12:00 AM
if RT=0 implies either R=0 or T=0 right...I see no reason for that condition given only R+T=1

how can the probability of photon being reflected and transmitted at the same time be not 0??
P(reflec) + P(transmit) = 1
P(reflect and transmit) = P(reflect)P(transmit)
= RT
shouldnt it be = 0?
they both are independent events

2 hours later…
2:11 AM
Are Jaan Kalda's ipho study materials one of a kind
Because they are absolutely brilliant
But sadly not all topics are covered

2:25 AM
[Random]
What does it mean for something to have no future:
Consider an observer O with some light come J. Let J+ be the future half of the light come
Normally, general relativity requires all dynamics to be deterministic, thus any future event p of O is already determined by the past light cone J-(p)
((Except for very pathological cases as detailed in 0celo's blog Einstein and The Evidence
)
But consider the topology of spacetime has a structure such that no event in $J^+(O)$ is connected to their respective future light cones $\{J^+\}$, then we have these events not the cause of anything in the future including $J^+(O)$ and thus they effectively disconnected with their future
For any such event $q$ such that for any event $p$ in $J^+(O)$ where $|p-q| < \epsilon$, $J^-(p) = \varnothing$
Likewise for $q$, $J^+(q) = \varnothing$
$q$ is thus an example of an event that has no future and the ps that mentioned earlier are examples of events that has no past
These are the basic units you need in order to construct a spacetime such that there are non deterministic regions of dynamics, without the spacetime being eternally expanding or inside a black hole to generate a Cauchy horizon
It is unclear what will be observed by any observer locating at $q$, will they be erased from existence some infinitesimal proper time later, or something else
I think I would love to learn more about coordinate free versions of GR, this is so convenient

3:27 AM
@MohammadAreebSiddiqui the events "the incident photon is reflected" and "the incident photon is not reflected" are certainly not independent.
It's really not different than the situation for a coin flip. If P(H)=P(T)=1/2, then P(H)P(T)=1/4 is not zero. But H and T are not independent outcomes, so P(H and T) != P(H)P(T)
(Or, even more simply: Mutually exclusive events are not independent.)

3:48 AM
@Secret I think there's an issue of is such a causal structure allowed

@enumaris Well, I don't see how it cannot be done by making the point p in question to be topologically disconnected from it's future or past light cone by e.g. removing a small annular region near p

Well, a manifold in GR is a differentiable one...it has to be locally Minkowski. If you start removing regions from the manifold willy nilly, I feel like you will create boundaries along which the "manifold" is no longer a manifold in that sense
I have no proof of this, but I think some care should be taken

Ah right, yes, once sections start to be removed, it will ceased to be a manifold. hmm...

I don't think there's any problem removing bits from a manifold. It's just that you will now have incomplete geodesics.

yeah, from a physical perspective, it would be like you couldn't look left all of a sudden cus you ran up to a boundary or something...how odd would that be...lol
well I think there could be ways in which you can remove bits from a manifold...but the way in which they are removed probably has to be special

4:03 AM
A black hole is an example where geodesics are future incomplete and a white hole is an example where geodesics are past incomplete.

well they are incomplete because of the presence of the singularity
the singularity itself is not generally thought of as part of the manifold

Yes, so we have a manifold with a bit missing.

a singularity missing...and the singularity itself is...theoretically...difficult

@JohnRennie I have to ask, how'd you find that?

One should not remove parts of a manifold without care is what I'm getting at. You would have to prove that removing these pieces makes mathematical and physical sense.

