/sci/ - Science & Math » Thread #11261198

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2 questions for 1 thread limited proposal!

Anonymous Sun 29 Dec 08:45:24 2019 No.11261198 View Reply Original Report

Quoted By: >>11261233 >>11261531 >>11261628 >>11261822 >>11263426 >>11263641

1. Have you studied (especially as a part of your uni course taught by your lecturer) Qunatum Physics: an Introduction Based on Photons?

Seems interesting but I can't find it anywhere for free and don't really wanna pay insane dallaz

How is it?

2. What books on mathematical physics are recommended by your teachers? As I understand, mathematical physics itself has different definitions in Russian speaking and English speaking segments of academia.

Specifically, I'd like something that explains operators and their eigen functions, generalized functions - particularly Greene's and Delta functions in depth (and generally all the shit you need to solve electrodynamics and QM problems).

Anonymous

Anonymous Sun 29 Dec 2019 09:04:44 No.11261228 Report

Quoted By: >>11261233

I never get why does non-commuting operators mean that you can't measure both of the responsive values at the same time?

Anonymous

Anonymous Sun 29 Dec 2019 09:08:10 No.11261233 Report

Quoted By: >>11261235 >>11261293 >>11261566

>>11261198
Springer books are typically trash.
>>11261228
It doesn't mean you can't measure them at the same time: you definitely can. It does mean that you can't know both results with arbitrary precision. More precision in one measurement must mean less in the other. That's the uncertainty principle.

Anonymous

Anonymous Sun 29 Dec 2019 09:12:23 No.11261235 Report

Quoted By: >>11261240 >>11261262

>>11261233
So the uncertainty principle in general is applicable to any pair of parameters whose operators commute ?

Anonymous

Anonymous Sun 29 Dec 2019 09:17:17 No.11261240 Report

Quoted By: >>11261269

>>11261235
>whose operators *don't* commute
fixed it

Anonymous

Anonymous Sun 29 Dec 2019 09:35:43 No.11261262 Report

Quoted By: >>11261269 >>11261615

>>11261235
It depends on what the commutator is, but essentially yes. The inequality in Heisenberg's uncertainty principle depends on the expectation value of the commutator, and is not necessarily h-bar /2 for a pair of operators.

Anonymous

Anonymous Sun 29 Dec 2019 09:43:40 No.11261269 Report

Quoted By:

>>11261240
Yeah that's what I meant.

>>11261262
Are there are other commutators than AB - BA?

What's expectation value?

Anonymous

Anonymous Sun 29 Dec 2019 10:02:45 No.11261293 Report

Quoted By:

>>11261233
>Springer books are typically trash.
Not necessarily true, they publish a lot of translations of good books as well. And this author is from Russia - although the original textbook is in English.

Anonymous

Anonymous Sun 29 Dec 2019 10:49:00 No.11261363 Report

Quoted By:

I hate that they never tell you WHY do you have to use operators in QM.
They just throw you all the shit and then at some point much later they'll let you in on the secret.

It's like the ultimate cold open.

Anonymous

Anonymous Sun 29 Dec 2019 12:22:47 No.11261462 Report

Quoted By: >>11263048

I wonder if in US they teach physics before they teach the math that's required to use that physics

Anonymous

Anonymous Sun 29 Dec 2019 13:17:42 No.11261531 Report

Quoted By:

>>11261198
I wish we could have threads like this but apparently there's just not enough people here

Anonymous

Anonymous Sun 29 Dec 2019 13:47:24 No.11261566 Report

Quoted By: >>11263671

>>11261233
Hmm is there a formula for this, for two random parameters? Cause I know how to derive Uncertainty Principle by integrating wave pack and considering expressions for fluctuations but it's typically just t, x, p, E and w.

Anonymous

Anonymous Sun 29 Dec 2019 14:22:01 No.11261615 Report

Quoted By: >>11261625

>>11261262
wait, is time an operator?

Anonymous

Anonymous Sun 29 Dec 2019 14:30:24 No.11261625 Report

Quoted By: >>11261634

>>11261615
>wait, is time an operator?
Good question, but no. You can define a time uncertainty principle but it is a little different in form from the others

Anonymous

Anonymous Sun 29 Dec 2019 14:35:27 No.11261628 Report

Quoted By: >>11261991

>>11261198
>and generally all the shit you need to solve electrodynamics and QM problems
You don't need rigorous functional analysis proofs to actually solve problems in electrodynamics and QM. If you are interested in generalized functions open up any book on functional analysis and read the definitions, but don't blame most physics books for not containing this material

Anonymous

Anonymous Sun 29 Dec 2019 14:41:15 No.11261634 Report

Quoted By: >>11261639

>>11261625
Well I know that through Fourier transform integrating wave pack with E in exponent you can get delta(E)*delta(t) = h_bar/2 which is exactly like for x and p_x.

