The h Bar

6:00 PM

@Blue As per the note, we consider some observables which can't be expressed with x and p as irrelevant

Anonymous

In physics, complementarity is both a theoretical and an experimental result of quantum mechanics, also referred to as principle of complementarity. It holds that objects have certain pairs of complementary properties which cannot all be observed or measured simultaneously. The complementarity principle was formulated by Niels Bohr, a leading founder of quantum mechanics. Examples of complementary properties that Bohr considered: Position and momentum Energy and duration Spin on different axes Wave and particle-related properties Value of a field and its change (at a certain position) Entanglement...

Anonymous

> A crucial difference between classical quantities and quantum mechanical observables is that the latter may not be simultaneously measurable, a property referred to as complementarity. This is mathematically expressed by non-commutativity of the corresponding operators, to the effect that the commutator

Anonymous

@taritgoswami I don't think that statement makes any sense

bolbteppa

The point of scattering is to ask how the particle would deviate from an otherwise well-known trajectory, the simplest thing to ask is - how would a free particle at constant velocity deviate due to a scattering center, an experiment like Rutherford is trying to fire electrons in such a way so that they behave as free particles at constant velocity once they are shot and then end up acting as free particles again once scattered so that we can focus only on the scattering properties

Anonymous

Maybe ask your prof to clarify

tarit goswami

6:02 PM

That means there are some observables that can't be expressed using x and p, though we don't need them as using x and p we can understand any dynamics fully

curiosity

@taritgoswami there are very important oberservables that cant be expressed in terms of x and p: the spin degree of freedom

bolbteppa

What's the difference between a particle in the narrow region missing the scattering center and a particle outside the narrow region (in a model where we're not constrained by the size of the electron gun size) missing the scattering center if all we care about are the particles that do scatter

Anonymous

"That means there are some observables that can't be expressed using x and p"

Anonymous

And I'm asking for an example of such an observable

tarit goswami

@Blue let me check the notes again tomorrow. :p

curiosity

6:04 PM

a particle cant miss the target is the beam is narrow enough so that the wavefunction is spread over the space where the potential is nonzero

tarit goswami

@Blue as @curiosity stated, spin is one such

Anonymous

@taritgoswami How do you define spin in classical mechanics?

bolbteppa

In Rutherford's scattering experiment most of the particles do miss the target

curiosity

yea but those go straight through, at $\theta =0$

bolbteppa

It sounds to me like you're saying we can make the beam so small we can be sure the electron follows a certain path

curiosity

6:06 PM

they dont hit the detector at other angles

we can make to beam so small that the wavefunction will interact with the whole potential, but still be narrow enough so that it wont reach the detector for other angles but zero degrees

bolbteppa

How do you know they go straight through

curiosity

because i can put the detector at $\theta =0$ and measure the scattering amplitude there

tarit goswami

@Blue what I have understood, in classical mechanics spin is irrelevant then.. As the note stating that "any observables we can't express with x and p are irrelevant in classical mechanics", though it is relevant in QM

curiosity

then i know how many will go through

bolbteppa

So the detector has infinitely small size

curiosity

6:08 PM

practically

bolbteppa

That's impossible, non-zero angles are always possible no matter how small, you have no idea

We can't tell if some fired off at a tiny angle

Anonymous

@taritgoswami It's not irrelevant, it's non-existent. There's no notion of spin in classical mechanics similar to that in quantum mechanics.

curiosity

sure but i neglect the error of the angle

its neglectable in comparison to an angle of $\frac{\pi}{4}$ , say

tarit goswami

@Blue Ooh, we don't have notion of spin in classical mechanics, right

curiosity

which is perfectly relevant in a scattering experiment

bolbteppa

6:10 PM

So we're already doing approximations, I see, and what's the difference between using a wave function where a free particle is constrained to definitely hit the scattering center, and a wave function where the free particle could hit the scattering center or miss it

Semiclassical

Half-integer spin, at least

curiosity

the difference is whether we say that the beam can reach the detector at large angles like $\frac{\pi}{4}$

tarit goswami

@Blue It will be good to see examples of classical observables that can't expressed using x and p

curiosity

which it cant do in experiments like the one from rutherford

Anonymous

@taritgoswami There's none

Semiclassical

6:12 PM

you can perhapd make sense of integer-valued spin in some classical way

Anonymous

@Semiclassical Like angular momentum?

