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

s3a

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

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.

Attached Thumbnails

 

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

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

I am aware that this problem neglects the vertical direction.

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.

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 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.

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.

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.

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 agree. :D

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/teachin...es/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.)