## QM - Time Ordering Operator

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### QM - Time Ordering Operator

QM was the first lecture I've ever heard, where just about everything the professor said was (strictly speaking) wrong, and I've experienced some difficulties trying to learn this stuff by heart or work with it. Still, most things gave the impression that they could be well-defined, if one would bother to learn the appropriate mathematical tools. But some definitions exceed my suspension of disbelief, like this.

The time ordering Operator T is defined as
T \{{A(t_1), A(t_2)\}} = \begin{cases} A(t_1)A(t_2) & t_1 >= t_2 \\ A(t_2)A(t_1) & t_1 < t_2 \end{cases}

Where A(t) is a time-dependent Operator on H (the "Hilbert" space QM operates on), i.e. a continuous function R->End(H).
That's as strange a definition as I've ever seen, and it looks like this should be defined on some product space. But in the first example of it's application it gets used like this:
T\int_0^t\int_0^t \mathrm dt' \mathrm dt'' A(t')A(t'') = \int_0^t\int_0^t \mathrm dt' \mathrm dt'' TA(t')A(t'') = \int_0^t\int_0^{t'} \mathrm dt' \mathrm dt'' A(t')A(t'') + \int_0^t\int_{t'}^t \mathrm dt' \mathrm dt'' A(t'')A(t') = 2 \int_0^t\int_0^{t'} \mathrm dt' \mathrm dt'' A(t')A(t'')

While the last equation certainly holds, the time ordering operator seems to work more like an "I'll do whatever you wish" wonder operator. It gets an ordinary linear operator, which is in no way time dependent, writes it as an integral, enters the integral and writes the integrand as a product, then does it's magic on the product. As I understand it, that's not at all well defined (I can easily write the same Operator T works on as a different integral or just a single product and obtain a different result). Is there any way to do this right?

And while we're at it, is there a mathematical definition of path integrals? What we did in the lecture was more than fishy and wikipedia is not much better.
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Mindworm

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### Re: QM - Time Ordering Operator

I'm not a mathematician (but a physicist!), so I don't know if I can help you that much. But here's a few quick points:

1) In my lectures, I always learned that there is indeed a mathematical definition for path integrals. I never looked it up though. The problem is, that if you do things rigorously, you're dealing with pretty badly behaved functions. So it's not trivial to do properly in the context of a physics lecture. From what I remember, the correct mathematical definition is in the context of statistics and Markov Chains. If you have done some statistics, you should be able to find more.

2)
Mindworm wrote:That's as strange a definition as I've ever seen, and it looks like this should be defined on some product space.

Well. From the definition it's a bilinear map from the space of operators to the space of operators. Which is not that special. Simplest example is the crossproduct in 3-dim. vector calculus. Take two vectors, get one vector. Linear in all arguments.

3) Time ordering is used extensively in quantum field theory. While qft is certainly not taught with full mathematical rigor (since it's mathematics are still far from fully understood), you might find nicer/alternate definitions looking at those texts. Here you are actually quite often dealing with contour integrals in the complex plane. Then, the time ordering is related to the choice of integration contour. It's a bit of a long detour if you're taking a quantum mechanics class. But if you want to take it, see e.g. Peskin & Schroeder, page 31 and chapter 4.2.

4) Your question concerning the example is basically, why the time ordering operator commutes with integration (which is what's happening in the first equality). I'm not sure if I can answer this. But if you assume that they commute, the remaining part of the calculation has some a nice way of thinking about it (again in Peskin, page 85). If you think of the two time variables as an R^2 (a real plane), you find that the integral on the right covers a triangle, while the integral on the far left gives a square. It happens to be double the area of the triangle.
Last edited by tooyoo on Sun Jul 22, 2012 3:31 pm UTC, edited 4 times in total.
tooyoo

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### Re: QM - Time Ordering Operator

Maybe one more point. When I was taking quantum mechanics, I always hated it when lecturers were saying: "Well, this is an operator on some Hilbert space. To deal with this properly, you need to take a lecture on functional analysis, but for now, just imagine it to be an infinite dimensional matrix." Eventually I took some lectures on Banach and Hilbert spaces. I can tell you: It was a bit of a letdown. Not that it isn't interesting, but it didn't really improve my immediate understanding of QM at all. Not a bit. The unfortunate truth is, that you can really think of operators as infinite-dimensional matrices. It will probably matter if you look into series over operators and wonder about convergence. So if you're really serious about your perturbation theory, you'll have to take a look at that. Otherwise just think of linear algebra.
tooyoo

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### Re: QM - Time Ordering Operator

Well. From the definition it's a bilinear map from the space of operators to the space of operators. Which is not that special.