Semiclassical

right

Anonymous

But for that we'd have to deal with a system of particles

Anonymous

Single particles are point particles in CM, which are completely defined by $x$ and $p$

Semiclassical

Sure

Anonymous

6:14 PM

And if you define spin using a system of particles, even then it will be defined by $(x_1,x_2,...,p_1,p_2,...)$ plus constraints

Anonymous

Tarit was asking for an example of an observable in CM which cannot be defined by $\mathbf{x},\mathbf{p}$

Anonymous

I'm pretty sure there can't be any

bolbteppa

@curiosity not sure what you mean, but the model where you begin with incoming plane waves of course derives the Rutherford cross section

curiosity

what are you not sure about?

what part?

bolbteppa

How are all the books doing something wrong if it rederives this result

curiosity

6:16 PM

i dont claim they are wrong

im saying i dont understand why they are right

i dont see why a model where the incoming particle beam is as broad as the whole laboratory would predict the right scattering amlitudes

since in the real experiment the beam is narrow

so narrow that it cant reach other angles but zero (approximately)

only if it was scattered

bolbteppa

They are right because they model the incoming particle as an asymptotically free particle that scatters off a scattering center into another asymptotically free particle and an experiment like Rutherford gives exactly this behavior even though you think we have to model the incoming electrons as constrained in a certain region, there is no experimentally creatable electron gun that could constrain them to fire in a narrow enough region, there is always uncertainty in the electron direction

curiosity

but this uncertainty is very small compared to an angle of $\frac{\pi}{4}$

Betrieb einer Elektronenstrahlröhre mit Ablenkung durch E-Feld

►

this is how broad the beam spreads

not more

tarit goswami

@Blue Probably non-existance of such observables is stated as irrelevant

bolbteppa

But they travel such a long distance you can't ignore it, there is no way you can make them all fire at the angle $\theta = 0$, even a tiny angle will add up as it moves along the path

Semiclassical

I mean, if the argument is that the scattering theory calculations on a standard QM book are not sufficiently realistic to be applied directly to an actual system

curiosity

6:24 PM

thats just not true, you can make the beam straight enough so that the error of the angle is negligible in comparison to something like pi

Anonymous

@taritgoswami Possibly

curiosity

@sem

@Semiclassical im talking about that fact that we equate two asymptotic expressions of the wavefunction

Semiclassical

Well, that’s probably true. But the solution is presumably to look at a book that talks about actual detector geometry

curiosity

where in one of the expression there is a plane wave term

that should not exist in my opinion

bolbteppa

Not only the detector geometry, also the electron-gun geometry and the scattering center geometry lol

curiosity

6:26 PM

$ \psi(\vec x) \sim \sum_l \frac{A_l}{kr} \sin(kr - l\pi/2 + \delta_l) P_l(\cos \theta) $

Semiclassical

Well, there’s definitely one problem with a plane wave solution : a real scatterer has a finite width in position space

tarit goswami

@Blue Can you read the 1st paragraph ocw.mit.edu/courses/physics/8-04-quantum-physics-i-spring-2013/…

bolbteppa

In $\Psi = e^{ikz} + f(\theta) \frac{e^{ikr}}{r}$ the argument is that $e^{ikz}$ is wrong because the electron gun has a finite width and constrains the incoming particles to a finite region while $e^{ikz}$ allows for any incoming region

curiosity

and $\psi(\vec{x}) \sim e^{ikz} + f(\theta) \frac{e^{ikr}}{r}$

enumaris

need 150 more rep...

Anonymous

6:27 PM

@taritgoswami Done

Semiclassical

there’s a pertinent reference in Griffiths as far as the 1D case goes, let me track it down

tarit goswami

@Blue I'm left with only 6%charge :p

Anonymous

Okay, we can talk later then

tarit goswami

Bye for now

Anonymous

Bbye

tarit goswami

6:29 PM

K

Semiclassical

The other related problem is that, if you send a single particle towards a scattering center, this should presumably be a time-dependent process

