The h Bar - 2018-03-19 (page 1 of 4)

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1:38 AM

0

Q: Why is it intuitively unreasonable for this transition probability to grow quadratically in $t$?

In Sakurai's "Modern Quantum Mechanics" section 5.6, there is a seemingly simple statement made that I do not understand the logic of. The author is considering a physical situation in which we "turn-on" a time-independent potential at $t=0$, and ask what the relevant transition probabilities are...

quantum-mechanics

hmm...

2:04 AM

@DanielSank it's met the threshold =)

3 hours later…

4:45 AM

Thank you blue for the site

2 hours later…

6:44 AM

@JohnRennie Hi ! Good morning :)

Morning :-)

Dawood ibn Kareem

6:56 AM

@Abcd I think the whole idea is that the part of the chain that's not in free-fall is the part that's reached the floor already, which has a mass of $M x / L$. The force it exerts on the floor is just its weight, which is $M x g / L$.

Either that or I've misunderstood the question disastrously.

Anonymous

@DawoodibnKareem There's also a force due to the sudden momentum change

Anonymous

$F=\frac{dp}{dt}$

Anonymous

Basically the sum of those two forces

Dawood ibn Kareem

But you don't have a $dt$ in the question, so you can't work this out.

Anonymous

@DawoodibnKareem You don't need it ;)

Anonymous

7:00 AM

There's a clever trick

Anonymous

But I want abcd to try it out first

Dawood ibn Kareem

Oh, yes, I see.

The question is quite difficult to understand though.

Anonymous

Looks like it's from Irodov. He's an expert in cryptography

2

Anonymous

:P

Anonymous

But yeah, it needs some acquaintance with similar type of problems

7:14 AM

@Blue Irodov an expert in cryptography! Yes, that's a pretty good summary :-)

Anonymous

:D

8:22 AM

Can anyone guide me how to post a picture here....?

Mithrandir24601

@NehalSamee Press the little 'upload' button next to the 'send' button and proceed to find the file on your computer, or type/copy in a url in the 'from the web' section

Man...I'm using Android ..There's no upload button...

you can't do it from the phone app, yes

,

you'll have to upload it elsewhere and post the link

8:25 AM

😥😥😥

OK then ... I'm trying to post the question here...

A spring is hanging at 20m height from the ground . A ball of mass 0.2kg is shot towards the spring with velocity 49m/s such that extension of spring is 3m . What will be the rebound velocity of ball on the ground ?

I found v at height 20m and then its kinetic energy ... This kinetic energy is equal to stored potential energy of spring . Then I found k equating the energies and consequently F...

Dawood ibn Kareem

I suspect the spring is a red herring. Any energy it absorbs is going to get given back to the ball.

The acceleration a plus gravity is used in $v²=u²+2(a+g).20$ ...

But I also tried another way ...

I tried to use conservation of energy ...

At height 23 M , I tried to say that total energy is gravitational and spring potential , which I equated with kinetic energy of ball ... But the answers are different ...

Where I went wrong ...?

I think the energy equation should be gravitational = spring + kinetic?

8:40 AM

Why so ?

Since it is the ball the causes the spring to be compressed in the first place?

But ... mmm...The terms do not match...Can you put numerical values...?

Negative energy outpops...

Is laws of conservation of energy right

If so how energy was created for the first time

Is everyone sleeping?

9:00 AM

Talking about another topic...?

@Akash.B a lot of energy comes from changing around bonds in chemical reactions. The energy that is created in this case was already present as stored chemical potential energy

(I'm not sure if that is exactly what you are looking for?)

how did energy formed in space?

@Akash.B what energy?

is the spring pointing towards the ground or to the sky but placed 20m above the ground?

9:16 AM

energy produced at the core of the sun

Dawood ibn Kareem

9:41 AM

How did I miss this? "I think if the president walked across the Potomac, the media would report that he can't swim" - Sarah Huckabee Sanders.

10:01 AM

hi,

the composition of the velocities in the RR gives a Galilean referencial (for a well-chosen transformation), is this known?

RR?

Relativity

also who is Sarah Huckabee Sanders, is she the fusion of Sarah Palin, Mike Huckabee and Bernie Sanders?

3

Relativity is only one R

sorry in french we say "relativité restreinte"

Yo, anybody got the 3D Fourier transform of a parabola handy?

10:09 AM

Is a parabola a $L^2$ function?

Yeah

real(sqrt(1-r^2))

I'll trade you a shinny Fourier transform of an infinite double helix

Hello chat

I think the trick is to convert to cylindrical coordinates...

I guess someone here most be familiar with Buoyancy ?

