« first day (2775 days earlier)      last day (2153 days later) » 

12:24 AM
given rest energy of electron $E_e = m_e \cdot c^2$, its compton wavelength $\lambda = \dfrac{\hbar}{m_e\cdot c}$ and fine structure constant $\alpha = \dfrac{e^2}{\hbar\cdot c}$. The associated radii for quantized energy levels for this electron is given by $r_n = \dfrac{n^2\cdot \hbar^2}{m_e\cdot e^2\cdot c^2}$. How would I write $r_n$ in terms of only $n, E_e, \lambda$ and $\alpha$ ?
I reached at $r_n = \dfrac{n^2\cdot \hbar\cdot c}{\alpha^2\cdot E_e}$
not sure what to do with that $\hbar c$ in the numerator
correction: $r_n = \dfrac{n^2\cdot \hbar^2}{m_e\cdot e^2\cdot}$
update: i reached at $r_n = \dfrac{n^2\cdot \lambda}{\alpha}$
is this correct for an electron with potential $V = \dfrac{-e^2}{r}$ ?
 
12:51 AM
nvm i calculated quantized energy levels using this $r_n$ and it worked!
 
1:43 AM
@MohammadAreebSiddiqui To learn about building quantum computers, you'd probably want to study experimental physics at university, specializing in something like low-temperature atomic physics.
 
i'd prolly have to get a double bachelors b/c where i live i can't live off a degree in physics :P
 
Sure, that might be an option. Or you could minor in physics, if that's available to you. Or just take some physics classes for fun while you're getting a degree in something else. Or you could move elsewhere.
Or, don't forget, getting a degree in physics doesn't mean getting a job doing physics.
 
im currently doing quantum physics from the opencourseware of mit
i took classical mechanics before
its fun doing physics
sadly getting a degree in physics wont get me a good job to get set for my masters
 
It sounds like you already have a career plan for yourself which does not involve quantum computing
 
2:04 AM
it actually does not, im planning to save money after graduating and getting a job in CS. then ill move to get a degree in physics and start my quantum computing path after that
hideous plan but i have no option
 
2:14 AM
Q. A photon collides with a free electron. Explain why the photon cannot be completely absorbed. (havent been taught photoelectric effect or compton scattering)

My answer: as soon as the photon collides and starts to transfer energy, electron gains energy and moves away stopping further transfer of energy from photon. (if true, why didnt photon transfer all energy at once given it is just a packet of energy?)
 
Photons do transfer all their energy at once. (At least, best to assume so unless you are getting into some very detailed theoretical research)
 
so is it because the electron didn't get the exact amount of energy it needed to liberate or jump to a higher energy state? but this was cause no absorption at all. what causes partial absorption?
 
You're asking about a free electron, right?
Do you know what energy states are available to a free electron?
 
3:24 AM
@Slereah That looks like the equation that describe predator prey cycles
 
3:58 AM
@MohammadAreebSiddiqui the interaction has to conserve both momentum and energy. If you work through it you will find this makes it impossible for he photon to be completely absorbed by the electron.
 
4:24 AM
@vzn Well that taught me one thing: I will only talk to her face to face
Unlike cyberspace and any fictional places, you cannot block anyone in real life
(not without killing one or the other or more mildly, some kind of protection restriction law thingy)
This is why the best answers to the most difficult questions often came from real life, because reality is such a bitch, its immense presence will ensure anyone will regurgitate the answers to you
(and sometimes, a combination of exploding internal organs, massive stroke, black outs and etc. due to the immense pressure)
(Lol I am joking, nobody will explode when being confronted, otherwise all of us would have died already just by meeting with each other)
which brings me to a strong argument that unlike other people, I actually like real life alot. Real life is the ONLY realm which ghosting and blocking and trolls have no control of
So yeah, if humanity ever figured out how to upload their consciousness, it will be a massive disaster
(NB Putting trolls there just to show reality is brutal but fair, it will not discriminate against blockers and ghosters, but their antithesis trolls as well)
 
Anonymous
4:48 AM
@MohammadAreebSiddiqui Advertisement time: Quantum Computing SE. Ask your QC questions there. :P
 
Anonymous
If you're interested in architecture that's good, because currently, we don't have many people interested in that area of QC other than Daniel Sank and agaitaarino. Maybe @Mithrandir24601 too.
 