Except that that doesn't seem to be enough. A(t) and A(t') are both time-independent operators, so T wouldn't know the order in which it should multiply them. You need t and t' for that, which the space of operators doesn't contain. So you need t and t' seperately. But then you can't even use T on the integral in the example, which by the way you couldn't even do if it were just bilinear.
I'm not sure if I can answer this. But if you assume that they commute, the remaining part of the calculation has some a nice way of thinking about it (again in Peskin, page 85). If you think of the two time variables as an R^2 (a real plane), you find that the integral on the right covers a triangle, while the integral on the far left gives a square. It happens to be double the area of the triangle.

As I said, the last equation I understand (and this is indeed how I thought about it). But the integral shouldn't even be a proper operand for this operator, let alone commute. Even if it did, the integrand does not have a unique decomposition into a product, so one isn't allowed to use it.
To illustrate my thoughts, consider
A: R-> End(H)
t -> sin(t)*x+cos(t)*p
Since [x, p]=ih A(0) and A(pi) do not commute.
Periodicity gives A(0)A(pi)=A(2pi)A(pi)
Using T on both sides we get TA(0)A(pi)=TA(2pi)A(pi)
thus A(pi)A(0)=A(2pi)A(pi) which is a contradiction.

The unfortunate truth is, that you can really think of operators as infinite-dimensional matrices. It will probably matter if you look into series over operators and wonder about convergence. So if you're really serious about your perturbation theory, you'll have to take a look at that. Otherwise just think of linear algebra.

I will take functional analysis next semester and I know it will help my understanding of QM since we already started a bit on Sobolev spaces, of which H seems to be a union (funny enough, it turns out not to be a Hilbert space). So at least I know what space we were operating on all the time. And thinking about operators as infinitely dimensional matrices is a bit unhelpful, since it suggests a few things about them that are not true. Not that it keeps our professor from using them. Two commuting operators do not have a common base of eigenvectors. A single operator does not always have one. It holds for Hermitean (?) operators, but without a scalar product, which H does not have, I don't know how to define that. Even with it, I don't (in linear algebra we learned that conjugation on infinitely dimensional vector spaces is more complicated).
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Mindworm

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### Re: QM - Time Ordering Operator

Ah, interesting.

A simple example: suppose you had an operator A(t), and an operator B(t) = A(-t).

Then A(1)A(2) = B(-1)B(-2), but if you time-order both sides, T{A(1)A(2)} = A(2)A(1) while T{B(-1)B(-2)} = B(-1)B(-2). Our equality just became unequal. Magical.

One way to salvage this would be if A(1)A(2) =/= B(-1)B(-2), even if B(t) = A(-t). But I think that makes the integral stuff even more magical.

Also, "Hermitian."
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Charlie!

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### Re: QM - Time Ordering Operator

Alright. Apologies if this is a bit patronizing, but let's turn this not into a debate with examples and counterexamples. Time ordering has been around for a long time, works well in physics and is also part of the parts of physics that have been understood mathematically. So I think while it's nice to construct counterexamples, the best course of action is simply that one of us heads out to the library and takes a look how the thing is defined properly. A starting point might be
- Deligne "Quantum fields and strings" volume 1
- Zeidler's "Quantum field theory I: Basica in Mathematics and Physics"
Yeah. They're QFT books. There's probably also some stuff in the qm literature. one of the more eminent series on mathematical physics was written by Walter Thirring. Check volumes 3&4. I'm a bit too busy for this right now, so I'm not sure if I can get round to it. Since both of you seem to care considerably more than I do about the mathematical foundations of physics, it might also make more sense for one of you to sort this out.

Still, I can try to give a few comments. Mindworm's counterexample is flawed. (I'll tell you in a second.) However, Charlie!'s does capture the problematic part, and I don't have the answer.

Mindworm wrote:Except that that doesn't seem to be enough. A(t) and A(t') are both time-independent operators, so T wouldn't know the order in which it should multiply them.

