, but rather highlight how it connects to everything else.

Here, I'll be providing you with the necessary ingredients, so to speak. But a full treatment of Levitita deserves its own video. Usually you'd see the levita defined in terms of cyclic permutations of 1 2 3. So this might be familiar to you. For example, if you take epsilon 1 2 3 that's just one.

But if you switch the three around, you pick up a minus sign. And if you switch it around again, you pick up another minus sign. And so that's just one. And if you repeat any two indices, you get zero. Now, I'd like you to see how we define this tensor in geometric algebra.

What I'm going to do is explain the terms in this equation and we're going to check if it works. So here we're taking the outer product EI with EJ with E K and then we're multiplying by the reverse of the pseudo skater given by the dagger symbol here. So instead of E1 wedge E2 wedge E3, we have E3 E2 E1. And the outer product is simply the geometric product because all of these are orthogonal. Now let's see if this works for the simplest case where we have two repeating indices.

So say epsilon 112 and as we discussed before there is no area between E1 and itself. So the wedge product vanishes and epsilon 112 goes to zero. For the case of cyclic permutations say epsilon 1 2 3 first we write out the reverse pseudo scaler then all these wedge products become geometric products. Now you see the E3 E3 in the middle. Well that just becomes one since it's just the inner product between the two.

And you can see the pattern for the E2 E2 and then the E1 E1. So epsilon 1 2 3 is just one. Now for the anti-yclic permutations say epsilon 132. Again we expand the pseudo scaler and drop all these wedges. But uh now I must reorder this e2 e3 in the middle for things to cancel out.

And if I do, I pick up a minus sign because the wedge product of these two vectors is anti-ymmetric. And now we can cancel out the same as before. And here we end up with a minus one. So you see we retrieve the old definition using the geometric algebra definition. Now that we have a handle on the algebra, we are almost in a position to revisit the cross product.

But first, I'd like to make a stop at the outer product to showcase the simplicity of geometric algebra. We can first rewrite A and B in terms of the basis vectors and then simply expand. From here, all we need to do is distribute. Now, any basis vector wedged with itself goes to zero. And that takes care of all these diagonal terms.

And I'd like to collect terms with opposite wedge products. So, I'll shuffle things around. Now let's bring these scalers up to the front and flip these outer products at the cost of a minus sign. Bringing all these scalers together, we get this. These coefficients look familiar.

Focusing on the E1 wedge E2 term for just a moment, that is just the geometric product of the two. Now I can multiply by E3 E3 because that's just one and I can shuffle twice picking up two minus signs. Now I multiply by epsilon 23 because that's just one. And remembering the formula we had for the cross productduct the epsilon 123 E3 is just E1 cross E2. And this E1 E2 E3 is just our sudo scalar I.

I'm going to use this relation to rewrite all the wedge products up there. Now we know how to take the cross productduct between any of our basis vectors. So we can write those vectors instead. And notice all those coefficients along with this minus sign over here. This is just the cross productduct with a factor of i to the side.

So going back to the cross productduct, we know that a cross b lies in the plane perpendicular to both a and b. But uh why not this vector? Well, it can't be because for a wedge b we need to start curling our hands from a towards b. So our thumb points up. Now as for the magnitude well that's given by the area of the parallelogram a wedge B which by the way also happens to be AB sin theta and that's what we learned early on. So it's really the outer product that is fundamental and you can see that because the crossroduct exists only in three dimensions.

One thing we haven't talked about though is this pseudo scalar. It seems to relate the plane a wedge B to an orthogonal vector. Now you might have many questions but let's tackle just a simple question. What is the action of this pseudoscaler like say in another dimension say in 2D? So let's consider E1. What is perpendicular to E1? Well, we know it's E2.

Now, I'm going to multiply by E1 E1, which is just one over here. Then I'm going to switch getting a minus sign. Then switch again. But this term is just the pseudoscaler in two dimensions. H what about E2? Well, let's copy this.

then multiply by the pseudoscaler on both sides. Now here I have I squared. This I^2 is just E1 E2 multiplied by itself. So again we do a switch here and pick up a minus sign. The E1 E1 both go away and so do the E2 E2 and whoa we have I^2 is minus one.