Does anyone know where I might be able to find a derivation of Einstein's field equations in the frame like formulation?
 
vzn
@bolbteppa holy @#%& rethinking my pov on her, she might be just a little )( crazy! check out her numerous music videos, esp luv the mismatched socks lol o_O :P youtube.com/watch?v=m47mJVUJqrw
 
vzn
5:21 AM
@Secret (as usual, not entirely sure what youre talking about...) ghosting? try caspering... :P dailymail.co.uk/femail/article-5736691/…
 
6:00 AM
> Instead of ignoring someone, you’re honest about how you feel, and let them down gently before disappearing from their lives.
> The general consensus on how to Casper someone is saying something nice before blaming your lack of compatibility. For example: “You’re great, but in all honesty I don’t think the spark is there. It’s been great getting to know you though.
That's not so bad. At least there is information here because of the honesty, thus that information can be used to prevent a future caspering to occur again in a new relationship
> Most people are familiar with terms like ghosting (when you casually ignore someone’s existence and messages). But what about ghostbusting, when you force them to reply?
Haha, I am queen of ghostbusting then
In fact that iluminati sounding The Plan is all about mangling the society so much that it ghostbusts reality itself, so that everyone will eventually lost their ability to ghost
> Freeclimbing is what happens when you go on a dangerous fact-finding mission about someone you’ve matched with. First you Google their name, then scale the results without the safety net of an undo button to save you. For even if you unlike their graduation photo from 10 years ago, they still get the notification.
I don't use that for dating (because I have no dates), but I do use that alot on ghostbusting.
In some case, had I had an FBI friend, I will find excuse to tie a ghoster to a crime case so that the FBI will torment him/her until the ghostbust occurs
So yeah, of all that modern dating lingo, ghosting is the worst
and not only I have zero tolerance on ghosting, I have negative tolerance on ghosting, that I will use every and any single laws of physics, social dynamics, screwing up the ecosystem if necessary, screwing up peoples life if necessary, screwng up politics, peace etc. etc. etc., just to ensure the ability for this world to demonstrate the ghosting phenomenon to be erased forever
and that's is the degree of resentment I have on the ghosting phenomenon. If I am Kim Jong Un, I won't hesitate to MAD just to ensure ghosting never exist
(Except Kim won't do that since he is not crazy)
 
Anonymous
6:47 AM
@Secret Sometimes it may be necessary though...
 
That's true, especially when the other partner is a troll. ghosting is one of the most effective antidote to trolls to be honest, I cannot deny that
 
Anonymous
And sometimes you're just so busy with your life that you don't even remember someone's existence. That happens a lot.
 
Anonymous
@Secret Not only for trolls, but also for people whom you don't find interesting (speaking from the other perspective here)
 
@Blue I actually try to counter that my having a notebook of names and significant events related to them, but yeah, keeping track of 300+ names is not easy
 
Anonymous
If you're being ghosted the best option is to move on rather than ghost-busting tbh :P
 
6:52 AM
@Blue For that, I will often state clearly I am not interested, and then if they persists, I will then ignore. But I usually won't ignore people from the start. That might be just how my morality to demand everyone including myself to be at least accountable works though...
@Blue easier said than done, given my personality trait to require almost everything and every phenomenon to be traceable back to its roots (interestingly, this trait is also what give me my research ability in academia)
 
Anonymous
@Secret While that is good, the majority of humans are quite different. Ignoring people you don't find interesting is more of an evolutionary trait I guess. I am not commenting on whether it is morally right or wrong.
 
Yeah, I don't know how common moral system think of that really
 
Anonymous
It would be interesting to see how much this phenomenon persists in the animal and insect kingdom.
 
7:19 AM
https://www.psychologytoday.com/us/blog/evolution-the-self/201604/the-curse-apathy-sources-and-solutions
Still looking for evolutionary roots
 
7:29 AM
0
Q: What is the evolutionary or biological origin of apathy and indifference. Is it uniquely a human thing?

SecretWe knew that social apathy is one of the main reasons why changes cannot be brought about easily due to the general disinterest of the public, and has ties with many mental diseases But does the act of indifference and apathy actually has an evolutionary advantage and more curiously, does anima...

nah, google is not helping here, posted a BSE on that now
 
Anonymous
@Secret Yeah, I looked a bit. Couldn't find much. Upvoted your question. It's quite interesting and I'd like to know the answer too :)
 
Yeah, it never come across me to the thought that indifference may be biological, thus this make it interesting intrinsically speaking
esepcially in light on how there are many evolutionary evidence that altruism is a byproduct of selfishness
 
7:44 AM
@Blue - can I nab you for a moment in the Classical channel?
 
 
3 hours later…
10:19 AM
I have edited your post, partly to correct the English, and in particular the misuse of the word disinterest, which means lack of bias from having an interest (financial etc.) in something. I have also changed "We knew" (presumably meant as We know) to "I believe". I believe it is bad practice to make unsupported assertions of this sort, attributing them to others, rather than taking responsibility for them yourself. I personally neither know what you assert, nor do I believe it. — David 56 mins ago
Huh, I thought that apathy resisting change is a well known thing in progressives?
 
@Secret it's rather pompous but in the interests of getting an answer I'd probably let it go
 
Well, I am just curious, no harm is done by David
 
But I suspect you're looking at this the wrong way.
 
