Understanding Heisenberg's Uncertainty Principle

[s3a]

Homework Statement

The problem and its solution are attached as TheProblemAndSolution.jpg.

Homework Equations

Δx = λ/sinθ
Δp_x = 2h/λ sinθ

The Attempt at a Solution

In the equation given in the problem (Δx = λ/sinθ), why is the uncertainty of the ELECTRON written as a function of the PHOTON's wavelength (and the microscope's angle)? Similarly, in the equation given in the solution (Δp_x = 2h/λ sinθ), why is the uncertainty in the x component of the momentum of the ELECTRON written, again, as a function of the PHOTON's wavelength (and, again, the microscope's angle)? If it wasn't clear from the above, I am placing emphasis on the fact that Heisenberg's Uncertainty Principle is all about being able to measure, with great certainty, either the position or momentum of the ELECTRON while the other factor remains highly uncertain BY USING CHARACTERISTIC VALUES OF A PHOTON (RATHER THAN AN ELECTRON). I would really appreciate it if someone could tell me why this is the case.

P.S.
The involvement of the sine of the angle of the microscope is not baffling or confusing to me, it's that I don't understand why information about the photon is used to describe uncertainty of a characteristic of an electron.

P.P.S
I am not looking for an overly complex answer (unless it is inherently so) so, please keep the answer as simple as possible.

 

[BruceW]

Is this for homework or do you just want the answer?

 

[s3a]

I double-posted by accident. Check the next post.

 

[s3a]

This is not homework. (I don't know if you noticed but, the solution is included in my main post.)

I'm just trying to understand this problem inside-out. I'd appreciate it if I can have an answer that simply addresses my concerns and nothing more (in order to not confuse me or add unnecessary difficulty).

 

[BruceW]

the size of an object which you want to take a picture of must be at least as large as the wavelength of light which you use to take the picture. This is a law of optics. (You can work it out using classical wave theory). Therefore, the uncertainty in the electron's position will scale with the wavelength of the photon.

And the uncertainty of the momentum of the electron is the same as the uncertainty in the momentum of the photon. This is because momentum is conserved in the collision, so if the photon had an uncertainty in its momentum, then after the collision, the electron will have an uncertainty in its momentum.

 

[s3a]

Before I ask my other questions, I'd like to ask: Is Δx the right arrow from the collision point to the top of the arrow (touching the lens)?

 

[BruceW]

no, the arrow from the collision to the lens is meant to show one of the possible paths of the photon.

Δx itself will depend on the experiment, the example doesn't give enough detail really. It is meant to be an instructive example, not something to actually calculate.

 

[s3a]

Therefore, the uncertainty in the electron's position will scale with the wavelength of the photon.

It makes sense to me when you said that “the size of an object which you want to take a picture of must be at least as large as the wavelength of light which you use to take the picture.” but, could you help me intuitively understand where the Δx = λ/sinθ equation comes from?

This is because momentum is conserved in the collision

Could you please also explain why the conservation of momentum in the collision implies that the uncertainty of the momentum and the photon are equivalent?

 

[s3a]

Therefore, the uncertainty in the electron's position will scale with the wavelength of the photon.

It makes sense to me when you said that “the size of an object which you want to take a picture of must be at least as large as the wavelength of light which you use to take the picture.” but, could you please help me intuitively understand where the entire Δx = λ/sinθ equation comes from?

This is because momentum is conserved in the collision.

Could you please also explain why the conservation of momentum in the collision implies that the uncertainty of the momentum of the electron and that of the photon are equivalent?

 

[BruceW]

It makes sense to me when you said that “the size of an object which you want to take a picture of must be at least as large as the wavelength of light which you use to take the picture.” but, could you help me intuitively understand where the Δx = λ/sinθ equation comes from?