Anonymous

Anonymous Sun 29 Dec 2019 14:44:07 No.11261639 Report

Quoted By: >>11261662

>>11261634
And how do you define delta(t) in that case? The thing is the definition of uncertainty of t is not quite the same as x,p,E, etc

Anonymous

Anonymous Sun 29 Dec 2019 15:08:16 No.11261662 Report

Quoted By: >>11261695

>>11261639
It's the average life span of system with energy E.

You integrate a wave pack in respect to k after Taylor expanding w with respect to k ignoring terms after second and plugging in. You get sin(ksi)/ksi where ksi is 1/2*delta(k)(x - w'_0*t) and if you take delta(ksi) to be pi which corresponds to most of the area by setting x = const you can get the expression.

Anonymous

Anonymous Sun 29 Dec 2019 15:25:13 No.11261695 Report

Quoted By: >>11261698

>>11261662
In ordinary QM we usually deal with a conserved number of particles, often just one. Doesn't make a whole lot of sense to talk about life spans of particles there.

Anyway when I asked
>And how do you define delta(t) in that case?
it was a rhetorical question. I know you can write down something like an energy-time uncertainty principle but it is not the same as those involving two non-commuting operators like position and momentum. In effect you get a different energy-time uncertainty principle for every operator that doesn't commute with the Hamiltonian.

Anonymous

Anonymous Sun 29 Dec 2019 15:28:10 No.11261698 Report

Quoted By: >>11261704

>>11261695
Particles can change energy levels, so life span of system with a given energy *state* seems fairly relevant.

Anonymous

Anonymous Sun 29 Dec 2019 15:33:36 No.11261704 Report

Quoted By: >>11261717

>>11261698
>Particles can change energy levels
That's where you're wrong. Energy is conserved

Anonymous

Anonymous Sun 29 Dec 2019 15:42:16 No.11261717 Report

Quoted By: >>11261729

>>11261704
It's conserved when accounting for photons emitted or absorbed but if you just look at an atom or charged particle - because it's not an isolated system - the energy can change.

Anonymous

Anonymous Sun 29 Dec 2019 15:50:29 No.11261729 Report

Quoted By: >>11261801

>>11261717
>In ordinary QM we usually deal with a conserved number of particles, often just one. Doesn't make a whole lot of sense to talk about life spans of particles there.

Do you see why restricting to one isolated particle like we do in a basic QM course changes things? I am trying to answer your question about the uncertainty principle and its relation to operators. I am telling you that time is not an operator and the energy-time uncertainty principle, in terms of lifetimes or whatever you want, is not quite of the same form as the momentum-position one in ordinary quantum mechanics. If the only way you can think to define time uncertainty is by extending the system to quantum electrodynamics then that just proves my point

Anonymous

Anonymous Sun 29 Dec 2019 16:41:53 No.11261801 Report

Quoted By: >>11263190

>>11261729
But what is the physical reason time can't be an operator? In Relativity it's equivalent to spatial coordinates but here it seems to be separate.

Anonymous

Anonymous Sun 29 Dec 2019 16:52:17 No.11261822 Report

Quoted By: >>11261827 >>11261966

>>11261198
Introduction to Quantum Mechanics by Griffiths to start. Modern Quantum Mechanics by Sakurai afterwards

Anonymous

Anonymous Sun 29 Dec 2019 16:56:40 No.11261827 Report

Quoted By: >>11263026

>>11261822
What about Feynmann and Landaushiftz?

Anonymous

Anonymous Sun 29 Dec 2019 17:16:51 No.11261849 Report

Quoted By:

Anyone familiar with Albert Messiah?

Anonymous

Anonymous Sun 29 Dec 2019 18:37:34 No.11261966 Report

Quoted By:

>>11261822
> Sakurai
94 or 10?

Anonymous

Anonymous Sun 29 Dec 2019 18:46:15 No.11261991 Report

Quoted By:

>>11261628
Follow-up question: what's good on functional analysis ?

Anonymous

Anonymous Mon 30 Dec 2019 02:25:13 No.11263026 Report

Quoted By: >>11266201

>>11261827
>Landaushiftz
kek im gonna use this

Anonymous

Anonymous Mon 30 Dec 2019 02:34:31 No.11263048 Report

Quoted By: >>11263729

>>11261462
The instance of this I experienced was my required 3rd year fluid mechanics course requiring solving PDEs for a major that doesn't even require taking PDE.