@Secret No, certainly not all "progressives" agree that "apathy" is the main reason why society doesn't change. Which, actually, is a very weird thing to say because society changes all the time, just compare today to e.g. the 1950s
 
It's more a case of there being no evolutionary disadvantage to having little interest in people outside of your own tribe
 
10:27 AM
hmm... maybe I don't understood social apathy enough, because I often get those information from random news articles on the internet (and sometimes on TV and magazines)
But before today's discussion, I never had even thought about the possiblity of apathy's evolutionary ties, thus the discussion so far is quite intriguing even from a purely academic perspective
 
Anonymous
@JohnRennie That's a nice viewpoint. But I can imagine lots of possible situations where some people would be rather uninterested in most people belonging to their own tribe and be more interested people belonging to another tribe. However, I think you're talking more in the perspective of "being protective of your own tribe", which is reasonable
 
I guess that's an unknown unknown to me that is just made known
 
Anonymous
Interest in someone usually arises from awe and/or the fact that the person I'm interested me can provide me something which I want.
 
I'm not sure the lack of interest in societal change at large is usefully characterized as "apathy". Apathy - as in the article you link to - means an unusual lack of emotions, or unusually weak emotions, when compared to a "baseline human". But certainly, if this is a large-scale phenomenon that resists societal change, then it is a feature of the baseline human rather than a deviation from it.
 
Anonymous
@ACuriousMind Yeah, "apathy" is not the right word there. I agree
 
10:33 AM
The question you seem to want to ask is - "Why is it that so many people are not very interested in supporting large-scale societal change?". Even granting that indeed the lack of interest is the main hurdle here - and not, say, people disagreeing about what exactly such change should look like - it is not clear that there needs to be any other explanation than that it is hard for most humans to empathize with - and thus care about - people they have never met and never will.
 
That's certainly a valid explanation, though Blue's claim intrigues me on whether there is also a biological connection to the behaviour, similar to how altruism is born from selfishness at least evolutionary speaking based on an issue of NewScientist (and some of the journal articles it links to)
More broadly, do we know which social phenomenon has evolutionary roots and which is not. It is a question I never pondered before thus I don't really know what my stance on this
 
@Secret all of our behaviour has "evolutionary roots" in the sense that we are a product of evolution. I don't understand the question.
 
10:49 AM
I think I am thinking about something more conspicious like the relation between altruism and selfishness, which can be traced all the way back to the chimpanzees and other humans. Lt me try to find a citation to support my point:
That's a computer model study on how bacteria make the host more selfless
and this is one of the first in line of the kin selection studies
I am not sure if things like happiness, lack of interest or many other social phenomenon has such deep roots back in the evolutionary timeline. This is what I mean by "evolutionary roots"
 
What you actually mean seems to be whether or not a particular social trait - like altruism - is evolutionarily advantageous.
All traits are rooted in evolutionary history, but not all are advantageous, or actually selected for. For many traits there's just not enough evolutionary pressure to get rid of them after some mutation introduced them.
 
yeah, as that is often used to justify how it can persist for long periods of time evolutionary speaking
hmm...
Right, thus Blue's discussion makes me wonder of the evolutionary advantage of lack of interest of something, which Johnrennie seemed to touched upon some aspect of that question above
 
Anonymous
11:10 AM
Changing topic a bit. @ACuriousMind could you explain this sentence: "The straightforward application of the CBH formula shows that by judiciously turning on and off Hamiltonians, the experimenter can drive the system along any time-evolution corresponding to the unitary operator $U=e^{iAt}$" where $A$ lies in the algebra $\mathcal{A}$ generated from the set $\{\tilde{H}_1,...,\tilde{H}_l\}$ by commutation"? Can't understand what they mean in the part "generated...by commutation"
 
Anonymous
PDF here. It's from the second page
 
@Blue It's the algebra whose generators are $H_1,\dots,H_n$, all their commutators $C_{ij} = [H_i,H_j]$, all their second-order commutators $C_{ijk} = [[H_i, H_j],H_k]$ and so on
 
Anonymous
@ACuriousMind Oh, okay. Thanks!
 
12:04 PM
holy snail, what a terribly bad answer upvoted like hell at physics.stackexchange.com/questions/195016/… more than 33 upvotes for claiming that Ohm's law is not derivable because it breaks down for most, if not all, conductors. Of course it breaks down, but this does not imply it cannot be derived.... What the heck is this . I wish this answer was either removed or downvoted down to negative infinity votes
 
@ofhe_iAgDWolbuuTZO_5X1L6uuwfVP Easy there. That answer explains how, historically, Ohm's law came about. The other answers show a modern derivation. You're free to think the distribution of votes should be different, but please stick to constructive criticism instead of aggressive hyperbole.
 
It gives a wrong answer to the question asked by the person who asked it.
Which by the way did not mention at all how it originally appeared.
It lacks logical connection and makes a wrong statement.
This leads me to apply a subjective connotation of "terribly bad answer".
 
I think it's a valid answer to the question, but could be clearer about there being modern derivations of it. The author even acknowledges that in the comments, so it's not that they're unaware of it
 
The comments are worthless anyway, and they contradict what is written in his answer
shall he edit his answer with what he wrote in his comments, the answer would be worth to be upvoted
but this isn't the case
I have examples of questions where the top answer is either totally wrong or doesn't even adress the question and is upvoted like hell while a much better, concise and real answer is near 0 upvotes.
 
