Question

In Heisenberg's uncertainty principle, it is assumed that the uncertainity of the momentum of the electron agrees with that of the photon.

Assuming this is true (tell me if it's not), why is this the case?

If more information is needed, just ask.

Answer 1

Heisenberg's uncertainty principle is only applicable to Micro-Particles

Answer 2

since both particles are extremely (almost infinitely) small, the assumptions hold equally well for both, as @Yahoo! has mentioned. Uncertainty starts to diminish as the object being observed approaches the size of a few atoms, and by the time we get to objects as large as a speck of dust or an apple, uncertainty just about disappears from relevant measurements

Answer 3

Here is the problem I am doing.: http://i.imgur.com/St5yW.jpg

Answer 4

scattering = collision.

Answer 5

So, what does "scattering = collision" mean? I feel it somehow is the reason why the uncertainty in the momentum of the electron is the same as that of the photon.

Answer 6

Also, wouldn't Heisenberg's uncertainty principle apply even more to electrons since they're smaller than microparticles?

Answer 7

the change in an electron's momentum is due to its collision with a photon... it's like a billiard ball collision.

Answer 8

I'm not sure what the problem is asking... I'm not really sure what you're asking either... the idea is that attempting to 'cheat' the uncertainty doesn't succeed ...

Answer 9

So wait, are you saying that change_in_momentum_vector = electron_momentum_vector + photon_momentum_vector and also that change_in_momentum_vector = electron_momentum_uncertainty = photon_momentum_uncertainty or did I say something irrelevant?

Answer 10

no, the change in momentum isn't the uncertainty in the momentum.

Answer 11

The problem is just asking to show that if position uncertainty decreases, momentum uncertainty increases.

What I am asking for is a reason to justify that Δp_x = Δp_x_electron = Δp_x_photon.

Answer 12

In other words, why is the electron's uncertainty the same as that of the photon's?

Answer 13

yeah, the electrons have a well-defined momentum prior to the collision, so the idea is, if you can measure their position precisely in some clever way, there'd be a violation of the uncertainty principle...

Answer 14

I don't agree with that; to my knowledge, you can measure position precisely but at the expense of the momentum's certainty.

Also, I'm simply asking about why both particles' uncertainty is the same and not about the entire problem.

Answer 15

As in, knowing one of the two is not a violation of the uncertainty principle.

Answer 16

" you can measure position precisely but at the expense of the momentum's certainty."
yep, and vice versa

Answer 17

Yes but, that's not what I am asking.

The whole point of Heisenberg's Uncertainty Principle is to have either the momentum or the position OF THE ELECTRON be known to a great degree of accuracy.

However, when finding the momentum uncertainty, we use information about the PHOTON.

Which means that the momentum uncertainty of the photon is equivalent to that of the electron (at least for this problem) but I cannot justify why this is the case so I am asking you guys: why is this the case?

Answer 18

the idea here, I guess, is that uncertainty in photon position is dependent on lambda, so you'd think that by choosing photons with a very very small lambda, you might be able to violate the uncertainty relation. But small lambda means large frequency... and since momentum depends on frequency (and thus uncertainty in momentum depends on frequency), you can't violate the relation ...

Answer 19

you're missing the point a bit... the experiment is about having electrons with very well-defined momentum (much less uncertainty than hbar/2)... then, measuring their position accurately with photons by using that set-up, you should know momentum and position more accurately than the uncertainty relation allows...

Answer 20

but you can't...

Answer 21

I am not confused about why the uncertainty principle cannot be violated.

Sorry if I am not expressing myself well but, I just want to know why we can say that the uncertainty in the momentum of the photon is the same as the uncertainty in the momentum of the electron.

Answer 22

and where is that said?

Answer 23

the photons are for measuring electron position...

Answer 24

it's laid out clearly here; the relation between photon uncertainty in position and uncertainty in momentum is even more explicit, because they define momentum in terms of wavelength (lambda)
http://en.wikipedia.org/wiki/Heisenberg%27s_microscope

Answer 25

Okay, perhaps, I was thinking about it wrongly.

Now I'm guessing that it's not that we are dealing with the uncertainty in the photon's momentum and saying it's equal to that of the electron but rather that the electron momentum uncertainty is a a function of photon properties.

So basically, I am confused as to how the uncertainty in the electron's momentum is a function of photon properties (like the wavelength). By the way, I noticed that this also applies to the position as given in the question.

I have to go urgently but I will be back later to check what you answered.

Answer 26

There is no practical uncertainty in electron momentum. That's the point. You know momentum precisely... if a photon collides and travels through the lens, the idea is you will know where the collision took place. So by measuring the photon's position and momentum, (which should be easy, since it went through the lens and was focused to a point, and the momentum it gained or lost should be easy to find), then you've localized an electron that had a well-defined momentum...
think of the same experiment with classical particles (like billiard balls): if you shoot two billiard balls of well known momenta at each other, and they collide so that one of them rolls right through a hole you've made, then you know everything about the collision just from the fact that a ball with a easily measurable momentum rolled through the hole after the collision.
For any number of reasons, this doesn't work though, with unclassical particles... if you decrease wavelength of the photons in order to decrease uncertainty in their position, then you increase uncertainty in momentum...
if you increase the size of the lens to focus the photons better, you decrease uncertainty in position, but uncertainty in momentum increases...