Wait A(t) and A(t') are time-independent? Maybe I misunderstand, but is the " t " and the " t' " not time? And if they were independent, it simply means that you leave the order as is

Mindworm wrote:To illustrate my thoughts, consider
A: R-> End(H)
t -> sin(t)*x+cos(t)*p
Since [x, p]=ih A(0) and A(pi) do not commute.
Periodicity gives A(0)A(pi)=A(2pi)A(pi)
Using T on both sides we get TA(0)A(pi)=TA(2pi)A(pi)
thus A(pi)A(0)=A(2pi)A(pi) which is a contradiction.

I think your example is a bit flawed. You have two choices: (a) Either you do regular quantum mechanics. Then time is not an operator, but just a real variable. space and momentum can be. And they do not commute. However, your map t-> sin(t) x + cos(t) p is then not a map from the reals to the reals, but from the reals to the space of operators. Then, A(t) is not dependent on time, but on operators, and time ordering is not sensible. As an aside, concluding that A(0) and A(pi) do not commute because they depend on t is not clear either (trivial example: A(t) = any c-number).

(b) You consider quantum field theory. Here, both time and space are just numbers, parametrizing a set of operators (the quantized fields). Then, you won't need the commutation relation.

Charlie! wrote:A simple example: suppose you had an operator A(t), and an operator B(t) = A(-t).

Then A(1)A(2) = B(-1)B(-2), but if you time-order both sides, T{A(1)A(2)} = A(2)A(1) while T{B(-1)B(-2)} = B(-1)B(-2). Our equality just became unequal. Magical.

One way to salvage this would be if A(1)A(2) =/= B(-1)B(-2), even if B(t) = A(-t). But I think that makes the integral stuff even more magical.

Also, "Hermitian."

Yeah. It's a nice example. It somehow distills the formal issues "Mindworm" was getting into. I'll let you know when I can think of something smart.
tooyoo

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### Re: QM - Time Ordering Operator

Its best to think of time-ordering as simply a notation, not an operator. If you don't use the time-ordering notation, the equations for path integrals and certain perturbative expansions are longer and uglier, but you can write them down. Its just a compact way of writing down the equations that show up all over the place. I can write down a bunch of integrals over various time regions with different ordering, or I can use the 'Time ordered integral' notation.

Also, some path integrals have nice mathematical formulations (Ito calculus/Wiender Processes/Levy processes), but unfortunately lots of the physically relevant situations simply aren't rigorous.
SU3SU2U1

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### Re: QM - Time Ordering Operator

Wait A(t) and A(t') are time-independent? Maybe I misunderstand, but is the " t " and the " t' " not time? And if they were independent, it simply means that you leave the order as is

Maybe I should use less description and more formulas so the idea becomes clearer:
When I say "A is a time-dependent operator", I mean A is a continuous function R->End(H)
A(t) and A(t') (t, t' in R) are therefore Elements in End(H) (what we call "time-independent operator" or simply "operator").
Lots of physicists don't distinguish between an element of a space and a function into that space, which leads to miscommunications like this one.
Then time is not an operator, but just a real variable. space and momentum can be. And they do not commute. However, your map t-> sin(t) x + cos(t) p is then not a map from the reals to the reals, but from the reals to the space of operators.

That's what I was doing and yes, A is a map from the reals to the space of operators, which is a time-dependent operator. Unless I'm very much mistaken.
Then, A(t) is not dependent on time, but on operators, and time ordering is not sensible.

I don't understand. A(t) is an operator and does not depend on anything.

Its best to think of time-ordering as simply a notation, not an operator.

So T would be a function on formal strings into formal strings. That's not at all what my professor said, but it makes sense. Thank you.

Also, some path integrals have nice mathematical formulations (Ito calculus/Wiender Processes/Levy processes), but unfortunately lots of the physically relevant situations simply aren't rigorous.

Does that mean "the" path integral isn't defined, but there are some specific situations in which it can be?
I don't suppose there's a book or website which explains where and how path integrals can be formulated.
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Mindworm

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### Re: QM - Time Ordering Operator

For what it's worth, physics texts also sometimes define things via Heaviside's theta function
T[A, B](t_1, t_2) = \theta(t_1 - t_2) A(t_1) B(t_2) + \theta(t_2 - t_1) B(t_2) A(t_1),

but I'm pretty sure you won't be happy with that either.

As I wrote a few times, and as you wrote in your last comment, you'll really have to take a look at the mathematical physics literature. There might be a very little on the web, but I doubt it. The only text I can think of is Thirring:
Walter Thirring, here's an Amazon link
but if you take a look into the "also viewed" section at amazon, you'll find a number of texts like this, or this, or this, or this.

