# Why does an observer affect the electron?

1. Jun 3, 2006

### quddusaliquddus

Hi all,
In quantum physics (i.e. the double slit experiment with electron), why does mere act of observing the electron affect the fact of whether its a wave or particle?

Please answer in laymens terms if possible as i am no physicist!

2. Jun 3, 2006

### mathman

It is not observing as such, but the experiment. For the double slit, the electron exhibits wavelike behavior. I don't think anyone can completely explain it, but it comes out of quantum theory.

3. Jun 3, 2006

### pallidin

The issue involves "finality"
Observation collapses the probability distribution by virtue of intervention.

4. Jun 3, 2006

### pallidin

Think of an expanding balloon. That is your "electron" or "photon" going through the 2-slits.

Now, take a needle and poke the balloon. The "needle" is "observation"

Stop. Think about this: Though you can poke the balloon from many possible directions, the balloon will collapse essentially the same way but will alter in other ways depending on where(observation) you "poked" it.

5. Jun 4, 2006

### hanlin_sg2002

I think, to put it simply, similar to a photon, an electron can behave both as a wave and as a particle.

6. Jun 4, 2006

### Perturbation

Think of it in terms of the diffraction of water at a slit. If I let the water pass through the slit it will diffract, assuming the width of the slit is much smaller than the wavelength of the water. If I stick a bucket or something to collect the diffracting water in front of the slit and then poor it back it's not going to continue displaying diffraction.

The water is analogous to the electron wave-packet and the bucket analogous to the detector. It's the slit that causes the diffraction. If I detect an electron at a slit it's now in a definite position (in our analogy the water is definitely in the bucket) and when it continues there's no more diffraction because in detecting it I've "destroyed" the diffraction behaviour like I did with the water.

Last edited: Jun 4, 2006
7. Jun 4, 2006

### quddusaliquddus

Sorry guys. I still dont understand. Does the observation do something "physical" to the electron e.g. hit the electron with a photon?

8. Jun 4, 2006

### quddusaliquddus

So, how is the electron observed? Im assuming its not by shining a light on it.

9. Jun 4, 2006

### El Hombre Invisible

It's a question of confinement. If a particle is entirely unconfined, i.e. a free particle, its wavefunction is a simple sinusoidal function extending infinitely into space in the direction of its motion. You cannot directly observe a particle in such a state because its probability function is essentially zero everywhere. To observe the particle directly, you have to narrow it down to a smaller region of space. The smaller the confinement (i.e. the more you narrow down where the particle is) the more particle-like it will be. The less confined it is, the more wave-like it will be.

10. Jun 4, 2006

### RandallB

Well in laymen’s terms as you requested, yes it is.

Most of the answers you’ve gotten are about how to describe the solution; your question is more about finding the problem that needs explaining.

It works like this: With electrons going though two slits you set a “trap” at just one slit. Basically you shine a light across that one slit and look for the shadow of an electron going though.
Now to be sure you only count the shadows passing that match detections at the screen (The screen that looks for interference or no interference).
You can do this because you are sending just one electron at a time.

Important part of your test – you run it with, and without, the trap turned on, while the other slit is closed.
You test for constant counts on the screen to be sure your trap does not hurt the transmission of electrons making it through the one slit.
Assume you pick an interval that gives a consistent 10,000 electrons detected at the screen in a normal single slit pattern, with or without the “shadow” being created and when you do count the shadows it matches with 10,000.

Now you continue the experiment with both slits open, for the same time interval you now see 20,000 electrons hit the screen as expected due to two slits. And still when you count the shadows when the light trap is turned on only 10,000 shadows counted at the one slit as expected.
Thus, with the shadow counter working you can isolated the pattern just for the electrons at the screen that match with electron shadows at the first slit.
The remaining detections couldn’t come from slit one so they are from electrons that went through slit two.
And what kind of patterns do we see isolated for each side? normal single slit patterns.
AND we still see the same for the whole pattern with all 20,000 electrons when we ignore the shadow counter and can not divide the electrons.
BUT; when we turn the light off that allowed the shadow to be made for the shadow counter to use, the double slit interference pattern appears for the 20,000 electrons!

Now when we were hitting the electrons going through slit one with light we may expect we would lose the pattern for them.
But we went though extra steps in setting up the test to guarantee the second slit was not to be touched by our light beam so why doesn’t the other half of the electrons retain an interference pattern? We didn’t touch those electrons.
Conclusion – In classical laymen’s terms some part of the electrons going through slit two must also need to sneak through the tested slit one, and are ‘hurt’ by the light, yet without creating a shadow!

Now you are onto the follow-up question – what in laymen’s terms does that other part of the electron look like?
There is no classical laymen’s answer to that, only mystical descriptions.

You can use QM superposition, BM guide-waves, MWI and many other proposed theories to describe what might be happening within the rules of each theory.
But until we have a GUT or TOE those are tentative descriptions not definitive explanations.

