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antimony

01:23

my goodness that frequency energy question has created a mess. as a student the different answers has really muddled my understanding

the post has now been edited 'The electromagnetic spectrum's wavelengths all travel at the same speed, c'

so now its wavelengths not particles

antimony

01:39

even if OP meant classical i don't understand why people are talking about wiggling electrons......

if all wavelengths have the same velocity, surely we're talking about a vacuum. are there any materials with no dispersion at all frequencies?

OR do people mean that a wave (or photon) in pure vacuum doesn't have energy until it interacts with something to impart that energy to? I never heard of that before???

5 hours later…

John Rennie

07:03

Part of the problem I that the original question is unclear. At first glance I read it as asking why E = hf, in which case the highest voted answer was completely wrong. However on a second read it isn't clear that was what was being asked.

When any question reaches the Hot Network Questions list answers tend to be upvoted on the basis of plausibility rather than correctness since most of the people voting are not physicists. That's just life I'm afraid.

What is annoying is that no-one has posted the real explanation i.e. the modes of the electromagnetic field behave like simple harmonic oscillators. I've toyed with writing an answer along these lines, but given the lack of clarity in the question I wasn't sure it was worth the effort.

123

07:36

Hi All..

Hello @JohnRennie Sir

What is the reason when system can not split into two parts, not able to change its state of rest or uniform motion?

Sir Cumference

07:54

honestly love this book lol

2 hours later…

Ryder Rude

09:29

@SirCumference lolol Its inventor thought it deserved a cool name

@ACuriousMind But I think the C* algbera is of observables. And observables are generators. Then why do you say that the algebra is of the group of exponentials of generators?

ACuriousMind

@RyderRude because the unboundedness of $x$ and $p$ is annoying

Ryder Rude

@ACuriousMind i don't know what unboundedness means. Is it related to topology of the algebra?

ACuriousMind

technically you have to take the algebra of actual projectors associated with $x$ and $p$ as observables if you want a bounded algebra

Ryder Rude

Yeah, the projection operators will do as observables

ACuriousMind

@RyderRude "unbounded" means that there are arbitrarily high values in the spectrum. In particular both $x$ and $p$ not only are unbounded but also have continuous spectrum

this makes them mathematically very ugly operators and hard to deal with

Ryder Rude

09:33

But then again, none od the projection operators satisfy the heisenberg algebra

And we start the theory by postulating the heisenberg algebra

ACuriousMind

so you don't really want to work with them directly - just look at the SvN theorem, while we do have uniqueness as along as the exponentiated CCR hold, we don't have uniqueness just for $[x,p] = \mathrm{i}$

Ryder Rude

@ACuriousMind that makes sense

@ACuriousMind oh

Would u say that the clifford algebra of pauli matrices counts as a C* algebra

Maybe it needs additional structure like a norm and adjoint

ACuriousMind

I don't know what the "Clifford algebra of Pauli matrices" is

Pauli matrices are a 3d Clifford algebra

Ryder Rude

Yeha

So, in abstract algebra, you will define the quantum theory of spin by writing a clifford lagbera

The 3D one

ACuriousMind

but sure, the algebra generated by the Pauli matrices is a $C^\ast$-algebra

Ryder Rude

09:37

With additional structure defined on, like a norm and self adjoint

ACuriousMind

every subalgebra of operators on a Hilbert space is a $C^\ast$-algebra and since the Pauli matrices are just operators on $\mathbb{C}^2$, there isn't any problem here

Ryder Rude

Yeah

But what about in abstract, like defining the spin theory without reference to hilbert space

Then we would just write the clifford algebra, right?

The clifford algebra would be analogous to heisenberg algbera here

But with additional structure

Because clifford algebra's definition does not have an adjoint operation

ACuriousMind

I feel this is a distinction without a difference - the norm/adjoint you need to impose is exactly the one you get from the faithful representation on $\mathbb{C}^2$

otherwise what are you even doing :P

in fact every $C^\ast$-algebra is isomorphic to the algebra of bounded operators on some Hilbert space

i.e. there aren't any "abstract" algebras that you can't represent just as operators on a concrete space. A lot of the $C^\ast$-algebraic approach to QM is just showing the whole machinery is overkill if you just want to do QM :P

Ryder Rude

I have never cared about the algebra either

In fact, the adjoint structure of the algebra is pretty much stolen from matrix adjoint

That structure makes sure that we're talking about matrices

Still, there is some beauty in describing the theory in a minimalistic way

Like in algebras

1 hour later…

Physics Meta

11:07

Q: Flag declined even thought it was (possibly) correct

I've flagged this post with the 'other' option for flags, and described it as 'Homework question'. The post was closed as a homework question, and my flag was declined. The justification was that I did not use a standard flag, and was referred to What is flagging?. In this page it is not specifie...

An_Elephant

11:34

@ACuriousMind Hi. Like mathjax works on main site, does it work on rooms as well? I have seen people typing in mathjax commands but these commands do not appear as they should look to me?

ACuriousMind

@An_Elephant see the room description for how to run MathJax in chat

An_Elephant

I am sorry but how do you see room description?(I can't find anything like this :(

Oops sorry , I get it.

3 hours later…

Daniel Underwood

14:28

Maybe I'm just getting old and turning cynical, but I'm surprised he hasn't tried to rebrand Science as Wolfram at this point wolfram-media.com/products/…

ACuriousMind

14:53

I have nothing to say about Wolfram that Feynman didn't already say in his oft-quoted letter to him :P