The h Bar - 2019-02-17

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2:01 AM

532nm might be most commonly recognized as the second harmonic of the typical output Nd:YAG laser.

output of a Nd:YAG*

2:34 AM

10 hours ago, by PM 2Ring

We lost the documentation on quantum mechanics. You'll have to decode the regexes yourself.

I am starting to suspect whatever "code" that wrote the universe, is nothing we can conceive. Just look at the patterns of scientific breakthrough, there is no way to encode that

user351417

3:11 AM

0

A: Centripetal force at the pole is different from the equator

I think whats confusing everyone is that if there is no centri. force at the north and south pole, then people should be sinking into the earth in those places because of the force of gravity, unless gravity doesn't exist.

user351417

Gorgeous answer.

Anonymous

10:00 AM

user image

2

Anonymous

Hah, this is fine still works!

Anonymous

What Easter Eggs do the chat sites have?

user351417

that is fine

10:55 AM

hey all

Anonymous

@Reign Holla.

:D

user351417

@Blue Nobody spells 'hola' with 2 ls...

user351417

Eek

user351417

(this is the perfect opportunity for you to pull a this is fine :P)

Anonymous

11:02 AM

@Chair Sure they do.

Anonymous

Urban Dictionary agrees with me too. Beat that.

Anonymous

user image

user351417

That's just weird.

Anonymous

Hmmm, Shor's paper is way clearer than the Wikipedia exposition. It's a pleasure to read well-written and well-typeset papers (without abrupt logical jumps). Grover's original paper, on the other hand, sucks, as far as typesetting goes.

Anonymous

This number theory stuff is really cool tho. Wish I knew more.

11:11 AM

user image

Noice

Captain Bohemian

speaking is easier than listening.

typing is easier than reading.

2 hours later…

1:01 PM

hello everyone

Anonymous

@Student404Mus Hello onebody.

Could somone explain for me this paragraph :
"Suppose now that we measure the
X operator. The state |0> is equal to a uniform superposition of |+> and |−>.
The measurement postulate then states that we get the state |+> or |−> with
equal probability after performing this measurement. If we then measure the Z
operator again, the result is completely random" ?

Anonymous

Ugh. Use MathJax! :)

Anonymous

27

A: Any chance of MathJax in chat?

As a workaround while this request is pending, there exist several client-side workarounds that can be used to enable LaTeX rendering in chat, including: ChatJax, a set of bookmarklets by robjohn to enable dynamic MathJax support in chat. Commonly used in the Mathematics chat room. An altern...

i think it is clear. no?

1:04 PM

@Student404Mus What exactly do you want to have explained?

i didn't understand when he says "If we then measure the Z
operator again"

did he mean : Z|0> or ZX|0> ?

literally not clear

Neither. Measurement of an operator on a state does not mean applying the operator to a state

Anonymous

@Student404Mus We have a habit of looking down on users who don't use MathJax. So it's better if you get acquainted with it soon. ;)

@Blue ok np

Anonymous

Also, typeset text looks nice! Why use the ugly formatting when you have something much nicer at your disposal.

1:07 PM

@Blue That's...a very unkind way to state it. I much prefer if users use MathJax, but I don't "look down" on users who don't (I just might not go to the trouble to parse their equations :P)

Anonymous

@ACuriousMind Eh, I was just joking. :P

@ACuriousMind Could you figure it out ?

i was thought that measurement means applying operator on states!

@Student404Mus The text is saying that if you first measure X, and then measure Z on the result, the outcome of the Z measurement is 50-50

i see

@Student404Mus No, it doesn't! "measuring an operator" means that you perform a mysterious (literally, look up "measurement problem") operation on the quantum state whose result is an eigenstate of the operator measured.

Just applying an operator to a state does not give you an eigenstate, nor would it be probabilistic

The Born rule tells you what outcomes of the measurement are likely with which probability.

1:11 PM

ok. i think this was clear for me

the physicial outcomes are the eigenvalues of that operator

Anonymous

In short, "measuring the $X$ operator" in this context means, measuring in the $X$ basis i.e. $\{|+\rangle, |-\rangle\}$ basis. Similarly, "measuring the $Z$ operator" means measuring in the $Z$ basis i.e. $\{|0\rangle, |1\rangle\}$ basis.

