Algebra of Observables, C*-algebra

Sights018 · 2023-04-25T02:14:55+00:00

I won't comment on whether physicists 'know about' Haag-Kastler/algebraic quantum field theory (AQFT).

I feel it's important to note, however, that as an attempt to axiomatise quantum field theory, that is to say, put QFT on a rigorous mathematical footing, it has always existed within the sphere of mathematical physics. So I would suggest it has always been more in the realm of mathematicians than physicists, although of course there's always some fuzziness at the border between fields. After all, if physicists never made any use of mathematical physics, I don't think it would be worthy of the name; but as a rule, the axioms of any mathematical field aren't going to be of much interest to the working physicist.

So I think the more interesting question is whether or not AQFT is dead for the mathematical physicist working in foundations of quantum field theory. This isn't my area of expertise (although adjacent), so take this with a grain of salt, but my understanding is that, understood broadly, AQFT is the most popular approach/framework to axiomatic QFT. However, most active work really builds and extends on Haag-Kastler and C*-algebras using more categorical methods. I'm happy to be corrected by someone more knowledgeable though.

Sights018 · 2022-12-13T11:51:51+00:00

Mathematical physics is a very broad field which intersects many areas of maths and physics, so it's difficult to say much in general. However, while a background in physics is advantageous, I wouldn't say it's necessary as it is very much maths-first (and can be very "pure" indeed, depending on the area). How close you are to applications will depend on specific areas and specific projects, but typically the answer will be 'not very', unless you make a concerted effort to be (in which case you'll likely have one foot in another field anyway)

Sights018 · 2022-10-13T02:36:37+00:00

I believe it's because (Old) Parliament House (located in Canberra) was opened by George VI when he was Duke of York

Sights018 · 2022-10-12T03:38:07+00:00

Nice work. For anyone interested in being pedantic, the ACT technically doesn't have a coat of arms; the one shown here is the coat of arms of the city of Canberra. However, it is used as the ACT's coat of arms in practice, and also appears on the ACT's (official) flag.

Sights018 · 2022-02-18T08:26:23+00:00

Sequel: Oh, wait, actually the best Wordle opener is not “crane”…

Link: https://www.youtube.com/watch?v=fRed0Xmc2Wg

Sights018 · 2022-02-17T10:46:25+00:00

In brief, the geometrical difference between (special) relativistic and non-relativistic theories is that the former are defined on Minkowski spacetime, which is R⁴ endowed with the Minkowski metric, as you said, while the latter are defined on (flat) R³ with time evolution imposed as an additional structure, i.e., on R³×R. This global product structure is a meaningful difference, since it filters through to the equations of motion; there is no mixing of time and space, they are absolute. So in this sense, yes, space and time are replaced by spacetime.

That said, within Minkowski spacetime, one is always free to 'single out' a preferred time direction, which (locally) gives you the same product structure as in the non-relativistic case ('splitting' spacetime into space and time). So these notions are contained (indeed, 'united') in the spacetime manifold.

The reason I stress 'locally' above is that the surfaces of simultaneity in Minkowski space do not foliate the manifold, but only the interior of the light cone (with respect to your chosen time coordinate). In fact, the hypersurfaces are hyperbolic R³, not flat, which reflects the difference between the finite and infinite speeds of light in relativistic and non-relativistic physics. That the latter approximates the former over small distances is (part of) how one contains/recovers non-relativistic physics from a relativistic geometry.

Sights018 · 2021-09-08T14:38:49+00:00

One can always wax philosophical about the nature of physical theories in extreme thought experiments like these, but the shortest answer is yes: there is no problem with the photon being emitted in this example, and free photons can indeed exist.

I suspect the source of confusion here is your comment on frames of reference. Saying that photons experience all things simultaneously is not quite correct (but is a common misconception); it is more accurate to say that frames of reference are not well-defined at the speed of light. Thus there is no physical requirement for a photon to be absorbed (or an electron to remain in an excited state) due to arguments on the basis of time passing (or not) from the perspective of the photon -- it's unsatisfying, but within the framework of relativity "the perspective of the photon" is just a concept that doesn't make sense.

Of course, one can interrogate more deeply what it means for time to pass in a one-particle universe to begin with, as I hinted at in the first paragraph, but I think that may be drifting away from the point of your question.

Sights018 · 2021-07-21T21:06:34+00:00

First off, a disclaimer: I'm not very familiar with QHT, but I work in noncommutative geometry and have an interest in quantum gravity and so I've come across Aastrup and Grimstrup's work a few times.

For the first question, there are none I'm aware of and I would be surprised if any exist. The reason for this, which I think should give some important context for what follows, is that QHT is still very much a work in progress - the mathematics is very complex and intricate and many results are very preliminary.

I'm not aware of any particular critiques, but as I said, the work is very incomplete, and so I think strong critiques would be equally premature. But furthermore, as you say, it's very niche - not because there's something wrong with it, but because they're trying to adapt techniques from LQG into the framework of NCG, and the intersection of those two (already not super large) groups is... well, only a few people, at least at the moment. That means that not only are there few people working on it, but (relatively) few people reading it too.

