July 21, 2012 9:42 AM Subscribe

Morton and Vicary on the Categorified Heisenberg Algebra - "In quantum mechanics, position times momentum does not equal momentum times position! This sounds weird, but it's connected to a very simple fact. Suppose you have a box with some balls in it, and you have the magical ability to create and annihilate balls. Then there's one more way to create a ball and then annihilate one, than to annihilate one and then create one. Huh? Yes: if there are, say, 3 balls in the box to start with, there are 4 balls you can choose to annihilate after you've created one but only 3 before you create one..."
posted by kliuless (78 comments total)
18 users marked this as a favorite

This insight, that funny facts about quantum mechanics are related to simple facts about balls in boxes, allows us to 'categorify' a chunk of quantum mechanics—including the Heisenberg algebra, which is the algebra generated by the position and momentum operators. Now is not the time to explain what that means. The important thing is that Mikhail Khovanov figured out a seemingly quite different way to categorify the same chunk of quantum mechanics, so there was a big puzzle about how his approach relates to the one I just described. Jeffrey Morton and Jamie Vicary have solved that puzzle.btw John Baez sez: "Don't even look at it unless you're an expert in mathematical physics, but it's important - and I think someday it could change our understanding of quantum mechanics." :P (I didn't; I know next to nothing about this stuff, but I found it interesting so I thought I'd share ;)

The big spinoff is this. Khovanov's approach showed that in categorified quantum mechanics, a bunch of new equations are true! These equations are 'higher analogues' of Heisenberg's famous formula

pq − qp = −i

So, these new equations should be important! ... Now Jeffrey and Jamie have shown how to get these equations just by thinking about balls in boxes... The important thing is this: those equations are not something we get to choose, or make up. They are what they are, and they're just sitting there waiting for us to discover them.

When they say "In quantum mechanics, position times momentum does not equal momentum times position!", my first instinct is to say "what kind of moron includes arbitrary increments and decrements of terms in their definition of multiplication?".

In other words, they are doing math, but they are not doing the math the terms would lead you to believe they are using.

So how is my grumpy knee-jerk ignorance wrong here?

posted by idiopath at 9:57 AM on July 21, 2012 [2 favorites]

In other words, they are doing math, but they are not doing the math the terms would lead you to believe they are using.

So how is my grumpy knee-jerk ignorance wrong here?

posted by idiopath at 9:57 AM on July 21, 2012 [2 favorites]

Wait, what exactly does this have to do with Breaking Bad?

posted by item at 10:00 AM on July 21, 2012 [2 favorites]

I find that when attempting to explain these things it helps to use extremely clear and precise English. That paragraph about the balls just made me go "wat", and I have a degree in physics. I wish some of these sciencey types would spend a bit of time improving their English. The best science communicators have made that effort, and so should anyone else who wants to make difficult ideas intelligible to a broad audience.

posted by Decani at 10:01 AM on July 21, 2012 [4 favorites]

posted by Decani at 10:01 AM on July 21, 2012 [4 favorites]

Which gets us to ... could somebody please explain centrifugal force to me?

Please, I'm an artist. It says so on that degree thing at the bottom of one of my file boxes.

posted by philip-random at 10:04 AM on July 21, 2012

Far be it from

And that's why I'm an ecologist.

posted by Jimbob at 10:05 AM on July 21, 2012 [2 favorites]

Not knee-jerk, not ignorant. It's a no-win situation for math people: you find a way to combine two items of a certain kind (like matrices) to get a new one. It looks like multiplication except that it does not commute (pq is not equal to qp). What to do? You can either make up a new name like combinification and no one will guess what it means or you can use an old one that is at least accurate by analogy--but may lead to confusion when the analogy breaks down.

posted by TreeRooster at 10:06 AM on July 21, 2012 [4 favorites]

posted by TreeRooster at 10:06 AM on July 21, 2012 [4 favorites]

(Actually, isn't this pretty much related to what us common folks in the low sciences call "degrees of freedom"?)

posted by Jimbob at 10:08 AM on July 21, 2012

posted by Jimbob at 10:08 AM on July 21, 2012

Look, they're just misusing terms and deliberately omitting context to make this sound mysterious. "xp" does not represent "position **times** momentum", because that isn't multiplication, and "x" and "p" are not (in general) real numbers, but operators. Yes, momentum and position operators do not commute. That does not mean "zomg multiplication of real numbers is different in QM".

posted by Salvor Hardin at 10:15 AM on July 21, 2012 [3 favorites]

posted by Salvor Hardin at 10:15 AM on July 21, 2012 [3 favorites]

This post was written on a web site that's mostly aimed at researchers working in mathematics and mathematical physics. As such, I don't think it's fair to complain that it doesn't do a good job explaining the results to a lay audience -- that's not its aim.

Also, the first main quote of the post is from John Baez (who incidentally had possibly the first ever blog, This Week's Finds in Mathematical Physics) and not from Morton and Vicary, in case the layout of the post caused anyone confusion on that point. For 17 years, Baez wrote a weekly column explaining various aspects of mathematical physics in wonderful, accessible (to researchers) and clear ways. I have a lot of respect for his exposition.

posted by louigi at 10:16 AM on July 21, 2012

Also, the first main quote of the post is from John Baez (who incidentally had possibly the first ever blog, This Week's Finds in Mathematical Physics) and not from Morton and Vicary, in case the layout of the post caused anyone confusion on that point. For 17 years, Baez wrote a weekly column explaining various aspects of mathematical physics in wonderful, accessible (to researchers) and clear ways. I have a lot of respect for his exposition.

posted by louigi at 10:16 AM on July 21, 2012

TreeRooster: isn't that what composition of operations means? I understand that composition and multiplication are related, but yeah, what Salvador Hardin said.

posted by idiopath at 10:16 AM on July 21, 2012

posted by idiopath at 10:16 AM on July 21, 2012

Metafilter: the magical ability to create and annihilate balls.

posted by cmoj at 10:18 AM on July 21, 2012 [1 favorite]

posted by cmoj at 10:18 AM on July 21, 2012 [1 favorite]

Incidentally, John Baez may also read MetaFilter; I say this because of this Google+ post of his, at the bottom of which he credits this MetaFilter post about obscure music at Ubuweb.

posted by louigi at 10:20 AM on July 21, 2012 [1 favorite]

posted by louigi at 10:20 AM on July 21, 2012 [1 favorite]

Am I the only person who, on first reading, was wondering why Joan Baez had such a keen interest in mathematical physics?

posted by spectrevsrector at 10:23 AM on July 21, 2012 [11 favorites]

posted by spectrevsrector at 10:23 AM on July 21, 2012 [11 favorites]

spectrevsrector, no, you are not the only one.

posted by mr_crash_davis at 10:26 AM on July 21, 2012

posted by mr_crash_davis at 10:26 AM on July 21, 2012

spectrevsrector, they're cousins, as a matter of fact.

posted by Lemurrhea at 10:32 AM on July 21, 2012 [2 favorites]

posted by Lemurrhea at 10:32 AM on July 21, 2012 [2 favorites]

Oh great, so now I have a box filled with an unknown number of multiplying balls as well as a possibly dead cat. Thank you quantum physicists, really, thank you. It's just as well that I live in Hilbert's hotel so I have somewhere to put it all!

posted by fallingbadgers at 10:41 AM on July 21, 2012 [7 favorites]

posted by fallingbadgers at 10:41 AM on July 21, 2012 [7 favorites]

Although I'm not an expert on mathematical physics, my research interests have recently been drifting towards quantum algebra and category theory, and from my point of view this paper is interesting but not the kind of paper that people outside of the math/physics community really would be especially excited about. The idea of "categorification," for example, is not a new one and has been one of the main paths of quantum algebra for some time now (see, eg, this MathOverflow question), and picking out this particular paper for a lay audience occludes the long history of categorified algebra.