4:05 AM
@SirCumference someone else mentioned it :-)

there is great care in dealing with the singularities present in physics. And a lot of work has gone into trying to hide singularities behind event horizons...
if you just remove a chunk of the manifold...that's like...pretty catastrophic imo T_T

tbh, I really have no idea what happens when something enters a nondeterministic region, because any of the following can happen:
1. The object gets erased from existence because it "fell out of spacetime"
2. The object's present state cannot be inferred from its past states once a threshold is crossed (that's the case for Cauchy horizons)
3. The object disappears and then reappeared elsewhere

I think when you start to get to issues as esoteric as like "falling out of spacetime", you should start to evaluate whether your theory still has predictive power in that regime.
I wouldn't trust a mathematical result like that without some physical evidence.
It's not to say such weird events can not occur

This is why I often found these results strange

Just because an object's future trajectory isn't integrable doesn't mean we need resort to the supernatural :-)

4:17 AM
Yeah I saw that news a few weeks ago
I'm not well versed in GR enough to comment either way
at this point it seems to be more of a math question imo

Sometime later I need to revise just how many nondeterministic scenarios of spacetime can occur because they are quite interesting
Although likely to not be physically relevant to us alot

well there are "spacetimes" that have closed timelike curves
like the Godel spacetime
it doesn't surprise me that some other caustics can occur in GR
if global hyperbolicity has to be taken as a postulate of GR, I'm fine with that :P
the question would then be reduced to physically finding a situation where global hyperbolicity breaks down
I'm...not actually sure if the presence of singularities breaks down global hyperbolicity
intuitively it would seem to be yes but I'm not sure

1 hour later…
5:42 AM
howdy, i got some elementary questions on bra-ket notation, but reading the wiki article didn't make it obvious so I was wondering if i could ask here
Does $\langle a | b | c \rangle = \int a \circ b \circ c$ ?

@frogeyedpeas What does the $\circ$ operator mean here?

You don't have the integral sign unless you express the (co)vectors a,c and linear operator b in some continuous basis

one of the issues with learning Bra-ket notation from a source like Griffiths that starts from a "physical" perspective is that one is often stuck in viewing things in terms of the position (or momentum) representation. A book like Ballentine that approaches things from much more of a mathematical perspective will be better for that issue. However...there is something to be said about keeping an eye on the physical implications of a theory and not getting too bogged down in the math.

I like dirac's book the best for bra-ket(after all he created it)
Except the lack of problems, that book is the most clear

yeah, bra-kets are equally fine in finite dimensional spaces, which is why Susskind introduce them in the context of qubits

5:53 AM
interestingly I have not read Dirac's treatment lol
there are a great many books that I should read that I have not...
but right now I have to finish this book on Reinforcement Learning as that has a direct effect on my livelihood
(sorta)

@frogeyedpeas hello?

heh, this convo is a bit...4 sided...
needs to be 5 sided though

Why stop at 4? If I were you I will go full on $\mathfrak{c}$ :P
::finiteness is soooooooo overrated::

well there were 4 users who responded lol

Anonymous
6:22 AM
@enumaris Name of the book?

which one?
Ballentine?

Anonymous
@enumaris The RL which supposedly has a direct influence on your livelihood. :)

oh
Barto and Sutton Reinforcement Learning
I'm reading a pre-print that they released
for edition 2
edition 1 is pretty old

Anonymous
Thanks. I don't have much experience in RL. But have been studying some deep learning stuff in the past few weeks. What's the basic concept in reinforcement learning though? A reward and punishment system...sort of (in order to reduce the error)?

Anonymous
Gotta pick up RL sometime

6:26 AM
it's almost exclusively formulated in terms of a markov decision process
state -> action -> reward (or punishment) -> new state
there's an environment which an agent interacts with. The agent has a policy that selects an action given a state, the feedback form the environment is the reward and the next state

Anonymous
But how do you decide the reward/punishment? Are you given the ideal case ?

so you have an string of S-A-R-S-A-R-S-...
oh, the reward and punishment should be set by the environment
e.g. if you play space invaders the reward would be your points
it's not set by the learning agent
that would defeat the purpose
but there is some reward engineering you can do depending on your task

Anonymous
Okay. Interesting, I see. So it's not a gradient descent type of thing.