As a final note: The difference between physics and mathematics is - among other things - that you can work with things that you haven't defined rigorously. You do actually quite often. The simple reason being, that you always have experiment to check whether what you're doing makes sense. So when you learn physics, you simply take definitions such as the above, run with it and get a feel for it. It is rather common for maths students to find this disturbing.
tooyoo

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### Re: QM - Time Ordering Operator

That sort of attitude irks me though. I am firmly a physicist, but we need to be sure the emperor actually has some clothes on underneath it all. If you are making a physical approximation (eg, "no air resistance") you are generally upfront about it. When you play fast and loose with mathematics, this must be done with some awareness and not a purely cavalier attitude that eventually an experimentalist will come and clean up after you.

But with time ordering, I do think it makes the most sense as notation rather than an operator, or in QFT as a prescription for a contour integral.
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### Re: QM - Time Ordering Operator

If you are doing theoretical particle physics, you definitely need to make sure the math makes sense on your own, because the experiments might be a long, long time away.
chenille

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### Re: QM - Time Ordering Operator

doogly wrote:That sort of attitude irks me though. I am firmly a physicist, but we need to be sure the emperor actually has some clothes on underneath it all.

To preserve my sanity I am working under the assumption that most things could theoretically be well defined, if one was willing to get at least a bachelor degree in maths before doing any physics (like the space QM happens in. And in what way Dirac distributions can be made a base of it, seeing as there are also functions in it). Otherwise I don't know why it should produce results at all. This theory gains a bit of credence from the definitions of manifold and Lie-group I had to endure in that lecture (a group that depends continuously upon a parameter.. what?), since I know what these things actually are.
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Mindworm

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### Re: QM - Time Ordering Operator

Yeah that does not sound like a lie group at all. I feel for you.
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### Re: QM - Time Ordering Operator

Me too, but while it's an awful definition, it does make some sense. A Lie group is one that is also a differentiable manifold, so the elements can be specified by coordinates according to some continuous function. If you were foolish enough to read the "Pressures" thread, you can say you've seen worse ways to define coordinate systems.
chenille

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### Re: QM - Time Ordering Operator

a group that depends continuously upon a parameter.. what?

This is a decent 'working-man's' definition of a Lie group- and if you've worked with more mathematical formalisms for Lie group's you should be able to work out why (can you put coordinates on the manifold?)

In general what a physicist cares about is 'how can I tell if I can represent this physical operation with a Lie group?' and having a definition along the lines of the one above works well for that.

And remember- there is a lot of physics that simply hasn't been made rigorous yet. You can't panic everytime you see something you don't know how to make rigorous.
SU3SU2U1

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### Re: QM - Time Ordering Operator

doogly wrote:That sort of attitude irks me though. I am firmly a physicist, but we need to be sure the emperor actually has some clothes on underneath it all. If you are making a physical approximation (eg, "no air resistance") you are generally upfront about it. When you play fast and loose with mathematics, this must be done with some awareness and not a purely cavalier attitude that eventually an experimentalist will come and clean up after you.

Ah - I didn't mean that you throw everything overboard and simply go for it. So thanks for emphasizing that. But these days there's certainly a continuous spectrum from pure mathematics with full mathematical rigor to experimental physics where you have the certainty of your data.

So while I think that I agree with you in principle - you really have to do some cleaning up and don't just leave it to others, I have encountered so many maths students complaining about lack of rigor on physics, that I think it's important to point out that physics is not maths. To add a little to this:

1) It's not realistic to try to teach students always fully rigorous mathematics when learning physics. Otherwise you'll never get anywhere. Those students I know who focussed on mathematics too much know a bit too little of experiments, and those who did too little maths have some inadequacies when rigor is important. In quite a few areas of high energy theory you need to have some idea of algebraic geometry and topology, differential geometry and topology, partial differential equations, functional analysis, etc. before doing any research whatsoever - and that on top of having a firm founding in actual physics and experiment. So while the above definition of a Lie group is awful, you see it really often. And while I hate it, I have to admit that it works really well. So I think you have to learn some pure maths when studying physics, so you know the right places to look things up, so that you know how to formulate and understand a proof and what it means to be rigorous.