12:24 PM
greetings gents
 
Anonymous
@ofhe_iAgDWolbuuTZO_5X1L6uuwfVP Do you have a solution in mind?
 
Anonymous
@Slereah sexist!
 
Slereah dies not exist
 
@Blue other than posting in the chat and expecting people to downvote/upvote more accurately, no. Posting comments below a wrong answer and claiming it is wrong does not work (and I was suggested to do that in physics meta). comments can be deleted annonymously and leaves 0 trace. While messages of the chat are here forever
 
and no, there is no cake
because the cake is a lie
 
12:28 PM
@ofhe_iAgDWolbuuTZO_5X1L6uuwfVP Comments pointing out that an answer is wrong will typically not be deleted.
At least not unless there's an a different reason to delete them.
 
@ACuriousMind you deleted a comment I had posted below an incomplete (but not wrong) answer, claiming it was incomplete
 
@Blue Are there any ladies on the chat
 
And while chat is there "forever", only a tiny fraction of users read chat, and no one reads chat logs from more than a few days ago.
 
Anonymous
Dec 26 '17 at 7:19, by Secret
@JohnRennie My 13 year old self is a boy, my mature self is a girl. I am currently somewhere in between
 
If you want a sizable fraction of people to see your claim that an answer is wrong and react to it, then comments are the way to go
 
12:30 PM
but anyway, I hold what I claim. I remember once when I did that (claimed the answer was wrong), the dude realized it was true and instead of editing his answer, he edited the question asked, to match his own answer. The edit got approved (really!). so in the end I looked like a fool with an obsolete comment
and all the mods said I took it too personally (which is true) and that it wasn't a big deal (I still cannot swallow that)
 
Anonymous
Quality control on a site like Physics SE is indeed quite difficult. There's really nothing much we can do about it...
 
I agree
I try to make it better though, I really try
Here the top answer does not really answer the original question, but nevertheless states an interesting fact. physics.stackexchange.com/questions/404703/… Nonetheless I believe V.F.'s answer is right to the point (I hadn't seen his answer when I wrote mine) I wish his answer was more upvoted than the current one
the answer basically boils down to "because not every monotonely increasing function must diverge"
that's the real answer, IMO. spotted by V.F.
and btw this question is a duplicate that nobody noticed yet
 
I think you are perhaps looking a bit too zealously for "the" correct answer. One answer focuses on constant current, another on constant voltage. Why does it matter so much where in the answer ranking they are? We allow multiple answers to questions for a reason - there is no requirement that any single answer must be perfect and consider the question from all angles.
 
I agree but the top voted answer does not answer the question
even with a constant current, the melting of the wire might not be reached
something the current top voted answer does not address. but it's not a bad answer
it's just that there are other better answer(s), the one of V.F. for example
So I agree that the current top voted answer should NOT be downvoted, since it gives an interesting viewpoint
 
Anonymous
The voting system is broken (quite natural on a site where there are much more beginners than experts). Everyone knows that. Just collect whatever you can from all the answers combined
 
12:44 PM
but I do not agree that it should be on the top when there are better answer(s)
 
I don't see how you can claim any of them is "better" than the other: The question does not specify whether there's a constant voltage or a constant current supply. One answer explains why it doesn't happen for constant voltage, the other why it doesn't happen for constant current.
 
I agree @Blue
the question was general, so the answer provided by V.F. which applies for both constant current or voltage, is better
it simply is better, giving the information that an increasing function need not diverge
and that's exactly what the person who asked the question was missing
 
Ah, I see what you mean.
 
yeah I'm not saying the current top answer is bad. it should be upvoted, but I think V.F. answer should be the most upvoted in an ideal world
 
Well, that's the FGITW problem: That answer came over a day later than those higher up, meaning less people have seen it
 
12:48 PM
yes. I left a comment under VF's answer to hopefully slowly reverse the tides in his favor
 
When you post an answer to a question that's older than a few hours and already has somewhat satisfactory answers, you need to be prepared for a looooooong wait until you climb over the answers which had a head start
 
Anonymous
1:04 PM
@SirCumference How's life?
 
1:47 PM
What are the recommended books in standard model for students who have master's degree?
Please help
 
Anonymous
@Student404Mus Wikipedia has a lot of book recommendations. Scroll down
 
Anonymous
I'm certainly not an expert and can't suggest anything
 
Anonymous
2:01 PM
@ACuriousMind On page 3 [here](https://arxiv.org/pdf/1710.03599.pdf) they say "let us define a set of $M$ unitary operators $\{\mathcal U_k\}$ acting on a register of $N+1$ qubits according to:


$$\mathcal{U}_k := |0\rangle\langle 0| \otimes \Bbb I \ + |1\rangle\langle 1|\otimes e^{-i|x^{(k)}\rangle\langle x^{(k)}|\Delta t}$$

for a small time $\Delta t$"

I'm a bit confused here, because the dimension of the $\mathcal{U}_k$ matrix operator (which seems to be of dimension $2d\times 2d$ to me) doesn't seem to be compatible to act on a state vector of $N+1$ qubits (which we can consider to b
 
Anonymous
$\Bbb I$ is the $d$-dimensional identity matrix
 
What's $\lvert x^{k}\rangle$?
 