One of the difficulties for lay readers of this sort of mathematics is that although it's called "quantum" algebra, the connections to actual physics and in particular quantum physics are tenuous at best. I've been thinking a lot about mathematical objects called quantum groups recently, but "quantum groups" is a term for a large and hazily-defined collection of mathematical objects that in their full generality goes far beyond standard ideas in physics. (For those that know a bit of math, for many people the term "quantum groups" even applies to any group object in the category of schemes, which is an extremely broad category of objects).

Also, I should say that the idea of pq not equaling qp is perfectly standard and even generally expected in mathematics and physics. It's one of the cornerstones of quantum theory, now an elder branch of physics, that operations occur in a noncommutative ring; and noncommutative rings are central in many branches of math, so I think the focus of this fpp is perhaps a bit off by focusing on this by now totally normal aspect of math and physics.

posted by Frobenius Twist at 10:45 AM on July 21, 2012 [4 favorites]

One of the difficulties for lay readers of this sort of mathematics is that although it's called "quantum" algebra, the connections to actual physics and in particular quantum physics are tenuous at best. I've been thinking a lot about mathematical objects called quantum groups recently, but "quantum groups" is a term for a large and hazily-defined collection of mathematical objects that in their full generality goes far beyond standard ideas in physics. (For those that know a bit of math, for many people the term "quantum groups" even applies to any group object in the category of schemes, which is an extremely broad category of objects).

Also, I should say that the idea of pq not equaling qp is perfectly standard and even generally expected in mathematics and physics. It's one of the cornerstones of quantum theory, now an elder branch of physics, that operations occur in a noncommutative ring; and noncommutative rings are central in many branches of math, so I think the focus of this fpp is perhaps a bit off by focusing on this by now totally normal aspect of math and physics.

posted by Frobenius Twist at 10:45 AM on July 21, 2012 [4 favorites]

I think this comment/question and the followups on John Baez's more in-depth blog post is helpful:

posted by crayz at 10:46 AM on July 21, 2012 [2 favorites]

So for example, if you have 4 $10 bills in your wallet, then there are 4 distinct ways to remove one, because each bill is an individual entity with its own distinguishing characteristics. But the dollars in your bank account are fungible, so if you have $40 in your account there aren’t 4 distinct ways to withdraw $10 from it to reduce it to $30."I agree this is weird. The sense I have, after playing with this for a few years on and off, is that this analogy isn’t physically meaningless, but it’s not obvious how literally to take it."

So what’s going on here? Is it just a misleading coincidence that the same mathematical relation holds for the creation and annihiliation operators as holds for balls and dollar bills? Or does this relation tell us that quanta of energy should be thought of – quite counter-intuitively – as having distinct identities?

posted by crayz at 10:46 AM on July 21, 2012 [2 favorites]

Premise 1: Things move straight unless acted upon.

Premise 2: Forces in different directions add to (or subtract from) each other, creating a resultant force.

Swing a bucket on a rope. The bucket is constrained in a circular path by the tension the rope exerts from where you're swinging. The bucket would go straight forward without this tension (let go of the rope and see which way the bucket goes: tangentially to the circular path it's constrained to. That's why that sling you played with when you were 10 never heaved the rock where you thought it would). The force of the bucket's tendancy to go straight and the force of the rope's tension add up to an apparent "force" pulling the bucket straight out along the line of the rope. That's centrifugal force, and that's why it's called "fictional."

posted by tspae at 10:50 AM on July 21, 2012 [5 favorites]

(I just want to pop in here and say how happy Metafilter makes me when I can go from rubber-band-exploding watermelons to Categorified Heisenberg Algebra! Thanks y'all!)

posted by Ron Thanagar at 10:54 AM on July 21, 2012

posted by Ron Thanagar at 10:54 AM on July 21, 2012

Beavis: Heh, heh, he said "balls".

Butthead: Heh, heh, and "functor".

Beavis: Heh, heh, and he said Khovanov's equations arose from the representation theory of symmetric groups, when more properly they arose from representations of the corresponding braid groups.

Butthead: Yeah. What a douche.

posted by twoleftfeet at 11:06 AM on July 21, 2012 [8 favorites]

Butthead: Heh, heh, and "functor".

Beavis: Heh, heh, and he said Khovanov's equations arose from the representation theory of symmetric groups, when more properly they arose from representations of the corresponding braid groups.

Butthead: Yeah. What a douche.

posted by twoleftfeet at 11:06 AM on July 21, 2012 [8 favorites]

I follow the n-category cafe (I have a background in mathematical physics), in fact MeFi, the n-category cafe, and the Guardian are the 3 websistes I look at every day. But I *never* expected to see anything from the n-category cafe make it to the front page here! It's really only intended for researchers in the field. You are expected to know what things like "braided monoidal category" and "Khovanov homology" mean. If not, it's probably not the greatest place to dive in ... it's the deep end of the pool!!

But John Baez is wonderful. In addition to his long-standing educational projects, he's also taking on global warming ... his homepage is a treasure-trove of interesting stuff .... and more approachable

than the Vicary/Morton paper!

posted by crazy_yeti at 11:10 AM on July 21, 2012 [2 favorites]

But John Baez is wonderful. In addition to his long-standing educational projects, he's also taking on global warming ... his homepage is a treasure-trove of interesting stuff .... and more approachable

than the Vicary/Morton paper!

posted by crazy_yeti at 11:10 AM on July 21, 2012 [2 favorites]

tspae has it down. There does exist such a thing as "centripetal force," which pulls a body in motion towards the center of a curved path, but what people commonly refer to as "centrifugal force" is in actuality just an application of inertia, which is a tendency, not a force. A body in motion in a straight line wants to continue on that vector, and when constrained on a circular path will at every instant tend along the straight-line-vector of that instant.