Now, that should look familiar, but let's sweep it under the rug for just a second and move on. So, up there, we're going to have a minus sign. Now, let's expand the pseudo scaler and do a switch here on the left, picking up a minus sign. Now, this term is the reverse of the pseudo scaler. The minus sign goes away on both sides and we get this very interesting this sudo scaler.

But what about in higher dimensions? To illustrate why this actually has to generalize, let's attempt to construct an orthogonal vector ourselves. So say we want a basis vector EJ that is orthogonal to all the other elements in our algebra. In in other words this the pseudoscaler of our algebra can now have n terms for n dimensions. And what we can do is take the outer product of all the elements in our algebra. Then multiply by the reverse of the pseudo scalar to not let this all reduce to plus or minus one.

We exclude the term ej from our product. Now this looks awfully reminiscent of the levichita definition we had before except we're missing a term. So to compensate for that, we need a sign factor. And here its job is to preserve the handedness of our set. But let's ignore that for now.

Let's just see if this works. So let's go to three dimensions as an example and look for E2. The sign term here becomes minus1. From our formula, we must wedge E1 with E3 and then multiply with the reversed pseudoscaler with a sign factor. We expect to get something orthogonal to both E1 and E3.

Either this vector or that vector. Expanding the pseudoscaler and dropping the wedge, we get this. the E3 E3 go away and we do a switch here and the E1 E1 go away and so we get E2. So you might think this is redundant, you know, we we're looking for a vector that is like E2 in the sense that it's orthogonal to E1 and E3. We say a dual to E2, but we ended up right back at E2.

So you might think that a vector and its dual are in general related by this equation. But that's not always the case. To illustrate this, let's consider a different algebra. The fourdimensional space-time geometric algebra given by the familiar manowski metric. Here I'll be using the mostly minus version for this algebra.

We have one time like and three space-like basis vectors that form an orthonormal set. This orthogonality can be written as this. Now here the inner product is defined according to the metric. So for our signature taking the space-like gamma 2 inner product with itself we get minus one. This is important because we previously defined duality using this relation.

So the inner product of gamma 2 with its dual has to be one. But in order for both of these to be true, we can't simply say that gamma 2 is equal to its dual. The only way to satisfy both relations is if the dual of gamma 2 is equal to negative gamma 2. But this is not the case for all basis vectors. For the time like gamma 0, the inner product is one.

And you can use a similar line of reasoning to see that gamma 0 is equal to its dual. So it's really the metric that defines the inner product and in turn the relations between vectors and their duels. In the language of tensors, we say that the metric tensor is used to raise and lower the indices. Now I'd like to explore another layer of complexity for the space-time geometric algebra. Given our basis vectors, we can only form grade two elements by vectors using either of these combinations.

Here I denotes that I'm using space-like vectors with indices 1 2 or three. And I'd like to focus on this set of by vectors sigma i. From this set, we can form scalers by squaring sigma i. That is just gamma i gamma 0 multiplied by itself. If I do a switch, I pick up a minus sign.

But the space log gamma i squared to minus1. So we just end up with one. The pseudo scalar for this algebra is then sigma 1 sigma 2 sigma 3 and the lvich vita accordingly is written as sigma i wedge sigma j wedge sigma k multiply by the reverse of the pseudoscaler these wedges again go away and I'm going to multiply by the pseudoscaler on the right this i dagger I when expanded we can see that using the property above the sigma 1 sigma 1 go away and so do the sigma 2's and the sigma 3es so I daggerize just one using this result up there this goes away and I'm going to multiply by sigma k on the right the sigma k sigma k go away and we get this because of the anti-ymmetry of the levich vita the sigma i sigma j is negative sigma j sigma i. So I'm going to add two sigma i sigma j's together and then flip the order of these picking up a minus sign. But this that's just the commutator.

Substituting a result from the right side, we get this. And I would like to pause a little bit here and ponder. I have not made any mention of physics or quantum mechanics whatsoever. Yet, these are precisely the commutator relations defining the poly algebra. But you might ask, how did these get in here? But I'd like to postpone this question for now and instead take a look at what else we can do with these sigma by vectors by first making a correspondence with quantum mechanics.