Anonymous
It is a state vector of a $d$-level quantum system. $k$ is a just a label indicating the number of the training sample
 
$d$? You mean $N$? :P
 
Anonymous
 
Anonymous
2:07 PM
$N=[\log_2(d)]$
 
Anonymous
I'm getting confused now, lol
 
Anonymous
$N$ is the number of qubits required to represent a d-dimensional vector $\mathbf{x}$
 
Anonymous
Where $\mathbf{x}=\{x_1,x_2,...,x_d\}^{T}$
 
Anonymous
and $x_i\in\{+1,-1\}$
 
Hm, in that case, I agree that $U_k$ as written is not an operator on $N+1$ qubits.
Rather, it acts on a single qubit times a state of that $d$-level system.
However, I suspect the authors mean that you should take the $N$-qubit state that represents $\lvert x^k \rangle$ and use that in that formula.
 
Anonymous
2:13 PM
@ACuriousMind Which formula?
 
@Blue The one you wrote down for $U_k$.
 
Anonymous
But still $|x^k\rangle$ will be ~$2^N$ dimensional, isn't it?
 
Anonymous
I mean the state vector
 
Why "still"?
Anyway, yes, that vector is then $2^N$ dimensional. What's the problem?
 
Anonymous
I couldn't really understand what you meant by "it acts on a single qubit state times a state of that $d$-level system"
 
Anonymous
2:15 PM
Could you elaborate on that a bit?
 
Well, I deleted that message for a reason :P
 
Anonymous
5 mins ago, by ACuriousMind
Rather, it acts on a single qubit times a state of that $d$-level system.
 
Anonymous
It's still here ^ :P
 
Ah, that, sorry
Well, "times" is meant in the sense of the tensor product here. When you take $\lvert x\rangle$ as a $d$-level state, then $\lvert x\rangle\langle x\rvert$ acts on the space of such states (it's a projector). If you now take its exponential, it still acts on that space. If you tensor that with $\lvert 1\rangle \langle 1\rvert$, then the resulting operator acts on the tensor product of the $d$-level space and the 1-qubit space
 
Anonymous
Well, yes, in that case $|x\rangle \langle x|$ is $d\times d$ dimensional and $|1\rangle \langle 1|$ is $2\times 2$ dimensional. The tensor product is $(2d)\times (2d)$ dimensional. This operator (matrix) can act on a $(2d)\times 1$ dimensional state vector (as you say "tensor product of a $d$-level space and a $1$-qubit space"). But in this context what is $\mathcal{U}_k$ acting on?
 
Anonymous
2:24 PM
They say $N+1$ register of qubits
 
Anonymous
Say perhaps is acting on the tensor product of a state of $N$ qubits and $1$ additional qubit. Does that sound right?
 
Anonymous
State of $N$ qubits would be a $d$-dimensional vector and the state of $1$ additional qubit would be a $2$-dimensional vector. So their tensor product can be written as a $2d$ dimensional column vector
 
@Student404Mus the end of Peskin and Schroeder's QFT, the end of Schwartz QFT and the Standard Model
 
@Blue No, the state of $N$ qubits is only $d$-dimensional if you didn't have to round when taking the log.
 
Anonymous
@ACuriousMind Okay, okay, leaving that approximation aside :P
 
Anonymous
2:30 PM
Otherwise is it okay/makes sense?
 
@SigmaAlpha maybe section 2.4 of damtp.cam.ac.uk/research/gr/members/gibbons/… ?
 
Anonymous
Let's just consider for simplicity that $d$ is a power of $2$
 
@Blue Yes
 
@vzn (note the author) it's actually pretty good:
 
Anonymous
@ACuriousMind Thanks :)
 
vzn
2:46 PM
> 2:05 Since there is only one string theory, it seems reasonable to hope that it might have only one quantum vacuum describing our universe and leading to definite predictions for the values of all the dimensionless natural constants. ---Witten
 
@bolbteppa thanks for help mat
 
2:58 PM
what about standard model
does it have a specific books or it's supplementary part of particle physics?
 
Anonymous
 
Anonymous
$\mu_j(\rho)$ and $|v_j(\rho)\rangle$ are the eigenvalues and eigenvectors of the density matrix $\rho$. They are preparing a composite state $\sum_{t=1}^{2^T}|t\rangle \otimes |\psi\rangle$ (where $|\psi\rangle$ is a state vector of $N$ qubits). Clearly $\sum_{t=1}^{2^T}|t\rangle \otimes |\psi\rangle$ is not $2d$ dimensional (i.e. it is not a tensor product of a $N$-qubit state and a single qubit state).
 
Anonymous
Then I'm not sure how it makes sense to apply $U_t$ on the composite state. Can't understand on what they're applying $U_t$ to end up with $\sum_j \beta_j|\tilde{\mu}_j(\rho)\rangle \otimes |v_j(\rho)\rangle$.
 
