Tag Archives: linear algebra

The Representation of group G on vector space V is really a left action of the group on the vector space

Say what? What does this have to do with quantum mechanics? Quite a bit. Practically everything in fact. Most chemists learn quantum mechanics because they want to see where atomic orbitals come from. So they stagger through the solution of the Schrodinger equation where the quantum numbers appear as solution of recursion equations for power series solutions of the Schrodinger equation.

Forget the Schrodinger equation (for now), quantum mechanics is really written in the language of linear algebra. Feynman warned us not to consider ‘how it can be like that’, but at least you can understand the ‘that’ — e.g. linear algebra. In fact, the instructor in a graduate course in abstract algebra I audited opened the linear algebra section with the remark that the only functions mathematicians really understand are the linear ones.

The definitions used (vector space, inner product, matrix multiplication, Hermitian operator) are obscure and strange. You can memorize them and mumble them as incantations when needed, or you can understand why they are the way they are and where they come from. So if you are a bit rusty on your linear algebra I’ve written a series of 9 posts on the subject — here’s a link to the first https://luysii.wordpress.com/2010/01/04/linear-algebra-survival-guide-for-quantum-mechanics-i/– just follow the links after that.

Just to whet your appetite, all of quantum mechanics consists of manipulation of a particular vector space called Hilbert space. Yes all of it.

Representations are a combination of abstract algebra and linear algebra, and are crucial in elementary particle physics. In fact elementary particles are representations of abstract symmetry groups.

So in what follows, I’ll assume you know what vector spaces, linear transformations of them, their matrix representation. I’m not going to explain what a group is, but it isn’t terribly complicated. So if you don’t know about them quit. The Wiki article is too detailed for what you need to know.

The title of the post really threw me, and understanding requires significant unpacking of the definitions, but you need to know this if you want to proceed further in physics.

So we’ll start with a Group G, its operation * and its identity element.

Next we have a set called X — just that a bunch of elements (called x, y, . . .), with no further structure imposed — you can’t add elements, you can’t mutiply them by real numbers. If you could with a few more details you’d have a vector space (see the survival guide)

Definition of Left Action (LA) of G on set X

LA : G x X –> X

LA : ( g, x ) |–> (g . x)

Such that the following two properties hold

l. For all x in X LA : (e, x) |–> (e.x) = x

2. For all g1 and g2 in G LA ( g1 * g2), x ) |–> ( g1 . (g2 . x )

Given vector space V define GL(V) the set of invertible linear transformations of vector space. GL(V) becomes a group if you let composition of linear transformations become its operation (it’s all in the survival guide.

Now for the definition of representation of Group G on vector space V

It is a function

rho: G –> GL(V)

rho g |–> LTg : V –> V linear

The representation rho defines a left group action on V

LA : (g, v) |–> LTg (V) — this satisfies the two properties above of a left action given above — think about it.

Now you’re ready for some serious study of quantum mechanics. When you read that the representation is acting on some vector space, you’ll know what they are talking about.

The pleasures of enough time

One of the joys of retirement is the ability to take the time to fully understand the math behind statistical mechanics and thermodynamics (on which large parts of chemistry are based — cellular biophysics as well). I’m going through some biophysics this year reading “Physical Biology of the Cell” 2nd Edition and “Molecular Driving Forces” 2nd Edition. Back in the day, what with other courses, research, career plans and hormones to contend with, there just wasn’t enough time.

To really understand the derivation of the Boltzmann equation, you must understand Lagrange multipliers, which requires an understanding of the gradient and where it comes from. To understand the partition function you must understand change of variables in an integral, and to understand that you must understand why the determinant of the Jacobian matrix of a set of independent vectors is the volume multiplier you need.

These were all math tools whose use was fairly simple and which didn’t require any understanding of where they came from. What a great preparation for a career in medicine, where we understood very little of why we did the things we did, not because of lack of time but because the deep understanding of the systems we were mucking about with simply didn’t (and doesn’t) exist. It was intellectually unsatisfying, but you couldn’t argue with the importance of what we were doing. Things are better now with the accretion of knowledge, but if we really understood things perfectly we’d have effective treatments for cancer and Alzheimer’s. We don’t.

But in the pure world of math, whether a human creation or existing outside of us all, this need not be accepted.

I’m not going to put page after page of derivation of the topics mentioned in the second paragraph, but mention a few things to know which might help you when you’re trying learn about them, and point you to books (with page numbers) that I’ve found helpful.

Let’s start with the gradient. If you remember it at all, you know that it’s a way of taking a continuous real valued function of several variables and making a vector of it. The vector has the miraculous property of pointing in the direction of greatest change in the function. How did this happen?

The most helpful derivation I’ve found was in Thomas’ textbook of calculus (9th Edition pp. 957–> ). Yes Thomas — the same book I used as a freshman 6o years ago ! Like most living things that have aged, it’s become fat. Thomas is now up to the 13th edition.