You can use gradient descent methods with it
that's if you use function approximation
or if you use policy gradients

Anonymous

Anonymous
6:30 AM
Thanks for the overview

yeah np :D
you can integrate deep learning into RL
then you get deep-RL
which is what Alphago did to beat the world champ

Anonymous
Haha. Sounds cool

Basically the function approximation part you use a deep neural net
so you have an intermediate stage S->f(S)->A where f(S) is some function approximation of S so you don't have to worry about learning a separate value for every state S
f(S) allows you to generalize results from one S to many other states
so S = image of arcade game, f(S) = convnet(image)
instead of say S=every pixel and then choosing an action directly based on the information from each individual pixel
bam, you got the 30 second overview of Deep-RL. There's a ton of other techniques you gotta use to actually make it effective though...XD

@jo
@JohnRennie its operator composition, $a,b,c$ are functions of functions of the domain of the integral. Example $a = \psi(x) \times \text{arg}$ $b = x \times \text{arg}$ and $c = \psi^*(x)$ then by my definition $\lange a | b | c \rangle = \int \psi(x) x \psi^*(x)$

That...makes no sense... D:

6:47 AM
I don't like my syntax actually now that I think about it, i'm trying to be very abstract and expressive but its more probably more confusing than it needs to be

In $\langle a | \hat{b} | c \rangle$ a and c are wavefunctions and $\hat{b}$ is a hermitian operator, so I'm not sure what you mean by $\psi(x) \times arg$ means.

@JohnRennie what would be the literal definition of that expression in terms of sums or integrals?
I would like to start from a clean slate

$\langle a | \hat{b} | c \rangle = \int a \hat{b}(c)$

okay cool, so thats what i thought but wasn't totally sure. Thank you! :)

I've used $\hat{b}(c)$ to make it clear the operator $\hat{b}$ is acting on the function $c$. Normally we'd just write $\int a \hat{b} c$

6:49 AM
if the position-representation of $|a\rangle=\psi(x)$ and the position representation of $|c\rangle=\phi(x)$ then $\langle c|b|a\rangle = \int \phi^*(x)b_x\psi(x)dx$ where $b_x$ is the position representation of $b$

hmm so @enumaris I noticed your definition and @JohnRennie differ since you have complex conjugation in your integral, which he doesn't

I should

Oops, yes, my mistake. It should be $\int a^* \hat{b} c$

I missed the bra ket on the a and c sorry
see my updated post for more clarity

ah now i see, cool and good :)

6:52 AM
but note that the position representation is just one of many possible representations
so it's not always to your benefit to think of the object $\langle c|b|a\rangle$ in terms of a specific representation

Is the rule of complex conjugation specific to a representation?

I would say no. The complex conjugation part arises when you turn a ket into a bra
the duality between a state $a|a\rangle$ for some complex number $a$ is $a^*\langle a| \leftrightarrow a|a\rangle$
I feel like...the more I explain...the more there will be to explain...
lol
It might be best to refer to some lecture notes or refer to a textbook
e.g. Ballentine

Anonymous
@frogeyedpeas The corresponding member of a ket vector in the dual space is the bra vector which turns out to be the Hermitian conjugate of the ket. The interesting fact is that each vector of a Hilbert space has a one-to-one correspondence to members of the dual space (which is unique).

yes...
but now you have to explain what a dual space is...and what a Hilbert space is...
T_T
it never ends...

Anonymous
You always have Wikipedia :P

7:00 AM
I find wikipedia to be really good as a reference or as a refresher if you've already learned a subject before...but very rarely is it a good first source
the explanations tend to be too technical

Anonymous
Wikipedia + Stack Exchange is good combination :)

Anonymous

Anonymous
Or else, of course there are textbooks

Anonymous
The best thing about Wikipedia is that it does not dilute the details (most of the times)

yeah I suppose. I find that in my experience, for all these issues of Hilbert spaces and the like Ballentine was the best source.

Anonymous
7:03 AM
There are over a 1000 introductory QM books I guess :P

I am digesting this all slowly, I think I will get a copy of ballentine though if it is highly recommended

the trade-off though is that Ballentine, at least for the first several chapters, reads like a math text rather than a physics one
It doesn't really get to a physical problem for a long while...it spends a lot of time developing the mathematical formalism of QM

I would love something like math text devoid of physical meaning, and then a follow up book that is much more about intuition.

well then, Ballentine might be the one for you :D

actually the dream would be to just learn QFT as a mathematical framework directly

7:04 AM
QFT is a different beast...

