Why does an observer affect the electron?

[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!


Thanks in advance

 

[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.

 

[pallidin]

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

 

[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.

 

[hanlin_sg2002]

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

 

[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.

 

[quddusaliquddus]

Sorry guys. I still don't understand. Does the observation do something "physical" to the electron e.g. hit the electron with a photon?

 

[quddusaliquddus]

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

 

[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.

 

[RandallB]

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

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.

 

[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!

 

[JesseM]

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

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!

 

[silver-rose]

this is Heisenberg uncertainty principle no?

 

[quddusaliquddus]

thanks guys. now i think i understand it better.

 

[ZapperZ]

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!

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.

 

[jtbell]

[The HUP] 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.

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.

 

[ZapperZ]

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.

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.

 

[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.

 

[ZapperZ]

ZapperZ,
I think your differentiation between 'measurement uncertainty' and uncertainty measured by other means(e.g. the single slit experiment) is a bit 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.

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.

 

[gptejms]

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.

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.

 

[ZapperZ]

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

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.

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.

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

Zz.

 

[gptejms]

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)?

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.

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.

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).

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.

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.

 

[jtbell]

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.

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.

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).

 

[ZapperZ]

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.

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?

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

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

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.

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.

 

[JesseM]

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.

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.

 

[ZapperZ]

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.

I don't understand why it is unjustified. A measurement uncertainty is due to our technique. We improve our technique, we improve our uncertainty. We do this all the time in experimental physics - just look at how more certain we know the mass of the top quark now when compared to when it was discovered. But is this the HUP? Nope!

I still want someone to show me how one could get the HUP relation from ONE single measurement, with or without measurement uncertainty. I measure the position of an electron, and THEN, measure its momentum. Where is my "HUP" here?

Zz.

 

[JesseM]

I don't understand why it is unjustified. A measurement uncertainty is due to our technique. We improve our technique, we improve our uncertainty.

I suppose it's just a question of terminology--I wasn't aware that the term "measurement uncertainty" had the accepted meaning "uncertainty due to measurement technique" (not saying it doesn't, just that if it does I wasn't aware of this). Would you say that "intrinsic measurement uncertainty due to the fundamental laws of physics" is an oxymoron, for example?

Googling "measurement", "uncertainty" and "heisenberg" I found this abstract which says "At the same time, this relation represents an instance of a Heisenberg uncertainty relation for measurement imprecisions." Would you disagree with their use of "measurement imprecisions" as well?

I still want someone to show me how one could get the HUP relation from ONE single measurement, with or without measurement uncertainty. I measure the position of an electron, and THEN, measure its momentum. Where is my "HUP" here?

The uncertainty is only in the simultaneous value of two different variables. But are you saying there is something in the accepted definition of "measurement uncertainty" that implies it only deals with the uncertainty in single measurements?

 

[ZapperZ]

I suppose it's just a question of terminology--I wasn't aware that the term "measurement uncertainty" had the accepted meaning "uncertainty due to measurement technique" (not saying it doesn't, just that if it does I wasn't aware of this). Would you say that "intrinsic measurement uncertainty due to the fundamental laws of physics" is an oxymoron, for example?

I don't know if it is an oxymoron, but it is certainly puzzling. The HUP has nothing to do with experimental/measurement uncertainty. A measurement can certainly contains experimental uncertainty. But when you have already MADE a measurement, you know the value VERY up to the degree of uncertainty of your measurement technique, equiment, etc. So if you have large pixels, the best you can do is the actual size of the pixel itself.

But is this the HUP? It is not.

Googling "measurement", "uncertainty" and "heisenberg" I found this abstract which says "At the same time, this relation represents an instance of a Heisenberg uncertainty relation for measurement imprecisions." Would you disagree with their use of "measurement imprecisions" as well? The uncertainty is only in the simultaneous value of two different variables. But are you saying there is something in the accepted definition of "measurement uncertainty" that implies it only deals with the uncertainty in single measurements?