2) To give an example: Take textbooks on general relativity. Quite a few are somehow differential geometry textbooks in disguise. So while you learn pretty good definitions of what a manifold is, it's sometimes a bit obscured why GR is set up that way in the first place. I think the same might happen if you give a quantum mechanics lecture to undergraduates wondering about spectral theory of linear operators. They might learn all the maths, but it might escape them why we use this definition and not another one. If you take Weinberg's "Gravitation and Cosmology", it's a really nice point that he always let's the physics take precedence over the maths.

3) Similarly, if you work in theory, you need to have an idea when you have to be rigorous and when you don't. So I think you can justify your calculations in two ways: Either, you know the maths is 100% correct. Or you know that it agrees nicely with experimental data and the general intuition you get from theory. Ideally you have both, but quite often you only get one (or a mixture of the two). And I don't think that's so bad. After all, if you get a good model of actual physics, there's still time to think how you can make things rigorous.

4) Also this:
SU3SU2U1 wrote:And remember - there is a lot of physics that simply hasn't been made rigorous yet. You can't panic everytime you see something you don't know how to make rigorous.
tooyoo

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### Re: QM - Time Ordering Operator

SU3SU2U1 wrote:
a group that depends continuously upon a parameter.. what?

This is a decent 'working-man's' definition of a Lie group

It sounds like it would imply all Lie Groups are connected though.
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### Re: QM - Time Ordering Operator

It sounds like it would imply all Lie Groups are connected though.

You can start from the continuous-parameter definition, and start defining your tools (generators, Lie brackets, etc) and show the formulations are equivalent. It seems somewhat backwards to mathematicians but its probably the more common approach to teaching physicists. I think of the Lorentz group (or maybe Poincare group) as the most common disconnected Lie group in physics but when I teach QFT I always start from the transformations that depend on continuous parameters (boosts, translations, rotations). I find this approach works for the way physicists think (or at least, how I think, and so how I want them to think). Start with the physical operation, and then build the math structure.

I once taught a QFT for mathematicians at a summer school and this approach was a disaster. I had to go back and define the structure, define the Poincare group as isometries in Minkowski space, and do everything 'backwards' from my normal approach. In the end I had developed these generators and showed how we could interpret them as various operations. Both approaches work, and are complementary.
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### Re: QM - Time Ordering Operator

@Charlie! and the OP:
There is no contradiction; the time ordering operator is not a map of operators, but rather a map that takes a tuple of time-dependent operators A(t), B(t), C(t)... AND a tuple of times t1,t2,t3... and yields an operator (clearly not time dependent). Ergo, one cannot just time order the product A(1)A(2) because it is just a time independent operator; What one means by

T(A(1) A(2))

is actually T[A(..),A(..)](1,2) (read as "the time ordered product of the time dependent operator A(..) with itself evaluated at times (t1,t2)" ). The equality

A(t1)A(t2)=B(-t1)B(-t2)

for arbitrary t1 and t2 is just "numerical" equality of time independent operators and not an equality of tuples [A(..),A(..)](t1,t2) = [B(..),B(..)](-t1,-t2) since B(..)!=A(..) -- note that the fact that B(-t)=A(t) is entirely irrelevant as far as the tuples are concerned.

The definition given by tooyoo using Heaviside's theta is perfectly rigorous, and can easily be generalized to arbitrary finite products of operators.

Now the interesting reason there's an apparent paradox in using time ordering naively is that B(t)=A(-t) is time-orientation reversed with respect to A(t), as an operator-valued curve, and time ordering "detects" this of course -- otherwise it would not really be time "ordering" at all! Look at http://en.wikipedia.org/wiki/Differential_geometry_of_curves#Reparametrization_and_equivalence_relation for details.

Anyway, in my experience time ordered products in physics appear in exactly two contexts in my experience:

1) When solving for the propagator of an equation
\dot{\Phi}(t)=A(t) \Phi(t)\ and \ \Phi(0)=1
where A and Phi are linear operators, the solution can be expressed as a time-ordered exponential,
\Phi(t)=T\exp \left(\int A(t) \ dt\right)

2) Inside correlation functions

where one does not need to consider such subtleties.

