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Why is information about the electron described with information of a photon [...]?

by s3a
Tags: electron, information, photon
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BruceW
#19
Nov10-12, 08:51 AM
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p.s. sorry I've taken a while to reply, I have been under the weather.
s3a
#20
Nov22-12, 11:26 AM
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Okay, I think I'm beginning to get it more.

Right now, I need to understand two more things to my knowledge.:

1) I'm beginning to understand the Δx = λ/sinθ more. Basically, a larger wavelength takes more physical space and needs a larger slit so it can pass through. Also, sinθ will yield a value between 0 (excluded) and 1 (included) meaning that the slit will be larger or equal to the wavelength of the photon and that the photon is somewhere within the slit. In other words, it makes sense to me why Δx and λ are proportional. What doesn't make sense to me is how the sinθ part is there. I would like your help to dive into the single-slit diffraction non-formal “proof” to figure out how to get the sinθ part if you don't mind.

2) I would like to understand how the momentum uncertainty is the Δp_x = 2h/λ sinθ equation. I understand it geometrically so, what I am asking is basically when you said
We know that adding the same constant to each data point doesn't change the spread of the values, therefore the spread is the same.
, how do we know that this constant spread is (always) equal to 2h/λ sinθ? Or is that not what's going on? Assuming what I said so far is correct for #2, can you show that to me mathematically please? I believe an explanation using vector summation would be the one I am looking for.

QUOTE]p.s. sorry I've taken a while to reply, I have been under the weather.[/QUOTE]
That's fine, I'm not rusing you. In fact, I am very grateful for your answers. Also, sorry for taking long to respond myself. I've seen your post for a while now but, I was either too busy or too tired to post and it's pointless to post if I haven't thought through what I am writing (since this thread is about understanding a physical concept rather than just responding to a regular email). Technically, I am tired now too but, I'm still well-rested enough to be able to understand what's going on if I think about it slowly.
s3a
#21
Nov28-12, 03:17 PM
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Actually, I think I get the theory now.

The only minor confusion I have now is simply: "Shouldn't Δx ~ λ/sinθ be Δx ~ λ/sinθ - -λ/sinθ = 2λ/sinθ instead such that the final answer is Δx Δp_x ~ 8πħ?"
BruceW
#22
Dec1-12, 07:28 PM
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Quote Quote by s3a View Post
Actually, I think I get the theory now.

The only minor confusion I have now is simply: "Shouldn't Δx ~ λ/sinθ be Δx ~ λ/sinθ - -λ/sinθ = 2λ/sinθ instead such that the final answer is Δx Δp_x ~ 8πħ?"
Good point. 2λ/sinθ is the width of the central peak, but I think they used Δx ~ λ/sinθ because the intensity within the central peak is not uniform, so the spread is not as great as you might first think.
s3a
#23
Dec5-12, 05:53 PM
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Can you relate what you said in your most recent post (the one before this one that I am writing now) to the concept of a Diffraction-limited system ( http://en.wikipedia.org/wiki/Diffraction-limited_system ) for me please since it seems to be strongly related to what I am asking?

Based on the equation from the Wikipedia link I'm now posting, it seems that Δx ~ λ/(2nsinθ) - -λ/(2nsinθ) = 2λ/(2nsinθ) = λ/(nsinθ) which is λ/sinθ when n = 1 for the problem that is the focus of this thread.
BruceW
#24
Dec6-12, 05:18 AM
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yeah, definitely. The equation Δx ~ λ/sinθ comes from the theory of far-field diffraction. And in this problem, we are saying that our uncertainty of the position of the electron is due to the equation for diffraction.

In the wikipedia page, they are saying the smallest resolution size is due to diffraction. This is really the same thing as our problem, where the resolution size is the uncertainty of the position of the electron.
s3a
#25
Dec6-12, 12:13 PM
P: 559
Okay now, I need to discuss the index of refraction. There exists negative indices of refraction which would just make Δx negative (which doesn't change anything meaningful since it just means that we are referring to Δx as the distance from point B to point A rather than from point A to point B) but what if the index of refraction is a positive value that is not 1? That would affect the uncertainty of the photon's position (and by extension, the electron's).

