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6:50 AM

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Q: This is a closed question. How to ask this question?

I have a state |0,α⟩ i.e. One coherent state in the initial state. What is the evolved state of |0,α⟩ if the $H=a^{†}b+b^{†}a$ which is time-independent Hamiltonian. $$|0,α⟩(t)=e^{−iHt/ℏ}|0,α⟩$$ $$=cos(tH/ℏ)−isin(tH/ℏ)|0,α⟩$$ $$=cos(t/ℏ)H−isin(t/ℏ)H|0,α⟩$$ $$=cos(t/ℏ)(a^{†}b+b^{†}a)|0,α⟩−isin(t/ℏ...

discussion

Adil Mohammed

7:29 AM

Whats the avg value of drift velocity? 10^5? or 10^-4?

Or speed of light???

John Rennie

As I recall the drift velocity is around a millimetre per second for a typical current.

Don't confuse this with the speed that signals propagate along a wire, which is around the speed of light.

Adil Mohammed

8:18 AM

Ah yes you are right

Drift velocity is 10^-4 m/s and thermal velocity is 10^5 and propagation speed is c

So the reason why the electric bulb lightens as soon as we switch on, is because the propagation speed is c right?

Wolgwang

8:43 AM

@AdilMohammed Nope

ACuriousMind

@AdilMohammed note that John said around the speed of light, not equal to the speed of light.

for rule-of-thumb calculations it's fine to take it as equal to, but you shouldn't come away thinking that the propagation speed of electric signals is the speed of light.

More Anonymous

9:21 AM

How does one find the Lagrangian if the potential is both velocity and time dependent? Like $\dot p = V(x,\dot x , t)$ ?

Where $p$ is moment, $V$ is potential, $x$ is position and $t$ is time

ACuriousMind

if it's velocity-dependent, it's not a potential

More Anonymous

@ACuriousMind Yea ... But I think question still remains how do I get the Lagrangian (even if it is not a potential)

P.S I do have a specific example in time

ACuriousMind

in general, there is no Lagrangian (or at least no purely variational) formalism for arbitrary dissipative systems, but see physics.stackexchange.com/q/20929/50583, physics.stackexchange.com/q/147341/50583

Prithu biswas

If a professor told us how to do an answer of an physics assignment(which is a part of an active exam) and if I believe that there is a flaw in the logic of the professors answer and if everyone is writing professors answer for their assignment (except me) then:
(i)Should I write down professor answer in my own assignment even though I believe that there is a flaw.
(ii)Can I ask question about the flaw on this forum?

More Anonymous

Well if your in India:

1. Yes
2. No that would be a humiliation for your prof

@ACuriousMind Is there an unconventional Lagragian for the E.O.M I ask for? (See my latest asked question)

ACuriousMind

9:34 AM

@MoreAnonymous well, have you tried to apply any of the tricks Qmechanic enumerates in the answer to the second question I linked?

I'm afraid I don't have a magical ability to see Lagrangians :P

More Anonymous

@ACuriousMind I'll give it a shot

I honestly think the top comment is super misleading

ACuriousMind

what top comment?

More Anonymous

The whole $L=T-V$ can't be applied here

ACuriousMind

you can click on the timestamp of a comment to get a link that works well when pasted in chat, you don't need to take a screenshot (I have no idea where that comment is or how it relates to the discussion at hand)

More Anonymous

Oh I was just mentioning that while digressing

ACuriousMind

9:39 AM

@Prithubiswas 1. Depends on whether you place more importance on passing the exam or on being right :P 2. In principle sure, but be careful - if you just ask about how to correctly solve an exercise you will likely run afoul of our homework-and-exercises policy and your question will be closed.

Nihar Karve

10:03 AM

@MoreAnonymous I don't know if you can see this or not, but people are piling close votes on your question for being "homework-like"

maybe edit it so that it looks less like an exercise at first glance

More Anonymous

@NiharKarve I did. But I don't think it's homework like :/ ... I suspect the reason is they think it's a simple L= T-V as the top comment suggests

@NiharKarve Feel free to suggest any edits

Nihar Karve

I don't think it's homework-like either. But from experience, many people tend to close questions if a) they even look like an exercise (while in your case it's just clear formatting) and b) they are classical mechanics or equivalent. Perhaps you should include a bit about what you have tried apropos the solution

satan 29

10:19 AM

@ACuriousMind why though?

i mean, how exactly are you defining "potential" ?

ACuriousMind

@satan29 because "potential" usually means "the force $F(x,t)$ is the gradient of the potential $V(x,t)$"

satan 29

so cant we have the gradient of $V(x,\dot{x},t)= F(x,\dot{x},t)$ ?