Anonymous
My favourite was Shankar. But if you're an absolute beginner I very highly recommend not learning QM directly before classical mechanics. The other alternative might be to go down the quantum computing path (which is finite-dimensional) and much easier to grasp.

Ballentine is normal QM
QFT can suck it

I've been down the quantum computing path, my particular interest is in a (currently ill defined) question that inviolves heisenbergs uncertainty principle and relativity

along with every single QFT textbook I've ever read

*i've been down very briefly
*i can only factor some shit, and explain some basic quantum algos and circuits, i think i spoke a bit more pompously than i should have

7:06 AM
morning
@frogeyedpeas just read PCT then :p

Anonymous
What's PCT?

@Slereah "https://mathoverflow.net/questions/27701/proof-of-pct-theorem-for-haag-kastler-‌​nets-in-qft">
?
i'm not confident if that is a book, a theorem, an article or something else, do you have a link @Slereah?

Anonymous
16

I want to learn QFT, because I have heard of its applications in mathematics, I am not interested in scattering cross sections and such. Where can I start to learn? Only books I found are either way above my level or too physics oriented. I know undergraduate math level and have had courses in QM...

why not just learn it normally
you won't get any motivation

Anonymous
I guess he doesn't care about the physics :P

7:10 AM
learning QFT from a math perspective the first time around will do you no help
because if you're a math person you'll already know most of the math covered or if you don't you won't like how it's written and if you're in physics you won't learn any physics
just learn the physics QFT and then learn the math separately

hmm this is a fair point you bring up
i guess i do want to learn the physics, but (now this may be just plain wrong) I would like to leverage my math background to push towards QFT as quickly as possible
as that seems to be the only setting in which the types of questions i'm interested can be well posed and studied

I gotta say though, that in all my learnings of QFT, the tie back with QM has been extremely subpar
it's like you just learn something completely different and you spend all your time calculating S-matrices and cross sections...-.-

here is question actually: it's possible to study the "mechanics" that classical E&M creates (which is special relativity) without ever once talking about electricity and magnetism itself...
Similarly you can start playing around with QM without any real science (ex: deriving the uncertainty principles using fourier analysis for arbitrary wave functions). Is there a way to do relativistic QM without actually spending too much time on "real physics"? I.e. does it make sense to talk about the underlying mechanics independent of the actual fields and particles we observe in nature?
since my interests are mostly mathematical artifacts i think

you can certainly study math without the physics
that's what most math departments do...applied math...is like...almost all math lol
Probably you can study representation theory? There's a lot of connections between particles/fields and the representations of the poincare group

7:17 AM
@frogeyedpeas are you sure about that?

and then gauge theories if you want to get into that...

@Eulb sure about what?, I stated a lot of things which could be worth questioning in that paragraph

i replied to the first sentence
doing e&m and basic SR without the physics is just doing vector and tensor calc more or less
but what's the point of that because you can just learn it from a math book since the math used it basically a done deal by now

yea but you can still derive time dilation and mass energy equivalence
its possible to show those things and be absolutely clueless how E&M works

that's because you'd lack physical intuition by avoiding any mention of physical concepts

7:23 AM
15 mins ago, by frogeyedpeas
i'm not confident if that is a book, a theorem, an article or something else, do you have a link @Slereah?
lol x 1

@Eulb well my point is my general theme of questions are about notions of uncertainty and higher statistical moments in a relativistic framework. The ideas i'm playing with don't directly invoke say the Strong Force or E&M, so if there's a way for me to learn the bare minimum so i can make my questions well posed, and answer them, I would probably like to do that
I don't think thats a good thing, ideally, i would learn everything, but i'm also lazy and i'll be honest in that regard, that if i can get away with it, a lazy route would be nice :)

what are you trying to do?
you're not making sense to me.

Boguliubov also did a math QFT book

delol x 1
(ridiculum success)
Anyway...