Anonymous

@Student404Mus I don't really like the way that paragraph is written. It's quite vague. Which textbook is it?

from Quantum information theory by Mark M.Wilde

Anonymous

Hmmm. If you're learning quantum computing better pick up something standard like Nielsen and Chuang.

yes i read from both

Anonymous

1:18 PM

Oh, and this is a good time to advertise Quantum Computing. :P

Anonymous

Ask all your QC and QI doubts there. ;)

thank you !....
but take a look, measuring Z again, mathematically how we could write it?

Anonymous

$$|0\rangle = |+\rangle + |-\rangle$$

Anonymous

If you measure this in the $X$ basis, you have $50\%$ chance of getting $|+\rangle$.

Anonymous

Now, after you've gotten $|+\rangle$, you measure it again in the $Z$ basis. Again, you have a $50\%$ chance of getting $|0\rangle$ and another $50\%$ chance of getting $|1\rangle$. The fact that the initial state was $|0\rangle$ doesn't have any effect on the final measurement. This is presumably what they mean by "random".

Anonymous

1:25 PM

@Student404Mus What you mean by "how could we write it"?

applying operators over their basis: X|+>=|+> , X|->=-|->

for Z|0>=|0> and Z|1>=-|1>

Anonymous

You can't represent measurement with a unitary operation like that.

Measurement is not an operator, at least not unless you include the state of the measurement apparatus into your system.

i see

i think i missed something or we didn't agree on something
ok look at this quotation "Suppose that we prepare the state |0>. If we measure this
state in the Z basis (Isn't Z|0>=|0> ?), the result is that we always obtain the state |0> because the prepared state is a definite Z eigenstate"

Anonymous

Sure. $|0\rangle$ is one of the basis states of the $\{|0\rangle, |1\rangle\}$ basis ($Z$ basis).

Anonymous

1:39 PM

If you're confused, write it as $1|0\rangle + 0 |1\rangle$. The probability of getting $|0\rangle$ is $|1|^2=1$ whereas the probability of getting $|1\rangle$ is $|0|^2=0$.

Anonymous

BTW I don't think you should represent measurement in the $Z$ basis as $Z|\Psi\rangle$. That's misleading.

alright

finally latex is displayed !

Anonymous

@Student404Mus See! It's much nicer. We're no longer in the age of Yahoo Answers. :P

1:55 PM

lol

i wasn't intalled chrome extension for SE chat

Anonymous

I use robjohn's bookmarlet on Firefox.

Anonymous

Chrome extension should be fine too.

yes it works fine now

The author adds "The Z measurement result is $|0\rangle$ or $|1\rangle$ with equal probability if the result of the X measurement is $|+\rangle$ and the same outcome occurs if the result of the X measurement is $|-\rangle$. This argument demonstrates that the measurement of the X operator “throws off” the measurement of the Z operator. The Stern–Gerlach experiment was one of the earliest to validate the predictions of the quantum theory". "Throws off" means the randomness right?

Anonymous

Use |1\rangle to get $|1\rangle$. Use \langle 1| to get $\langle1|$.

Anonymous

MathJax doesn't have the \bra and \ket commands defined.

2:10 PM

:( unable to edit it again

Anonymous

There you go. :P

Anonymous

Anyways, I'll go out for a walk now. Good luck with your question.

yes see you later

thank you for your help

@ACuriousMind do you see any way to unblock myself in physics.stackexchange?

2:25 PM

@Student404Mus You are affected by the automatic question ban, see e.g. this meta post and its linked posts. Your options are a) to wait (it will allow you to ask a new question at some point) b) to edit your old, downvoted posts to make them better so that they are no longer negatively scored.

Anonymous

209

Q: What can I do when getting "We are no longer accepting questions/answers from this account"?

Do not repost the question you were about to ask until you have READ EVERYTHING WE ARE ABOUT TO TELL YOU. While trying to ask a question, one could get: We are no longer accepting questions from this account. See the Help Center to learn more. Likewise, for answers: We are no longe...

support faq asking-questions post-ban

Anonymous

The most important point there is that deleting your negatively voted questions won't help. And even if you delete your account and re-create a new one you'll be restricted to 1 question per week. So it's better if you try to improve your existing questions.

but what if my questions are not in their right place?

Anonymous

If you've already deleted some questions, undelete them soon.

Anonymous

@Student404Mus What do you mean?

2:30 PM

i've been banned a long time

Anonymous

Yeah, but there's not much to do about it. One option to get out the question ban would be to write up some really good quality answers.

i mean they cannot be replaced with good ones. like homework questions or they are unclear even for me...

so, answering questions rise my reputation hence it may get me out of banned users list

Anonymous

@Student404Mus Yes. So do that. But as I said, write good answers.

