Camp... Camp... Camp...

Torch light, sticks, bats & balls, tumbler & plates, umbrella... What then? :D

@ManishEarth: Whenever I read my old answers, I hate my English (historical...)

Anything that I can do....

(I hate the whole answer - I mean, I wanna delete those)

uh, fix it?

But, you can't do that to 100 of 'em... :/

(always bump it front)

Slowly then :)

If someone notices it, they'd be angry with my revisions...

There's a possibility that many people can think that I'm doing these revisions just because my answer want to get some attention... -_-

This makes me feel awkward...

meh, if you're improving grammar, its ok

k... I'll decide it when I return from my camp... Maybe, I should give myself some rest :)

yep

physics.stackexchange.com/questions/68940/… If this was physicsforums, this post would be "initiating flame war in 3...2...1...."

@twistor59 Hmm, looks innocuous to me :)

Why would it start a flamewar?

(I think I have an idea, but not sure)

No, it's fine, it's just that on PF everyone gets hot under the collar about the ontological status (LOL) of virtual particles

ah

Like e.g. this Phys.SE post and links therein, which spurred dozens of comments.

ah

Hey @ManishEarth @Qmechanic : i have a quick question

fire away

its on very basic of q mechanic which i am just learnin... but a bit of a wild one

Consider a wavefunction$\psi(x) = \exp(-kx)$ for $x\in (0,L)$ and $0$ otherwise. for get about time dependence of the wavefunction, we just concentrate on the static case for now.

@Manish did you get till now

yeah

The wavefunc is invalid, it's discontinuous

at 0 and L

Thats ok, I am still at basic stuff now. I am not going to introduce any time dependence or shcrodinger, just basic. it may not conform to any real world situation but a conceptual one. ok?

No, not even a conceptual one

You can't have a $\Psi$ like that.

its written in most books

hm?

at introduction

in most books it's $\exp(-ikx)$ or $\sin kx$ something

just to explain probability

that's different

sorry its actally -ikx, i missed i

Yeah. OK, continue

It basically shows that the particle position is uniformly distributed on (0,L) right

yep

at it momentum is concentrated around $k$

I'm not sure though, this may be invalid too as the imaginary component won't be continuous

@RajeshD yeah

$\hbar k$ to be exact

ok, i am not going to deal with continuity later, just assume for now

ok hbar, the room we are in :-)

I mean I'll try to deal with conitnuity later

@RajeshD We can't assume that here, it's obvious that it's discont. Which means that paradoxes may arise. But nvm, we'll ignore it

okay

Now $k$ could be anything right?

I mean how ever large or how ever small

in this case, yes, because we are neglecting boundary conditions

now assume the $hbar k$ is very very small som[pared to the dimension $L$, I mean let assume $hbar k$ is sufficiently small

kk

What I wonder is a particle with very low velocity/momentum, can it be found with equal probability in an entire space (0,L), wouldn't it sound starnge and may be tooo random?

@RajeshD Like I said, apparent paradoxes will arise if you neglect the continuity boundary condition

Also, note that HUP is an inequality

I can have a particle with infinite position and momentum uncertainty

ok assume continuity with a smooth cutoof function

it shouldnt matter now

@RajeshD then $k$ becomes dependent on $L$

Like I said:

If not, you have an extra equation and that will quantize or restrict the allowed values of $k$

we can make smooth cutoff function arbitrarily approximate to square function we have such functions in math, then $k$ essentially approximately reamins same

This has nothing to do with the continuity of the cutoff

This has to do with the fact that you ignored continuity initially

given any arbitrarily small $k$ getting a sharp cutoff function wouldn't be a problem

Which means that we dropped an equation -- the boundary condition

If you don't want to ignore continuity, even with a sharp cutoff, you have to include the continuity BC

@RajeshD Anyway all of this is besides the point, you can still choose a $k$ of a chosen order of magnitude, just that it's quantized

okay fine but almost uniform right. just a bit I'll introduce you to functions with smooth cut off to you in a min, which i think would clear the air

The cutoff doesn't matter: The choice here was "Are we neglecting the continuity constraints?"