It is simple singe-slit diffraction. I.e. if the light comes from a place of width d, then it will end up moving in a direction where theta is less than the theta given by the formula (most of the time). As I said, this is classical physics. So the question is kind of using a mix of classical and quantum reasoning. This question is not doing the calculations very precisely, the point of the question is to give you a feeling of the uncertainty principle, not working knowledge of it.

Could you please also explain why the conservation of momentum in the collision implies that the uncertainty of the momentum and the photon are equivalent?

Well, I am assuming it is the photon which initially has most of the momentum in the system, and its momentum is initially well-defined, so the total momentum of the system is well-defined. And then after collision, we know that the momentum of the photon and electron must add up to this initial value. We can repeat this experiment many times to find out how the photon's momentum after the collision varies from its average. And we know that each time we do this experiment, the momentum of the electron + that of the photon must add up to a constant value. Therefore, the way the electron's momentum varies over many experiments must be exactly the same as the way the photon's momentum varies. Therefore, the uncertainty in their momentum must be the same.

EDIT: in other words, imagine doing 10 experiments. For each experiment, you will obtain different values for the final momentum of the photon. We know momentum is conserved for each experiment, so we can write down the final momentum of the electron in each experiment. If we calculate the spread in the values of electron and photon momentum, then hey-presto, they will be the same, because in each experiment, the electron momentum is simply the same constant (initial momentum), minus the photon momentum. 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.

 

[s3a]

I am not satisfied in “just getting” the idea that you can only know either the momentum or the position of the electron with great certainty. Not knowing everything I need to know to make intuitive sense of it ultimately confuses me.

(1) The formula for the uncertainty of the electron's position is starting to make more sense to me now. It is assuming m = 1 (first order), right?

(2) You say that the total momentum is well-defined because the initial momentum of the photon is well-defined (and I assuming you're just taking the word of the problem for the electron's momentum being well-defined). But, how, exactly, is the electron's momentum well-defined?

(3) What logic or evidence is your assumption about the photon always having a greater initial momentum than the electron based on? Does this even matter? If it is the case, that the electron's initial momentum is larger, everything you said about the uncertainties of their momenta being the same still stands, right?

(4) Despite “just knowing” that a greater uncertainty about the momentum of the electron or its position will imply a smaller uncertainty in the other factor, I don't see how the position is related to momentum in the first place. I get that momentum is the product of mass times velocity and the velocity is the rate of change of displacement which is a net change in position but, beyond that, I don't see how the two are related. What's their “initial relationship” such that Heisenberg could even think about writing down that Δx Δp_x = constant (or an inequality instead of the equality)?

Also, I now get how their uncertainty in momentum is the same (thanks to you). :)

If anything I said is unclear, just ask me to rephrase it in a better way.

Sorry for being for pedantic.

 

[BruceW]

I am not satisfied in “just getting” the idea that you can only know either the momentum or the position of the electron with great certainty. Not knowing everything I need to know to make intuitive sense of it ultimately confuses me.

Yep, for the same reason, this problem confused me a lot when I first saw it. It is probably best to try not to think on this problem for too long. It is best to start learning quantum mechanics. (for example, particle in a square well, particle in free space, hydrogen atom, etc. These problems are better to get an understanding of the basics of quantum mechanics).

(1) The formula for the uncertainty of the electron's position is starting to make more sense to me now. It is assuming m = 1 (first order), right?

I'm not sure what you mean by this.

(2) You say that the total momentum is well-defined because the initial momentum of the photon is well-defined (and I assuming you're just taking the word of the problem for the electron's momentum being well-defined). But, how, exactly, is the electron's momentum well-defined?

(3) What logic or evidence is your assumption about the photon always having a greater initial momentum than the electron based on? Does this even matter? If it is the case, that the electron's initial momentum is larger, everything you said about the uncertainties of their momenta being the same still stands, right?

Hmm. The reason that the question assumes the initial momentum of the electron is well defined is to make the problem as simple as possible. This can be done by making a beam of electrons which is sufficiently spread out throughout space. And no, in this problem, it doesn't matter which particle has the most momentum.