But again, how does one get the uncertainty relation of a single measurement of position followed by the momentum? Plug this in into the HUP relation, and what do you get?

Zz.

 

[JesseM]

I don't know if it is an oxymoron, but it is certainly puzzling. The HUP has nothing to do with experimental/measurement uncertainty. A measurement can certainly contains experimental uncertainty. But when you have already MADE a measurement, you know the value VERY up to the degree of uncertainty of your measurement technique, equiment, etc. So if you have large pixels, the best you can do is the actual size of the pixel itself.

But in your answer you are simply assuming that your notion of the meaning of the term "measurement uncertainty" is the correct one (and part of your notion of the meaning seems to be that 'measurement uncertainty' and 'experimental uncertainty' are interchangeable, which I don't agree with). To me "measurement uncertainty" just means that there is a limit on the precision you can pin down the value of some variable or variables, it doesn't say anything about whether this limit is due to the particular experimental technique you are using to measure the value or whether it's due to fundamental physics. Surely you'd agree that if my notion of the meaning of the term "measurement uncertainty" was the commonly-accepted one, then the HUP would be a form of measurement uncertainty according to this definition? If so, then this debate is really over the sociological question of how most scientists would understand the term. My quick google search didn't turn up many examples I could see of people using "measurement uncertainty" in the way I think of it, so that would lend support to your version of the meaning of the term. On the other hand, I did find that one example of a scientific abstract which referred to the HUP as dealing with "measurement imprecisions", which suggests that at least some scientists would not see "imprecision in measurement" as synonymous with "imprecision in experimental accuracy" as you seem to do.

But again, how does one get the uncertainty relation of a single measurement of position followed by the momentum? Plug this in into the HUP relation, and what do you get?

I'm not sure what you're asking. When dealing with successive measurements the HUP wouldn't say there are any limits on the precision to which you can measure each variable, although of course if you measure position first and momentum second then the probability distribution for position will be sharply changed by a precise measurement of the momentum. The HUP is a limit on how narrowly you can confine the probability distributions of position and momentum (or other noncommuting variables) at a single moment in time.

 

[ZapperZ]

But in your answer you are simply assuming that your notion of the meaning of the term "measurement uncertainty" is the correct one (and part of your notion of the meaning seems to be that 'measurement uncertainty' and 'experimental uncertainty' are interchangeable, which I don't agree with). To me "measurement uncertainty" just means that there is a limit on the precision you can pin down the value of some variable or variables, it doesn't say anything about whether this limit is due to the particular experimental technique you are using to measure the value or whether it's due to fundamental physics. Surely you'd agree that if my notion of the meaning of the term "measurement uncertainty" was the commonly-accepted one, then the HUP would be a form of measurement uncertainty according to this definition?

But that is what I have been asking you to show. I make ONE measurement of the position of an electron. How is the uncertainty of that single measurement somehow is "according" to the HUP? Isn't the accuracy of your measurement of the position depends on your technique/equipment/etc? This is what I have been asking you to show. How are you able to reconcile the definition of the HUP as having a statistical spread over a large number of measurement over what you perceive to be a "measurement uncertainty" that is present whether you measure once, or a million times?

Zz.

 

[gptejms]

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?

I don't know what you mean by your question 'what momentum uncertainty'--the photon imparts momentum to the electron.The uncertainty in the momentum is due to the uncertainty on where exactly the photon enters the microscope after scattering(&. thereby what momentum it imparts to the electron).

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

What's \lambda=h/p?You have momentum on one side and wavelength on the other--is this not duality?

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.

Even in a laser you never have a 'single frequency' photon.There is a spread and that's why there's no contradiction between 1 & 2.

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.

If a single photon didn't have a frequency/wavelength,it wouldn't be able to cause the photoelectric effect i.e. liberate a photo-electron.

 

[JesseM]

To me "measurement uncertainty" just means that there is a limit on the precision you can pin down the value of some variable or variables, it doesn't say anything about whether this limit is due to the particular experimental technique you are using to measure the value or whether it's due to fundamental physics. Surely you'd agree that if my notion of the meaning of the term "measurement uncertainty" was the commonly-accepted one, then the HUP would be a form of measurement uncertainty according to this definition?