Whereas the standard calculation is time-independent

curiosity

yes in that case i think we have to treat it as a wave packet

i think we can approximate it with the plane wave though

but not with one that extents over a large angle

Semiclassical

Well, I suspect the out is that we’re only interested in scattering amplitudes

curiosity

yea

Semiclassical

We don’t care about time-of-flight measurements, at least not in this setup

curiosity

6:33 PM

but we should care about the beam being narrow

Semiclassical

Narrow relative to what, though

I don’t think an electron beam will typically be narrower than the scatterer

curiosity

yes not with respect to the scatterer but to the position of the detector

we dont want any incoming particles to reach the detector at any other angle but zero degrees

Semiclassical

I don’t think you can require that. Even in Rutherford scattering you have the possibility of back-scattering for instance

curiosity

yes sure scattered particles can reach any angle

Semiclassical

It may or may not be probable but I don’t see why it’d be forbidden

curiosity

6:37 PM

but the particles cant go out of the gun and directly into the detector

Semiclassical

Sure they can

bolbteppa

Maybe a way to say it is, in $\Psi = e^{ikz} + f(\theta) \frac{e^{kr}}{r}$ the term $e^{ikz}$ represents no scattering occurring while $ f(\theta) \frac{e^{kr}}{r}$ represents the scattering occurring, and just because only a small region may pass through the electron gun it doesn't mean that inside the electron gun we didn't have what amounts to free particles moving

Semiclassical

They’d just miss the scatterer

curiosity

not at angles other then zero

Semiclassical

They’d just miss the scattering center

curiosity

6:38 PM

yes but then they would only reach the detector if it was at $\theta =0$

Semiclassical

Eh. I’d want an actual geometry before I’d commit to that

bolbteppa

If we ignore reality and just allow $\theta = 0$ why is there a problem

Incoming

curiosity

@bolbteppa i agree, but only for $\theta =0$

the geometry can be seen here tcm.phy.cam.ac.uk/~bds10/aqp/lec20-21_compressed.pdf

bolbteppa

So if all the particles go towards the scattering center at $\theta = 0$, they don't all have to scatter, the ones that don't scatter are represented by $e^{ikz}$ and the ones that do by $f(\theta) \dots$

curiosity

yes

but when we consider the asymptotic form of the wavefunction at large angles, the literature still keeps the $exp(i k z)$ part

bolbteppa

6:42 PM

curiosity

yes thats exactly the picture i have in mind

in the asymptotic form, the $exp(i k z)$ part should only be kept for $\theta =0$ thats my claim

Semiclassical

I think the pertinent question is "which literature"

if we're talking intro QM books, I'd believe it

curiosity

griffiths

messiah

25

Q: Phase shifts in scattering theory

I have been studying scattering theory in Sakurai's quantum mechanics. The phase shift in scattering theory has been a major conceptual and computational stumbling block for me. How (if at all) does the phase shift relate to the scattering amplitude? What does it help you calculate? Also, any...

this answer

Semiclassical

sure. I'd chalk it up to them not trying to be completely realistic

curiosity

also my lecture at university

every literature that ive seen, frankly

Semiclassical

6:45 PM

presumably a more realistic treatment would account for the finite transverse width of the incoming electron beam

curiosity

thats hard to believe

for me

bolbteppa

So the question is, why do we keep $e^{ikz}$ in $\Psi = e^{ikz} + f(\theta) \frac{e^{ikr}}{r}$ if the particle scatters?

curiosity

yea

Semiclassical

because you're solving it as though the incoming beam has infinite transverse width

that's fine for an intro textbook

curiosity

in order to derive the form of the phase shift and the coefficients in $\psi(\vec x) \sim \sum_l \frac{A_l}{kr} \sin(kr - l\pi/2 + \delta_l) P_l(\cos \theta)$

we set this expression equal to $\Psi = e^{ikz} + f(\theta) \frac{e^{ikr}}{r}$

why dont we just set it equal to $\Psi = f(\theta) \frac{e^{ikr}}{r}$ for $\theta \neq 0$

and to the term plus plane wave for $\theta =0$

Semiclassical

6:55 PM

one thing I will note is that, if you want precise answers to this, you should probably ask an experimental particle physicist

bolbteppa

One point, you can explicitly derive $\Psi = e^{ikz} + f(\theta) \frac{e^{ikr}}{r}$ by approximating the Green function solution of the Schrodinger equation

curiosity

i know the derivation, we dont have to add the $exp(i k z)$ term, it just doesnt mess up the solution