(4) Despite “just knowing” that a greater uncertainty about the momentum of the electron or its position will imply a smaller uncertainty in the other factor, I don't see how the position is related to momentum in the first place. I get that momentum is the product of mass times velocity and the velocity is the rate of change of displacement which is a net change in position but, beyond that, I don't see how the two are related. What's their “initial relationship” such that Heisenberg could even think about writing down that Δx Δp_x = constant (or an inequality instead of the equality)?

Excellent question. Historically, quantum mechanics came about because people started to see places where the classical theory did not hold. For example, electromagnetic radiation seemed to be explained much better if it were absorbed and emitted in discrete 'quanta'. People also related the quantum-mechanical theory to electrons, and it turned out to be a good model for them, and people started diffracting electrons, similarly to how you would diffract light. So, we have a theory which explains much of the behaviour of both electrons and photons. The uncertainty principle is really self-evident, given the postulates of quantum mechanics. In other words, since we have a theory (quantum mechanics) which describes our world very well, then for the time being we assume its postulates are not dis-proven. Therefore the uncertainty principle is not dis-proven. (You could change the phrase 'not dis-proven' with 'true', but 'truth' sounds like quite an unscientific word, right?)

EDIT: To answer your last question in another way, we know that for electromagnetic radiation, wavelength and position are intimately linked, and this wave nature (which is more obvious for electromagnetic theory) is one 'facet' of quantum theory, which is a theory underlying photons and electrons. So it is not so surprising that for electrons, wavelength and position are also intimately linked. And also, the inverse of wavelength is proportional to momentum, therefore momentum and position are linked in a special way in quantum mechanics.

 

[s3a]

I'm not sure what you mean by this.

I was adapting the single-slit diffraction formula to the problem of this forum thread. Here is the formula (and more).: http://www.physics.uoguelph.ca/applets/Intro_physics/kisalev/java/slitdiffr/

'truth' sounds like quite an unscientific word, right?

I agree since we can never know for sure if a certain theory is a representation of the truth or a qualitatively good approximation of it until we find evidence disproving it.

1) How does you measure the final momentum of the photon bounced in real life? Do you analyze a single-slit diffraction image on a screen?

2) How do you measure the position of the photon? Do you at all or do you just say “I know it's somewhere in the slit” and then you attempt to make the slit as small as you can (by using light with as small a wavelength as possible) and then, since the uncertainty is small, you can say something like “screw the specifics, it's between this line here and that line there which are super close so we can just assume it's in the middle of the two lines”?

3) I get, mathematically (in the way that the solution of the problem in this thread shows), that if you decrease the wavelength of the light in an attempt to make the slit (and the electron's uncertainty in position) smaller, you will increase the uncertainty in the electron's momentum (which is what the solution of the problem in this thread is all about) however, I don't understand, scientifically, how this principle works. I know it's a fundamental limitation imposed by nature (rather than a limitation of our measurement tools). I'm looking for an answer that's on a sub-atomic particle level (that involves thinking of the collision and what happens after it).

4) In real life, do we measure the momentum of the electron that's prior to the collision by accelerating very spaced-out electrons using a potential difference and then shining light opposite to the electrons and, since light is unaffected by the potential difference due to its lack of charge, it should not introduce any new uncertainty when the collision occurs? What about the interaction of electrons between one another? That introduces new uncertainty but, is that uncertainty also part of the uncertainty principle? Do we just neglect that completely (since the interactions are minor since the electrons are spaced out heavily)?

As for the initial momentum of the light, nothing confuses me; it's simply choosing a wavelength and “shooting” it.

In short:
* I have minor doubts about how to get the initial momentum of the electron in real life that I need cleared.
* I know how to get the initial momentum of the photons in real life.
* I'm thinking that the analysis of the single-slit diffraction pattern is what allows a scientist to measure the momentum of the photon after the collision.

P.S.
Sorry if I said anything stupid or repeated myself because, I have a headache as I write this.