The simplest example of a continuous real valued function is a topographic map. Thomas starts with the directional derivative — which is how the function height(north, east) changes in the direction of a vector whose absolute value is 1. That’s the definition — to get something you can actually calculate, you need to know the chain rule, and how to put a path on the topo map. The derivative of the real valued function in the direction of a unit vector turns out to be the dot product of the gradient vector and any vector at that point whose absolute value is 1. The unit vector can point any direction but the value of the derivative (the dot product) will be greatest when the unit vector points in the direction of the gradient vector. That’s where the magic comes from. If you’re slightly shaky on linear algebra, vectors and dot products — here’s a (hopefully explanatory) link to some basic linear algebra — https://luysii.wordpress.com/2010/01/04/linear-algebra-survival-guide-for-quantum-mechanics-i/. This is the first in a series — just follow the links.

The discussion of Lagrange multipliers (which is essentially the relation between two gradients — one of a function, the other of a constraint in Dill pp.68 -> 72 is only fair, and I did a lot more work to understand it (which can’t be reproduced here).

For an excellent discussion of wedge product and why the volume multiplier in an integral must be the determinant of the Jacobian — see Callahan Advanced Calculus p. 41 and exercise 2.15 p. 61, the latter being the most important. It explains why things work this way in 2 dimensions. The exercise takes you through the derivation step by step asking you to fill in some fairly easy dots. Even better is  exercise 2.34 on p. 67 which proves the same thing for any collection of n independent vectors in R^n.

The Jacobian is just the determinant of a square matrix, something familiar from linear algebra. The numbers are just the coefficients of the vectors at a given point. But in integrals we’re changing dx and dy to something else — dr and dTheta when you go to polar coordinates. Why a matrix here? Because if differential calculus is about anything it is about linearization of nonLinear functions, which is why you can use a matrix of derivatives (the Jacobian matrix)  for dx and dy.

Why is this important for statistical mechanics. Because one of the integrals you must evaluate is of exp(-ax^2) from -infinity to + infinity, and the switch to polar coordinates is the way to do it. You also must evaluate integrals of this type to understand the kinetic theory of ideal gases.

Not necessary in this context, but one of the best discussions of the derivative in its geometric context I’ve ever seen is on pp. 105 –> 106 of Callahan’s bok

So these are some pointers and hints, not a full discussion — I hope it makes the road easier for you, should you choose to take it.


How to think about two tricky theorems and other matters

I’m continuing to plow through classic differential geometry en route to studying manifolds en route to understanding relativity. The following thoughts might help someone just starting out.

Derivatives of one variable functions are fairly easy to understand. Plot y = f(x) and measure the slope of the curve. That’s the derivative.

So why do you need a matrix to find the derivative for more than one variable? Imagine standing on the side of a mountain. The slope depends on the direction you look. So something giving you the slope(s) of a mountain just has to be more complicated. It must be something that operates on the direction you’re looking (e.g. a vector).

Another point to remember about derivatives is that they basically take something that looks bumpy (like a mountain), look very closely at it under a magnifying glass and flatten it out (e.g. linearize it). Anything linear comes under the rubric of linear algebra — about which I wrote a lot, because it underlies quantum mechanics — for details see the 9 articles I wrote about it in — https://luysii.wordpress.com/category/linear-algebra-survival-guide-for-quantum-mechanics/.

Any linear transformation of a vector (of which the direction of your gaze on the side of a mountain is but one) can be represented by a matrix of numbers, which is why to find a slope in the direction of a vector it must be multiplied by a matrix (the Jacobian if you want to get technical).

Now on to the two tricky theorems — the Inverse Function Theorem and the Implicit Function theorem. I’ve been plowing through a variety of books on differential geometry (Banchoff & Lovett, McInenery, DoCarmo, Kreyszig, Thorpe) and they all refer you for proofs of both to an analysis book. They are absolutely crucial to differential geometry, so it’s surprising that none of these books prove them. They all involve linear transformations (because derivatives are linear) from an arbitrary real vector space R^n — elements are ordered n-tuples of real numbers to to another real vector space R^m. So they must inherently involve matrices, which quickly gets rather technical.

To keep your eye on the ball let’s go back to y = f(x). Y and x are real numbers. They have the lovely property that between any two real numbers there lies another, and between those two another and another. So there is no smallest real number greater than 0. If there is a point x at which the derivative isn’t zero but some positive number a to keep it simple (but a negative number would work as well), then y is increasing at x. If the derivative is continuous at a (which it usually is) then there is a delta greater than zero such that the derivative is greater than zero in the open interval (x – delta, x + delta). This means that y = f(x) is always increasing over that interval. This means that there is a one to one function y = g(x) defined over the same interval. This is called an inverse function.

Now you’re ready for the inverse function theorem — but the conditions are the same — the derivative at a point should be greater than zero and continuous at that point — and an inverse function exists. The trickiness and the mountains of notation come from the fact that the function is from R^n to R^m where n and m are any positive integers.

It’s important to know that, although the inverse and implicit functions are shown logically to exist, almost never can they be written down explicitly. The implicit function theorem follows from the inverse function theorem with even more notation involved, but this is the basic idea behind them.