But that is what I have been asking you to show. I make ONE measurement of the position of an electron. How is the uncertainty of that single measurement somehow is "according" to the HUP?

I still don't understand your point. According to my definition above, measurement uncertainty is "a limit on the precision you can pin down the value of some variable or variables". So why do you keep emphasizing that I must show uncertainty in the value of a single variable? There is nothing in the definition that rules out forms of measurement uncertainty which only apply to the simultaneous value of pairs of variables.

Isn't the accuracy of your measurement of the position depends on your technique/equipment/etc?

That's one of the things it can depend on, yes. But there is nothing in the laws of physics that prevent us from creating techniques/equipment which measure the position of a particle to any desired degree of precision, at least until you get to the Planck scale.

This is what I have been asking you to show. How are you able to reconcile the definition of the HUP as having a statistical spread over a large number of measurement over what you perceive to be a "measurement uncertainty" that is present whether you measure once, or a million times?

I would understand "measurement uncertainty" to be like probability--you can talk about the probability an event will happen on a single trial, but under the frequentist interpretation of probability, what you really mean by that is the frequency that event would happen if you did a large number of identical trials. Similarly, if you look at a large number of trials where the momentum has been measured to be within a certain range, you could use the HUP to predict the range of positions found in a very precise position measurement immediately afterwards. Multiple trials are also needed to understand ordinary classical measurement uncertainty due to equipment--if you look at large number of trials where an imprecise precision-measuring device found the position to be at a certain pixel, and immediately afterwards a more precise position measurement was made, the uncertainty in the first measurement tells you the range of positions that will be found by the more precise measurement in the entire set of trials, given identical readings by the first device in each trial.

 

[ZapperZ]

Then I don't understand what we are arguing about, or why you were having problems with what I said originally.

Zz.

 

[ZapperZ]

I don't know what you mean by your question 'what momentum uncertainty'--the photon imparts momentum to the electron.The uncertainty in the momentum is due to the uncertainty on where exactly the photon enters the microscope after scattering(&. thereby what momentum it imparts to the electron).

And how is this "uncertainty" the same as the HUP?

What's \lambda=h/p?You have momentum on one side and wavelength on the other--is this not duality?

How is this "duality"? Just because you equate a momentum with a wavelength? This is how we define "duality"? Since when?

Open a classical E&M text (see Jackson, for example). You'll notice a discussion on "radiation pressure". There's no invokation of any QM here in any form since light is still described as EM radiation. Yet, you can still have "pressure", and thus, "force per unit area" and thus "momentum change" associated with that force. By your logic, classical E&M also have "duality" just because it has this association.

If this is true, then this "duality" shouldn't be such a revelation that it is now since it was common in classical E&M.

Even in a laser you never have a 'single frequency' photon.There is a spread and that's why there's no contradiction between 1 & 2.

But this "spread" is not found in the description of a SINGLE photon. You just wrote above the relationship between the momentum and the wavelength. There's no "spread" there because such a photon has only ONE wavelength. The "spread" in wavelength due to the variation in the state of transition is measured STATISTICALLY, i.e. you do this for more than one photon. You then measure a variation in wavelength of a number of photons, not the variation in EACH photon. THIS is the spread. This spread has nothing to do with your "wavepacket" the way you imagined. You were using this wavepacket as being the location of ONE photon. This makes no sense. I could, for example, cool down a gas and the "spread" in the wavelenth of light being emitted will be smaller. Does this mean that the wavepacket representing each of the emitted photon is now tighter, implying the "size" of a photon emitted by a cold gas can be affected by the thermal variation of the source?

If a single photon didn't have a frequency/wavelength,it wouldn't be able to cause the photoelectric effect i.e. liberate a photo-electron.