Answer 27

The bottom line is, if they were simply two particles, they'd act like billiard balls. But because they behave like wavicles (getting diffracted by the lens, showing a dependence of momentum on wavelength), you'll never be able to tell precisely how they collided or what paths they took as you would with billiard balls..

Answer 28

^waveicles, +1

Answer 29

I'm now confused because the question is saying if Δx is small, show that Δp_x is large but you said (and I agree) that p_x (electron momentum) is known because it is given to us by the problem.

However, since we know the exact value for p_x (electron momentum), that would mean that Δp_x = 0 (or super close to it) which would mean that Δp_x would be super large would contradicts what is trying to be shown.

What's wrong with my thinking?

Answer 30

nothing. that's exactly the point...

Answer 31

for one thing this thought experiment is pretty limited... it treats an electron of known momentum as a particle... which is exactly what it is least like when momentum is sharp.

Answer 32

brb

Answer 33

when momentum is sharp, it means the particle isn't localized... and as uncertainty in momentum ->0 uncertainty in x gets very large... ie the wave packet that describes the particle is spread across a wide range of x...

Answer 34

but this is the argument he used, and it is more or less correct... even if it's not modeled very accurately...

Answer 35

but it sounds like you might be confusing the electron with the photon...?

Answer 36

I still do have confusions with the electron vs photon, yes.

Answer 37

The real point he was trying to counter was that someone (Einstein maybe ... I don't recall the deets) said " well if we make electrons with well known momentum and then we look through a high quality lens at them, we will eventually 'see' one...(an electron will collide with a photon and send it off through the lens) .. so doesn't this contradict your principle prof. Heisenberg, since it will be seen to be at x and also with a well known momentum ??"

Answer 38

Yes, I know it will have moved by the time it's seen. :)

Answer 39

(If that's what you were getting at.)

Answer 40

So he responded by saying that the principle couldn't be circumvented this way, because the photon carrying the information is going to obey the uncertainty relation (and he showed how it would do that)

Answer 41

Does the electron fuse with the proton when one sees it?

Answer 42

I mean photon not proton.

Answer 43

naw the photon scatters off the electron... like two billiard balls colliding... but of course it isn't like two billiard balls colliding...

Answer 44

that's the simplistic idea anyway. If you want to know what 'really' happens, you'll have to mess around with Feynman diagrams...:)

Answer 45

Is the reason why high momentum of an electron causes high uncertainty in position because the impulsive force will be great in the collision and will therefore make the position change more and that large change will make it be far from where we see it to be when the photon hits our eye (or whatever)?

Answer 46

Or is classical thinking wrong in this case?

Answer 47

ΔP = FΔt

Answer 48

or dP = Fdt if you prefer.

Answer 49

that's pretty classical... the real reason is... and it's actually pretty simple but a bit weird...
that any particle is just a bunch of sinusoids added up in a certain way... don't ask what the sines and cosines are... it's not clear what they are. The point is that to get a particle to actually behave like a particle ( to be sharp in position) you have to add up a huge amount of waves, each of which represents a certain momentum.... and the inverse is true... to get a particle with sharp momentum you have to add a huge amount of waves in the 'x' space... so the particle isn't even a particle but a spread...

Answer 50

It has to do with Fourier series and transforms... if you have a highly localized particle.. (a spike at x=c) the transform is a sine wave, meaning an infinite number of momenta are needed to create that spike....

Answer 51

I didn't learn Fourier stuff yet.

Answer 52

Let me take a look back and try to see if I can bring this question back to basics because I doubt the confusion is so profound.

Answer 53

When an electron collides with a photon, do they both have the same amount of energy after the collision?

Semi-classically, I think of regular collision as the more massive ball has a lower velocity therefore, it has lower kinetic energy but more mass and mass is energy.

Am I correct in thinking this?

Answer 54

no, I mean, they could I suppose, but it's not relevant..

Answer 55

I was asking if it was always the case but anyways, moving on. Let me think of something else to ask.

Answer 56

the point of this experiment is really the photon... you'd like to get precise info about the electron's position from the photon it hits and sends flying at you... but you can't get the precise info...

Answer 57

Okay so, if the electron has very low momentum then it will hit the light and the light will have very low momentum and why does that mean the electron will be easy to find?

Answer 58

the photon*

Answer 59

scratch that

Answer 60

if momentum is low uncertainty in momentum is low...

Answer 61

Why?

Answer 62

for "if momentum is low uncertainty in momentum is low..."

Answer 63

so it might be 'hard to find'... (frequency is low, lambda is large, photon will have a large uncertainty in position)

Answer 64

because you're limiting the range of momenta...

Answer 65

Sorry. I don't know if I'm getting dumber right now because I've been at this (and other stuff) for long.

Answer 66

if you say it can't have a momentum greater than... whatever... then you're making momentum sharper...

Answer 67

like say you make tthe photon very low momentum... say the photon can't have a momentum greater than h bar /2 well... that means your uncertainty in position has to be greater than one... which is pretty large....

Answer 68

one or greater than one you mean, right?

Answer 69

sure.

Answer 70

I gotta head to class... maybe someone else can pick this up... I'll be back later this evening if you're around...

Answer 71

I'll be back tomorrow since I got exhausted. Thanks for your help (so far).