A few other points on differential geometry. Much of it involves surfaces, and they are defined 3 ways. The easiest way to understand two of them takes you back to the side of a mountain. Now you’re standing on it half way up and wondering which would be the best way to get to the top. So you whip out your topographic map which has lines of constant elevation on it. This brings to the first way to define a surface. Assume the mountain is given by the function z = f (x, y) — every point on the earth has a height above it where the land stops and the sky beings (z) — so the function is a parameterization of the surface. Another way to define a surface in space is by level sets: put z equal to some height — call it z’ and define the surface as the set of two dimensional points (x, y) such that f (x, y ) = z’. These are the lines of constant elevation (e.g. the contour lines) – on the mountain. Differential geometry takes a broad view of surfaces — yes a curve on f (x, y) is considered a surface, just as a surface of constant temperature around the sun is a level set on f(x,y,z). The third way to define a surface is by f (x1, x2, …, xn) = 0. This is where the implicit function theorem comes in if some variables are in fact functions of others.

Well, I hope this helps when you plunge into the actual details.

For the record — the best derivation of these theorems are in Apostol Mathematical Analysis 1957 third printing pp. 138 – 148. The development is leisurely and quite clear. I bought the book in 1960 for $10.50. The second edition came out in ’74 — you can now buy it for 76.00 from Amazon — proving you should never throw out your old math books.

An old year’s resolution

One of the things I thought I was going to do in 2012 was learn about relativity.   For why see https://luysii.wordpress.com/2012/09/11/why-math-is-hard-for-me-and-organic-chemistry-is-easy/.  Also my cousin’s new husband wrote a paper on a new way of looking at it.  I’ve been putting him off as I thought I should know the old way first.

I knew that general relativity involved lots of math such as manifolds and the curvature of space-time.  So rather than read verbal explanations, I thought I’d learn the math first.  I started reading John M. Lee’s two books on manifolds.  The first involves topological manifolds, the second involves manifolds with extra structure (smoothness) permitting calculus to be done on them.  Distance is not a topological concept, but is absolutely required for calculus — that’s what the smoothness is about.

I started with “Introduction to Topological Manifolds” (2nd. Edition) by John M. Lee.  I’ve got about 34 pages of notes on the first 95 pages (25% of the text), and made a list of the definitions I thought worth writing down — there are 170 of them. Eventually I got through a third of its 380 pages of text.  I thought that might be enough to help me read his second book “Introduction to Smooth Manifolds” but I only got through 100 of its 600 pages before I could see that I really needed to go back and completely go through the first book.

This seemed endless, and would probably take 2 more years.  This shouldn’t be taken as a criticism of Lee — his writing is clear as a bell.  One of the few criticisms of his books is that they are so clear, you think you understand what you are reading when you don’t.

So what to do?  A prof at one of the local colleges, James J. Callahan, wrote a book called “The Geometry of Spacetime” which concerns special and general relativity.  I asked if I could audit the course on it he’d been teaching there for decades.  Unfortunately he said “been there, done that” and had no plans ever to teach the course again.

Well, for the last month or so, I’ve been going through his book.  It’s excellent, with lots of diagrams and pictures, and wide margins for taking notes.  A symbol table would have been helpful, as would answers to the excellent (and fairly difficult) problems.

This also explains why there have been no posts in the past month.

The good news is that the only math you need for special relativity is calculus and linear algebra.  Really nothing more.  No manifolds.  At the end of the first third of the book (about 145 pages) you will have a clear understanding of

l. time dilation — why time slows down for moving objects

2. length contraction — why moving objects shrink

3. why two observers looking at the same event can see it happening at different times.

4. the Michelson Morley experiment — but the explanation of it in the Feynman lectures on physics 15-3, 15-4 is much better

5. The Kludge Lorentz used to make Maxwell’s equations obey the Galilean principle of relativity (e.g. Newton’s first law)

6. How Einstein derived Lorentz’s kludge purely by assuming the velocity of light was constant for all observers, never mind how they were moving relative to each other.  Reading how he did it, is like watching a master sculptor at work.

Well, I’ll never get through the rest of Callahan by the end of 2012, but I can see doing it in a few more months.  You could conceivably learn linear algebra by reading his book, but it would be tough.  I’ve written some fairly simplistic background linear algebra for quantum mechanics posts — you might find them useful. https://luysii.wordpress.com/category/linear-algebra-survival-guide-for-quantum-mechanics/

One of the nicest things was seeing clearly what it means for different matrices to represent the same transformation, and why you should care.  I’d seen this many times in linear algebra, but seeing how simple reflection through an arbitrary line through the origin can be when you (1) rotate the line to the x axis by tan(y/x) radians (2) change the y coordinate to – y  — by an incredibly simple matrix  (3) rotate it back to the original angle .

That’s why any two n x n matrices X and Y represent the same linear transformation if they are related by the invertible matrix Z in the following way  X = Z^-1 * Y * Z

Merry Christmas and Happy New Year (none of that Happy Holidays crap for me)