Ah, now you are in my territory. You see, a photon is defined having a clump of energy. In most cases, you never just get ONE photon, but rather a stream of photons that have a classical equivalent of light with a well-defined concept of "wavelength". This is where if you have an antenna, as you look at the E field of the stream of photon passing by your antenna, you see an oscillating E-field. But this requires MORE than one photon for you to detect such oscillation. This is where we make the connection between the E-field content (the frequency or wavelength of oscillation) with the energy in EACH photon. But you do not detect such an oscillation when you have only ONE photon even when you could, mathematically, assign an energy to one photon. The same argument can be made in talking about "polarization" of a photon - it is an equally meaningless concept with there aren't a bunch of them around.

And photoemission simply care that you have a sufficient energy for emission of an electron. It doesn't require only something with a "wavelength-type" of energy for the liberation of electrons. I can heat something up, or shoot it with classical electrons, and I could get electron emission. Photoelectric effect is simply consistent with the concept of a photon carrying some energy that is transferable to the solid. The fact that this energy can be associated with some "wavelength" when there's many of these photons doesn't affect the solid's ability to emit an electron.

Zz.

 

[gptejms]

And how is this "uncertainty" the same as the HUP?

Ok,call it MUP!(where the M stands for measurement).Now what you are trying to say is that MUP is different from HUP.The former is due to the limitations of our measuring apparatus/instrumentation whereas the latter is the 'real uncertainty'(and this is what Heisenberg really had in mind)--according to you.

I repeat this is an artificial demarcation.HUP is identical to MUP--in fact,QM talks only of meaurements.HUP tells us of meauremement uncertainties only.Even the single slit experiment is telling you of the measurement uncertainty--you try to 'measure' the position of an electron accurately by making it pass through a narrow slit(you have a small delta x),but as a result of this you see that you've got a diffraction pattern on the screen(so you end up getting a large delta p).All these are measurement uncertainties.There is no meaning to uncertainty without a measurement!

Open a classical E&M text (see Jackson, for example). You'll notice a discussion on "radiation pressure". There's no invokation of any QM here in any form since light is still described as EM radiation. Yet, you can still have "pressure", and thus, "force per unit area" and thus "momentum change" associated with that force. By your logic, classical E&M also have "duality" just because it has this association.

Ok,there's pressure,there's momentum change--but this momentum change is not quantum in nature.How can you make a conclusion of particle nature here?

But this "spread" is not found in the description of a SINGLE photon. You just wrote above the relationship between the momentum and the wavelength. There's no "spread" there because such a photon has only ONE wavelength. The "spread" in wavelength due to the variation in the state of transition is measured STATISTICALLY, i.e. you do this for more than one photon.

Don't agree with this.Tell me a single experiment which measures the photon to have a single wavelength.The spread in frequency(at least the homogeneous part of broadening) relates to a single photon.

 

[JesseM]

Then I don't understand what we are arguing about, or why you were having problems with what I said originally.

Well, I took it that you had a problem with what I said originally--you quoted me and said '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'. Would you agree that if "measurement uncertainty" is defined in the way I am doing it, then the HUP is a form of measurement uncertainty, just one that has to do with fundamental physical laws rather than technique?

 

[ZapperZ]

Well, I took it that you had a problem with what I said originally--you quoted me and said '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'. Would you agree that if "measurement uncertainty" is defined in the way I am doing it, then the HUP is a form of measurement uncertainty, just one that has to do with fundamental physical laws rather than technique?

Here's the problem I had with it from the way I understand it. You can have a "measurement uncertainty" even for the SINGLE measurement. But I have been asking how one gets the HUP out of such a thing, to prove that you can't, yet, you can still have a "measurement uncertainty". Even you have mentioned somewhere along the way that measurement uncertainty can contain the HUP plus other uncertainty, haven't you? So how can this and the HUP be the same thing?

The other problem that I have is that I don't think you have illustrated what you mean by such a thing clearly by using it for a particular example. You will notice that I keep trying to illustrate what I meant by using specific example, such as using a single slit case. I'd like to see how you extract such "measurement uncertainty", and how this is also the HUP.