That's all.

posted by Navelgazer at 11:12 AM on July 21, 2012 [1 favorite]

Is this something I would need a ~~TV~~ **PHD** to understand?

posted by idiopath at 11:12 AM on July 21, 2012

posted by idiopath at 11:12 AM on July 21, 2012

How about when you multiply 60 mph * 5 hours to determine that you would travel 300 miles in 5 hours at that speed? All kinds of applications of math to the real world involve multiplying displacements, velocities, momenta, etc.

posted by crazy_yeti at 11:13 AM on July 21, 2012

I understand that part crazy_yeti, the part I don't understand is saying:

"5 hours * 60 mph is different from 60 mph * 5 hours because a variable may be incremented or decremented and 4*60 is different from 59*5"

(this was probably a very bad paraphrase)

posted by idiopath at 11:16 AM on July 21, 2012

"5 hours * 60 mph is different from 60 mph * 5 hours because a variable may be incremented or decremented and 4*60 is different from 59*5"

(this was probably a very bad paraphrase)

posted by idiopath at 11:16 AM on July 21, 2012

The problem as I see it is that modern physics is not necessarily soluble in language. It lives in the space of advanced mathematics, and it's nigh impossible to accurately translate it into English.

Talking about physics without math requires metaphors. There are lots of really good metaphors for different situations, but they are fine-tuned. A single "but what if...?" and the metaphor falls apart, and you need to switch to a different metaphor, and the whole coherence of your explanation starts to fall apart. The smarter your student is, the quicker this happens.

Which isn't to say that people trying to popularize physics aren't doing god's work. But I think there's a fundamental limitation that we all have to accept where after a certain level of abstraction is peeled away the only remotely honest answer is "

posted by no regrets, coyote at 11:18 AM on July 21, 2012 [2 favorites]

Idiopath: Non-commutativity in QM has nothing to do with variables being incremented or decremented. I think this thread has just confused you.

Two standard ways to understand it: the Schroedinger picture (wave mechanics, operators) or the Heisenberg picture (matrices).

In the Schroedinger approach, the state of a particle, or system, is described by a complex function Ψ(x,t) where x represents the space component (let's just do this in 1-d for simplicity). This is the famous "wave function".

"Observables" are modeled by operators which act on the wave function.

The "position" operator (typically called 'q') is just the operator which multiplies Ψ by x.

The "momentum" operator (typically 'p') is a partial derivative with respect to x (times a factor of -i to make things come out right)

Now, if you apply the operators in one order, you get a different result than if you apply them in the other order. Truth is, we're not really 'multiplying' anything, but applying operators to a function. (It's only "multiplication" by analogy).

If you remember your calculus, you will recall the (Leibniz) rule for the

derivative of a product: (f*g)' = f' * g + f * g'

So, if we multiply Ψ by x, then take the x-derivative, we get

d/dx (x*Ψ ) = (d/dx x) * Ψ + x * d/dx Ψ

But (d/dx)x = 1, so

d/dx (xΨ ) = Ψ + x * dΨ/dx

On the other hand, if we take the derivative (momentum) first, then multiply by x, we just get

x * dΨ /dx

So, we've shown that pq is not equal to qp, in fact qp-pq=i, which is the "canonical commutation relation" of quantum mechanics.

Now, one can rewrite p and q in terms of "creation and annihilation" operators, and find the corresponding commutation relations for those, and this does correspond to the fact that "there are more different ways to add then remove a ball than there are ways to remove then add".

There's also the Heisenberg approach, which I won't get into, other than to say that in this case, instead of operators acting on a wave function, we've got matrices multiplying a state vector in some Hilbert space. And we all know matrix multiplication is not commutative. (Rotate a box around the x-axis then they y-axis. Repeat the motions, but in the other order. The results are different. And matrix multiplication is like rotation).

When Heisenberg invented matrix mechanics, he didn't know about matrices - they weren't as widely known then as they are now - and he came up with his own idea

about numbers which "do not commute". Later his advisor, Max Born, apparently a bit more knowledegable in mathematics, informed him that he had reinvented matrix multiplication.

posted by crazy_yeti at 11:37 AM on July 21, 2012 [7 favorites]

Two standard ways to understand it: the Schroedinger picture (wave mechanics, operators) or the Heisenberg picture (matrices).

In the Schroedinger approach, the state of a particle, or system, is described by a complex function Ψ(x,t) where x represents the space component (let's just do this in 1-d for simplicity). This is the famous "wave function".

"Observables" are modeled by operators which act on the wave function.

The "position" operator (typically called 'q') is just the operator which multiplies Ψ by x.

The "momentum" operator (typically 'p') is a partial derivative with respect to x (times a factor of -i to make things come out right)

Now, if you apply the operators in one order, you get a different result than if you apply them in the other order. Truth is, we're not really 'multiplying' anything, but applying operators to a function. (It's only "multiplication" by analogy).

If you remember your calculus, you will recall the (Leibniz) rule for the

derivative of a product: (f*g)' = f' * g + f * g'

So, if we multiply Ψ by x, then take the x-derivative, we get

d/dx (x*Ψ ) = (d/dx x) * Ψ + x * d/dx Ψ

But (d/dx)x = 1, so

d/dx (xΨ ) = Ψ + x * dΨ/dx

On the other hand, if we take the derivative (momentum) first, then multiply by x, we just get

x * dΨ /dx

So, we've shown that pq is not equal to qp, in fact qp-pq=i, which is the "canonical commutation relation" of quantum mechanics.

Now, one can rewrite p and q in terms of "creation and annihilation" operators, and find the corresponding commutation relations for those, and this does correspond to the fact that "there are more different ways to add then remove a ball than there are ways to remove then add".

There's also the Heisenberg approach, which I won't get into, other than to say that in this case, instead of operators acting on a wave function, we've got matrices multiplying a state vector in some Hilbert space. And we all know matrix multiplication is not commutative. (Rotate a box around the x-axis then they y-axis. Repeat the motions, but in the other order. The results are different. And matrix multiplication is like rotation).

When Heisenberg invented matrix mechanics, he didn't know about matrices - they weren't as widely known then as they are now - and he came up with his own idea

about numbers which "do not commute". Later his advisor, Max Born, apparently a bit more knowledegable in mathematics, informed him that he had reinvented matrix multiplication.

posted by crazy_yeti at 11:37 AM on July 21, 2012 [7 favorites]

Not necessarily a PhD but at least several years of concerted study. That's the problem with hard science - it's HARD.

posted by crazy_yeti at 11:40 AM on July 21, 2012

Which reminds me of one of my favorite comments on the financial crisis. Can't find the source right now, but some clever soul said something like "All those Wall Street finance wizards had backgrounds in mathematical physics - they should have known that you can't just create a bunch of money out of nothing without creating an equal amount of antimoney!". (If anybody knows the source for this, let me know).

posted by crazy_yeti at 11:45 AM on July 21, 2012 [5 favorites]

crazy_yeti:

thanks, I knew I didn't understand, thanks for helping me get a better map of my ignorance.