its a boundary condition that we're allowed to change

we have to solve
$(\Delta+k^2)\psi_k(\vec x)=U(r)\psi_k(\vec x) $

bolbteppa

Slide 39 and 40 tcm.phy.cam.ac.uk/~bds10/aqp/lec20-21_compressed.pdf you do need to add it, it's the homogeneous solution you need to include as part of solving an inhomogeneous linear differential equation

curiosity

$\psi_k(\vec x)=\psi_0(\vec x)+\int G(\vec x-\vec y)\psi_k(\vec y)d^3y$ does the trick, with or without the $\psi_0(\vec x)$

bolbteppa

So they call it Lippmann-Schwinger for some reason

curiosity

6:59 PM

$(\Delta+k^2)\psi_0=0 $

its a solution even without the $\psi_0$

bolbteppa

So you're asking why can't we just ignore the homogeneous solutions of a linear PDE

curiosity

yea

if the boundary conditions force us to do so

the BC are given by the experimental setup, which is a narrow beam

Semiclassical

One thing to note, I guess, is that you can go from a wide beam to a narrow beam by convolving the former with a gaussian

curiosity

if we could show that this convolution doesnt change the scattering amplitude, i would be happy

Semiclassical

well, I don't think you'll be able to avoid the following: If you change the transverse width of your electron beam, then the scattering distribution should change

the wide-angle portion should presumably be unchanged, but the small-angle part would change

curiosity

7:06 PM

yea there is no reason to expect that it wouldnt change the scattering distribution

Semiclassical

with the 'small-angle' regime being determined by the beam width and the size of the detector

curiosity

thats why i had the trouble with the plane wave in the first place

i guess i will ask my professor about that

seems we wont reach a conclusion now

but thanks for the discussion guys!

bolbteppa

I think this comes down to quantum mechanics

If the particle was incoming, you can't ignore the probability that it might end up as an outgoing particle without scattering

Semiclassical

whereas I think this comes down to considerations of experimental context

bolbteppa

I remember seeing a justification for why the end result is $\Psi = A [ e^{ikz} + f(\theta) \dots ]$ and not $\Psi = A e^{ikz} + B f(\theta) \dots$

The latter possibility is the only way you can possibly ignore the homogeneous solution

Semiclassical

7:10 PM

i mean, if you see a particle backscatter, then there really should be no probability that that occurred without the particle scattering

bolbteppa

I forget why now

curiosity

well just plug the psi into the schroedinger equation and you will see that we can have a solution without the plane wave

Semiclassical

I guess it's also worth noting that, in actual practice, what you'd probably do are two experiments

curiosity

with $(\Delta+k^2)G(\vec x)=\delta(\vec x) $

Semiclassical

1) do the scattering experiment without any scatterer

2) include the scatterer

curiosity

7:11 PM

$\psi_k(\vec x)=\int G(\vec x-\vec y)\psi_k(\vec y)d^3y$ is a solution

Semiclassical

the difference between the two is presumably the interesting part

bolbteppa

But in QM you need the most general solution by linearity/superposition principle

curiosity

not if you want to solve a particular problem

bolbteppa

You then need to use the general solution to argue why some terms are zero in a specific example

curiosity

yes and the picture you posted captures the argument

im out guys

bolbteppa

7:14 PM

The only way you could do that is if $\Psi = A e^{ikz} + B f(\theta) \dots$ were the wave function, but for some reason $\Psi = A [ e^{ikz} + f(\theta) \dots ]$ is actually the wave function, you can't ignore the $e^{ikz}$

curiosity

have a nice evening

Semiclassical

@bolbteppa Well, here's my problem with it: Suppose there wasn't a scatterer there at all.

Then you'd have $\Psi = Ae^{i k z}$. That's fine if you're within the beam width, but not outside of it

bolbteppa

You can rewrite this in terms of an S matrix where the identity part $I$ is $e^{ikz}$ with no scattering as well

Semiclassical

hmm

I mean, I can understand there being dispersion in the transverse direction as you move along the direction of the beam. but the beam should nevertheless have finite transverse width no matter how far you look, so long as no scattering actually happens

bolbteppa

I mean the beam is so big with respect to the electons atomic dimensions it can be treated as a plane wave, I think that's fine right?