@Secret lol what

7:30 AM
I am just a bit... abtruse today, that's all

Anonymous
How's Shankar's QFT book? @Eulb

condensed matter one?

Anonymous
Yeah

Anonymous
Is it mostly condensed matter?

Anonymous
Or does it introduce QFT well too?

7:31 AM
it's not QFT QFT
it's like stat mech/ condensed matter QFT

Anonymous
?

Anonymous
@Eulb Ah

if you want to learn QFT from a condensed matter perspective as a QFT kind of person you should read altland and simons
it does QFT proper more or less

Anonymous
@Eulb What are the pre-requisites though?

QM and basic knowledge of QFT and stat mech and some solid state physics
i didn't read it in full but I did use the symmetry breaking chapter as a reference once.

Anonymous
7:34 AM
I see. And what's a "physicists'" standard introduction to QFT? (for those who're not interested in condensed matter/solid state)

the standard in grad school seems to be peskin/schroeder (or srednicki or zee if your course does path integral quantization first) and then supplemented with weinberg for further reading.
the course I took used peskin/schroeder but that's not the best book for self learning

@Eulb i'm not able to send any messages, my wifi just dropped and reconnected. I'll ping u when get back to my home country and have a stabler setup :)

the best book for learning QFT as a beginner self learning would be schwartz @Blue

goddamnt
I lost like 3 comments, and then anotehr 10 spam messages of me complaining about losing the 3 comments
i lost that too

- "QFT and the standard model" by schwartz. it's very similar to peskin/schroeder in format

7:36 AM
I lost like 3 comments, and then anotehr 10 spam messages of me complaining about losing the 3 comments

Anonymous
@Eulb Okay, thanks! I'll check those out once I'm done with QM a bit more thoroughly

you can start learning QFT after like chapter 15 in shankar QM
skipping chapter 8
you don't need to learn QM scattering or perturbation theory or path integrals before learning it in QFT for most intro books
they cover it fully

Schwarz is a fine book

Anonymous
Gotcha, cool :) GR and QFT are the two things which are in my bucket list for the next couple of years

i'm actually going to read schwartz again this fall to refresh (basically re-learn at this point since it's been so long) for QFT 2 in the winter because i didn't end up taking that but want to. for now i'm just doing math though
@BalarkaSen sup

7:43 AM
\o

currently being strangled by topology

y so

because i'm not spending enough time on it

which is bad and i'm working on fixing that asap

7:45 AM
yes i am very angery at you for doing that

same

@Blue I thought you wanted to go into development?

i should kms

There isn't a huge call for QFT and GR in IT :-)

your learning velocity should be multiple km/s, yes

7:46 AM
i'm still plagued by a few remnants of mental illness

Anonymous
@JohnRennie software development? No, lol. If anything I'm interested in machine learning and quantum computing, if I'm to take up an industry job

Anonymous
I meant machine learning also involves some sort of software development but it's quite different

Anonymous
And I enjoy it

@Blue ah, OK, but there's not much demand for QFT and GR in that field either!

but at least now i'm trying to get my shit together while simultaneously taking topology and pdes to keep the pressure on. @BalarkaSen
midterms is in like 10 days and i'm freaking out already

Anonymous
7:48 AM
@JohnRennie Hehe. Physics and math will always remain hobbies for me :P

the current weather trend seems to indicate that i'll be learning everything the night before lol

Anonymous
They are enjoyable but they don't pay XD

@Eulb that aint good
start now

@Blue Don't misunderstand me - I approve :-) It's just that I thought the Indian university system didn't give you the spare time to indulge in side tracks like that.

@BalarkaSen discord

7:51 AM
@JohnRennie Well, people doing QKD are beginning to look at things like decoherence of single photon states (i.e. a non-classical quantum state) as a result of the gravitational field...

I'm unconvinced you need mastery of GR for that ...

@Eulb I'm working rn, i'll check later today

@JohnRennie Probably not 'mastery', but some knowledge, definitely

Anonymous
@JohnRennie That is true, but JEE preparation had made me so robotic during high school and I was so stripped off of my hobbies, that I seemed to have lost a personality of my own. Nowadays, I don't mind getting a bit lower grade if I am spending doing something I enjoy, instead.

