The h Bar

Moses

5:13 PM

@Secret Your it would make reading the less ambiguous if they did that (with regard to you operator hat comment)...

Mithrandir24601

@JaimeGallego :o I could swear it's not ham. Horrifically expensive, but so good :D

John Doe

@Mithrandir24601 I'm going to upload a pic of the definition of resolving power from the text I am using.

Mithrandir24601

5:36 PM

@JohnDoe OK, thanks - I can't guarantee you that I'm right, but this seems to be along the lines I was thinking of - I've got a few images and things that are perfect for this, so I think I'll go and make an answer (for once)

Moses

@ACuriousMind If say we know the position of a particle exactly, does this then imply that variance of the momentum would be infinite (using the uncertainty principle) and hence the energy would be infinite as well?

Mithrandir24601

5:50 PM

@JohnDoe I take it that you're taking that from "Quantum Measurement Theory and its Applications"?

ACuriousMind

@Moses Yes, and that's why an exactly localized particle is not a physically possible state.

(but it would not be the energy that's infinite - just its variance)

Slereah

Not if the particle is in a box!

Moses

@ACuriousMind Yeah I would expect the variance and not the energy. This applies to all systems where the Hamiltonians are functions of the momentum operator right?

Slereah

Well as said

Particle in a box can have well defined positions

ACuriousMind

@Slereah In a box you have issues defining a good momentum operator in the first place, though :P

Slereah

5:57 PM

what is the issue

the boundaries?

ACuriousMind

yup

You can make it well-behaved with periodic boundary conditions, but then you have a particle on a ring, not in a box

Slereah

Eh, it's probably the realistic case, anyway

Universe with periodic boundary conditions

ACuriousMind

@Moses What exactly do you mean by 'this' in that question?

In the case where we are in a box/ring, the uncertainty principle becomes a bit more subtle to apply - the eigenstates of position/momentum do not lie in the domain of definition of their commutator, so the naive uncertainty principle is not applicable (this would also hold in the infinite case, but there the eigenstates are not even states (not normalizable) to begin with, so, eh.)

Moses

@ACuriousMind The implication: infinite varaiance in momentum implies infinite variance in energy.

ACuriousMind

This question discusses that.

@Moses Yeah, well, you see, "infinite variance" is not actually a thing that exists if we are careful with our words

The variance is the difference of two expectation values, and expectation values are always finite

Moses

6:02 PM

@ACuriousMind I understand

ACuriousMind

The "infinite variance" really tells you nothing more that something in the pseudo-mathematical machinery physicists use for QM broke

In this case it's both that the eigenstates of position/momentum are often not really states, and neglecting the domains of definitions of the operators involved in the uncertainty principle

So I would abstain from ascribing any specific significane to this "infinite variance" beyond that

Slereah

it's a good heuristic

Moses

@ACuriousMind I understand, that is the exact context in which the topic is discussed in the notes I am using.

John Doe

@Mithrandir24601 Great thanks. Yeah are you famaliar with it? It's quite hard going in that book. Going from Sakurai to that is quite tricky.

Mithrandir24601

@JohnDoe I've found a couple of sources that briefly mention it and I'm pretty sure I'm thinking of the right thing, in which case, it formed part of my MSc dissertation :)

I just didn't know that it has a name...

John Doe

6:16 PM

@Mithrandir24601 :) What are the chances?...I'm sure that question is explained somewhere in the book. Have you read through this book at all?

Mithrandir24601

@JohnDoe Nope, but I've found it that page on Google books and it's the equation immediately after that made me reasonably sure that I'm thinking along the right lines - it seems to be an obscure/infrequently explained thing

John Doe

@Mithrandir24601 Yeah 'obscure/infrequently explained thing' sounds about right...I can send you a soft copy of the book, you can let me know.

Mithrandir24601

@JohnDoe Nah, I've got the pages around that area, thanks to Google. Thanks though

John Doe

@Mithrandir24601 Okay no prob

Moses

6:43 PM

@ACuriousMind If we consider $e^{x-c}$ as an operator $f(x)$ as we discussed earlier then do we consider $c$ as just a constant or is $c = cI$?

Mostafa

6:55 PM

Anyone knows what is a Mac Pizza?

Slereah

also why is the wavefunctional even $\Psi[\phi(x), t]$

Oh wait no

That's probably fine

but on the other hand

What is exactly the operator $\hat \Phi$

ACuriousMind

@Moses the latter

Slereah

It should be a distribution, so what is $\hat \Phi [f] \Psi[\phi(x),t]$

Is it just $$\int \phi(x) f(x) \Psi$$

Moses

@ACuriousMind In a text it says that a state $| \psi \rangle$ has a mean momentum of $0$. Would you assume that mean momentum means expectation value of momentum?

ACuriousMind

@Moses yes

Felix.C

7:13 PM

hey can someone please quickly explain to me what overtones are?
As far as I understand if we have a string of length $\delta$ fixed at both sides then there is the fundamental frequency $\nu$ then the wavelength $\lambda = 2*\delta$. If we now suppose an arbitrary standing wave we can write it as a fourrier series being the infinite harmonics, e.g. those frequencies being a integer multiple of the fundamental.

I heard from some sources I don't really trust, that the overtones are part of the set of the resonant frequencies, but I don't understand how adding a frequency not being a integer…

bolbteppa

8:18 PM

Is this right:

Lie algebra extension

In the theory of Lie groups, Lie algebras and their representation theory, a Lie algebra extension e is an enlargement of a given Lie algebra g by another Lie algebra h. Extensions arise in several ways. There is the trivial extension obtained by taking a direct sum of two Lie algebras. Other types are the split extension and the central extension. Extensions may arise naturally, for instance, when forming a Lie algebra from projective group representations. Such a Lie algebra will contain central charges. Starting with a polynomial loop algebra over finite-dimensional simple Lie algebra an...

$[G_1,G_2] = \frac{d}{dt} (g_1(t)g_2(t)g_1^{-1}(t)g_2^{-1}(t))|_{t=0}$?

Aspen Rand

9:55 PM

Hi

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