 

[BruceW]

I was adapting the single-slit diffraction formula to the problem of this forum thread. Here is the formula (and more).: http://www.physics.uoguelph.ca/applets/Intro_physics/kisalev/java/slitdiffr/

Yeah, that's right. Although I don't think 'first order' is the correct terminology. It's just that most of the light from single-slit diffraction falls within the first minimum.

 

[BruceW]

3) I get, mathematically (in the way that the solution of the problem in this thread shows), that if you decrease the wavelength of the light in an attempt to make the slit (and the electron's uncertainty in position) smaller, you will increase the uncertainty in the electron's momentum (which is what the solution of the problem in this thread is all about) however, I don't understand, scientifically, how this principle works. I know it's a fundamental limitation imposed by nature (rather than a limitation of our measurement tools). I'm looking for an answer that's on a sub-atomic particle level (that involves thinking of the collision and what happens after it).

This problem is fairly intuitive, if you think about the collision. (although not all problems in quantum mechanics are intuitive at all). Anyway, in this problem, If you decrease the wavelength of the light, you are increasing its frequency, and therefore increasing its momentum. When the photon collides with the electron, there is a certain probability that the electron will be 'bounced' away in a certain direction. And if the photon initially has more momentum, it will transfer more momentum to the electron.

So If we just think about what happens to the electron, after collision it gets bounced away in a random direction, and it gets bounced further if the photon had more momentum. So, after collision, the uncertainty in the momentum of the electron is greater when the photon has greater initial momentum.

 

[BruceW]

1) How does you measure the final momentum of the photon bounced in real life? Do you analyze a single-slit diffraction image on a screen?

I think it depends on the type of experiment you are doing... If the light is monochrome, then you could do a simple diffraction experiment to find the wavelength (and therefore momentum). If the light is made up of lots of different wavelengths, then maybe you'd need some kind of spectrometer, for example a ccd.

 

[BruceW]

2) How do you measure the position of the photon? Do you at all or do you just say “I know it's somewhere in the slit” and then you attempt to make the slit as small as you can (by using light with as small a wavelength as possible) and then, since the uncertainty is small, you can say something like “screw the specifics, it's between this line here and that line there which are super close so we can just assume it's in the middle of the two lines”?

the equation for diffraction does take into account the fact that the light could come from any point within the slit, but this is not a problem, since the slit is small compared to the distance the light travels from the slit. Does this answer your question?

 

[BruceW]

4) In real life, do we measure the momentum of the electron that's prior to the collision by accelerating very spaced-out electrons using a potential difference and then shining light opposite to the electrons and, since light is unaffected by the potential difference due to its lack of charge, it should not introduce any new uncertainty when the collision occurs? What about the interaction of electrons between one another? That introduces new uncertainty but, is that uncertainty also part of the uncertainty principle? Do we just neglect that completely (since the interactions are minor since the electrons are spaced out heavily)?

We could even accelerate the electrons, let some of them come out of the 'accelerating tube' (sorry that's probably not correct terminology), then turn on the beam of photons, so that the collisions would happen without the potential difference in the background. (This would all need to be within a vacuum, of course).
And yes, we can neglect the interaction between electrons, since we can make a beam in which they are not close to one another.

 

[BruceW]

p.s. sorry I've taken a while to reply, I have been under the weather.

 

[s3a]

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]

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]

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]

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]

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]

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=refractive+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]

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]

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]

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]

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) traveling through one medium, right?

 

[BruceW]

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.

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.

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) traveling through one medium, right?

Oh, yes definitely.

 

[s3a]

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]

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]

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]

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.

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]

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]

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.

 

[s3a]

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.

Replace "electron" with "photon" and that's what I was saying. I do get that the uncertainties of the photon are numerically equivalent to that of the electron but, that's indirectly determined.