Anonymous

Anonymous Mon 30 Dec 2019 03:35:35 No.11263190 Report

Quoted By: >>11263671 >>11264143

>>11261801
If you have an operator like a time operator that has a canonical commutation relation with the hamiltonian, that implies (by the Stone-von Neumann theorem) that the spectrum of energy is unbounded from below. But for the sake of stability we always want a lowest ground state.

If you want to introduce relativity, you are right, space and time need to be treated on an equal footing. But what actually happens is that you can't define good spatial position operators either. There are still momentum and energy operators, but there is no position operator that satisfies all the properties you would expect a position operator to have in relativity.

I should say, you can make some progress trying to keep position operators along with a time operator in a way that is similar to the way string theory is formulated, but as far as I know you can't actually do much with this approach as far as adding interactions is concerned.

Anonymous

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Anonymous Mon 30 Dec 2019 05:34:11 No.11263426 Report

Quoted By:

>>11261198
1. I have not.
2. The principal seminal volume of texts is Methods of Mathematical Physics by Simon and Reed, however this might be beyond what you are looking for. For exposition into the Green function method and distributions in general I'd suggest you looking into texts on PDEs and optimization, such as Schwarz or Brezis. Strocchi also has some good stuff on such things in his Non-Perturbative QFT text.
>As I understand, mathematical physics itself has different definitions
In general it means the study of the mathematical aspects of physical theories, but this itself already spans several if not all mathematical disciplines. It's hard for people who's been working on one end of math-phys to look into the opposite end.

Anonymous

Anonymous Mon 30 Dec 2019 07:40:56 No.11263641 Report

Quoted By: >>11263722

>>11261198
1. No but I'd be wary based on the title. The quantum nature of light is fucking complicated and very subtle. I'd daresay there's way too much nuisance in QED and quantum optics to warrant using it as an introduction to QM, you'll just end up learning half-truths and getting misconceptions built into your head without even realizing it.

There's a reason the physics community has a meme about "obtaining your photon license" (i.e. you shouldn't even be using the word "photon" at all until you've sufficiently mastered the subject).

2. Boas and Arfken are the standard undergrad and grad texts. While I'm not a huge fan, they certainly get the job done and you'll learn how to use and apply a wide range of math techniques.

If you want rigor, I can't recommend Reed and Simon hard enough. They're pretty expensive/rare I think, but you can get all the volumes off libgen probably. Fucking amazing books, in particular for anything functional analysis related (i.e. all of QM). These are bona fide math texts, however, and certainly not for beginners. They won't teach you computational tricks and methods. But if you want to do theory, you should at least be aware of R&S. You'll understand quantum theory so much better if you understand functional analysis rigorously.

P.S. there are some pretty good questions in this thread, I'm impressed it's not a schizo fest. If I have time tomorrow I'll try to answer some or expand a little more some others.

Anonymous

Anonymous Mon 30 Dec 2019 07:51:38 No.11263671 Report

Quoted By:

>>11263190
wonderful answer.
>>11261566
>Hmm is there a formula for this
$\displaystyle \sigma_A^2\sigma_B^2 \geq \frac{1}{4}\Big|<\Psi|[A,B]|\Psi>\Big|^2$
So it depends on the operators A and B, as well as the state $\Psi$

Anonymous

Anonymous Mon 30 Dec 2019 08:14:20 No.11263722 Report

Quoted By: >>11264479

>>11263641
> Boas and Arfken
What's the name though? Can't seem to find it on libgen

Anonymous

Anonymous Mon 30 Dec 2019 08:19:16 No.11263729 Report

Quoted By:

>>11263048
Are you an engineer? I'm in top uni studying physics and we don't have those. Just some brief mentions within mechanics and classical mechanics courses.

Anonymous

Anonymous Mon 30 Dec 2019 10:02:19 No.11263866 Report

Quoted By:

I wonder what's more credible: this https://www.ocf.berkeley.edu/~abhishek/chicphys.htm list or what my professors recommend (given that I'm in a Russian speaking country so at least some non-translated books are gonna be skipped).

Good thing I know the language I guess, some bachelors say there's nothing good on QM in Russian aside from Landau which is hard.

Anonymous

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Anonymous Mon 30 Dec 2019 10:56:39 No.11263922 Report

Quoted By:

So here's what I struggle with. This is ladder operators for angular momentum.