Zz.

 

[JesseM]

Here's the problem I had with it from the way I understand it. You can have a "measurement uncertainty" even for the SINGLE measurement.

Only in the same sense that you have a "probability" for a single event--it's a notion that implicitly depends on the frequency of different outcomes in a large set of identical trials.

But I have been asking how one gets the HUP out of such a thing, to prove that you can't, yet, you can still have a "measurement uncertainty". Even you have mentioned somewhere along the way that measurement uncertainty can contain the HUP plus other uncertainty, haven't you? So how can this and the HUP be the same thing?

I think you're misunderstanding me, I didn't say "measurement uncertainty" is just the same thing as the HUP. Rather, the HUP is a type of measurement uncertainty, but there are also other forms of measurement uncertainty due to purely classical considerations like the resolving power of your measuring apparatus.

The other problem that I have is that I don't think you have illustrated what you mean by such a thing clearly by using it for a particular example. You will notice that I keep trying to illustrate what I meant by using specific example, such as using a single slit case. I'd like to see how you extract such "measurement uncertainty", and how this is also the HUP.

How I extract measurement uncertainty from what? Your example, or any example? If we are talking about the HUP as an example of measurement uncertainty, then you'd illustrate it in exactly the same way that you illustrate the HUP. For example, you could make an inexact momentum measurement followed by a precise position measurement immediately afterwards, and repeat over a near-infinite number of trials; then looking just at the subset of trials where the inexact momentum measurement found the momentum to be within a certain range, you could look at the range of different precise positions found immediately after in this same subset of trials, and see that the range of the momentum vs. the range of position obeys the relation given by the HUP.

If you want to illustrate an example of measurement uncertainty that is not due to the HUP but due to the particular measurement technique you're using, the basic idea would be similar, as I suggested in a previous post:

if you look at large number of trials where an imprecise precision-measuring device found the position to be at a certain pixel, and immediately afterwards a more precise position measurement was made, the uncertainty in the first measurement tells you the range of positions that will be found by the more precise measurement in the entire set of trials, given identical readings by the first device in each trial.

In every case, the idea is that you can determine the uncertainty in a given type of measurement by taking a large number of trials where you made that type of measurement and it gave you a certain answer, then immediately afterwards you made a much more precise measurement, whether of the same variable or a different variable; the "uncertainty" is in your prediction of the result of the second precise measurement, given only knowledge of the result of the first measurement. The wider the range of possible results of the second measurement in a large set of trials where the first measurement gave the same result on each trial, the greater the "measurement uncertainty" associated with the first measurement's ability to predict the value of the variable measured by the second measurement.

 

[ZapperZ]

Only in the same sense that you have a "probability" for a single event--it's a notion that implicitly depends on the frequency of different outcomes in a large set of identical trials.

But how are you to know this after you have performed just ONE set of measurement?

I think you're misunderstanding me, I didn't say "measurement uncertainty" is just the same thing as the HUP. Rather, the HUP is a type of measurement uncertainty, but there are also other forms of measurement uncertainty due to purely classical considerations like the resolving power of your measuring apparatus.

And I say that the HUP is not a type of measurement uncertainty. You can have an absolutely perfect measurement instrument giving you an ideal zero uncertainty, and you can still have the HUP, which isn't really an uncertainty, but rather than inherent spread of a value measured repeatedly. This is why I distinguised the HUP from the "measurement uncertainty". They are not of the same specie. You can improve the accuracy of your measurement of any observable independent of each other. I can make my measurement of the position as accurately as I want, independent of the accuracy of how I determine the momentum of that particle afterwards. They are not related and not coupled together as described by the HUP. But this doesn't mean that I have the ability to predict the outcome of the NEXT measurement, even when I have equipment to accurately determine both position AND momentum.