Look at the attached image. What happens if the collision is not horizontally in the middle of the lens? Surely, we do not expect it to be there especially since, if we did, we would know the horizontal position to a 100% certainty. The way I had pictured the determination of the electron's position in my mind before thinking about horizontally translating the collision with respect to the lens was that we don't care about the vertical displacement and, we know that the photon and electron touched each other in order to collide (since photon's don't have a charge) so, the horizontal position must have been somewhere along the interval of the lens. However, when translating the point of collision, I am confused.

 

[BruceW]

Think about a real microscope. It only picks up light from a small region. If I put a spider in the middle point of a microscope, then I will just see a spider, even though there might be a fly just off-center.

That is the point of the lens. If you stand at the place on the screen where the diagram shows, you will mostly receive light made from collisions at the middle point.

Edit: of course, some photons will come from off-center collisions and still reach you. But the whole point of this problem is that we average over many collisions, so almost all of them will come from the middle.

 

[s3a]

That answer's logic doesn't satisfy my curiosity and, it also introduces additional uncertainty since there is no guarantee by nature that the average photon position is at the horizontal centre. Furthermore, even if the average photon position is at the horizontal centre, it gives me no assurance for a particular random scenario and, this (thought) experiment is done on a case-by-case/photon-by-photon basis.

Alternatively, I was thinking of an “If then” approach to the problem that is the focus of this thread such that if the photon is at the horizontal centre, this is how we would deal with the situation (where “this” is the method used by the solution in the image) however, again, if we knew that it would be in horizontal centre, then that makes the (theoretical) experiment unnecessary.

 

[BruceW]

Furthermore, even if the average photon position is at the horizontal centre, it gives me no assurance for a particular random scenario and, this (thought) experiment is done on a case-by-case/photon-by-photon basis.

The thought experiment says that the precision with which the electron is localised is Δx This is the same thing as the precision with which the collision is localised in the horizontal direction.

It gets a little confusing because we then say that Δx is equivalent to the size of a slit which we are letting light through classically. But it does sort of make sense, if you don't pay too much attention to the details.

As I said, this problem is a nice example for when you start to learn about quantum mechanics, but it has a sloppy mix of quantum and classical theory. So my advice is to move on to better qm problems, not to think for too long about this one. On the other hand, if you find this problem interesting, then there's nothing wrong with that.

 

[s3a]

The thought experiment says that the precision with which the electron is localised is Δx This is the same thing as the precision with which the collision is localised in the horizontal direction.

I am aware that this problem neglects the vertical direction.

It gets a little confusing because we then say that Δx is equivalent to the size of a slit which we are letting light through classically. But it does sort of make sense, if you don't pay too much attention to the details.

The details for “What if?”s are important to me even if the answers are approximations. Put differently, I like to have an intuitive explanation for everything even if it's not perfectly accurate.

As I said, this problem is a nice example for when you start to learn about quantum mechanics, but it has a sloppy mix of quantum and classical theory. So my advice is to move on to better qm problems, not to think for too long about this one. On the other hand, if you find this problem interesting, then there's nothing wrong with that.

I am covering other things simultaneously; I love knowing every single detail about which I have a question in my mind when doing a problem.

Basically, if the collision were translated horizontally the way I drew on the picture (except I should have labelled the angle something different such as alpha), I could see how the horizontal momentum could range between the same amount as the situation where the collision is in the horizontal middle of the lens and that, because of conservation of momentum, both sides of the equation need to have the same uncertainty term which means that the electron's momentum has the same uncertainty. What I cannot fully see is how the electron's position is uncertain by the amount of the single-slit diffraction/diffraction-limited system. The best I can think of is when thinking of light as a wave and that when the electron collides with the photon, we know that the electron was somewhere along the photon implying that the uncertainty is the wavelength but that doesn't take into account the angle that could increase the uncertainty significantly. Also, I've watched a video involving a single-slit experiment (not involving a lens) and, when the wavelength of the light is made small (and constant) and the slit's width is made small (and constant), the angle must get large and, basically, I'd like to involve the angle in my thought process somehow.