According to the problem that is the focus of this thread, the photon will be observed through a lens and, as mentioned here ( http://www.wolframalpha.com/input/?i...index+of+glass ), the index of refraction of the lens is a positive value that is not 1.

If it was just some slit with no glass, we could say that the index of refraction is 1 in a vacuum or very close to it in air. Is the problem saying “we're assuming the index of refraction is a positive value that's not 1 but since it's not important to the argument, we'll just ignore since it is an approximation after.”? By the way, is it correct for me to believe that the reason why a lens is being used is because it allows us to converge the (monochromatic) light at a specific point allowing us to treat it more like a particle than a wave when measuring it?
BruceW
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Dec7-12, 04:46 AM
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In this problem, they get this Δx ~ λ/sinθ because they are assuming the refractive index of the substance between the electron beam and the lens is roughly equal to one. They probably chose this just because it is the simplest option. Also, it would fit with the idea of the experiment being done in a vacuum.

And yes, the lens will have greater than 1 refractive index. This doesn't enter into the equation above, because after the light goes through the lens, it's wavefunction is no longer intimately linked to the wavefunction of the electron. Can you think why? And yes the lens is used to focus the light given off (just so that we can see it really).
s3a
#27
Dec7-12, 09:20 PM
P: 559
The wavefunction seems to require more knowledge that I do not have including knowledge of probability. Could you tell me the bare minimum I need to know about it to understand your previous point please?
BruceW
#28
Dec8-12, 06:10 AM
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The wavefunction of a particle (e.g. a photon) contains all the physical information we have about the particle (e.g. its momentum). And in the collision of photon and electron, we have certain rules about conservation of momentum, e.t.c.

Therefore, immediately after collision, the wavefunctions of the two particles are intimately related. And while the two particles are moving away after the collision, their wavefunctions are still related. But when the photon goes into the lens, the lens will interact with the photon, changing its wavefunction, so the wavefunction of the photon is no longer intimately related to the wavefunction of the electron.

I hope I explained that ok. Just ask if I didn't explain something. So anyway, the whole point of what I was trying to say in the last post, is that after collision, we get a simple equation for the angle within which the photon is most likely to go. But when the photon interacts with the lens, then the motion of the photon will also depend on what kind of lens we are using, right?
s3a
#29
Dec8-12, 12:26 PM
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The motion of the photon will also depend on what kind of lens we are using, right?
Yes. To my understanding, the index of refraction of the lens will change the angle where the photon exits the lens and, by consequence, will change the velocity and momentum as well.

This means that the wavefunction of the photon will change since the momentum is part of the data held by it and that will be changed therefore, the wavefunction will also change as a consequence.

In this problem, they get this Δx ~ λ/sinθ because they are assuming the refractive index of the substance between the electron beam and the lens is roughly equal to one. They probably chose this just because it is the simplest option. Also, it would fit with the idea of the experiment being done in a vacuum.
Okay so, based on what I'm quoting here, it seems that the equation for this diffraction-limited system holds up until and exluding the lens's position.

I was going to say that this confuses me further because, the experimenter is using the lens to focus the light at a point to detect it as a particle and that, this would introduce additional uncertainty but that is not the case since the refraction is measurable to a theoretical 0% uncertainty so we could “correct” the “screwed-up” wavefunction where we only have the theoretical uncertainty from the uncertainty principle, right?

And yes, the lens will have greater than 1 refractive index. This doesn't enter into the equation above, because after the light goes through the lens, it's wavefunction is no longer intimately linked to the wavefunction of the electron. Can you think why? And yes the lens is used to focus the light given off (just so that we can see it really).
I mentioned why I think this is so above (with the angle changing explanation and its consequences) but, additionally, this is so because the index of refraction changed from the vacuum to the lens and the diffraction-limited system's equation is just for light (or something else that's wave-like) travelling through one medium, right?
BruceW
#30
Dec8-12, 07:24 PM
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Quote Quote by s3a View Post
Yes. To my understanding, the index of refraction of the lens will change the angle where the photon exits the lens and, by consequence, will change the velocity and momentum as well.