This equation is famous in quantum mechanics. It's really reminiscent of Oilers's formula but using the sigma Z operator and you can find proofs in many many textbooks. Now I'd like to convince you that the geometric algebra analog is also true and to prove that we could go through the math but instead I'd like to remind you that our sigas square to one aka they are unitary and also there's something that I swept under the rug our sudo scaler it squares to minus1 acting exactly like the imaginary unit and And that's all we need to say that if this is true, then this too must be true. Now, you've probably also seen this come into play in a sandwich product in quantum mechanics. And all of this jargon simplifies to just a cosine theta and a sin theta term.

And again using similar argument, the geometric algebra equivalent also holds. Now in geometric algebra we call this a rotor because as you can see it takes sigma 1 and moves it along the unit circle in the sigma 1 sigma 2 plane. In fact this i sigma 3 term can be expanded and the sigma 3 sigma 3 go away leaving us with sigma 1 sigma 2 the plane of rotation. We can see this in action by substituting for theta. For example, when theta is p /4 or p /2.

In fact, the sandwich product enables us to move anywhere along the unit sphere. The math behind all this is something I'd like to elaborate on in a future video. But for now, let's focus on the visuals. Perhaps this sphere in quantum mechanics is familiar to you as the block sphere or in the context of quantum optics maybe the point sphere. describing polarization.

But this sphere should really be attributed to Reman who introduced complex geometry to the world. In quantum mechanics, a vector inscribed in the sphere is used to represent a two-level system, the cubit with two complex numbers V and W. But because these are complex numbers, shouldn't we be using two spheres? The key lies in global phase invariance. We say that scaling our state with a complex number lambda leaves it unchanged. So we can scale our state by 1 / v to get this where u is w over v.

But then we have a problem when this v in the denominator goes to zero. And if you've ever looked up quantum mechanics literature, we say it's all fine and good if we accept that when v goes to zero, u goes to infinity. But what does that mean? And what does that say about the vector space inhabited by our quantum state? But before we dive into these abstract spaces, let me just illustrate that this idea is not so bizarre. Take the function x^2 + 1 for example and imagine standing at the origin looking at it straight ahead. From your perspective it would seem that very far away the parabola is looking more like an ellipse.

As y goes to this point at infinity x goes to zero. This more so resembles our world because of how we see. But what about our quantum state? What vector space does it inhabit? To explain, I need to introduce you to the real projective line. Imagine an artist sitting at the origin with his easel sitting at y equals 1. To project objects in space onto his painting, he simply imagines lines connecting these objects to himself, then finds where these lines intersect his painting.

Now instead of the artist imagine a special camera in addition to projecting objects beyond the screen, this camera can also project objects between itself and the screen and it can also look behind and project objects from behind onto the screen. Mathematically to project any point we simply need to scale both x and y by some factor say lambda for the point at 42 this is simply 1/2 for the point at -4 minus 2 this would be minus a half and while this factor changes notice that the ratio between x and y does not for any point sitting on the same line. That's because these lines we drew pass through the origin having zero y intercept. So only the slope m is needed and scaling both y and x by some lambda does not change the slope. This is analogous to our situation with the quantum state, but instead of the complex number u, we have m.

Now, I'm going to raise the screen to illustrate another point. And I'm also going to have this blue dot here to track the projection of a point on a line. As we sweep across, the projection point moves to the left. But notice what happens when we get too far. As we approach the horizontal axis, the projection point goes very, very far away.

But as we exceed it, this projection point returns from said far far away. And even though these lines don't intersect at y equals z, the projection point wobbles back and forth from this point of no intersection, which corresponds with the point at infinity. And now this point at infinity is analogous to our situation with the complex numbers describing the cubit. Now one last thing I'd like to show here is take a look at this. We do a quarter rotation followed by another quarter rotation and that's one sweep up there.

Another half rotation and we get another sweep up there. Let's pause for a second. If you're familiar with quantum mechanics, this might raise suspicion for you. You know, a full rotation corresponding to only half a rotation. One last thing, U is a complex ratio defining our state.

So even though we have a ratio, it's a real ratio. To solve this last issue, we can extend the real projective line by going up a dimension to the real projective plane. Now instead of this circle we have many of these circles each corresponding to a screen and a parallel line through the origin. Collectively we have a sphere and two planes. Now in this space again only the ratio matters.

So we can scale our points arbitrarily say with 1 / y and then only the ratios x over y and z over y are needed to specify planes in our projective space. Our complex ratio u would correspond with two real ratios. So there is a correspondence between the complex projective line and the real projective plane. And the correspondence is made via a sphere. The reman sphere.