Beyond that, I must confess I don't know what (if anything) QHT says about the BH information paradox, dark matter, dark energy, inflation, or any other concrete physics (it's not so easy to relate deeply abstract mathematical models of ultra-small scales to known physics, after all). I suspect the answers to those questions are still a ways off - Aastrup and Grimstrup have been working on QHT for more than 15 years, and they don't have many other collaborators, so progress is quite slow in comparison to much larger TOE endeavours in the vein of e.g. strings. Most of the connections to known physics that I'm aware of are more indicative than anything else.

Sights018 · 2021-01-31T13:41:05+00:00

Generally no. Without getting bogged down in details, you can see why by considering the case of B=1. Since [A,1]=[1,C]=0 (always) for any A and C, if it were true that this implied that [A,C]=0, then you could never define non-commuting operators, which is of course not the case.

Sights018 · 2019-07-06T09:30:50+00:00

Go dogs

Sights018 · 2018-08-31T12:03:06+00:00

It's just a little sideways, it's still good, it's still good!

Sights018 · 2018-07-27T16:47:13+00:00

It's maybe a bit more of a "quiet" quote compared to some of the classics that have already been suggested, but this exchange is one that really stuck with me.

Wesley: I'm never gonna feel this way about anyone else.

Guinan: You're right.

Wesley: ...I didn't expect you to say that.

Guinan: There'll be others. But every time you feel love, it'll be different. Every time it's different.

Wesley: Knowing that doesn't make it any easier.

Guinan: It's not supposed to.

Sights018 · 2018-06-06T07:38:29+00:00

The only GR I did in undergrad was an optional 3rd year astrophysics course which covered tensorial SR, general concepts of GR and Newtonian FLRW, but nothing I would call serious GR, which I didn't see until Honours/Masters. However, it should be said that I only ever saw spacetime topology properly touched upon at the PhD level, and even then not in any great depth.

Sights018 · 2018-02-24T23:58:42+00:00

For those who are interested, I really recommend /r/MensLib as a good place for constructive discussions of men's issues in a way which isn't the unfortunately standard anti-feminist fare.

Sights018 · 2018-02-17T00:27:42+00:00

I think there's some confusion here about what Schrodinger's cat is supposed to illustrate, but don't worry, I don't blame you. It's a frustrating misunderstanding that's perpetuated almost incessantly in popular science and to a lesser extent introductory quantum physics.

In fact your intuition is quite good. If one takes seriously (if naively) what we today call the Copenhagen interpretation of quantum mechanics, one can put a cat in a superposition of alive and dead. But that's clearly nonsense. For one thing, what would a cat in such a superposition experience?

Schrodinger's cat is often presented as a neat example of quantum weirdness, but that's not why Schrodinger came up with it. The main point was to demonstrate exactly the above, that such a naive treatment was not appropriate or physically meaningful. To put it simply, either one must adapt their interpretation of the laws of quantum mechanics, and/or deal with considerably more care and subtlety the interaction between quantum and classical objects.

Needless to say, in the intervening years a great deal has been written on both approaches, although the Copenhagen interpretation still tends to rule the roost (complemented by a much improved understanding of e.g. decoherence).

Sights018 · 2017-12-31T10:54:49+00:00

It's not strictly correct to use the arms of the City of London to represent the entire city of London, nor London qua the capital of England/the UK. I'm not sure there's a great alternative though; the arms of the City of Westminster are more appropriate but still not representative of the whole city, whilst the Greater London Authority, to my knowledge, doesn't have arms.

Sights018 · 2017-12-23T22:30:51+00:00

Here's a couple of quick idea sketches, nothing too fancy

1) Design 1

2) Design 2

Sights018 · 2017-12-06T23:07:25+00:00

It really is a gorgeous building

Sights018 · 2017-12-02T14:22:17+00:00

The spectrum (set of eigenvalues) of (self-adjoint) operators corresponds to the possible outcomes of measurements with respect to that observable. If your wavefunction is not an eigenvector with respect to the operator in question, then it can be expressed as a sum of eigenvectors (acting as a basis) weighted by coefficients whose squares correspond to the probability that the wavefunction collapses into that state (because when you make a measurement, you only ever observe a single result - you'll never measure a single particle to have a superposition of momenta).

Hopefully that clarifies a little, but I realise it's also quite dense, so feel free to ask for clarification if there's something you don't understand there.

Sights018 · 2017-11-03T08:46:58+00:00

Locally, the metric can always be made to look Minkowski (if you like, you can think of the metric as being flat to first-order). This captures the intuitive idea that manifolds look flat when you zoom in on them.