The problem is that you have 6 downvoted, deleted and closed questions from a while ago and that only a single one of your non-deleted questions has a score higher than 0

oops !

2:35 PM

And then, two months ago, you posted this after the system warned you that you were close to a question ban and you just never edited the question in response to people telling you it was unclear and let it be closed and deleted.

2:58 PM

@AvnishKabaj Astro?

3:44 PM

Hello hbar.

Anonymous

Hello, phone number.

I expected better from Physics chatroom. :P

It isn't a phone number.

:P

Anonymous

The speed of light could very well be ~~your phone number~~ you.

@Blue :)

Imagine some company selling that number for a hefty price to a dorky customer who would ask for it :)

Nevertheless, I am back here in this chatroom for another reason.

For shameless publicity.

For anyone who feels/wishes to chip in:

4

Q: Is this physical model exactly solvable?

There exists a popular model in the Physics of heavy quark bound systems, called the Cornell potential model, in which the inter-quark potential is modeled to vary with radial distance $r$ as $$V(r) = - \frac{\kappa}{r} + \frac{r}{a^2}$$ The mathematical problem is reduced to solving the radi...

ordinary-differential-equations physics mathematical-physics mathematical-modeling singularity

Anonymous

Boy, that's one fine example of answering in comments. :)

3:53 PM

@Blue I concur. I hope he/she develops it further, and nails it completely.

Like I'd mentioned there, it would make a lot of people look silly. Numerical solutions are very abundantly found for this.

But if an exact solution exists, why would HEP phenomenology blokes bother solving it numerically?

I feel something's gotta give, in this context.

@299792458 Do they use it for anything where an exact solution would be needed?

@ACuriousMind Hello ACM. Long time. I do know at least one instance in this context, where any potential inaccuracy in the wavefunction can make perceptible difference. It is in QCD Stark Effect, the leading order energy level shift depends on the momentum-space integral of $ \vert \partial^2 \psi/\partial k^2 \vert$.

Let me see if I can find a reference.

@299792458 Aha - "leading order", so you're doing an approximation anyways, so it's not a big deal if your wavefunction is only numerical

It seems plausible to me that just no one ever invested much energy in trying to find an exact solution because the numerical solution is good enough

@ACuriousMind That is certainly plausible. But if the NLO terms are also wavefunction dependent (through similar integrals), do we have any a priori reason to believe that adding the NLO terms would counterbalance the effect of the inaccuracy. What if the total term picks up some overall factor?

4:08 PM

I'm not saying that any terms "counterbalance". I'm just saying that it's not a big deal if you use a numerical solution for something that's an approximation anyway.

Correct.

Modern numerics are rather cheap and precise, so the error you incur by using a numerical solution won't be large in any case

(unless your equation is ill-behaved, but yours looks pretty tame to me)

But we can still need the analytic solution, if it exists, to gauge how close/far we are from the the same term evaluated using analytic vs numerical wavefunctions.

You don't need to have the exact solution to estimate the error for most numerical methods.

But when statements are made, based on numerical wavefunctions, but with no mention of errors, they would convey the impression of being exact. Even if we aren't trying to legitimize them, knowledge of analytic wavefunctions would help. I mean, if you have an elegant solution at your disposal (through a differential equation solution), why bother with a numerical solution?

4:22 PM

Sure, there's no reason to use the numerical solution if you have an analytic one. I'm just explaining why I think that not much effort was spent on trying to find the analytic one.

I would not dispute that.

Like, you say that the existence of an analytic solution would make the people using the numerical one look "silly". I disagree, since the numerical one seems to have sufficed for their purposes.

Suppose you try to perturb around this Hamiltonian, for a theoretical calculation, you would need the exact eigenfunctions of the unperturbed Hamiltonian, i.e. the Cornell model. With numerical approximations to the same, you can still evaluate, but the flavor of work changes totally.

@ACuriousMind Alright.

Anyhow, I was a little surprised when the Math.SE guys mentioned it (at least) seems to be an easy differential equation to solve, since even the guys who proposed the model originally, did only a numerical solution.

The second amusing part of the story is, the singular part of the potential is just the Coulombic term, which is clearly an exactly solvable form.

The linear part won't be expected be differ comprehensively from a quadratic term (i.e. a harmonic oscillator, another exactly solvable model) over a finite length, since max. probability would only occur at a specified distance from the origin. So, one would guess this is very well behaved differential equation to solve.

5:28 PM

@JohnRennie

do you understand what this means?

user image

Or anybody with knowledge of QM

i don’t understand really what it’s asking

Anonymous

They're talking about the translation operator.