11. Jun 4, 2006

### JesseM

It might help to point out that it doesn't need to be a person observing the particle, any interaction with macroscopic systems that leaves some sort of record of its path, even transiently, will do. For example, if you do a double-slit experiment with an electron rather than a photon, you must do it in a vacuum, because in open air the electron's interactions with air molecules would be enough to destroy the interference pattern, even though it would be nearly impossible for humans to reconstruct the electron's path by measuring the air molecules.

As far as I know, the boundary between the "macroscopic" classical world and the quantum world is not perfectly understood, although there has been progress due to a better understanding of things like decoherence. The question of how this boundary works is sometimes called the measurement problem, and it still seems to be an active area of research. So if you're confused, don't worry, scientists are somewhat confused by it as well!

12. Jun 4, 2006

### JesseM

Actually you can measure the position of an electron by bouncing light off it. As explained on this page (in the section 'Watching Electrons in the Double-Slit Experiment'), to resolve the electron's position with greater accuracy, you need to use light with a smaller wavelength, which means the photons will have more momentum (using DeBroglie's formula for the relationship between wavelength and momentum), and thus can impart more of their momentum to the electron. It turns out that the interference pattern is destroyed if the uncertainty in each electron's momentum is too large, so there's a minimum wavelength of light you can shine on the electrons and still get an interference pattern. When you actually calculate this minimum wavelength using the uncertainty principle, it turns out to be exactly equal to the distance between the slits...but to actually know which of the two slits it went through, you'd need a wavelength smaller than the distance between the two slits!

Last edited: Jun 4, 2006
13. Jun 7, 2006

### silver-rose

this is heisenberg uncertainty principle no?

14. Jun 9, 2006

### quddusaliquddus

thanks guys. now i think i understand it better.

15. Jun 9, 2006

### ZapperZ

Staff Emeritus
Again, there is a miconception here as implied by this post there the uncertainly principle is a "measurement" uncertainty, i.e. due to our technique. We need a shorter wavelength of light to probe the electron, and thus, we are using higher energy photons that cause it to be blasted away so much so that we cannot be certain of its momentum.

This is incorrect and is not about the HUP.

There are two important things to remember about the HUP:

1. It tells you the spread in an observable as you make repeated, indentical measurement. This is something crucial to keep in mind. It has nothing to do with the accuracy of a single measurement - this is the instrumentation accuracy and not the HUP. The fact that the uncertainty of the measurement of the outcome of operator A is defined as

$$\Delta(A) = \sqrt{<A^2> - <A>^2>}$$

should immediately tells you that this is meaningless for a single measurement (i.e. it is zero). If something has a WIDE spread in repeated measurement, even for an identically prepared and measured system, then your ability to predict the next measured value will be low. If the spread is very small, then your ability to make the same prediction improves since you know that most of the outcome will be confined to a small range of values. THIS is what is meant by the HUP.

2. I don't have to shoot high energy photons simply to know the location of an electron. If I have a beam of electron, I can simply pass it through a narrow slit. The instant an electron made it through the slit, I can immediately say that at that instant, the electron was in such an such transverse location. The uncertainty in its transverse position depends on how wide I make that slit. If I want to be more certain, I make the slit narrower. I didn't blast away at the electron with anything here.

However, here's where the HUP kicks in. At some point, as the slit gets narrower, its lateral momentum starts to acquire a larger possible range of values. If I shoot one electron through the slit and measure its momentum after it passes through that hole, then the NEXT electron that I measure will can have a widely different value of momentum for no apparent reason, even when it was prepared identically. The smaller I make the slit, the more difficult for me to anticipate what the NEXT momentum being measured would be.

Note that this has NOTHING to do with the uncertainty of a SINGLE measurement. I can measure as accurately as I want the momentum of a single electron that went through the slit. I can determine this as accurately as technologically possible, maybe it depends on the number of pixel per square inch on my CCD. The HUP doesn't play a role here at all.

I think the HUP, along with "superposition", as described within the standard treatment of QM, are the two concept that has the most misconception attached to it. This is something that needs to be cleared first before one can tackle issues surrounding them.

Zz.

Last edited: Jul 21, 2006
16. Jun 9, 2006

### Staff: Mentor

Also, these repeated, identical measurements are not to be performed on the same system, one after another. You have to imagine a large ensemble (or collection) of identically prepared systems, and making the same measurement once (at corresponding times, of course) on each system in the collection.

17. Jun 9, 2006

### ZapperZ

Staff Emeritus
I'm not sure if that would matter.

In EPR-type experiments, you can send 'single photons', one at a time, through the splitter. There's nothing here to say that one isn't getting identical systems, one at a time. I can do the same with electron sources also (in fact, it is easier with electron sources than with photons). So I can generate practically identical electrons heading towards a slit, one at a time, and repeat the measurement several times. This is essentially a repeated measurement of an identical system.