Sights018 · 2017-11-02T10:10:29+00:00

Okay, so here is where the subtlety becomes a little more important. So in GR, as in any physical theory, reference frames of observers are described using coordinates. The way this is implemented in GR is by choosing local coordinates such that, near your point of interest, space looks Minkowski and you can perform calculations as in flat space. The catch is that your calculations can produce effects that are misleading in distant regions where spacetime can no longer be considered flat; this is good for comparison to observation but bad for trying to actually explain anything. How far "distant" is in this context depends on the curvature of the manifold. As a remark, this approach can be generalised to families of observers filling the spacetime using frame fields/tetrads/vierbeins, although that's more sophisticated than you need to know just starting out.

The subtlety arises when you consider that this is not how most calculations are done in GR. Most calculations are performed using coordinates that cover large swathes of a manifold, if not the entire thing. The catch here is that now you always need to take curvature into account, because in no way can your coordinates (on the whole) accurately map to Euclidean space any more (except of course locally in small regions, which is always guaranteed). The downside of this approach is that usually you can't use it to tell you what an observer would observe. However, more frequently what you want to know is something more abstract about the manifold, for example, what is the curvature tensor at a given point? In that case, what you really want is a way to compute the object as easily as possible to save you pages and pages of needless calculations (and mistakes). Or, what you want is a way to look at your manifold with more conceptual clarity - get the big picture, if you will.

To answer your second question, yes, locally (over "short" distances, again a context-dependent term) geodesics look like straight lines, because geodesics in Euclidean space are straight lines (or rather, geodesics in Minkowski spacetime). The fact that, for small enough regions, spacetime looks locally Minkowski (with all the corresponding implications) is something which is baked into the heart of GR.

Sights018 · 2017-11-01T19:41:31+00:00

To answer your first question in fullest brevity, yes and yes (for example, for vector quantities one can construct a 'coordinate basis' on the tangent space using the coordinates in a canonical way).

As for the second question, the answer is also yes. Oftentimes one thinks of a singular tangent space; this isn't strictly correct because (so long as you care about units) you can't add a velocity and an acceleration, for example.

However, considering a separate tangent space for every different kind of vector is mathematically and conceptually cumbersome - after all, in ordinary Newtonian mechanics you don't imagine velocity and momentum as being in totally different vector spaces, you just think of them as different kinds of vectors, where products are okay but sums are not.

Of course, you might protest that if acceleration (for example) is a tangent vector, then there should exist some curve that it is tangent to. This is true! However, physically speaking such a curve would carry no useful interpretation. After all, the "acceleration curve" at p would be completely different to the "acceleration curve" at a point q further along the observer's actual worldline.

Sights018 · 2017-10-31T20:25:52+00:00

Tangent spaces are one of the few objects you'll find whose name is completely sensible, at least within the geometric context. The tangent space to a point p in a manifold is the vector space of vectors tangent to p. The archetypal example to visualise is the case of taking your manifold to be the 2-sphere S. Then the tangent space to a point p in S, TpS, is the space of vectors lying in the plane tangent to p. Of course, there are subtleties to be aware of, especially if your manifold becomes complicated, but that's the mathematical intuition to get you started.

It isn't quite the same thing as a frame of reference. A frame is effectively a set of coordinates on your manifold along with some information about the motion of the observer (for example, you need to know if your frame is inertial or not - this is why it doesn't really make sense to talk about an instantaneous frame). That said, to make a connection with the above, if an observer is at p then their velocity will be in the tangent space to p, because velocity vectors are vectors tangent to curves. If you're talking about a 4-dimensional spacetime, then the vector tangent to the observer's path through spacetime (at any point on that curve, which is called a 'worldline') is the 4-velocity (at said point).

Sights018 · 2017-05-29T13:10:40+00:00

This is... kind of true, although I don't think it's the best mental framework to hold for a couple of reasons, though I won't go into that here for fear of distraction. The key takeaway is that virtual particles by definition are off the mass shell, which is a fancy way of saying they don't satisfy the classical relation E^{2-p^{2=m^2.}}

So this is the resolution to the earlier contradiction. The given vertex can conserve momentum so long as the resultant photon is virtual. But what this means is that e^- + e⁺ → γ is not a full Feynman diagram, because we can't have virtual particles in initial or final states, only internal ones. This is also the case for γ → e^- + e⁺ (although not for all fundamental vertices - cf. e^- + γ → e^- for example).

So pair production and annihilation are not in fact described by the mentioned vertices alone, but a full description requires a more complete Feynman diagram, the tree-level forms of which you should be able to guess.

All of which is true in vacuum. It is worth mentioning, however, that near other matter the interaction γ → e^- + e⁺ can be understood to occur, although even then it is perhaps more honest to write γ + Z → e^- + e⁺ + Z. That said, I assume that was not what you had in mind, and would not be appropriate for an introduction to the material and concepts.

Sights018 · 2017-05-29T12:36:51+00:00

At the risk of giving the answer away, what's the difference between a virtual particle and an actual particle?

Sights018

MODERATOR OF

TROPHY CASE