My "PHD to understand" comment was a riff on an old cliche around here (in case you didn't know).

I think it is great that advanced stuff is being shared here, I like jumping in over my head every once in a while. I assumed (I hope not incorrectly), that if something this out of the ordinary were presented here, in such a general forum, beginners questions would be OK.

posted by idiopath at 11:45 AM on July 21, 2012

thanks, I knew I didn't understand, thanks for helping me get a better map of my ignorance.

My "PHD to understand" comment was a riff on an old cliche around here (in case you didn't know).

I think it is great that advanced stuff is being shared here, I like jumping in over my head every once in a while. I assumed (I hope not incorrectly), that if something this out of the ordinary were presented here, in such a general forum, beginners questions would be OK.

posted by idiopath at 11:45 AM on July 21, 2012

Absolutely! And I am doing the best I can to answer them. I just didn't get the PhD joke, sorry. And I hope that nothing I'm writing comes across as impatient or arrogant, it's great that folks are interested in this topic!

posted by crazy_yeti at 11:49 AM on July 21, 2012

Okay, so how does this apply to the politics of crowds?

Half-seriously.

The reason I raised the centrifugal question is that I was recently witness to a fairly absurd (ie: stoned) political discussion in which the LOUDER of the discussers suddenly started speaking of centrifugal forces (in a very positive way). Nobody had a clue what he was talking about, except him, of course. He clearly had a sense of something, just couldn't quite grasp it, turn it into words that actually communicated meaning ...

posted by philip-random at 11:51 AM on July 21, 2012

metaphors you articulate poorly in order to communicate a thought you don't quite grasp using concepts you never properly understood is one of the traditions of marijuana use

posted by idiopath at 11:53 AM on July 21, 2012 [7 favorites]

posted by idiopath at 11:53 AM on July 21, 2012 [7 favorites]

Heh, I'm working in the same group as Jamie Vicary. This stuff is really cool, but I agree that you shouldn't dive in to the deep end on the n-category cafe.

For a introduction to the diagrams, try Coecke's Quantum Picturalism (particularly section 3.1). For more fun stuff, try Baez's FUN stuff page.

posted by katrielalex at 12:01 PM on July 21, 2012 [3 favorites]

For a introduction to the diagrams, try Coecke's Quantum Picturalism (particularly section 3.1). For more fun stuff, try Baez's FUN stuff page.

posted by katrielalex at 12:01 PM on July 21, 2012 [3 favorites]

Maybe he was just using it in a very general, literal sense to mean social or spiritual forces that cause people to flee from the center? (Or maybe it actually made no sense, of course.)

posted by hattifattener at 12:01 PM on July 21, 2012

posted by hattifattener at 12:01 PM on July 21, 2012

Interesting link. Not how I was expecting to spend my Saturday morning. I hope to read it later and I hope even more to understand at least 5% of it.

posted by benito.strauss at 12:06 PM on July 21, 2012

posted by benito.strauss at 12:06 PM on July 21, 2012

This can't be true, because all knowledge is soluble in language. Or at least all

The symbols that mathematicians and physicist use make their knowledge look arcane and impenetrable, but each one of those symbols has an unpacked explanation in ordinary human language. The reason symbols are so helpful is because they can contain an arbitrary amount of knowledge (infinite compression if you will), and the reason certain formulas seem so impenetrable is because researchers do their best to compress as much knowledge into these symbols as possible, which in turn makes complicated ideas easier in some sense (but only if you already know what all the symbols mean).

The reason certain metaphors fail when trying to explain complicated mathematical or physical ideas is because they deliberately avoid trying to unpack all the meaning inherent in the symbols and therefor yield necessarily inaccurate comparisons. But that doesn't imply that the understanding itself is beyond human language. You just need

posted by grog at 12:24 PM on July 21, 2012 [5 favorites]

> each one of those symbols has an unpacked explanation in ordinary human language (grog)

Exactly! If you go to university to study maths, right near the beginning someone will give you a definition of a group. It's not a particularly difficult concept -- a set together with a way of combining elements of that set to make more elements -- but the axioms take a little space to write. You'll then spend the remainder of the term proving theorems in group theory i.e. facts that are true of groups.

If, later in your mathematical career, somebody says "... and the Xs form a group", that's just shorthand for "... and there is a way of combining two Xs to make another X so that this and that and the other." But you both know what a group is, so that's implicit. And you've played with groups before and know the main theorems, so a whole host of useful data comes from that one little phrase.

Later, you'll learn the definition of a field. Later, a vector space. Later, a category. And so on.

The stuff Vicary etc. is (and to a much lesser extent I am) working on, as with most advanced-level mathematics, is built up on a foundation of this kind of shorthand, and expresses concepts that would take*hours* to explain if we had to write them out in full each time. Not all of it is inherently difficult to understand -- indeed, I think that the graphical calculus for categorical quantum mechanics is so good precisely because it is built on intuitively obvious foundations. But to understand the results you do need to go away and learn all the definitions.

posted by katrielalex at 12:38 PM on July 21, 2012 [2 favorites]

Exactly! If you go to university to study maths, right near the beginning someone will give you a definition of a group. It's not a particularly difficult concept -- a set together with a way of combining elements of that set to make more elements -- but the axioms take a little space to write. You'll then spend the remainder of the term proving theorems in group theory i.e. facts that are true of groups.

If, later in your mathematical career, somebody says "... and the Xs form a group", that's just shorthand for "... and there is a way of combining two Xs to make another X so that this and that and the other." But you both know what a group is, so that's implicit. And you've played with groups before and know the main theorems, so a whole host of useful data comes from that one little phrase.

Later, you'll learn the definition of a field. Later, a vector space. Later, a category. And so on.

The stuff Vicary etc. is (and to a much lesser extent I am) working on, as with most advanced-level mathematics, is built up on a foundation of this kind of shorthand, and expresses concepts that would take

posted by katrielalex at 12:38 PM on July 21, 2012 [2 favorites]

Well, ordinary human language is full of ambiguities and leaves a lot of room for interpretation, which is why we have poetry and literature ... Mathematical/technical language is designed to get around this. E.g. when you say "if" do you really mean "if and only if"? Is "or" to be taken in the inclusive or exclusive sense? Technical languages (math, legalese, computer language) have to avoid these ambiguities. It's why we program computers in Python (or C++ or Ruby or Java or what-have-you) instead of trying to develop plain-English programs.

posted by crazy_yeti at 12:45 PM on July 21, 2012 [1 favorite]

Going back a step for those who would like: mathematicians spend much of their time taking a more basic or concrete idea and generalizing it somehow, by trying to think what the really important properties are that kind of define that basic thing. So take basic arithmetic operations. We've got some set of numbers (whole numbers, integers, real numbers, or whatever). As a kid, you get taught that the numbers 0, 1, 2, 3, ... just *are*, and that "addition" and "multiplication" are these things that we can do to take two numbers and get a new number. And if you went to a good school, maybe your teacher explained what this *means* in some sense, eg. how multiplication is related to addition. If you went to a really good school, or if you were just into this sort of stuff as a kid, maybe you ended up sitting down with your addition or multiplication tables and finding a bunch of patterns (very basic example: multiples of 5 end either in a 0 or in a 5).