How I extract measurement uncertainty from what? Your example, or any example? If we are talking about the HUP as an example of measurement uncertainty, then you'd illustrate it in exactly the same way that you illustrate the HUP. For example, you could make an inexact momentum measurement followed by a precise position measurement immediately afterwards, and repeat over a near-infinite number of trials; then looking just at the subset of trials where the inexact momentum measurement found the momentum to be within a certain range, you could look at the range of different precise positions found immediately after in this same subset of trials, and see that the range of the momentum vs. the range of position obeys the relation given by the HUP.

No, in a single measurement, show me how you determine such "measurement uncertainty". Use ANY specific example as you wish. I happened to use the single slit. If you wish, you can use that. Show me exactly where, when single particle passes through the slit, you'd determine all the types of "uncertainty", and how such a thing would lead to the HUP.

If you want to illustrate an example of measurement uncertainty that is not due to the HUP but due to the particular measurement technique you're using, the basic idea would be similar, as I suggested in a previous post: In every case, the idea is that you can determine the uncertainty in a given type of measurement by taking a large number of trials where you made that type of measurement and it gave you a certain answer, then immediately afterwards you made a much more precise measurement, whether of the same variable or a different variable; the "uncertainty" is in your prediction of the result of the second precise measurement, given only knowledge of the result of the first measurement. The wider the range of possible results of the second measurement in a large set of trials where the first measurement gave the same result on each trial, the greater the "measurement uncertainty" associated with the first measurement's ability to predict the value of the variable measured by the second measurement.

But isn't this what I said way early on in this thread? I brought up the statistical nature of the outcome and clearly stated that the SPREAD in the value of the observable (and NOT the value of the uncertainty of a single outcome) is what goes into the HUP.

I make a position measurement x_1, and in that single measurement, I have a measurement uncertainty \delta(x_1). I put it to you that this is not equal to, nor is what goes into the HUP, i.e. \Delta(x_1).

I then make several more measurement, giving me x_2, x_3, x_4,..., each of them giving me a measurement uncertainty \delta(x_1), \delta(x_2), \delta(x_3), ...

Now what exactly is involved when I have to find \Delta(x) that goes into the HUP? The \delta 's? Nope! It is all the values of x_2, x_3, x_4,.... This is where you get the average value and where the notion of an "average" value of anything is meaningful. That is why I said that even when you have an ideal measuring device that gives you ZERO measurement uncertainty, you will still have a spread in ALL of your measurement (not in a single measurement since we have zero measurement uncertainty in our ideal setup). It is this spread that is intrinsic to QM and consequently, intrinsic to our world.

Zz.

 

[JesseM]

But how are you to know this after you have performed just ONE set of measurement?

You can't, nor have I every claimed that you could. So why do you keep insisting on this condition, when it is not part of my definition of "measurement uncertainty"?

And I say that the HUP is not a type of measurement uncertainty.

Using your definition of "measurement uncertainty" or mine?

You can have an absolutely perfect measurement instrument giving you an ideal zero uncertainty

Not if you're talking about the uncertainty in two variables like position and momentum. I agree that you can measure either one individually with arbitrary precision, but I said before that my definition of "measurement uncertainty" can apply to pairs of variables rather than single variables.

you have and you can still have the HUP, which isn't really an uncertainty, but rather than inherent spread of a value measured repeatedly.

Of course it's an uncertainty, because you are uncertain about what the result of a precise measurement of variable #2 will be, given only knowledge of the result of a measurement of variable #1. Likewise, even if you measure variable #1 with great precision, if you immediately measure variable #2 and then immediately measure variable #1 again, you are uncertain about what the result of the second measurement of variable #1 will be, even if the time interval between the successive measurements is made arbitrarily small, which in classical mechanics would imply that the difference between the two measurements of the same variable should become arbitrarily small as well (on the other hand, if you measured variable #1 with great precision twice in succession without measuring variable #2 in between, then even in QM your uncertainty about the second measurement can be made as small as you wish).

This is why I distinguised the HUP from the "measurement uncertainty". They are not of the same specie.