(to be clear, this is a GR, not non-relativistic effect)

7:53 AM
@BalarkaSen nvm then it's not that important. just wanted to elaborate what i've been going through recently and how i'm fixing it.
for any math stuff i'll just stick to the SE chat

@Mithrandir24601 I think if you've got to the point of studying QFT on a curved background then you already have more than enough maths to do GR.

@JohnRennie Single photon states do require knowledge of quantum optics if you want to do it really properly and that is (often) done in second quantisation
So if you want to do it properly, you'd need knowledge somewhere (between QM & QFT) + GR [brackets used to clarify meaning of notation :P]

Dammit, where's my delivery?

@JohnRennie Only 8 stops away :)

That's nearly two hours the delivery man has taken to get to stop 1. What's he doing? Walking there?

8:02 AM
Maybe stop 1 is in Scotland and they're coming from London?

are all the roads in the uk squiggly like that
lol

@Eulb You urgently need to get back to that topology :-)

you too?
well darn i suppose i can't run away from it wherever i go it seems

I have a new mofo CPU on the way as part of my ongoing project to see how much CPU power I can cram into how small a computer before everything catches fire and blows up.

eee it's almost 5am though i better sleep or i'll miss class tomorrow which i probably won't even go to anyway
gnight
also me irl @BalarkaSen
6

8:11 AM
lol
go to slep

In $(2 z - \frac{\partial }{\partial z} )^n = \sum_{k=0} { n \choose k} (-1)^k \frac{d^k}{dz^k}(2z)^{n-k}$ what is the sum up to? It's supposed to be up to $\lfloor \frac{n}{2} \rfloor$ but why :\

@bolbteppa It looks like there should be extra terms... Set $n=2$ and summing to $\lfloor \frac{n}{2} \rfloor=1$ doesn't look right :/
I mean, even better, set $n=1$ and $2 z - \frac{\partial }{\partial z}\neq 2z$

It's an operator acting on $1$, $(2z - \frac{d}{dz})^n (1)$, so that for $n = 1$ you have it equal $2z$
In mathematics, the Hermite polynomials are a classical orthogonal polynomial sequence. The polynomials arise in: probability, such as the Edgeworth series; in combinatorics, as an example of an Appell sequence, obeying the umbral calculus; in numerical analysis as Gaussian quadrature; in physics, where they give rise to the eigenstates of the quantum harmonic oscillator; in systems theory in connection with nonlinear operations on Gaussian noise. in random matrix theory in Wigner–Dyson ensembles. Hermite polynomials were defined by Pierre-Simon Laplace in 1810 though in scarcely recognizable form...

8:27 AM
@bolbteppa Ah, then it makes more sense - the terms greater than $\lfloor \frac{n}{2} \rfloor$ consist of $\left(\frac{d}{dz}\right)^j\left(1\right)=0$ terms

Ah yeah

8:40 AM
\begin{align}
(2 z - \frac{\partial}{\partial z} )^n &= \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} { n \choose k} (-1)^k \frac{d^k}{dz^k}(2z)^{n-k} \\
&= \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} { n \choose k} (-1)^k (n-k)(n-k-1)\dots(n-2k+1)(2z)^{n-2k} \\
&= \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k { n \choose k} \frac{(n-k)!}{(n-2k)!} (2z)^{n-2k} \\
&= \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k \frac{n!}{k!(n-2k)!} (2z)^{n-2k} \\
&= \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k { n \choose k} \frac{(n-k)!}{(n-2k)!} (2z)^{n-2k} \\
Fun times

@bolbteppa At least it doesn't go on for pages and pages

QM Harmonic oscillator gives the Hermite equation $y'' - 2zy' + 2n y = 0$, series solution method gives the series expansion above just before the $|_{t=0}$ business starts, the ladder method on the Hamiltonian gives the $(2z - \frac{d}{dz})^n$ thing as part of the solution, and the Laplace method phys.uconn.edu/~rozman/Courses/P2400_16S/downloads/…
gives the Rodrigues $(-1)^ne^{z^2}\frac{d^n}{dz^n} e^{-z^2}$ thing as your starting point, still need to look at the Heisenberg and path integral methods properly :\
(Though I think the ladder method is basically the Heisenberg method, not sure :)