Now fast forward to your first algebra class. For a lot of North Americans, this was a bunch of arcane mumbo-jumbo, and was the point where math started getting hard and/or uninteresting. There's a reason for this: in addition to issues with the quality of grade school mathematics teaching, math changes a little bit at algebra. In algebra, we start thinking about numbers and operations a little differently. Pre-algebra was where I learned about the number line, for example, and started thinking about numbers as an abstract concept, rather than representing specific, physical quantities. Arithmetic operations on numbers then become a little more abstract, too: just some thing that we define, where we input two numbers, and it outputs a third number. Addition and multiplication of integers still have their original interpretation in terms of combining quantities of actual physical objects, but now we've kind of extended the way that we think about them. And often times (in a good algebra class at least), you'll have to start thinking about the properties of these arithmetic operations a little bit more carefully or formally/abstractly as well.

One property of the operation of multiplication is that it has an "identity": a number (the number 1, specifically), such that when we multiply any other number by 1, we get our original number back (3 times 1 = 3, 296 times 1 = 296, etc.). Another property is associativity: if a, b, and c are any three real numbers, then a(bc) = (ab)c; that is, it doesn't matter how we group numbers that we multiply. For multiplication of real numbers, we also have the commutative property: ab = ba for any two real numbers a and b.

If we are multiplying real numbers, then another property of multiplication is that we have "multiplicative inverses" -- if we start with a number, a, then it's inverse is a number that "undoes" the operation in the sense that you get back to the (additive or multiplicative or whatever operation you're using) identity. So the additive inverse of a is -a (because a + (-a) = 0, the additive identity), and the multiplicative inverse of a is 1/a (because a(1/a) = 1, the multiplicative identity). (So subtraction is just adding the additive inverse, and division is just multiplying by the multiplicative inverse.) Now instead of multiplying real numbers, think of multiplying whole numbers or integers. If a is an integer, then 1/a is a fraction not an integer. So multiplication on the real numbers has the property of having inverses; but multiplication on the integers does not have this property! So when we're talking about operations, like multiplication, we also have to specific the set of objects that we're defining the operation on.

For various reasons (having to do with how multiplication of matrices and of functions and similar objects are defined), mathematicians don't care quite as much about having the property of commutativity. Any operation on any set of mathematical objects that has an identity element in that set and that is associative is considered to be analogous to multiplication, so is often called multiplication on [whatever set it is that we're defining multiplication on: sets of numbers, sets of matrices, sets of functions, sets of operators on other sets, etc.]. Then we talk about commutative multiplications or noncommutative multiplications depending on if the operation has the property of commutativity or not.

(On preview, that's the more detailed basic version of what katrielalex is talking about with defining groups, etc.)

posted by eviemath at 12:48 PM on July 21, 2012

Now fast forward to your first algebra class. For a lot of North Americans, this was a bunch of arcane mumbo-jumbo, and was the point where math started getting hard and/or uninteresting. There's a reason for this: in addition to issues with the quality of grade school mathematics teaching, math changes a little bit at algebra. In algebra, we start thinking about numbers and operations a little differently. Pre-algebra was where I learned about the number line, for example, and started thinking about numbers as an abstract concept, rather than representing specific, physical quantities. Arithmetic operations on numbers then become a little more abstract, too: just some thing that we define, where we input two numbers, and it outputs a third number. Addition and multiplication of integers still have their original interpretation in terms of combining quantities of actual physical objects, but now we've kind of extended the way that we think about them. And often times (in a good algebra class at least), you'll have to start thinking about the properties of these arithmetic operations a little bit more carefully or formally/abstractly as well.

One property of the operation of multiplication is that it has an "identity": a number (the number 1, specifically), such that when we multiply any other number by 1, we get our original number back (3 times 1 = 3, 296 times 1 = 296, etc.). Another property is associativity: if a, b, and c are any three real numbers, then a(bc) = (ab)c; that is, it doesn't matter how we group numbers that we multiply. For multiplication of real numbers, we also have the commutative property: ab = ba for any two real numbers a and b.

If we are multiplying real numbers, then another property of multiplication is that we have "multiplicative inverses" -- if we start with a number, a, then it's inverse is a number that "undoes" the operation in the sense that you get back to the (additive or multiplicative or whatever operation you're using) identity. So the additive inverse of a is -a (because a + (-a) = 0, the additive identity), and the multiplicative inverse of a is 1/a (because a(1/a) = 1, the multiplicative identity). (So subtraction is just adding the additive inverse, and division is just multiplying by the multiplicative inverse.) Now instead of multiplying real numbers, think of multiplying whole numbers or integers. If a is an integer, then 1/a is a fraction not an integer. So multiplication on the real numbers has the property of having inverses; but multiplication on the integers does not have this property! So when we're talking about operations, like multiplication, we also have to specific the set of objects that we're defining the operation on.

For various reasons (having to do with how multiplication of matrices and of functions and similar objects are defined), mathematicians don't care quite as much about having the property of commutativity. Any operation on any set of mathematical objects that has an identity element in that set and that is associative is considered to be analogous to multiplication, so is often called multiplication on [whatever set it is that we're defining multiplication on: sets of numbers, sets of matrices, sets of functions, sets of operators on other sets, etc.]. Then we talk about commutative multiplications or noncommutative multiplications depending on if the operation has the property of commutativity or not.

(On preview, that's the more detailed basic version of what katrielalex is talking about with defining groups, etc.)

posted by eviemath at 12:48 PM on July 21, 2012

Mimesis?

Pictures?

Direct Mystic Transmission?

posted by crazy_yeti at 1:08 PM on July 21, 2012

You can translate advanced mathematics into English! It's kind of ugly, though. Basically, first you use English to teach your listener advanced mathematics. Then you use the new language that you've defined on top of English to explain whatever you want to explain.

posted by madcaptenor at 1:33 PM on July 21, 2012 [3 favorites]

all shared knowledge must be language-izable, because that's the only mode we have in which to share it.

Dance!

posted by benito.strauss at 2:38 PM on July 21, 2012 [2 favorites]

Dance!

posted by benito.strauss at 2:38 PM on July 21, 2012 [2 favorites]

Surely I'm not the only one who read John Baez as Joan Baez, and got really confused, really quickly?