Wait, because you are working with only one homogeneous solution and then the inhomogeneous term, why do you need separate coefficients or to form linear combinations

Why would you just randomly ignore the homogeneous solutions of an inhomogeneous linear pde

Abstractly one would expect the homogeneous solution to be a linear combination of a bunch of things and then adding the inhomogeneous solution before trying to fix the coefficients in that linear combination right

No you need to justify why the coefficients are what they are, hmm

Semiclassical

8:32 PM

@bolbteppa it's not the same thing, but I imagine you could pose a similar question as far as classical scattering by EM waves

bolbteppa

Most sources I remember seeing just skip explaining this, even ones that get it from 'Lippmann-Schwinger'

I vaguely remember it being a good reason :(

Semiclassical

well, how many sources are intended for particle physics experimentalists?

bolbteppa

0 :p

Semiclassical

8:48 PM

well, there you go :P

this looks like a more exhaustive treatment: books.google.com/…

enumaris

9:06 PM

do I dare to write a post about energy in GR...do I...

Semiclassical

do you dare disturb the universe

enumaris

hmmm

Reminds me of J. Alfred Prufrock

CooperCape

This might be a really dumb question but if $\vec v_A(t)=f_i(t)\hat i+f_j(t)\hat j$ and $\vec v_B(t)=g_i(t)\hat i+g_j(t)\hat j$ where $v_x$ is the velocity of $x$, then is the relative velocity of $B$ w.r.t $A$ just given by $\vec v_{B/A} = (g_i(t)-f_i(t))\hat i+(g_j(t)-f_j(t))\hat j\ ?$

Anonymous

@CooperCape Yes

CooperCape

And the same holds for $r_{A/B}$, where $r$ is displacement right?

Semiclassical

9:09 PM

@enumaris as it should

Anonymous

@CooperCape Right

CooperCape

I'm trying to prove that for two objects at their point of closest approach $\vec r_{B/A}\cdot\vec v_{B/A}=0$ but it's not working out

enumaris

At least non-relativistically it's right :D

CooperCape

At least i got that part correct

Trust I'm no where near to the point of being able to lorentz transform this

enumaris

wait, maybe it's also correct relativistically if by "w.r.t." you don't mean moving to the frame of A but just the separation speed within the original frame...

too lazy to think that one through

Anonymous

9:12 PM

1

Q: Why must closest approach occur when relative velocity is perpendicular to motion?

The first part i) I can solve correctly, but I need some advice and intuition on how to solve the second part ii). Here is the mark-scheme for the question: But for part ii) I do not understand their logic as shown in red i need to know why this must be the case for closest approach. Coul...

geometry trigonometry intuition

CooperCape

I mean they're measured to be moving from the perspective of one stationary inertial frame of reference

@Blue I did see that but I kind of wanted a more algebraic approach

Although maybe that's not plausible

Anonymous

Everything is possible algebraically. I'm a bit too sleepy now though

Anonymous

I'll see tomorrow morning

CooperCape

I'm gonna try and crack on with it tonight - thanks.

@Blue Actually I didn't look at the second answer on that post which gives a better explanation I think - might just use that but I'll still have a go with it algebraically.

Avantgarde

@CooperCape I thought you said "I'm gonna try and have crack tonight"

CooperCape

9:23 PM

@Avantgarde In Manchester - anything is possible

(Definitely neither true or quotable or said by anyone, ever)

Avantgarde

-CooperCape

circa 2018

CooperCape

I like this.

Oh I did it

It was actually quite easy

I just needed to not be lazy - classic

Anonymous

@CooperCape Yes, either $r_A=r_B$ or $d|\vec{r}_{A/B}|/dt = 0$ at minimum approach, isn't it?

CooperCape

Yeah that was what I did

Which is just repeatedly using product/chain rule

And after expanding horribly the part that isn't in the $(h(t))^{-1/2})$ (i.e the only part that can equal $0$) comes out to be equal to $r\cdot v$.

enumaris

9:52 PM

still debating...