@JakeRose It's a bit oddly worded, but I think you're supposed to a) find the operator $O$ that acts on $\psi(x)$ as $O\psi(x) = \psi(x-x_0)$ and express $O$ in the momentum representation b) show that the momentum-space wavefunction remains normalized under the action of $O$, too (which is an odd thing to ask since the Fourier transform is unitary, but whatever)

Oh it wants an operator that translates $\psi$ but keep the operator in terms of momentum?

Yeah, it just wants you to express how that operator would act on a momentum space wavefunction $\psi(p)$

5:44 PM

@ACuriousMind does that $\psi(p)$ tell you about position probability?

I don't understand the question

Does $|\psi(p)|^2 dp$ give the probability of finding the particle in a space dx?

or with a given momentum?

@JakeRose The latter

Why does the momentum distribution change if I move the particle?

to a different location that is

@JakeRose Who said that the momentum distrubtion changes? ;)

5:48 PM

then wouldn’t the momentum operator remain fixed?

$\psi(p)$ changes, sure, but nobody said anything about $\lvert \psi(p)\rvert^2$.

I’m confused

That's normal. :P How about you just go ahead and compute what happens to $\phi(p)$ when you shift $\psi(x)$, and then think about whether the result makes sense or not?

Can we say $\psi(x)$ give sposition and $\phi(p)$ gives momentum to avoid confusion

sure

5:51 PM

these two are related by Fourier transforms to be absolutely clear right?

Exactly

The mathematical question is what happens to the Fourier transform of a function if you shift the function by a constant.

Okay cool writing it out now

6:02 PM

@ACuriousMind gets multiplied by $exp(\frac{ipx_0}{\hbar})$?

@JakeRose yup

what does that mean though?

why does the momentum operator even change?

Well, for one, you can see that $\lvert \psi(p)\rvert^2$ remains unaffected

@JakeRose It's not "the momentum operator", it's the momentum space wavefunction

And if it didn't change, then you would lose information when switching between position and momentum space

how would information be lost?

Because if it didn't change, then you couldn't distinguish the momentum space representations of $\psi(x)$ and $\psi(x+x_0)$!

6:06 PM

is there any other reason?

why do we have to be able to distinguish between them?

@JakeRose A wavefunction is supposed to represent a unique state of the particle. If there were two different position space wavefunctions for a single momentum space wavefunction, then clearly the momentum space wavefunction doesn't represent a unique state.

Also, the Fourier transform is supposed to be reversible. You wouldn't be able to reverse it if it wasn't bijective.

Cool

how do I show that this is still normalized?

im not seeing it immediately

oh wait

nevermind

I always forget to take modulus

Anonymous

At this point you might also want to ask why momentum space and position space are related by the Fourier transform. :P

@Blue will get back to you just in a coffee shop that’s about to close

6:27 PM

@Blue This...is a difficult question with no straightforward (or rigorously justified) answer, AFAIK

To trigger you, I'll quote Allan Adams: "I'm going to declare it!"

@GodotMisogi The rigorous answer is the Stone-von Neumann theorem: Every pair of operators that obeys the canonical commutation relations is unitarily equivalent to the pair of $x$ and $\partial_x$ on $L^2(\mathbb{R})$.

And that $x$ and $\partial_x$ are related through the Fourier transform is a straightforward computation

Oh, I understand that part. But postulating the CCR itself is non-trivial as a physical question

You get the CCR algebra naturally from $x$ and $p$ on classical phase space with their Poisson bracket

Sure, you have to take some things as axiomatically given, but I see no mysterious "difficult question" here

Anonymous

Jul 3 '18 at 21:35, by ACuriousMind

@Blue There is no fully rigorous formulation, due to the Groenewold-van Howe no-go theorem - you cannot directly map all Poisson brackets to commutators in a naive and desirable fashion. Rigorous formulations are e.g. deformation quantization, mapping a deformed classical Poisson bracket - e.g. the Moyal bracket - to the commutator.

I think the point Blue was driving at was why such a relation should exist in the first place (like the mechanism of upgrading Poisson brackets to commutators), so I was answering with relevance to that

...and he beat me to it

6:36 PM

Then you're just asking "Why quantum mechanics?" and you should be upfront about it :P

It's not the Fourier transform that's mysterious, it's the idea of quantization as such

I understand. I skipped a couple of steps, and was too focused on triggering Blue :p

Anonymous

Tbh, I don't know/remember the whole story. And it's a long story...