I'm not sure what notation in the black rectangle means.
The sub (lambda mu, lambda mu) can't be matrix' element number, cause they aren't natural. Does it just mean that we choose the row that corresponds to the column containing mu or lambda (which should mean that the operator we're working with is already diagonalized)?

So as far as I understand what this is - is we take matrix of the operator (in this case I^2 - I^2_z), we use it on some general eigenvector ket that's common to both I^2 and I^2_z.

This ket is a column from U matrix which is used to diagonalize the operator matrix (it's not from [I^2 - I^2_z] but rather from both I^2 and I^2_z in this case). The result is obvious given the spectral equations for I^2 and I^2_z.

It turns matrix that acts on it to a ket. Then the inner product (or matrix product) of bra on the left (which in this case is that same column from U transposed - or a column from U^+) and the resulting ket gives a number.

Now... What are we actually doing here? I^2 and I_z are conjugates. That means we can diagonalize them at once (how is it done?). But here it's the difference between I^2 and I^2_z. What's the physical or even linear algebraic meaning behind this?

And generally speaking when you have <b | A | a> where A is the operator and a, b are it's eigenvectors what are you doing? Does A | a > give you some column from diagonalized matrix and then multiplying that by < b | give you some element from that column which is zero if b != a and eigenvalue if a = b?

Anonymous

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Anonymous Mon 30 Dec 2019 11:05:09 No.11263931 Report

Quoted By:

Also.

Why is there coefficient C(lambda*mu)? If you compare what we get here in the first line with the spectral equation of I_z for (mu + 1), then I^+(psi_lambda_mu) should be just psi_lambda_mu+1.

Anonymous

Anonymous Mon 30 Dec 2019 13:37:53 No.11264143 Report

Quoted By: >>11264221

>>11263190
So just how good and encompassing is quantum mechanics + STR such as that described by Dirac's equations ?

Anonymous

Anonymous Mon 30 Dec 2019 14:49:00 No.11264221 Report

Quoted By:

>>11264143
If you are dealing with a single fermion in a background electromagnetic field, Dirac's equation works fine. The modern interpretation is that the function in Dirac's equation is really a matrix amplitude in a quantum field theory. The only problem is when you try to interpret the function as a wave function in the same sense as in ordinary quantum mechanics.

Dirac's equation incorporates relativity, but like I said you can't define a good position operator in it. What I mean by a good position operator is one that has orthonormal eigenstates |x> that transform like position should under Lorentz transformations. Either the states |x> corresponding to different positions aren't orthogonal, or they don't transform right, you can't have it both ways.

And if we can't define a good position operator, it doesn't quite make sense to treat the function in the Dirac equation as a wave function of a particle. Such concerns don't really matter much on scales bigger than the Compton wavelength of the fermion though (since the states |x> are approximately orthogonal at scales bigger than this).

Anonymous

Anonymous Mon 30 Dec 2019 15:11:20 No.11264259 Report

Quoted By: >>11266208

Is quantum loop gravity still a thing or is it just the SM now?

Anonymous

Anonymous Mon 30 Dec 2019 17:10:30 No.11264479 Report

Quoted By: >>11264935

>>11263722
Sorry, they're 2 separate books. Boas is for undergrads, Arfken for grads. They're both called mathematical methods for physicists or something generic like that.

Anonymous

Anonymous Mon 30 Dec 2019 21:06:26 No.11264935 Report

Quoted By:

>>11264479
Mathematical Methods in the Physical Sciences

Anonymous

Anonymous Tue 31 Dec 2019 08:33:23 No.11266201 Report

Quoted By:

>>11263026
It's actually a super common name for it in Russian speaking countries.

Anonymous

Anonymous Tue 31 Dec 2019 08:36:41 No.11266208 Report

Quoted By:

>>11264259
still meme

Anonymous

Anonymous Tue 31 Dec 2019 09:54:29 No.11266287 Report

Quoted By:

r8 my study plan for QM:

1. Feynman lectures/Susskind video lectures
2. Griffin/my class lectures - parallel to that I do extra hard problems we get
3. Landau Short Course, vol. II (for the fuck of me can't find the original "long course" where QM is III and then QED is IV I think)

Also how long should this take given I've gotten the hang of wave packs and Schrodinger equation in simple cases already? I only need the first semester's worth mind you.

Anonymous

Anonymous Tue 31 Dec 2019 11:30:04 No.11266358 Report

Quoted By:

Hey can anybody check this? It seems like for any book I try to download from libgen I get some html file leading to the very page of downloading that pdf/djvu. Today is the first time I experience this.

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