If you use your definition, maybe not. If you use my definition, it's true that "measurement uncertainty due to equipment" is not of the same species as "measurement uncertainty due to HUP", but they are both members of the same genus, "measurement uncertainty". I don't see why you act as if there is something incoherent about my definition, or why your definition is the only one possible. Words and terms have no intrinsic meaning, if I wished I could use the term "measurement uncertainty" to mean the same thing as "tree", in which case a birch tree would be a type of "measurement uncertainty". As I said before, there is the sociological question of whether my use of "measurement uncertainty" is totally at odds with the way the term is used by scientists, but I did quote that abstract which suggests at least some scientists would describe the HUP as a form of "measurement imprecision", which is essentially the same term.

You can improve the accuracy of your measurement of any observable independent of each other. I can make my measurement of the position as accurately as I want, independent of the accuracy of how I determine the momentum of that particle afterwards. They are not related and not coupled together as described by the HUP. But this doesn't mean that I have the ability to predict the outcome of the NEXT measurement, even when I have equipment to accurately determine both position AND momentum.

But if we are to describe ordinary classical measurement uncertainty in operational terms (as opposed to taking a God's-eye-view where we always know the 'true' value of the position and can compare it to the result of a position measurement with a given apparatus), we must also talk about pairs of measurements, not single measurements. To determine the measurement uncertainty in the readings of a particular device, we need to have a more precise way of measuring the same variable, so that we can take two measurements in quick succession, the first using the device and the second using the more precise form of measurement. That way, we can look at the subset of trials where the device gave a certain reading, and then look at the spread of values of the more precise measurement among these trials--that will tell us the "measurement uncertainty" in the readings of the first device.

If you disagree, please explain how we can quantify plain old classical measurement uncertainty of an imprecise measuring-device, using only a SINGLE measurement with that device alone.

No, in a single measurement, show me how you determine such "measurement uncertainty".

Why do you think I need to show this, when it is not part of my definition of "measurement uncertainty" that it only applies to SINGLE measurements? That may be part of your definition, but it isn't part of mine.

But isn't this what I said way early on in this thread? I brought up the statistical nature of the outcome and clearly stated that the SPREAD in the value of the observable (and NOT the value of the uncertainty of a single outcome) is what goes into the HUP.

I don't understand what you mean by "NOT the value of the uncertainty of a single outcome" here. The value of the "uncertainty" of a single outcome also depends implicitly on the notion of a spread over multiple trials, just like the value of the "probability" of a single outcome depends implicitly on the frequency of that outcome in multiple trials. This is true of classical measurement uncertainty too, the "uncertainty" in a single reading of a given device corresponds to the spread of more precise values for that variable (whether measured by a more precise method immediately afterwards, or taking a God's-eye-view where the precise value of that variable is just known) in a large set of trials where the device returned that same reading. Of course, the HUP differs in that it's a type of uncertainty that only arises when you try to measure two different variables simultaneously (or in the limit as the time between measurements of different variables goes to zero), not a single variable; and, of course, it's also different in that it arises from fundamental laws of physics rather than the particular details of your measuring-device. But my definition of "measurement uncertainty" would subsume both types of uncertainty.

I make a position measurement x_1, and in that single measurement, I have a measurement uncertainty \delta(x_1). I put it to you that this is not equal to, nor is what goes into the HUP, i.e. \Delta(x_1).

I agree, but then nothing in what I have said implies they must be equal, they are two very different types of "measurement uncertainty" according to my definition.

Now what exactly is involved when I have to find \Delta(x) that goes into the HUP? The \delta 's? Nope! It is all the values of x_2, x_3, x_4,....

Likewise, when you talk about the classical uncertainty \delta(x) due to the imprecision of your equipment, according to a frequentist you can only really undestand the meaning of this by looking at a large set of trials where your equipment returned that same reading, and then considering the more precise value of the position x_1, x_2, x_3, ... on each of these trials (whether obtained by a more precise device which made a measurement immediately afterwards, or simply assuming we have omniscient knowledge of what the 'true' position was on each measurement). Without this implicit understanding the notion of classical "measurement uncertainty" would be meaningless, according to a frequentist.