You should cut down on your drinking

Anonymous

I'll have to read up the quantization thing properly someday.

Anonymous

@GodotMisogi Stop projecting. :P

6:42 PM

I haven't had a drink since this afternoon! But back to research, I guess

6:59 PM

1

Q: Wrongly marked duplicate about why decay is random

I am not sure how to raise this issue, but I was looking for an answer to the question "How do we know that nuclear decay is truly random and spontaneous?" This is an important question because we don't just think it is random, but it has been proven quite thoroughly that it MUST BE random by Be...

discussion answers duplicates

Anonymous

7:16 PM

@GodotMisogi What's your current research on BTW?

@Blue Currently I'm trying to work on optimization of designing amphibious aircraft. So I've immediately put myself into the trouble of dealing with nonlinear theories and numerical methods in fluid dynamics.

Anonymous

Huh, wow. Sounds like cool stuff. :)

Anonymous

I rode an amphibious aircraft once (a small one which could land on water). It was quite fun, but the capacity was very low (it only had seats for 10 people).

Anonymous

It would be nice to have larger ones (ideally all aircrafts should be amphibious).

Dec 11 '17 at 17:09, by Emilio Pisanty

Hitler actor Bruno Ganz interview about Youtube Downfall Parodies

Bruno Ganz, who played Hitler in Downfall, dies aged 77

2

$\Huge 😢$

8:15 PM

@Blue Lucky. Wish I could've. I've only ridden on a 12-seater turboprop once

@EmilioPisanty I'm waiting for "Hitler reacts to Bruno Ganz's death"

1 hour later…

Anonymous

9:21 PM

Hah, this GIF speed changer seems pretty handy. I was wondering how to slow this down.

10:38 PM

Hey all .. Should I post in secret Labs if I want some perspective if I want perspective if my question is well posed?

Uh...what secret labs?

the chatroom

SecretLabs (SE Branch)

The Labs, where the ideas are organised. Occassional attempt a...

Oh, that's Secret's chat room. I guess you should mainly post there if you want Secret's feedback on something ;)

ah ... would it be fine to post it here then? In math.stackexchange there is a room for this

(forget its name)

We don't have any "official" rooms besides this one.

10:42 PM

ah .. Well I wanted to ask if this was well posed (made sense?)

0

Q: Fluid in spatially enclosed universe and some additional assumptions?

Background In general relativity I was wondering about the following Universe: Imagine a spatially closed universe containing only a perfect fluid: $$x'_i =x_i + L_i $$ where $i =1,2,3$ one of the spatial components. Now, the perfect fluid in thermal equilibrium has the stress energy tensor: ...

thermodynamics general-relativity stress-energy-momentum-tensor

@MoreAnonymous I'm not really sure what you want as an answer here. 1. The equation of state relates pressure, temperature and volume, so it seems obvious pressure can be expressed as a function of volume and temperature. What makes you doubt this in this case? 2. The Lagrangian for a perfect fluid can be easily found by googling. Why do you think this standard Lagrangian doesn't apply to your case?

3. I don't understand the question about the line element at all - the "line element" is just a function of the metric, why would it care about the fluid?

I dont think the standard answer applies cause fluid's pressure is a function of the volume of the universe ... So when I perturb the metric this will also be perturbed (for the Laragian) and thus I dont expect the standard answer to apply ...

@ACuriousMind

10:59 PM

I don't think you should be treating the pressure as a function of the volume/metric in the Lagrangian. You only get that dependence from the equation of state, which should be a consequence of the Lagrangian as an on-shell relation, not an off-shell input to it.

I mean, there's nothing special about your situation - you always have an equation of state that relates pressure, temperature and volume, regardless of whether you're in a closed universe or not, and you don't treat pressure as a function of volume a priori in these other cases either, do you?

Im not sure ... I think one could equally well perturb the metric in such a way that the temperature changes ... but I wasn't interested in that

Ignore what I said above ...

I wanted to the 3 volume to be finite so I chose a closed universe ... But yes regardless Iit shouldn't matter if I have an open or closed universe ..

I thought it was an approximation when one didn't consider pressure as a function of volume ..

(the standard solution)

11:51 PM

@ACuriousMind you free for another QM question

?

If we have some particle and we take a measurement of position with uncertainty $\Delta x$ and then some measurement of momentum with uncertainty $\Delta p$ Immediately after. Will the order of the measurements affect the uncertainty? I.e if we then retook the measurement but with p first would we still get back the same uncertainties?

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