Yes? Ok then. Back to my Pre-Calc homework.

posted by spinifex23 at 3:47 PM on July 21, 2012

Yes? Ok then. Back to my Pre-Calc homework.

posted by spinifex23 at 3:47 PM on July 21, 2012

Isn't this just basic linear algebra and the well established properties of matrix multiplication?

posted by humanfont at 4:47 PM on July 21, 2012

posted by humanfont at 4:47 PM on July 21, 2012

Yes, but matrices are really counterintuitive when you're used to numbers.

posted by madcaptenor at 5:13 PM on July 21, 2012

posted by madcaptenor at 5:13 PM on July 21, 2012

I like to imaginge a Twilight Zone episode where an alien race discovered matrices first, and then there's some young upstart researcher who finds a sub-ring that's commutative. As they drive him away in the padded car he yells "Order doesn't matter! Do you hear me? Order doesn't maaaatttttteerrrrrrrr!".

posted by benito.strauss at 6:59 PM on July 21, 2012 [5 favorites]

posted by benito.strauss at 6:59 PM on July 21, 2012 [5 favorites]

"Matrix multiplication" isn't really multiplication (in the familiar sense). It's a composition of operators, which is why it doesn't commute. Matrix composition happens to have some things in common with multiplication though, and its convenient to represent and refer to it as "matrix multiplication" sometimes. Now I don't physics, but I gather from crazy_yeti's explanation that a similar confusion is going on with p & q.

posted by yeolcoatl at 8:06 PM on July 21, 2012

posted by yeolcoatl at 8:06 PM on July 21, 2012

I feel like I've learned something here.

posted by humanfont at 8:27 PM on July 21, 2012 [1 favorite]

posted by humanfont at 8:27 PM on July 21, 2012 [1 favorite]

Umm... composition means peforming two operations in sequence, one after the other. As linear transformations, yeah matrix multiplication is a composition, but perhaps people who are weirded out by the idea of noncommutative multiplication due to not having much experience with higher or more abstract math would be more familiar with matrix multiplication as a mechanistic process, if they are familiar with it at all?

posted by eviemath at 9:00 PM on July 21, 2012

posted by eviemath at 9:00 PM on July 21, 2012

"Matrix multiplication" isn't really multiplication (in the familiar sense)

A lot of education in formal mathematics teaches you to stop thinking in the familiar senses, and to just see where your assumptions take you. My favorite question for seeing how much formal math training someone has had is "Suppose every cat has seven legs and there are negative three cats in your yard. How many cat legs are there in your yard?".

posted by benito.strauss at 11:07 PM on July 21, 2012 [2 favorites]

A lot of education in formal mathematics teaches you to stop thinking in the familiar senses, and to just see where your assumptions take you. My favorite question for seeing how much formal math training someone has had is "Suppose every cat has seven legs and there are negative three cats in your yard. How many cat legs are there in your yard?".

posted by benito.strauss at 11:07 PM on July 21, 2012 [2 favorites]

Centrifugal force is a false force because the object is being brought into a curved trajectory by some other force, like the tension of the handle when you swing a bucket around. Centripedal force is just the name we sometimes use for the force of that tension, whatever it is; it's not an extra force that arises from the situation. Thus many different kinds of forces can serve as a centripedal force, like the tension of a rope, or the gravity that keeps a planet in orbit, but centrifugal force is always a result of inertia.

posted by JHarris at 12:03 AM on July 22, 2012

posted by JHarris at 2:44 AM on July 22, 2012 [2 favorites]

But everyone is familiar with the idea of operators that commute vs.

operators that don't commute.

I can put on my shirt and then my pants, or vice versa, and when I leave the house I won't look like an absent-minded professor. But putting on socks and shoes doesn't commute - if I put on my shoes and then my socks I probably won't pass as human, and I will also need new socks.

posted by madcaptenor at 4:27 AM on July 22, 2012 [5 favorites]

operators that don't commute.

I can put on my shirt and then my pants, or vice versa, and when I leave the house I won't look like an absent-minded professor. But putting on socks and shoes doesn't commute - if I put on my shoes and then my socks I probably won't pass as human, and I will also need new socks.

posted by madcaptenor at 4:27 AM on July 22, 2012 [5 favorites]

I've got a PhD in physics (but am an experimentalist, not a theorist) and would just like to express the opinion that if you can't explain what you're trying to say without using math, then you're really doing math, not physics.

This is forgivable. You can get a long way in science by*just* doing math. You can get most of the homework problems you're assigned, all the way up through grad school, correct without ever understanding the physics of them, if you understand the math. You can even do research this way -- don't worry about what the symbols stand for, just make sure they obey the symbol-manipulation rules and see if you can't get them to express a relationship you didn't know before in a way that follows from something you did.

And I've had professors who think that's enough, who think that you've understood the world when you've described it in terms of math, and balk at any demand for a more physical, visual, mechanical depiction of the processes the mathematical relations describe. For instance, I was taught thermodynamics this way; as a bunch of theorems and proofs, provable from a few initial assumptions. And that's how thermodynamics were discovered, too, by people doing math exercises. And asking "But what is entropy, really?" got me, from that professor, just symbols for an answer.

But there is another answer. Entropy a measure of the number of different ways you can arrange the components of something without changing the nature of the thing in any way you care about. You can move all the molecules of air in your room to different spots, but the air pressure, temperature, volume, etc, will be unaffected, and you won't notice, for instance.

I'd argue that the people who discovered the mathematical laws of thermodynamics weren't wrong, but the people who put those mathematical laws on the physical foundation of statistical mechanics were*more* right.

Anyway, my not-very-firmly held opinion is that people who talk about creation and annhilation of particles too much are guilty of taking the math too literally in the first place. Sure, mathematically, you can describe things in terms of these "creation and annihilation operators," but you can't ever take your box full of particles, "do an annihilation" in real life, and then have less particles. "Creation and annihilation" are mathematical operations, not physical ones, as far as I'm concerned.

The non-commutation of position and momentum makes perfect sense if instead you think in terms of waves. Naturally a wave pattern which repeats itself over and over again over miles, which therefore has a well defined wavelength, can't have a well defined location. And naturally a little temporary vibrating lump in the surface of the pond which is trapped between some rocks and therefor has a nicely defined location, can't have a very well defined wavelength. And quantum mechanics says that everything is a wave, and moreover, that the momentum of anything (how much force it can exert on you) is given by its wavelength, and so... It already all makes perfect sense. To me, anyway.

But I'm speaking for myself, here. There are a lot of mathematical physicists out there who disagree with my approach to the subject. (I'm just an experimentalist, after all.) Heck, I barely believe in particles at all, these days...

posted by OnceUponATime at 4:55 AM on July 22, 2012 [5 favorites]

This is forgivable. You can get a long way in science by

And I've had professors who think that's enough, who think that you've understood the world when you've described it in terms of math, and balk at any demand for a more physical, visual, mechanical depiction of the processes the mathematical relations describe. For instance, I was taught thermodynamics this way; as a bunch of theorems and proofs, provable from a few initial assumptions. And that's how thermodynamics were discovered, too, by people doing math exercises. And asking "But what is entropy, really?" got me, from that professor, just symbols for an answer.