I hope I'm not getting on your nerves with my great focus on little details.

 

[BruceW]

I am covering other things simultaneously; I love knowing every single detail about which I have a question in my mind when doing a problem.

I know what you mean :) me too, especially when it is a subject I find interesting. I guess we wouldn't be on this website if this weren't true of us.

Basically, if the collision were translated horizontally the way I drew on the picture (except I should have labelled the angle something different such as alpha), I could see how the horizontal momentum could range between the same amount as the situation where the collision is in the horizontal middle of the lens and that, because of conservation of momentum, both sides of the equation need to have the same uncertainty term which means that the electron's momentum has the same uncertainty.

Agreed. The uncertainties in the outgoing trajectories of the photon and electron will not depend on the horizontal position of where the collision occurred. And agreed that the photon and electron's uncertainty in momentum are linked because they both came from the same collision.

What I cannot fully see is how the electron's position is uncertain by the amount of the single-slit diffraction/diffraction-limited system.

Suppose that the position of the collision was limited to Δx around the centre. Then if we assume we can use classical physics, we get Δx ~ λ/sinθ. So how do we make the position of the collision to be limited to Δx around the centre? That's what the lens is for. The lens will mostly pick up light from around the centre, which is where we get Δx from.

I remember you were saying that the experiment is on a case-by-case basis. But I think it is much more likely that the experiment is done by allowing many collisions to occur. This is much more close to the idea of a microscope, where we let an image appear after many collisions have occurred.

Also, I've watched a video involving a single-slit experiment (not involving a lens) and, when the wavelength of the light is made small (and constant) and the slit's width is made small (and constant), the angle must get large and, basically, I'd like to involve the angle in my thought process somehow.

You mean in the video, the light was monochrome (just one wavelength)? If the slit was made small at the same rate as the wavelength, then the angle wouldn't change... Maybe something weird would happen when you got to the level where second quantisation is important.

In our problem, we also have just one wavelength and slit of width Δx. And we are using classical physics. I should say Δx ~ λ/sinθ Is only true for small angles, such that sinθ ~ θ In other words, the screen is much bigger than the slit, and the slit is far from the screen. The equation is pretty intuitive, if you think about different cases. First, the case when the wavelength is much smaller than the slit, so then the angle gets very small. This is because the light has no trouble squeezing through the slit, so it continues in the same direction it came from. And the other case, is when the slit starts to get to the same order of magnitude of size as the wavelength. Now the light has trouble squeezing through the slit, and when it comes out on the other side it gets diffracted, so the angle is larger than it was for a large slit.

 

[s3a]

I know what you mean :) me too, especially when it is a subject I find interesting. I guess we wouldn't be on this website if this weren't true of us.

I agree. :D

You mean in the video, the light was monochrome (just one wavelength)? If the slit was made small at the same rate as the wavelength, then the angle wouldn't change... Maybe something weird would happen when you got to the level where second quantisation is important.

The wavelength was monochromatic/held constant.

By luck, my eye caught something crucial I found on a random website on the Internet ( http://farside.ph.utexas.edu/teaching/qmech/lectures/node26.html ) which said something which I found acceptable to satisfy my “inner questions” for this particular problem.

Basically, the focus of the lens is where the resolving power is at a minimum meaning that, that's the point where the uncertainty is the greatest hence why it is acceptable to just deal with that one case when dealing with the calculations.

I'm now fully satisfied with what I know about this problem.

Thanks for everything ... Batman ( http://en.wikipedia.org/wiki/Bruce_Wayne ). ;)

 

[BruceW]

haha, no problem, I'm glad I could help. But maybe I am Bruce Willis ?

 

[s3a]

"I hadn't considered that." (Sheldon voice - from the Big Bang Theory). :P (If you're not obsessed with that show, you probably won't get it.)

(I know every episode of that show by heart so well that I laugh at the jokes even before they're made lol.)