8:58 AM
Never heard of the Heisenberg method

As in, solving it in the Heisenberg picture

Oh okay
This ladder method, can it applied to any problem or is there some restriction on the kind of potentials it can be applied on?
I ask this method because it involves this way to 'split' an operator into two operators , creation and annihilation. Also is this way of splitting it into two operators, is it unique?

Second quantization, also referred to as occupation number representation, is a formalism used to describe and analyze quantum many-body systems. In quantum field theory, it is known as canonical quantization, in which the fields (typically as the wave functions of matter) are thought of as field operators, in a manner similar to how the physical quantities (position, momentum, etc.) are thought of as operators in first quantization. The key ideas of this method were introduced in 1927 by Paul Dirac, and were developed, most notably, by Vladimir Fock and Pascual Jordan later. In this approach,...
Basically, the whole creation and annihilation method is called second quantization
It's supposed to apply to multi-particle QM problems, but sometimes it's applied to single-particle QM problems too, e.g. the Harmonic oscillator, so it's a bit confusing what's going on tbh

9:27 AM
Ahh I see

10:21 AM
@bolbteppa eh, they've stolen our site logo
but, that said

4

There are examples of time evolution of quantum dynamics with history dependence, such as these quantum random walk examples which make use of a memory parameter to influence the distribution of the random walk. I am wondering whether the rules of quantum mechanics allow the construction of a ve...

boy, oh boy, can you imagine the mess that this one would've become if our resident Angry Person was in a position to post?
@JohnRennie @ACuriousMind
@bolbteppa yes

revisiting an old question: I wonder if all linear systems failed to replicate (to be checked later)

look behind the PHYSICS

10:25 AM
@EmilioPisanty It will be a disaster that is big enough to generate 100 higgs bosons

@Secret would be
if
but no
thankfully
because
> The suspension period ends on Apr 18 '19 at 5:29.

> This account is temporarily suspended because of low-quality contributions
O, that's a new suspension reason

@EmilioPisanty :-)
Apparently he has his own web site now.

@JohnRennie does he, now?
good for him

@EmilioPisanty this is it

10:33 AM
hopefully he'll stop trying to use us as a soapbox once the timer runs out again

@Secret hmmmm. I thought that the previous suspension used the same reason but apparently it didn't
but there's some evidence of it from 2014 if not earlier
if you take this at face value then all the way back to 2009
this is pretty funny though

1 hour later…
12:04 PM
@EmilioPisanty Yes, that's always been one of the suspension reasons. We haven't done many manual ones since the process was automated for some common cases.

12:22 PM
Some people thought my question was unclear, and at least the revised version looks clear to me. How can I find out what's unclear about it?

https://physics.stackexchange.com/questions/410314/is-this-mental-concept-of-photons-wrong

@rob what's the moderator-team procedure for imposing a manual ban due to low-quality contributions? What are the criteria and how is the decision taking?

12:55 PM
Stupid factors of two off, also messed up on the operator definitions :'( So much simpler to note $(y - \frac{\partial}{\partial y})\cdot I = (- e^{\frac{1}{2}y^2} \frac{d}{dy} e^{-\frac{1}{2}y^2})\cdot I$
Can completely skip that computation

1:38 PM
0

There is an automatic answer ban mechanism that stops users from posting many low-quality answers. However, I'm pretty sure users automatically get immunity from the answer ban if their reputation is sufficiently high. Is this true? If so, it seems to me to be a uniquely bad decision for this pa...

2:15 PM
@JohnRennie where've I seen this before?

2:26 PM
@EmilioPisanty reminds me of this Onion video, though the context (US cable news) is different:

Anonymous
what's happenin

2:43 PM

Anonymous
@bolbteppa nice

Vaguely similar one for the cubic possible