But there is another answer. Entropy a measure of the number of different ways you can arrange the components of something without changing the nature of the thing in any way you care about. You can move all the molecules of air in your room to different spots, but the air pressure, temperature, volume, etc, will be unaffected, and you won't notice, for instance.

I'd argue that the people who discovered the mathematical laws of thermodynamics weren't wrong, but the people who put those mathematical laws on the physical foundation of statistical mechanics were

Anyway, my not-very-firmly held opinion is that people who talk about creation and annhilation of particles too much are guilty of taking the math too literally in the first place. Sure, mathematically, you can describe things in terms of these "creation and annihilation operators," but you can't ever take your box full of particles, "do an annihilation" in real life, and then have less particles. "Creation and annihilation" are mathematical operations, not physical ones, as far as I'm concerned.

The non-commutation of position and momentum makes perfect sense if instead you think in terms of waves. Naturally a wave pattern which repeats itself over and over again over miles, which therefore has a well defined wavelength, can't have a well defined location. And naturally a little temporary vibrating lump in the surface of the pond which is trapped between some rocks and therefor has a nicely defined location, can't have a very well defined wavelength. And quantum mechanics says that everything is a wave, and moreover, that the momentum of anything (how much force it can exert on you) is given by its wavelength, and so... It already all makes perfect sense. To me, anyway.

But I'm speaking for myself, here. There are a lot of mathematical physicists out there who disagree with my approach to the subject. (I'm just an experimentalist, after all.) Heck, I barely believe in particles at all, these days...

posted by OnceUponATime at 4:55 AM on July 22, 2012 [5 favorites]

Well, it's the logarithm of that number (basically). So some math creeps in....

How do you get the relation qp - pq = iħ out of that kind of handwaving?

But it happens

posted by crazy_yeti at 6:48 AM on July 22, 2012

The thing is, math is language. Language is the representation of the world in a set of cues so that we can communicate with one another; math is much the same. There are ideas implicit in our use of language - implication, logical constructions - that math plays with in a way that everyday language does not. But it's a difference of degree, not a difference of form.

The main reason it's tough to 'translate' into plain speech is that most of the important concepts involved are things that we don't think about or interact with on a day to day basis. Things like these creation and annihilation operators become quite natural to think about after spending quite a number of years thinking about physics, but they aren't the idioms in which we encode our everyday experience. And that makes it hard to convey, just as it's difficult to describe (for example) the taste of cheese to people from a lactose-free culture.

posted by kaibutsu at 9:04 AM on July 22, 2012 [1 favorite]

So I started multiplying ℂ by e

Next I thought I'd step up my game and throw down a general conformal map, but that left me feeling a bit queasy. At this point I was tired with the whole lot, so I embedded ℂ into ℝ

posted by benito.strauss at 12:02 PM on July 22, 2012 [3 favorites]

Two ways to answer that. The more general answer is: equations generally are more precise than physical descriptions. I don't do physics using only words either. I'm just saying that if you are going to insist that there

The more specific answer is: If you believe that things are waves and are described in terms of wavefunctions, then you can get to the "d/dx" definition of momentum without too much mind bending, and from "x = x" and "p = d/dx" you can almost get the real commutator. Or in other words: the same way Schroedinger did it. (Okay, the "i" still comes as a surprise. That's why Schroedinger was a genius.) But the general idea that position and momentum can't be known at the same time, or measured in arbitrary order, does follow naturally and immediately from the "handwavy" picture, rather than arising as a counterintuitive consequence of an otherwise arbitrary mathematical axiom.

As for the creation and annihilation operations happening "all the time," well, to be sure, a lot of people do describe lasers and particle collisions in that way. But to my mind they'd be better off sticking to the description of the fields, and probability distributions of field amplitudes and phases and energy states, than thinking in terms of creation and annihilation of particles. Lasers in particular are my area, and I know you don't

I keep typing and then deleting too-long and too-technical explanations of how I think about stimulated emission and absorption (which is semi-classically, as far as I can get away with it, and thinking about the energies of the quantized field, when I must, in a way more similar to the energy levels of a harmonic oscillator than to a number of particles).

If I had my way, "particle physics" would be renamed "field physics" anyway. The fields are far more fundamental, even strictly mathematically, than the particles, which are just excitations of the fields. When fields interact at high energy and are excited in new and different ways, I don't see what's really helpful about calling those excitations "new particles" and saying that they've been "annhilated" once they die down...

But okay, I know there are a lot of people who really like to think about photons as little particles that get created and destroyed, rather than as probability distributions on the amplitude of a plane-wave mode of the electromagnetic field. As I said, it's my not-very-strongly-held opinion that this is a too-literal interpretation of the role of "creation operators" etc in the mathematical description, but I recognize that it is a very common approach, and I just happen to think it's not as physically meaningful as the description in terms of fields and wavefunctions.

The part of my comment after "anyway" was meant to be specific to this post, and I don't feel as strongly about the question of whether creation operators etc. count as an example of "math, not physics," (it's borderline, for me) as I do about the general question of whether un-translatable mathematical descriptions in general should count as an explanation of a physical process. (I'm saying no, more emphatically.)

posted by OnceUponATime at 12:14 PM on July 22, 2012 [1 favorite]

But in mathematics this isn't seen as an always bad thing. One of the connotations of "calculus" is that one can carry out computations

The word "calculus" refers to a small stone used for counting, another computing tool. Leibniz's notation for functional calculus lets you find areas without having to worry about what they look like. And the lambda calculus is a reduction of computation to symbol manipulation (called, exotically, α reduction, β reduction, and η reduction) and is a big part of why we have computers today.

I'm jealous of you for having that physical intuition that I totally lack (ask about whether I'm supposed to look at tension and ignore friction, or look at friction and ignore tension to solve a ropes, weights, and pulleys problem and I'll scream). I agree that the symbolic formalisms are just tools and you really should have a deeper understanding behind them. But sometimes having the right tool can change the way you look at the world.

posted by benito.strauss at 12:20 PM on July 22, 2012 [2 favorites]

On preview, I'd love to hear your field-centric description of lasers, but it sounds like something best done over beer IRL.

posted by benito.strauss at 12:23 PM on July 22, 2012

posted by benito.strauss at 12:23 PM on July 22, 2012

benito -- I agree with everything in your post, and having a beer and discussing lasers sounds like exactly my idea of a good time.

posted by OnceUponATime at 1:33 PM on July 22, 2012

posted by OnceUponATime at 1:33 PM on July 22, 2012

crazy_yeti, thanks for that quote!

posted by IAmBroom at 6:46 PM on July 22, 2012

Pictures?

Well done, 2/3 of crazy_yeti & benito.strauss!

posted by IAmBroom at 6:49 PM on July 22, 2012

I like this definition. Thanks,

posted by IAmBroom at 6:53 PM on July 22, 2012

Lasers require three energy states in the lasing medium - excitation, metastable decay, and ground. The decay from the excitation state to the metastable decay state is an entropic heat loss to the system. Lasing occurs from that level down to ground.