Anonymous
Umm, wait $U_t = (\mathcal{U}_1...\mathcal{U}_M)^n$
 
Anonymous
But then even it's dimension should remain same as $\mathcal{U}_k$ if I'm not wrong
 
Anonymous
3:13 PM
Don't see how $U_t$ can be applied to something like $\sum_{t=1}^{2^T}|t\rangle \otimes |\psi\rangle$
 
Anonymous
@ACuriousMind Any idea? ^
 
@Blue Why not?
Or rather, what's space does $\lvert t\rangle$ live in therE?
 
Anonymous
@ACuriousMind $|t\rangle$ lives in a $2^{T}$ dimensional space ? It represents a state of $T$ qubits, no?
 
@Blue I don't know, I'm not really in the mood to read through a paper to understand what you're talking about :P
 
Anonymous
@ACuriousMind It's mentioned in the image I posted. But sure, I can understand. No worries, I'll ask it on the main site then
 
4:15 PM
@Blue Tough and stressful, but at least its summer :)
 
Anonymous
@SirCumference Vacations?
 
@Blue Nah, summer classes
Mostly to get humanities done with
 
Anonymous
@SirCumference Oh, on what?
 
@Blue Comedic storytelling. They're teaching me how to be funny
 
Anonymous
@SirCumference Aah, are humanities classes compulsory?
 
Anonymous
4:17 PM
Strange
 
Anonymous
@SirCumference Wtf...lol
 
@Blue Yeah, here we have to do 3 humanities classes, 3 social science, and 4 writing intensive
 
Anonymous
@SirCumference Pheeew
 
That's a common practice with universities ;-;
You don't have that?
 
Anonymous
Nope. We were done with humanities in the first semester
 
Anonymous
4:19 PM
You're graduating next year, right?
 
Two years
 
Anonymous
How long is it?
 
Anonymous
5?
 
I'm a junior now
4 years
Wait, I guess now I am graduating next year
This is why I'm not a math major
 
Anonymous
@SirCumference Umm?
 
Anonymous
4:21 PM
You were doing some double major thingy, no?
 
Can't subtract 1 from 4 :P
@Blue Well I'm considering whether I want to go through with it. Algebra seemed too abstract, but I still want to take analysis
How's life with you btw?
 
Anonymous
@SirCumference Not bad. Summer vacations going on and tearing my hairs over a few QC papers
 
A free electron doesn't completely absorb a photon when it collides. Is it because some energy of photon is used up as providing momentum to the electron, and the resultant total momentum is not 0?
 
Anonymous
@SirCumference How much does the analysis course cover?
 
$m_ec^2 + p_e^2c^2 = p_\gamma c$
hence, $m_ec = p_e^2c - p_\gamma$ ? the
 
4:27 PM
@Blue Well you can either take Real Analysis 1 and 2, or the honors versions of those
 
Anonymous
@SirCumference Sounds like a course based on Baby Rudin then
 
Is there a non-circular definition of work and kinetic energy?
 
@Blue Baby Rudin?
 
Anonymous
23
A: Free electron can't absorb a photon

Rob JeffriesCan't believe I can't find a duplicate. It is because energy and momentum cannot be simultaneously conserved if a free electron were to absorb a photon. If the electron is bound to an atom then the atom itself is able to act as a third body repository of energy and momentum. Details below: Cons...

 
Anonymous
@SirCumference "Principles of Mathematical Analysis"
 
4:37 PM
i cant believe these questions are given as assignment problems
:'(
 
@Blue Doesn't seem to be what we use :/
So far the textbooks I've seen for math courses have been pretty bad
 
Well, my great, great, ... great grandsprogs are doomed. Possibly.
And while I'm here, feast your eyes on this masterpiece:
Just ... amazing.
 
4:54 PM
@JohnRennie What's that?
 
is that dog food
 
@Slereah Everything is dog food. Just yesterday I watched a friend's dog trying to eat a blanket :P
 
not a good indication of what is dog food
for instance, many cats won't eat cat food
 
Well, cats have taste. And self-respect.
 
@ACuriousMind Ah, so chocolate is dog food? My suspicions confirmed
 
Anonymous
4:58 PM
@JohnRennie Is that...rice and chicken? Looks a bit like biryani with too much gravy
 
Anonymous
Nope that doesn't look like chicken
 
$\begin{pmatrix}\end{pmatrix}$
 
How's your new-ish job been @ACuriousMind?
 
Interesting, the behaviour described at tex.stackexchange.com/q/435813/71716 is not in Mathjax
@SirCumference Very good :) Keeps being interesting and not too stressful
 
Anonymous
Guess we all should forget our plans for grad school and join software COBOL development too. All the coders COBOL developers seem to be so satisfied with their jobs :P
 
5:04 PM
@ACuriousMind That's good to hear :)
 
COBOL is the goldmine
the frameworks come and go, but COBOL is forever
 
@SirCumference Just so you don't get confused, I don't actually write in COBOL, it's a running gag ;)
 
Anonymous
Don't believe him ^ @SirCumference
 
@ACuriousMind That would make more sense lol
 
Anonymous
This is something interesting:
 
Anonymous
5:13 PM
Grace Brewster Murray Hopper (née Murray; December 9, 1906 – January 1, 1992) was an American computer scientist and United States Navy rear admiral. One of the first programmers of the Harvard Mark I computer, she was a pioneer of computer programming who invented one of the first compiler related tools. She popularized the idea of machine-independent programming languages, which led to the development of COBOL, an early high-level programming language still in use today. Hopper had attempted to enlist in the Navy during World War II, but she was rejected by the military because she was 34 years...
 