\frac{\pi ^2 \left(J_0\left(\frac{1}{2} \sqrt{\text{kx}^2+\text{ky}^2}\right){}^2+J_1\left(\frac{1}{2} \sqrt{\text{kx}^2+\text{ky}^2}\right){}^2\right) (\sin (\text{kz})+i (\cos (\text{kz})-1))}{2 \text{kz}}

or that's what Mathematica tells me

10:16 AM

blue: h bar is very very strange today

Morning

warning: h bar is very strange today. There are a lot of incomprehensible spiritual sounding messages floating around

Anyway if i have solid object of some matter. It does float let say in water if it's density is smaller than waters density $ \rho_{object} < \rho_{water} \rightarrow \text{if true then it floats}$ ??

$$\frac{\rho_{object}}{\rho_{water}}$$ would be a multiplier which tells how well the object floats ? smaller value the better it floats ?

Strangeness #1:

1 hour ago, by Akash. B

Is everyone sleeping?

@Tuki yes

Dawood ibn Kareem

10:21 AM

@Slereah That's not as far from the truth as it could be.

Strangeness #2:

1 hour ago, by Akash. B

how did energy formed in space?

I wonder if i can use this ratio to determine how much volume is under water and how much volume is not in the liquid

@Tuki if the object displaces a volume of water $V$ then the upwards force on it is $g V \rho_\text{water}$. This is Archimedes principle.

$$ \frac{\rho_{object}}{\rho_{water}}\cdot 100 = \%$$ is the percentile of volume that is under water ?

And the downwards force is just $mg$ or $\rho_\text{object} V_\text{object} g$

Set the two forces equal to find the volume of water displaced.

10:24 AM

Yes they should be equal if it floats ?

@Tuki If the object is just floating i.e not moving up or down, then the net vertical force on the object must be zero. That means the upwards and downwards forces have to be equal and opposite.

But would the ratio between my object and water be equal to the amount of volume that is submerged ?

$$ \frac{\rho_{\text{object}}}{\rho_{\text{water}}}\cdot V_{\text{total}}=V_{\text{submerged}} $$

What's the function called that is composed of two bessel functions of the first kind added together in quadrature, aka ` z=[-10:0.1:10];plot(hypot(besselj(0,z),besselj(1,z)))` (MATLAB)?

If we set the two forces equal we get: $$ g V \rho_\text{water} = \rho_\text{object} V_\text{object} g $$ or: $$ V = \frac{\rho_\text{object}}{\rho_\text{water}} V_\text{object} $$

I wonder if this statement is false or true ?

10:29 AM

@Tuki and that's the same as your equation (give or take a few differences in notation)

I noticed

Sup dudes

I have a question on quantum mechanics, anyone good at that here?

maybe

OK well here goes

I heard the SE is useful for solving for wavefunctions

That is,

given a potential V (e.g. square well or something), and boundary conditions of wave function (e.g. psi(x,0)), one can solve for psi using the SE is that right/

where SE = Schrodinger equation

Sure, so whats the question? These are kinda like the modes.

10:34 AM

ok so my question is

@Kenshin Oh hey you're back

suppose instead of solving for the wavefunction

I am given a wavefunction instead

How can I calculate how the wavefunction evolves through time

Does the SE allow me to do this, or is the SE something used for solving for the wavefunction, rather than used to work out how a given wavefunction will evolve?

@Kenshin The propagator

aye, spin that bad boy up with some FDTD yo

hey @SirCumference how's it going

10:36 AM

$\hat U(t) |\Psi(0) \rangle = |\Psi(t) \rangle$

@Slereah how is the propogator found?

The cool part is that you can propagate it in momentum space easily under some approximations, like the born approximation. Except as an experimentalist the inverse problem, aka get the scattering potential sucks.

I sometimes wonder whether we can get U(t) by integrating the SE wrt time

A variety of ways, but usually $H U = -i\hbar \partial_t U$

Usually you have $U = e^{itH}$

or something like that

check for constants and signs, I don't remember everything by heart

cool thanks

so this is derived from the SE?

10:38 AM

You can derive it yes

ok next question

Am I correct to say that a wavefunction need not be a solution to the sE?

What do you mean by "a wavefunction"

well let's say I have a wavefunction, that I then perform a measurement on

@Kenshin You'd be wrong, it needs to be made of basis elements given by your particular box

Do you mean a wavefunction defined on $\mathbb R^3$, or as it evolve in time?