EDIT: @ the OP: Well about mathematical formulations of path integration... you might want to read A Modern Approach to Functional Integration by Klauder, where he gives a construction based on (his?) coherent-space path integral, which is a phase space path integral interpretation of the transition amplitude from one coherent state to another which solves the problem nicely --at least for QM in flat space-- for a rather large class of hamiltonians IIRC. Another related approach is expounded upon in Functional Integration: Action and Symmetries by DeWitt-Morette and Cartier but that book is much more difficult to read. There's also the book by Glimm and Jaffe but I have not read it yet. Finally, you may want to look into white noise analysis; one (possibly the only) textbook is Lectures on White Noise Functionals by Hida and Si Si which I have only started to read, but you should probably find introductory papers aiming at applying the results of the theory to QM and QFT.
akalexke

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### Re: QM - Time Ordering Operator

@akalexke:
That would be fine, except that's not how the time ordering operator is used(it usually doesn't get this much arguments). Look at your own example. I think using it as a simplification of notation is the right idea, since your definition, while rigorous, already implies that it can't be used for much else.

I will look into the books you mentioned for path integrals, they sound like they would be in my university's library.
@SU3SU2U1
The continuous parameter "definition" is at most a reminder or a help to imagine what it "looks like". It's certainly not equivalent to any definition of lie group for exactly the reasons doogly and others mentioned. If for no other reason, then because this definition does not establish any kind of relationship between the group action and the topology.

tooyoo wrote:1) It's not realistic to try to teach students always fully rigorous mathematics when learning physics. Otherwise you'll never get anywhere. Those students I know who focussed on mathematics too much know a bit too little of experiments, and those who did too little maths have some inadequacies when rigor is important.

I don't see why I can't have the cake and eat it, too. Learning formal definitions does not push physical interpretations out of my head.
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Mindworm

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### Re: QM - Time Ordering Operator

Mindworm wrote:I don't see why I can't have the cake and eat it, too.

Yeah, this is definitely how I like to do things.

I think it works very well when you understand the reasoning and history behind definitions. On the surface a definition can seem like it is the most arbitrary thing there is, because it is a starting point logically. A great example though is with a manifold, if you can explain what that extra nonsense about Hausdorf and second countable is doing tagged on to the end of the perfectly reasonable locally Euclidean story, it is helpful.

And oh, how much happier would QM students if they understood states as elements of Hilbert space rather than as functions defined on configuration space!
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### Re: QM - Time Ordering Operator

Mindworm wrote:I don't see why I can't have the cake and eat it, too. Learning formal definitions does not push physical interpretations out of my head.

And I think that that's the best approach you can take. There is no question that it is much better to understand both mathematics and physics.

What I tried to point out was that it is not necessary to be rigorous when you learn something for the first time. And when you're teaching a class, you might find that the majority of your students will profit from the omission of some rigor. This holds even in mathematics. A few years ago, I was the TA for a first semester analysis lecture. For some reason the professor thought it was necessary to discuss the axiom of choice in the first lecture. As a TA who needs to make sure that the students understand what's going on, you wonder if it's really necessary to discuss that with a bunch of first semester students who don't know how to write a simple proof.

It's similar with the Lie Group thing. If you are teaching QFT, it's eventually necessary for the students to know what a Lie algebra is. Eventually you will encounter problems in physics that require the global structure, but perturbative QFT is by and large a local thing. Unless your students are all exceptional or have way too much time, you might want to just do a quick introduction that gets you to generators and structure constants, maybe also some idea what a Killing form is - i.e. local properties - and then get on with the QFT lecture, which is difficult enough the first time round. And under those circumstances, it's absolutely perfect to use that approach. Of course, it's not a definition, but it's an approach.
tooyoo

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### Re: QM - Time Ordering Operator

Ugh.

OK, to scale it back, a great example for me is the cross product. It is not a vector, not really. This is something I do actually tell first year students learning angular momentum in my mechanics class. They often learn about vectors for the first time in a serious way in this same course. They are used to things pointing in a direction. Having some understanding that something is going on differently with angular momentum than with momentum, that there is something deep going on such that P points in the direction that a thing moves, and L points in a direction that a thing moves around, and the reason they are extremely different is not that your professor is delusional but that some mathematical structure is happening in a rich way, I think actually adds to their understanding.

I'm not saying give them a lecture on differential forms, but "cross products are different from regular vectors" is digestible and useful.
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### Re: QM - Time Ordering Operator

The continuous parameter "definition" is at most a reminder or a help to imagine what it "looks like". It's certainly not equivalent to any definition of lie group for exactly the reasons doogly and others mentioned. If for no other reason, then because this definition does not establish any kind of relationship between the group action and the topology.