In the ends, this would still give you a statistical ensemble - you just don't measure it all at once.

Zz.

18. Jun 9, 2006

### gptejms

ZapperZ,
I think your differentiation between 'measurement uncertainty' and uncertainty measured by other means as in the single slit experiment (which you seem to think is the 'real uncertainty') is artificial.Even in the case of a microscope,the resolution is related to the fact that the image of the object is really a diffraction pattern--and the width of the central bright fringe determines the resolving limit or the position uncertainty.

Last edited: Jun 9, 2006
19. Jun 9, 2006

### ZapperZ

Staff Emeritus
But that IS the "measurement uncertainty", similar to me using very large pixels on my CCD and thus my uncertainty on where the electron actually hit my detector. I don't have enough of a resolution! This isn't the HUP or else the HUP would be known before QM and would not be that strange. After all, applying the wave-nature of light gets you the same conclusion.

I can shoot a photon at an electron. Just because the electron momentum got changed so much after the collision tells me nothing about the HUP, because there's nothing to tell me that if I shoot another identical photon at an identically-prepared electron, that the outcome won't be the same. Why can't I destroy or alter the electron's momentum in the identically drastic way? There's nothing to say that I can't.

Zz.

20. Jun 9, 2006

### gptejms

Heisenberg discovered the uncertainty principle using his gamma ray microscope.In the case of microscope,it's the dual nature of light that leads to the uncertainty principle whereas in the case of single slit experiment the wave nature of the electron is the cause of the uncertainty relation.There is no fundamental difference between the two uncertainties.

Last edited: Jun 9, 2006
21. Jun 9, 2006

### ZapperZ

Staff Emeritus
Could you explain how the "dual" nature of light causes the uncertainty principle? I mean, in QM, there are no "duality" at all since I can derive all phenomena of light from one consistent description. So how does this "duality" got me the expression for $\Delta(x)\Delta(p)$?

Secondly, I'm not sure if it is historically correct to say that Heisenberg "discovered" the HUP from using a gamma ray microscope. He might have had the impetus to think about it using such a thing. Newton didn't discover the laws of gravity simply by observing a falling apple from a tree, even though it might have given him an inspiration for the idea.

Note that for your microscope to work, you never shoot just one photon at a time. Only then can you make any physical connection to the concept of "wavelength" as being a physical dimension.

This would be strange, since you appear to have different description for different things. Again, in QM, none of these are "different" things.

Zz.

22. Jun 9, 2006

### gptejms

The wave nature of light gives you the resolving limit of the microscope(i.e. position uncertainty) and the particle nature gives you the momentum uncertainty.

In QM you say there is no duality at all--if there were no duality there would be no QM in the first place.

Don't remember history,but from what I remember of what I've read,Heisenberg first proposed HUP using the microscope(and he was helped by Bohr in developing this idea).

What do you mean?You mean you need a lot of photons to relate it to the concept of wavelength?---this is not right!I would rather think of a single photon as a wavepacket with some dominant frequency/wavelength.

23. Jun 9, 2006

### Staff: Mentor

Sure. I didn't mean to imply that the the different systems in the ensemble had to be "running" simultaneously. My point was that you have to do the measurements on different photons or electrons or whatever, and not make repeated measurements of the same particle.

By "corresponding times" I meant that if you run the experiment serially on different particles, in general you need to make the measurement at the same elapsed time after preparation, for each particle (for systems in which the elapsed time affects the outcome).

24. Jun 9, 2006

### ZapperZ

Staff Emeritus
But what "momentum uncertainty"? You just said that the resolving power of a microscope depends on the wavelength of light. So where is the HUP here, i.e. where is the non-commuting operator partner in this scenario to allow you to derive the HUP?

Then could you please open a QM text and show me where the duality is?

Fine.

1. A photon as in a laser has a unique, SINGLE frequency.

2. Now, find me a 'wavepacket' in space. Now do a fourier transform of that wavepacket. Do you think you get a just ONE, single frequency? You don't!

Picture 1 and Picture 2 are self-contradictory. A photon is NOT a "wavepacket" simply by that reason alone.

If you think you can define a 'classical wavelength' when you have a single photon, I'd like to see it. Tell me how you could measure it as a start.

Zz.

25. Jun 9, 2006

### JesseM

I think that your "i.e." there is unjustified. You can describe the HUP as a measurement uncertainty while acknowledging that this uncertainty is built into the laws of physics rather than due to any particular measurement technique, and also without commiting to any sort of hidden-variables notion that the particle actually has a more precisely-defined position or momentum that we just can't measure (although Bohm's interpretation of QM shows that it's possible that this is true). Of course if you don't add these sorts of qualifications, I agree that calling it a measurement uncertainty is potentially misleading.