Question

(Question on problem about Heisenberg's Uncertainty Principle)

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.

Answer 1

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.

The problem and its solution:
http://i.imgur.com/jbweB.jpg

(Please give as simple an answer as possible and, involve advanced concepts as little as possible unless I request them specifically.)

Answer 2

If something is unclear, ask me to clarify it.

Answer 3

If this were a single-slit experiment, it would be cleared to me since it doesn't matter where the collision is but here with the lens it seems to matter.

Answer 4

This is not going to help you with the problem because I have an issue with the validity of the problem which you might take up with your instructor. That is the resolving power of any microscope is many time worse than the disturbance that the light photon will have on an electron. This I think invalidates the use of the optical criterion in this problem. This might be a good issue to discuss in this forum.

Answer 5

I would like to hear the instructors defense of the use of the classical optical criteria in this problem.

Answer 6

Isn't it *theoretically* possible to have a lens with a very good resolution.?:
http://en.wikipedia.org/wiki/Diffraction-limited_system

Answer 7

This is a thought experiment after all.

Answer 8

I'm not sure if the resolving power of a micro microscope would be given by the classical equation. Another issue is that the wavelength of the light photon is orders of magnitude larger than the apparent diameter of the electron. I am not sure this is in the spirit for the use of the Heisenberg uncertainty principle were we expect the observed entity to be influenced by a force by affecting it at some instant and receding with the position information only to have cause a change in its momentum.

Answer 9

I don't see how the resolving power of a microscope is relevant to this problem in a very quantitative manner.

As for the photon wavelength being larger, here's a video that talks about that.: http://www.youtube.com/watch?v=yrVi24pp_6I

I'm not sure I understood the last part but if I did understand you correctly, this problem does incorporate the concept of introducing additional uncertainty by the mere act of observation.

All i am really asking about is, what if the collision point was translated in the drawing I attached? Would that affect the uncertainty in the position of the electron compared to the collision point that was drawn by the textbook creators rather than what I digitally added in colours? If not (and, I am assuming that it will not since it makes no sense to assume that the collision is at the horizontal centre since that would mean there would be no need to try and figure out the horizontal position in the first place), how would the math change for example?

Answer 10

In a microscope you will see anything in a field of view defined by the position of the "eye" looking at the object. As the object moves horizontally the image does also in the opposite direction and eventually the image will become distorted and disappear because of the curvature and the diameter of the lens. The issue is how does the delta x change as it moves horizontally. It does not change as long as it stays in the distortion free field of view. For any lens Image size /object size is constant if the image and object distance are constant. you use the center in this case because the math is a little easier.

Answer 11

Just to say, your most recent post did provide valuable information to me.

1) Would the product of Δx and Δp_x be the same constant if the collision were translated horizontally?

2)
I can see how the uncertainty in the momentum of the electron would be the same if the collision were translated horizontally since momentum is conserved and if one side of the equation has an uncertainty term then the other side of the equation must have that same uncertainty term and the light must be seen and, if it's seen, it is assumed that the possible range for the values of the momentum of the photon covers all possible momentum values along the width of the lens.

I can also see how the uncertainty in position of the photon would remain the same since it would have to go through the lens to be seen. What I can't see is how the uncertainty in the position of the electron would remain the same. It must be linked somehow to the electron but I can't see how. (Keep in mind, I am talking about the horizontally translated collision.) Could you help me connect the dots please?

Answer 12

I'm not sure I understand your question. The uncertainty in the position of the electron is independent of its position and depends only on the uncertainly of the momentum.\[\Delta x \Delta p \ge h\] and delta p depends on the interaction between the "light" photon and the electron.. The microscope allows you to "see" how big delta x is. That delta x will be the same no matter where on the horizontal line the interaction takes place as long as the photon can enter the microscope.

Answer 13

What I'm not understanding is how I can say that the electron's position uncertainty is equivalent to that of the photon's in the case where the collision is not below some part of the range of the lens.

For example, in my drawing, the collision is not below the lens which means that I need to apply different reasoning than I was before. The reasoning I was applying before was that if the collision was horizontally within the lens and, if the photon was detected, then, that must mean that the collision happened somewhere within a horizontal range the size of the width of the slit.

In the case that the collision happens somewhere out of the horizontal range that is the width of the slit, I get that the position of the photon as it reaches the lens could be figured out with an uncertainty the width of the lens. What challenges my thinking is that, the horizontal translation away from the lens means that the collision will not be within the lens of the width and there is an additional uncertainty which is the horizontal value the collision has been translated by. In this case, shouldn't the uncertainty be Δx + amount_translated instead of just Δx?

P.S.
When I say in the horizontal range of the width of the slit, I mean that, the dot could be vertically translated to lie somewhere on the lens whereas in the part I drew, no vertical translation could place the collision dot on the lens since I translated it horizontally (out of range of the lens' width) as well.

P.P.S
If I'm not making perfect sense, tell me what specifically is unclear and I'll clear it up and/or rephrase it.

Answer 14

Aah Haa I think is finally see the issue. I have to leave this session for a few hours but I will try to address this issue later today. OK?

Answer 15

Alright. :)

Answer 16

This is as you said a thought experiment. It has presupposed that you had somehow localized it which would be very difficult to say the least. Having localized the electron say with its cross hairs your measurement will still be in error by delta x as given by the H.U. principle. How you localize it in the first place is not a consideration of the principle. Somehow you have to think you know where it is when you "see" it in the microscope. When you "see" it, it is not really where the microscope says because the photon interacted with the electron to move it after it bounces off.

Answer 17

Are you implying that it is THEORETICALLY possible to find the position of the electron right before the collision of the electron with the photon to a 100% certainty (if the microscope technology is sophisticated enough, etc)?

This would mean that the collision could (theoretically) always be at the horizontal centre of the lens (from the problem we're talking about).

Assuming all of the above is true, I'm a little confused as to the point of colliding the photon at all in the first place since the electron should be almost asymptotically close to where it was prior to the collision after the collision.