ACuriousMind

you might, but people usually don't call that a potential - potentials are associated with conservative forces

satan 29

Oh I see

Oh i vaguely remember you telling me that friction in LM can be dealt by introducing velocity dependent potentials...or something

John Rennie

11:16 AM

@AdilMohammed I think for a typical wire the propagation speed is in the range 0.1 to 0.5c.

Still fast enough for the light to appear to turn on instantly :-)

More Anonymous

11:32 AM

@NiharKarve Did something better ... figured out the answer

(posted it) :))

Nihar Karve

12:12 PM

nice :)

Adil Mohammed

12:48 PM

@ACuriousMind Oh i see, well my teacher said it was lol

Maybe i misheard him

@Wolgwang hey Wolgang arent you in 11th?

More Anonymous

1:03 PM

@NiharKarve After thinking about it abit longer I suspect my potential might be time asymmetric. Welp!

More Anonymous

3:40 PM

Is the probability of finding a particle in a region of phase space $\propto dx dy dz dp_x dp_y dp_z$ ?

Slereah

I mean it depends on the distribution

it's gonna be some integral over phase space, yeah

More Anonymous

I was thinking for a gas

Slereah

By using appropriate theorems on integral bounds you can say it's locally proportional to the phase space volume

More Anonymous

So what is the probability of finding 2 particles (with different momentum) in a region of volume space?

ACuriousMind

depends on your system!

More Anonymous

3:48 PM

A gas with temperature T?

Slereah

ideal gas?

More Anonymous

Yea

ACuriousMind

@MoreAnonymous if that volume is the entirety of the container of the gas, it's 1

Slereah

Ideal gas is about uniform I think?

ACuriousMind

your question is severely underspecified

More Anonymous

3:49 PM

@ACuriousMind I'm not thinking the entire volume of the container

I'm thinking of an arbitary volume in the container

ACuriousMind

well, in that case, for a single particle the probability should be uniform over the container, as Slereah says

More Anonymous

@ACuriousMind But isn't the momentum distribution given by the Maxwell distribution?

ACuriousMind

see, that's what I mean by "your question is severely underspecified" :P

are you asking "What is the probability density to find two particles with exactly the momenta $p_1,p_2$ in a given volume $V$?"

because that's a different question to asking "what's the probaiblity of finding at least 2 particles in a region of space"

More Anonymous

Yes!

ACuriousMind

position and momentum should be uncorrelated, so this should be $\frac{1}{V^2}f(v_1)f(v_2)\mathrm{d}x_1\mathrm{d}x_2\mathrm{d}v_1\mathrm{d}v_2$, where $V$ is the total volume and the $x_i, v_i$ are the positions/velocities of the two particles

More Anonymous

3:57 PM

where $f$ is the maxwell distribution right?

ACuriousMind

yes

More Anonymous

How did you get $V$ in the denominator?

ACuriousMind

because the probability to find a particle anywhere in the total volume must be 1

so the position space probability density is just $\frac{1}{V}\mathrm{d}x$

actually, should be $\frac{1}{V^2}$ up there due to 2 particles

More Anonymous

Thats what I was thinking

More Anonymous

6:01 PM

@ACuriousMind Why wont there be a factor of $^n C_2$ by the way (from permutations and combinations) ... Or is it because they are indistinguishable particles we dont care?

ACuriousMind

there probably should be, indeed

More Anonymous

Sigh .. sent the past hour trying to find reasons it wasnt there ... lol

@ACuriousMind But then will it remain $1/V^{2}$?

ACuriousMind

well, let's sanity check this - what should the probability to find 2 particles of arbitrary momentum inside $V$ be? Clearly 1, otherwise this isn't a probability, so the spatial part needs to be $\frac{1}{V^2}\mathrm{d}x_1\mathrm{d}x_2$

More Anonymous

Got it!

ACuriousMind

the velocity part is a bit trickier - what is the probability that two particles have velocities $v_1$ and $v_2$? the Maxwell-Boltzmann distribution tells us what we get if we pick an arbitrary particle and ask it what its velocity is, but you're asking "is there any particle among the $N$ with this velocity?"

More Anonymous

6:16 PM

Ah ... I see

ACuriousMind

my statistics knowledge is too long gone to immediately tell how to answer this question correctly :P

More Anonymous

@ACuriousMind Lol ... But I can use Maxwell Boltzmann to find N particles with velocity $v_1$ right?

In a volume V?

Ugh ... I'm terribly confused

Shing

11:17 PM

@MoreAnonymous the factor is due to particles number. For demonstration, consider a system of spins: $(up + down)^n$ (try n = 2) where the coefficients are the respective multiplicity of one particular state.

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