 

[ZapperZ]

Then I give up. I have no clue what we are arguing about anymore.

Zz.

 

[quddusaliquddus]

lol ... seems my seemingly innocent question has created quiet a stir in the PF community! :D

 

[Dr.Brain]

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!

can pls osmeone tell me how can we mathematically prove that minimum wavelength of photon = distance between slits

 

[masudr]

Classical optics: given the Fraunhofer condition, the intensity distribution is the Fourier transform of the transmission function at the slit. What you will find is that if the wavelength of the light is comparable to the slit size then you will get a diffraction pattern.

Quantum mechanics: given that the slit determines the position of the particle, and the detection on the screen measures the momentum of the particle as it left the slit, we relate the two via a Fourier transform (see the Marcella paper that ZapperZ has quoted over and over). Again you will find is that if the wavelength of the light is comparable to the slit size then you will get a diffraction pattern.

Hey presto, it's the same answer.

 

[Farsight]

quddusaliquddus: I'll try to give you a mental picture:

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

It doesn't.

An electron isn't something with a solid surface or a particular size. It's more like an electric standing wave. Imagine a pebble thrown into a pond. Plop! Now immediately freeze the image of the ripples, with a big peak in the centre. That's kind of what your electron usually looks like. Like a particle.

But fire it out of a gun and it spreads out like rubber onion rings. It now looks more like the ripples on the pond ten seconds after the stone went plop. Definitely like a wave. And remember there's no solid surface to it. It definitely isn't all in one place any more, and so it can be in two places at once.

But if any part of it snares on a detector or target screen and it stops, it snaps back into its original configuration and looks like a particle again.

 

[quantumcarl]

But if any part of it (an electron) snares on a detector or target screen and it stops, it snaps back into its original configuration and looks like a particle again.

Would the retina of the observer's eye be considered to be a target screen or detector?

If an electron is only a wave until it stops at your retina (and is then interpreted by the optic nerve and visual cortex as a solid) is this like saying that water isn't wet until you feel it?

 

[Farsight]

Would the retina of the observer's eye be considered to be a target screen or detector?

Yep.

If an electron is only a wave until it stops at your retina (and is then interpreted by the optic nerve and visual cortex as a solid) is this like saying that water isn't wet until you feel it?

Nope.

It's more like the situation where you close your eyes and feel a magnet's repulsion with another magnet. It feels kinda like there's something there, even though you ain't touching anything solid.

 

[masudr]

If an electron is only a wave until it stops at your retina (and is then interpreted by the optic nerve and visual cortex as a solid) is this like saying that water isn't wet until you feel it?

Electrons don't generally get to your retina. Photons do, though.

And, how much water does it take for it to feel wet?

 

[quantumcarl]

Yep.Nope.

It's more like the situation where you close your eyes and feel a magnet's repulsion with another magnet. It feels kinda like there's something there, even though you ain't touching anything solid.

Actually I don't know if I'm right to equate a retina with a detctor screen.

In the case of the detector... the observer observes the results of the electron becoming solid as it comes into contact with the screen. The observer's retina... as masudr points out, cannot even detect an electron on its own. Or, at least, we have not trained ourselves well enough to detect and electron hitting our retina.

So, with this being the case... it is the interpretation of the state of the electron given by the detector screen that the observer's eye is utilized to interpret.

So here we see that the eye is second party to the results of the wave collapsing (or whatever it does) and becoming, or behaving, like a solid.

In the case of a photon... again as masudr mentioned, I can see where the analogy of the eye being a detector plate or screen... because it is able to distinquish when the spectrum of em waves that is light is hitting it.

"Nature does nothing better than entertaining us." I must say.

 

[curious ron]

Ok, i have question. The double slit experiment is carried out in a vacuum so that foreign particles don't get in the way. So what about the so called quantum foam, particles and anti-particles appearing and disappearing, why do they not interference with the electron? or do they wipe each other out too quickly to make a difference.