But I'm not absolutely sure what you mean by these operators being commutative, so maybe I'm missing the point.

posted by IAmBroom at 6:56 PM on July 22, 2012

With regard to lasers - my training is in mathematical physics, not laser optics, so what I was taught about lasers is pretty theoretical! And it was also a long time ago. But I recall that the key observation was that once you have a number N of photons in a coherent state, the transition rate for additional photons to be created (emitted) in that state is higher than the transition rate for photons to be absorbed, and that this came down to the fact that the amplitude for the creation operator on an N-particle state is proportional to sqrt(N+1) while the amplitude for the annihilation operator is only sqrt(N), or

a|n>= √n |n-1>

a†|n> = √(n+1) |n+1>

which in turn can be derived from the commutation relations between a and a†.

Now, this is probably a horrible oversimplification of what's going on in a real laser. The above was taught to me in a QFT class (taught by a mathematician!) as an interesting application of raising and lowering operators (a.k.a. "creation and annihilation"), not as a detailed explanation of laser physics. Corrections are most welcome :-)

I've been looking for a reference on this, I believe it may be explained here: http://www.tcm.phy.cam.ac.uk/~bds10/aqp/handout_transitions.pdf

Sorry for any errors/inaccuacies. This has been a terrific discussion which I am enjoying tremendously.

posted by crazy_yeti at 7:42 AM on July 23, 2012

The operators by no means commute! I didn't say they commute, but that they

(which comes down to: more ways to add then remove an indistiguishable ball (photon) than ways to do it in the opposite order! Because if you add first, then you have more choices of which one to remove).

If the creation and annihilation operators commuted, all of this would fall apart.

posted by crazy_yeti at 7:47 AM on July 23, 2012

Yep, crazy_yeti, I suspect we're overlapping about as little as possible, while still both being knowledgeable about the topic.

But I recall that the key observation was that once you have a number N of photons in a coherent state, the transition rate for additional photons to be created (emitted) in that state is higher than the transition rate for photons to be absorbed,

[what's missing from your explanation is:]

*provided that the number of elevated electrons outnumbers the number of ground-level electrons*

... otherwise, there is a net absorption (Light Amplification through Stimulated Emission of Radiation < 1).

posted by IAmBroom at 11:13 AM on July 23, 2012

But I recall that the key observation was that once you have a number N of photons in a coherent state, the transition rate for additional photons to be created (emitted) in that state is higher than the transition rate for photons to be absorbed,

[what's missing from your explanation is:]

... otherwise, there is a net absorption (Light Amplification through Stimulated Emission of Radiation < 1).

posted by IAmBroom at 11:13 AM on July 23, 2012

... you are an optical systems engineer. I am a lapsed mathematical physicist. So our understandings of the laser are probably pretty different!

Makes me want to conduct an experiment where I throw you two together at high energies and see what happens.

posted by benito.strauss at 11:34 AM on July 23, 2012

Makes me want to conduct an experiment where I throw you two together at high energies and see what happens.

posted by benito.strauss at 11:34 AM on July 23, 2012

Well, if I we both survived the impact unharmed, I'd buy Broom a beer (or drink of his choice) and try to have a friendly chat and learn all about how lasers *really* work. Then I'd come looking for you and ask why you had to get all violent like that!! ;-)

posted by crazy_yeti at 12:05 PM on July 23, 2012

posted by crazy_yeti at 12:05 PM on July 23, 2012

But, but, but I thought that was how you did physics. I got the tension/friction thing wrong again, didn't I?

Seriously, I think you and I (math backgrounds) are just trying to lure Broom and Once into a beer-based working scientists seminar on lasers. Pity about the geography.

posted by benito.strauss at 12:26 PM on July 23, 2012

Seriously, I think you and I (math backgrounds) are just trying to lure Broom and Once into a beer-based working scientists seminar on lasers. Pity about the geography.

posted by benito.strauss at 12:26 PM on July 23, 2012

I'm game, provided either (1) crazy_yeti is a female physicist, or (2) crazy_yeti is holding a beer (I'm crazy-fast at stealing beers).

posted by IAmBroom at 12:38 PM on July 23, 2012 [1 favorite]

posted by IAmBroom at 12:38 PM on July 23, 2012 [1 favorite]

This is only sort of vaguely related to the topic of this post, but I've always wanted to get the opinion of some mathematical physicists on this paper, by a professor I once knew. It says that, at least for a single particle in a potential, you can write the Lagrangian in terms of the difference between the group velocity and the phase velocity of the particle, and the action, which is just the integral of that over the particle's path, is just the number of wavefronts that pass through the envelope. So when you're minimizing the action of a path, that's what you're minimizing.

This seems really deep to me, and hints at a physical reason for the principle of least action, which otherwise has always seemed like the*worst* kind of rule-of-thumbing to me, since I had no idea what the Lagrangian (unlike the total energy) was supposed to represent physically. (Though obviously it works). But I've never seen this point made in any class or textbook or anywhere outside of the context of this paper. Is it because everyone else thinks it's obvious (and minor?) or is it really a novel point and the paper's just obscure? Am I the only one whose mind is blown?

(BTW, I'm a female physicist, but I'm so female that I'm actually pregnant right now, so I won't actually be participating in any high energy collisions or enjoying any real life beer. Also BTW, props to crazy_yeti for the ASCII math. I agree with IAmBroom that there are not enough atoms or atomic energy levels in that description for it to make sense to me. Also, lasers don't put out N-particle states. The put out coherent states which superpositions with Poissonian distributions of photon number, and which are, oddly enough, eigenstates of the lowering operator. I haven't quite wrapped my head around that latter fact yet. I also just decided that I am much happier with those operators when they are called "raising and lowering" than when they are called "creation and annihlilation." So maybe I'm just being picky.)

posted by OnceUponATime at 8:56 PM on July 23, 2012

This seems really deep to me, and hints at a physical reason for the principle of least action, which otherwise has always seemed like the

(BTW, I'm a female physicist, but I'm so female that I'm actually pregnant right now, so I won't actually be participating in any high energy collisions or enjoying any real life beer. Also BTW, props to crazy_yeti for the ASCII math. I agree with IAmBroom that there are not enough atoms or atomic energy levels in that description for it to make sense to me. Also, lasers don't put out N-particle states. The put out coherent states which superpositions with Poissonian distributions of photon number, and which are, oddly enough, eigenstates of the lowering operator. I haven't quite wrapped my head around that latter fact yet. I also just decided that I am much happier with those operators when they are called "raising and lowering" than when they are called "creation and annihlilation." So maybe I'm just being picky.)

posted by OnceUponATime at 8:56 PM on July 23, 2012

"*They* put out coherent states which *are* superpositions..."

posted by OnceUponATime at 8:58 PM on July 23, 2012

posted by OnceUponATime at 8:58 PM on July 23, 2012

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