Mauricio Mendes

Hello people. Could anybody help me with the "car and bicycle rider" problem, please? Here's the link for the problem: ocw.mit.edu/courses/physics/8-01sc-classical-mechanics-fall-2016/assignments/MIT‌8_01F16_pset1_new.pdf

I tried to integrate the function to determine $v(t)$ and then integrate $v(t)$ to determine $x(t)$. $v(t)=b, 0 < t < t_1$ and $x(t)= bt, 0 < t < t_1$ Also, $v(t)=-c(\frac{t^2}{2}+t_1t), t_1 < t < t_2$ and $x(t)=-\frac{t^3c}{6} + \frac{ct^2t_1}{2}, t_1 < t < t_2$

enumaris

I get a page not found error

Mauricio Mendes

Try this one: ocw.mit.edu/courses/physics/…

Lozansky

@MauricioMendes So what's the problem?

You probably want a continuous velocity function

enumaris

You're solving part b?

or part a?

Mauricio Mendes

10:07 PM

@enumaris part a

Lozansky

So at $t = t_1$, your expressions for the velocity should match

Mauricio Mendes

@Lozansky i wanna know if my steps are correct

Lozansky

And same for the position function

enumaris

are you using $b$ as a proxy for $v_0$

Mauricio Mendes

@enumaris yup

@Lozansky when doing so, i got $t_1^2=\frac{b}{c}$

enumaris

10:12 PM

When you integrated to get $v(t)$ you need to remember to include the constant of integration..also it looks like there should be a minus sign between the two terms rather than a plus sign.

same for $x(t)$

Mauricio Mendes

@enumaris As for the constant of integration, should it be $v_0$?

enumaris

So it should be $v(t)=-c(t^2/2-t_1 t)+A$ where $A$ is some constant with units of velocity and then you have to use the initial condition $v(t=t_1)=v_0$ to determine what $A$ is

Don't think so, the constant of integration will be some function of $v_0$ and $c$

wait

right, the condition you have to use is $v(t=t_1)=v_0$ rather than $v(t=0)=v_0$ because your expression for $v(t)$ is only valid for $t_1<t<t_2$

Mauricio Mendes

@enumaris So $A=v_0 - \frac{ct_1^2}{2}$?

enumaris

looks right to me

Now do the same for $x(t)$

Anyone here good with python? I have a question about how best to pass variables...

Mauricio Mendes

10:28 PM

@enumaris Before getting $x(t)$, do I have to substitute $A$ in $v(t)$ and then integrate it?

enumaris

hmmm...actually I think I got it :D

@MauricioMendes yeah, but it's a constant so integrating it should be simple

Mauricio Mendes

@enumaris Ok, thanks!

enumaris

np

manooooh

11:18 PM

Hey everyone!

I would like to know why my question was put on hold:

0

Q: Find forces applied that supports a block

The pen of the figure hangs a block of mass $m = 1020~kg$. Find the modules of the forces applied in supports $A$ and $B$. I did the following, where $T_A$ is tension at point $A$, $T_B$ is the tension at point $B$ and $P=1020\cdot10=10200~N$ is weight (taking as reference the positive $x$-a...

forces vectors free-body-diagram

That question is about forces and I showed my work. I don't know why...

enumaris

"Check my work" kind of questions are off-topic for the main site

In order to make your post acceptable from the main site, you'd have to extract some conceptual question out of it and ask about that...you can ask your question here in chat though and probably someone will help you.

Anonymous

11:35 PM

@manooooh There's a dedicated Problem Solving room for such questions

enumaris

ugh, this last 150 rep is annoying to get lol

none of my recent answers are getting many upvotes...

Anonymous

-4

Q: How is spacetime created with reference to bosons and gravity

I have a theory that bosons, which confer gravity, are actually created by the decay of heavy elements. Hence a black hole is the biggest creator in the galaxy. This creates an outward flow of bosons from anything with mass. We observe this as a bending in spacetime as the bosons are in many ways...

gravity spacetime dark-matter dark-energy bosons

Anonymous

How this question pass through without getting closed as "non-mainstream" for 9 hours?

enumaris

I flagged it

dunno how long it takes mods to see it

Anonymous

ACM isn't around today

Anonymous

11:39 PM

Hmm

enumaris

-2

A: Is the dark matter present only around the galaxies?

Dark matter is a supersolid that fills 'empty' space, strongly interacts with ordinary matter and is displaced by ordinary matter. What is referred to geometrically as curved spacetime physically exists in nature as the state of displacement of the supersolid dark matter. The state of displacemen...