Anonymous
"Grace Brewster Murray Hopper (née Murray; December 9, 1906 – January 1, 1992) was an American computer scientist and United States Navy rear admiral.[1] One of the first programmers of the Harvard Mark I computer, she was a pioneer of computer programming who invented one of the first compiler related tools. She popularized the idea of machine-independent programming languages, which led to the development of COBOL, an early high-level programming language still in use today."
 
Anonymous
Computer scientist and rear admiral....too much talent!
 
@Blue Not to diminish her talent, but her navy rank was entirely due to her prowess as a computer scientist
 
Where's my admiralty
 
Hi, is some one of you present good in Navier-stokes equations?
 
Anonymous
5:29 PM
@ACuriousMind From a cursory reading it does seem so. They mention that she was a volunteer for WAVES and worked in the Naval Reserve but doesn't go into much details about what work she did there
 
vzn
@SigmaAlpha wondering, how is it different than whats here? en.wikipedia.org/wiki/Einstein_field_equations are you doing postgraduate work?
 
@JaydevSingh Hey there
 
@Slereah, @ACuriousMind Hi! Could you, please, verify that, in the section https://en.wikipedia.org/wiki/Simon%27s_problem, after I say "Let's now rewrite", the second expression is equivalent to the first?
 
Man I don't know quantum computing
 
5:43 PM
@Slereah It's just tensor products
It's just the mathematical part that you would need to check
No need to check that the concepts are correct
Just those two expressions, which are supposedly equivalent, but I am not sure, given my unfamiliarity with tensor product
 
@nbro I would put an additional pair of brackets around everything right of the sum sign in the second expression
I'm not sure you should be editing the current version of a Wiki article if you're unsure about what you're doing, though. How about working on a draft, and only changing the current version once you're confident in what you've written?
 
@ACuriousMind I am not really sure where you want me to add those parentheses. I added many parentheses already...
@ACuriousMind I am editing this article based on some notes by a professor, not just based on my knowledge, so, don't worry, I am not introducing "junk science"
 
Sorry, I wrote left where I meant right :D
 
Furthermore, the previous version was really bad...
 
I just want it to be clear that in the second equation the sum is not over $\lvert y\rangle$ alone but over $\lvert y\rangle \otimes\text{stuff}$
@nbro I'm not trying to discourage you from improving the article, just raising some doubts whether it wouldn't be better to have a draft version somewhere else and only copy it over once you're done
 
5:49 PM
So, are you saying that to add parentheses so that to have $\sum_{ y \in \{0, 1\}^n }
\left(
\left| y \right\rangle
\otimes
\left(
\frac{1}{ 2^n }
\sum_{x \in \{0, 1\}^n }
\left(
(-1)^{x \cdot y}
\left| f(x) \right\rangle
\right)
\right)
\right) $?
 
yup
although more parentheses do certainly not improve legibility, they do remove ambiguity :P
 
@ACuriousMind Yes, you're probably right. I should probably work on a draft first, before introducing 1000 edits. But maybe, if I work on a draft I may get discouraged...
Anyway, if you want to confirm, my edits are highly based on these notes: https://cs.uwaterloo.ca/~watrous/LectureNotes/CPSC519.Winter2006/06.pdf.
 
Hey @ACuriousMind, is there a non-circular definition of work and kinetic energy? I want to deduce the Work Energy Theorem but for this I would need to know already that K.E. is $1/2mv^2$ and work is $Fd$. How can I define those? I've only seen definitions using the WET itself which is a circular argument since the deduction of the WET requires knowing that K.E. is $1/2mv^2$.
 
But I added a few more explanations...
 
Anonymous
How does Wikipedia editing work btw? Anybody can edit anything without peer review? Or is the peer-review done periodically once in a while?
 
5:58 PM
@philmcole Hm, not quite sure what you mean. The work-energy theorem says that given the definitions of work as $W = \int F\mathrm{d}s$ where $F$ is the net force on an object and change in kinetic energy as $\Delta K = \frac12 m (v^2_\text{end} - v^2_\text{start})$, then $W = \Delta K$. Where's the circularity?
 
Yeah the circularity arises when I ask how we know that K.E. is $1/2 m v^2$ or $W = \int F ds$
Or should I just regard them as definitions (which I don't really want to do...)
 