Also, that doesn't matter either :P

You can still choose an arbitrarily large k. Just not an arbitrary k

Take the particle in an infinite box for example

$k$ is dependent on $L$, but also directly proportional to $n$

Smooth cutoff functions these cut off functions are smooth, they have derivatives of infinite order and they are still represent almost uniform probability

We can make $n$ as large as we wish

@RajeshD I know

That's besides the point

Manish I am not getting what you mean to point

lets make $k$ arbitrarily small

@RajeshD I'm saying that the cutoff thing is irrelevant

@RajeshD We can't do that

We can make k arbitrarily large

why?

lets get to that later

I lknow of it

I am not chaning $L$ so i am not messing with uncertainity m in momentum right?

@RajeshD See, regardless of the continuity of the function, in a physical situation, we need to include the continuity BC. So you've found a way to make the BC work for the flat function with a smooth cutoff. Fine. Still, the continuity BC will introduce quantization of $k$

hold on a bit

what do you mean by quantization of $k$

we have a sinc like function with sharp peak at $k$ in momentum space right?

which mean momentum is equal to $k$ with an u8ncertainity depending on $L$

Huh? The propability distribution of p is the same shape as that of position

The magnitude is top hat

$\int \Psi* \hat p \Psi = \hbar k\int\Psi*\Psi$ for $\exp(-ikx)$

$|\psi|$ is hat

picture the $|\psi|$ it is a square function which is uniform in (0,L) and zero else where

but $\psi(x)$ nis a complex sinusoid

$\sigma_p=p\sigma_x$ for $\Psi=\exp(-ikx)$

@RajeshD OK, so?

just consider only one k baba

ok

"only one k"?

Now I can have any desired $k$ and the momentum is $k$ with an uncertainity depending only on $L$, as if $L$ is quite high then this uncertainity is quite small irrespsctive of $k$

I mean 4k$ cghoosen

only one $k$

you are right

@RajeshD OK, so? Let the momentum be quite high

No I mean let it be too small, I ahe case momentum being very small, not highm talking about t

How are you sure you can make momentum arbitrarily small?

It could be quantized

It should be quantized

Whatever cutoff function you use, it will introduce a quantization

wait I dont get what you are saying, the momentum is always most probable $k$ with an uncertainity depending onlly on $L$, irrespective of $k$ choosen

(Also, the cutoff function itself will contribute to the uncertainty of momentum, because he momentum operator applied on a cutoff function gives an infinite value. Integrating that will create problems, and will probably create high peaks near the edges)

@RajeshD I mean that $k$ ought to be quantized

dont talk about quatization here, i dont know what you mean.....All I do is just picturize the Fourier transform of $\psi(x)$ which is $\phi(k)$ which gives the probability distribution of momentum right

@RajeshD Usually when you apply boundary conditions on a system, the parameters of that system get quantized

Now picture the Fourier transform , it is a convolution of fourier transform of cutoff function and the fourier transform of $\exp(-ikx)$

Wait a sec

dont talk of that, just look at fourier ributiontransform to get momentum probablity dist

alright, continue

Now te momentum distribution has a sharp peak at $k$ with dense concentartion around it and almost zero elsewhere

@RajeshD The cutoff funcs will create a problem

The spread of this concentartion around $k$ depends only on $L$ and if we fix $L$ to be at certain high valuehe this uncertainity is fixed at some low value

Okay, I'll picture a Fourier transform of cutoff function for you

$\langle \Psi |\hat p | \Psi\rangle$ has peaks at the edges

no issue, it may result in spurious other discrete moemta but with low probability

thats manageable

@ChrisWhite Well, that does create an issue -- if you can create an arbit (unbounded) low Delta p without changing Delta x then you can overthrow the uncertainty principle

Ok My question starts now

@Manish

...

assume $k$ is very very small, which means the most probable momentum/velocity is almost negligible, but still the particle is expected to found in (0,L) with almost equal probability right? I just wonder its kind of strange to believe

especially if we assume classical velocities, very very low compared to velocities of electrons/other particles

yes

Hmm

too much information removed I guess

@ManishEarth I dont want to open a new meta post now. But I think it is a pity that questions are looked adt by some mods and other people not to leand something nice and interesting about physics the OP potentially brings up, but exclusively with the aim to detect a flaw according to the present official close reasond. Being more generous and in favor of the interesting physics contained in a question and more focused on the superficial form of a post would help giving people the fealing back

@Dilaton And I don't want to join a discussion where you repeatedly ignore what I say and just barrel on with the same points.

that the site is physics oriented and take some of the heavy weight the policy and rules orianted atmosphere currently puts on physics SE.