All I can say is- sit down and work through it and you may find yourself surprised. The parameters supply you with coordinates, and you can use that to build up the topology, which is of course backwards from the usual way of doing things because mathematicians want to be as general as possible. I'm fairly certain the relatively new Stone and Goldbart mathematical physics book goes through the details of building up the usual Lie group structure starting from what I will call the more physicsy definition, but I'm not in my office to check. It might be worth having a look.

I don't see why I can't have the cake and eat it, too. Learning formal definitions does not push physical interpretations out of my head.

No, formalisms are often really, really helpful. But you have to be careful with this! The drive to formalize is great, but at this point in the history of science/mathematics the useful physical theories are out ahead of their formalization. We can't stop using perturbative QCD to do calculations just because no one knows how to make it rigorous- it consistently gives unambiguous, correct answers.

And oh, how much happier would QM students if they understood states as elements of Hilbert space rather than as functions defined on configuration space!

Isn't this how most of the relatively modern (post Von-Neumann even!) book teach it?

I'm not saying give them a lecture on differential forms, but "cross products are different from regular vectors" is digestible and useful.

I generally use the old term pseudo-vector, and its pretty easy to show it behaves differently under reflection. When I teach physics majors, I generally use this as an opportunity to talk about parity, and how spinning objects give you a chance to investigate parity violations
SU3SU2U1

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### Re: QM - Time Ordering Operator

All I can say is- sit down and work through it and you may find yourself surprised. The parameters supply you with coordinates, and you can use that to build up the topology, which is of course backwards from the usual way of doing things because mathematicians want to be as general as possible. I'm fairly certain the relatively new Stone and Goldbart mathematical physics book goes through the details of building up the usual Lie group structure starting from what I will call the more physicsy definition, but I'm not in my office to check. It might be worth having a look.

Indeed I know how manifolds can be defined by having a set and functions on open subsets of R^n into that set. I'm just saying that the"physical definition" only mentions one function, does not imply that said function is bijective or even locally bijective and does not mention the group action at all. Even if you take one function to mean an atlas and the functions to be bijective, which is not implied, you get at most a manifold which is also a group. Which does not a Lie group make.
We can't stop using perturbative QCD to do calculations just because no one knows how to make it rigorous- it consistently gives unambiguous, correct answers.

We could, but I see how it might not be helpful. But in my opinion, everything that could be well defined, should be. We lose no physical intuition by doing this. And who knows - it may even lead to using mathematical theorems for the structure you just defined, which point you in the direction of interesting new experiments. I don't actually know if that happened before (only the other way round), but I don't see why it couldn't.
The cake is a pie.
Mindworm

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### Re: QM - Time Ordering Operator

SU3SU2U1 wrote:
And oh, how much happier would QM students if they understood states as elements of Hilbert space rather than as functions defined on configuration space!

Isn't this how most of the relatively modern (post Von-Neumann even!) book teach it?

Perhaps you are familiar with a class of students that understands things the way books talk about them? May the sun never cease to smile on you. Or at least, they can talk about it this way, but fretting over EPR reveals that they are still thinking of the state as smeared out with a little bit over there and a little bit over here.
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### Re: QM - Time Ordering Operator

Mindworm wrote:We could, but I see how it might not be helpful.

I'm sorry, but this is simply a no. Physics is not mathematics. The two complement each other. Which is why it's so much fun.

I can agree with your other statements. It's indeed how maths and physics work. They provide ideas for each other. But if you always insist on formalism, you are simply robbing yourself of an approach to solve problems. As long as you don't pretend that something lacking it is rigorous, you're totally fine. And most people don't do that.

Concerning states in Hilbert spaces: I've seen both kinds of students. Depending simply on the level of the university. In general, you can expect that someone who has enjoyed a rigorous lecture in linear algebra will appreciate a state/operator formalism.
tooyoo

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### Re: QM - Time Ordering Operator

I would not read Mindworm's "you could" as a sincere granting of permission to scrap QCD if it suited you. This is one of those rhetorical techniques you hear so much about.

And maybe some comfort with linear algebra would help. My point is more that when the situation gets sticky (like EPR) the thinking reverts back to the simpler picture.
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### Re: QM - Time Ordering Operator

doogly wrote:I would not read Mindworm's "you could" as a sincere granting of permission to scrap QCD if it suited you. This is one of those rhetorical techniques you hear so much about.

Which is why I pointed out in the next line that I can agree with the remainder of his statement. Reading over things, I think at this point in the discussion I do agree with most of what all three of you are writing.
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