@philmcole But they are definitions
If you think they should be derived from something, then you need to provide an alternative definition from which to derive them
 
@Blue No peer review before applying changes, but you can follow a page and revert the changes that another editor did.
@Blue Since you're into CQ, if you're familiar with Simon's algorithm, feel free to have a look at that page. I would appreciate some help to make it better
 
Anonymous
@nbro I see, thanks
 
In particular, I would like to make the section "Simon's algorithm" (which should, in theory, be an overview of the algorithm) more consistent with the detail description of the algorithm, in the section "Detailed description of Simon's algorithm"
@Blue Btw, be careful, there are a few restrictions (like, do not remove tags at the beginning of articles without giving an explanation)
The section "Simon's algorithm" is based on prof. Varizani's video https://www.youtube.com/watch?v=VrlTdwBWc4g, whereas the other section is based on the notes I mentioned above.
 
Anonymous
6:10 PM
@nbro Okay, I'll check it out sometime this week and see if I can suggest some improvements. I personally find Wikipedia articles on quantum algorithms lacking a lot of details (and sometimes written using impenetrable language). It would be good if someone could use the standard textbooks like N&C and Vazirani/Watrous' lectures to improve the articles. What you're doing is a good initiative.
 
@ACuriousMind Okay thanks. I thought those were derived somehow. I will regard them as definitions from now on.
 
vzn
6:27 PM
> One memorable guest was Professor Stephen Hawking. “I asked him if he wanted to discuss the galaxy,” Stringfellow told an interviewer from The Times. “He types into his little box and out comes that synthesized voice: ‘No. I want girls.’ Ha ha ha! I WANT GIRLS!” thedailybeast.com/… thedailybeast.com/…
 
is it true that though all obervables can be expressed in terms of self-adjoint operators, not all self-adjoint operators are observables?
 
@CaptainBohemian Yes, this is the case if there are superselection rules, cf. e.g. this answer of mine
 
7:16 PM
Hello!
 
heyhey
 
I was stuck in learning a new concept rather constrain relation
I read somewhere for each part of the string where there is a velocity change use -v at that point if the velocity is decreasing its length and +v if its increasing
@ACuriousMind so i wrote my equation as v+vcos60+v_2 =0
As all the three velocities are increasing the lenghth of the string am i correct why or why not?
@ACuriousMind can you please help?
The other equation which I got verified by constrain relation is vcos60=-v_2 which gives me result I verified by constrain relation
 
Anonymous
7:48 PM
@Jasmine Use MathJax, please. Also, post a clearer image such that people don't have to rotate their heads by 90 degrees to make sense of it (the other alternative being: rotating the PC 90 degree clockwise)!
 
@Blue ok am sorry
I got $ v $ = $ 2 v_2 $ by constrain relation
 
Anonymous
@Jasmine Looks okay
 
But when i use the lenght increment decrement technique i get $ v $ = $ 3 v_2 /2 $
 
Anonymous
@Jasmine How?
 
According to the trick consider at all nodes if the velocity of the string is increasing the lenghth of strinģ or decreasing for increasing you take positive and for decrement you take negative
 
7:57 PM
@ACuriousMind Last week I asked a question with the Wikipedia page "superselection" about what "charged" means in the context of group theory. After that, I read that page repetitively with puzzles. Now I come back to that page and think I still don't quite understand what that page's explanation of superselection. Is superselection just a rule used to determine if two or more quantum states can be combined to form a quantum state?
 
I know why this logic works its like a result of constrain relation
I probably picked up wrong concept
 
Anonymous
@Jasmine 1. Define "node" 2. Define "velocity of string". It's not clear what you're saying. However, the simple logic is that the total length of the string must remain constant. So velocities pointing "out of string" must add up to and be equal to the sum of velocities pointing "into string". In this case quite simply $+v_2-v_1\cos(60)=0$.
 
Anonymous
You simply consider the velocities (of the ends) pointing "into" the string as negative and "out of" the string as positive.
 
Hmm now I see what that trick meant
I made the mistake of adding $ v_1 $ again but that we already added
 
8:13 PM
@ACuriousMind if a Hilbert space is irreducible, then there is no sueperselection rule, and then all self-adjoint operators on the Hilbert space are observables?
 
@CaptainBohemian 1. It makes no sense to say that a Hilbert space is irreducible. What you mean is that the representation of the algebra of observables on the Hilbert space is irreducible. 2. No, that does not follow. Irreducibilty is a necessary, but not sufficient requirement for "self-adjoints = observables", cf. Valter Moretti's answer to the question I linked: It also isn't true in the presence of non-Abelian gauge groups.
 
8:27 PM
@ACuriousMind you mean the long answer below your answer?
 
@CaptainBohemian Yup
 
8:40 PM
user image
2
@Blue ^That's my daughter's first haloween. Notice the rank and name.
The awards are also those that the admiral wore at her retirement modulo my sloppy hand with coloring pencils.
 
9:29 PM
in compton scattering, when we plot the graph of intensity of detected photon against their wavelengths when $\theta_0 = 0$ there is one peak which is justified as final wavelength is equal to the initial wavelength. As we increase the angle, why do we get the second peak? Why isn't the peak just shifted and instead another one is created?
 

« first day (2775 days earlier)      last day (2153 days later) »