This means that the wavefunction of the photon will change since the momentum is part of the data held by it and that will be changed therefore, the wavefunction will also change as a consequence.
Yep, that all sounds about right.


Quote Quote by s3a
Okay so, based on what I'm quoting here, it seems that the equation for this diffraction-limited system holds up until and exluding the lens's position.

I was going to say that this confuses me further because, the experimenter is using the lens to focus the light at a point to detect it as a particle and that, this would introduce additional uncertainty but that is not the case since the refraction is measurable to a theoretical 0% uncertainty so we could “correct” the “screwed-up” wavefunction where we only have the theoretical uncertainty from the uncertainty principle, right?
Yeah, up until the lens, we have that nice equation saying that most of the light will fall within an angle theta. Then when the light goes through the lens, it gets focused. The main idea of this problem is that before the light hits the lens, the equation (for diffraction-limited system) actually gives us an indication of the uncertainty principle, even though we are using classical physics.


Quote Quote by s3a
I mentioned why I think this is so above (with the angle changing explanation and its consequences) but, additionally, this is so because the index of refraction changed from the vacuum to the lens and the diffraction-limited system's equation is just for light (or something else that's wave-like) travelling through one medium, right?
Oh, yes definitely.
s3a
#31
Dec14-12, 08:27 PM
P: 559
even though we are using classical physics.
What is classical, specifically? Is it only the Abbe diffraction limit for a microscope and the dealing of light as a ray for the converging lens or is there more than that that is classical?
BruceW
#32
Dec15-12, 05:59 AM
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The entire problem is based around this equation Δx ~ λ/sinθ Which is classical physics. You could argue that the problem contains no quantum physics.

Edit: That's why this is a nice problem for people who are starting quantum physics, because it shows that with classical physics we can get behaviour which is similar to quantum physics.
s3a
#33
Dec15-12, 09:11 AM
P: 559
The entire problem is based around this equation Δx ~ λ/sinθ Which is classical physics. You could argue that the problem contains no quantum physics.

Edit: That's why this is a nice problem for people who are starting quantum physics, because it shows that with classical physics we can get behaviour which is similar to quantum physics.
Okay, that makes sense because, if I'm correct, that's the classical optics equation used with the idea of quantization of light (=photons).

Just to confirm though, the Abbe diffraction limit for a microscope and the dealing of light as a ray for the converging lens are both also classical, right?
BruceW
#34
Dec15-12, 10:03 AM
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Quote Quote by s3a View Post
Okay, that makes sense because, if I'm correct, that's the classical optics equation used with the idea of quantization of light (=photons).
I don't think that's correct. The equation Δx ~ λ/sinθ comes from the far-field approximation for diffraction of light. This doesn't take into consideration the 'particle-like' nature of light. In fact, I would say that any theory that does involve the 'particle-like' nature of light is non-classical.

Quote Quote by s3a
Just to confirm though, the Abbe diffraction limit for a microscope and the dealing of light as a ray for the converging lens are both also classical, right?
yep, that's right.
s3a
#35
Dec15-12, 10:27 AM
P: 559
I don't think that's correct. The equation Δx ~ λ/sinθ comes from the far-field approximation for diffraction of light. This doesn't take into consideration the 'particle-like' nature of light. In fact, I would say that any theory that does involve the 'particle-like' nature of light is non-classical.
Perhaps, we're miscommunicating. I didn't say that the equation takes into consideration the particle-like nature of light but, rather, that we're ("forcefully") combining the quantization of light/particle-nature of light/photon idea with the Δx ~ λ/sinθ equation and its underlying theory in order to obtain the uncertainty relation.
BruceW
#36
Dec15-12, 10:43 AM
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yeah, I think I see what you mean. The equation for diffraction from a slit is classical, but then when we say the size of the slit represents how 'spread out' the electron is, we are forcing a quantum concept into our problem.


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