10:40 AM

will the measurement always force the wavefunction to collapse to some solution of the SE?

$$ i \hbar \int_{t_0}^{t_1} \frac{\partial \lvert \psi (x,t)\rangle}{\partial t} dt = \int_{t_0}^{t_1} \hat{H} \frac{\partial \lvert \psi (x,t)\rangle}{\partial t}dt$$

In other words, do you mean $\psi(x)$ or $\psi(x,t)$

does this have meaning at all?

i'm talking psi(x,t)

Is there a firefox plugin for reading latex in chat?

10:41 AM

Then yes, not all such functions are solutions to the Schrodinger equation

@Mikhail yes, see this link

ok I have one last question guys, thanks for your help so far

the quesiton is,

@Kenshin So, the solutions to the SE gives you the "modes", every realizable signal needs to be made of those "modes" (however complex). You've introduced time, which made those "modes" more complicated compared to the time harmonic solutions (aka they also have a time variable).

If I introduce a new potential where it's peak is far from the peak of the current wavefunction

the new potential will take time to propagate through space (lmited by the speed of light)

parts of the wavefunction will thus be impacted by the new potential

but some parts of the wavefunction won't yet be aware of the new potential yet

Errr

10:44 AM

is there a way of dealing with such cases in QM?

you've done time haromics yourself

I'm not sure, because for a start, there's no limitation of speed with Schrodinger

is there a modified SE that can be applied

cos yes I see the SE assumes the potential impacts the particle immediately

Schrodinger is non-relativistic

is there a relativistic version, that's not QFT

10:45 AM

Sure, it's RQM

RQM has its problems, though

and it's really awful to deal with rigorous RQM

Sine-Gordon equation

The sine-Gordon equation is a nonlinear hyperbolic partial differential equation in 1 + 1 dimensions involving the d'Alembert operator and the sine of the unknown function. It was originally introduced by Edmond Bour (1862) in the course of study of surfaces of constant negative curvature as the Gauss–Codazzi equation for surfaces of curvature −1 in 3-space, and rediscovered by Frenkel and Kontorova (1939) in their study of crystal dislocations. This equation attracted a lot of attention in the 1970s due to the presence of soliton solutions. == Origin of the equation and its name == There are two...

might have got the wrong one

is the @Semiclassical around?

I need validation as a person and as a physicist =P

$Mathematics$ Mathematics

Associated with Math.SE; for both general discussion & math qu...

Not at the moment

@EmilioPisanty I'm afraid your crimes are too monstrous to ever be validated

@Slereah alas, it does seem that way

10:52 AM

Why the sun is red in the evening?

Can the time evolution operator be derived by integrating the SE wrt time?

hmm

$$i \hbar \int_{t_0}^{t_1} \frac{\partial \lvert \psi (x,t)\rangle}{\partial t} dt = \int_{t_0}^{t_1} \hat{H} \frac{\partial \lvert \psi (x,t)\rangle}{\partial t}dt$$

@Slereah this looks reasonable except somehow I am missing an $\mathbb{I}$

Guys, which of these two gambles would you prefer:

(1) 300,000 Guaranteed

(1)

10:55 AM

(2) 50% of nothing and 50% of 1,000,000

Because I like money

currency of your choice

Still (1)

ok

yep

Why is it even a choice

10:56 AM

well I now give you a seconc choice

and see if it is consistent

The two sums aren't even that far apart

(1) 99% of getting 200,000 and a 1% of getting 300,000

Constantine Black

Hi to all. May I ask something, if anyone has an idea he can drop it; Could there be any observational differences for gravitational waves coming from inflation if we defined differently the Stress-Energy tensor as given for example in the discussions of these posts:physics.stackexchange.com/questions/27048/… , physics.stackexchange.com/questions/119838/… .

(2) 99% of getting 200,000 and a 0.5% of getting 1,000,000 and 0.5% of nothing

does your choice now change Slereah?

Still (1) I'd say

10:58 AM

darn

come on guys I asked you a question

I choose (1) in the first round and (2) in the second round

Kenshin I have a gambling game for you

But of course if you keep chanching the ratios and sums, then I'm afraid we'll have to deal with

GAME THEORY

Which is unpleasant

(1) Guaranteed shit talk

10:58 AM

We'll have to whip out the utility functions

na it is just these two

(2) 50% of garbage philosophy and 50% of mindless rambling

Constantine Black

For example if working with the canonical SET coming from Noether's theorem and working with a SET as it is defined for GR( as in Hawking and Ellis, 1973), that is the Hilbert-Einstein SET.

Choose wisely

I choose (3)

10:59 AM

see JVN had 4 axioms from which he derived one should maximise expected utility

the 4th axiom is essentially that someone who chooses (1) in the first round will choose (1) in the second round

but psychological experiments show most people choose (1) and then (2) violating the 4th axiom of JVN

@Akash.B So don't bother asking questions here

because we are invisible to most people

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