Electrons and the Double Slit experiment questions

[tonyxon22]

First of all, I want to apologize ahead for three things:
1) Opening another tread about this experiment, with probably the same title than other 800 threads: I took a little time to read the other threads with similar titles and didn’t found this doubt in none of them, and also didn’t seem right to post this questions in any of those threads.
2) Giving the electron the property of “wave-particle duality”: I will not explicitly write down this term in the lines below because one of the things that I learned in many of the other threads is that this term causes misunderstanding since electrons are not particles nor waves, they are just “electrons”. However I think it could be implicitly understood in my expressions, especially because in most of the literature I have read the term subatomic PARTICLE is applied to all of them, photons, electrons, neutrinos, quarks, etc.. So I have great difficulty in accepting that an electron is not a “particle”, especially for my first question where the electron’s mass is involved into the problem.
3) Giving the electron a human characteristic by using the terms “it knows”: I am afraid this could be a terrible mistake but I find no other way to phrase my questions without it.
Second, the introduction:
As I understand, the experiment consist of 3 elements: (a) a source which shoots electrons. I hope it’s safe to say that it is capable of shooting one electron at the time when we push a button, so we know exactly when each one of them leave the source; (b) a wall with two small slits on it, and on top of each slit there is a detector that is capable to indicate whether a single electron or not passed through the slit; (c) a screen that will mark the position where the electron arrived.
Finally, the questions:
1) Electrons, which for the moment I will accept that are not particles nor waves, have mass. If that is the case, they cannot travel at the speed of light. In the case that we have the detectors ON over the slits, in order to determine which way the electron went, one should be able to calculate how much time did it take for the electron to go from the source to the screen. Knowing the distance between them we can then calculate the speed of the electron which should be slower than the speed of light. The question is: What happens when the detectors are OFF, so we do not know which way the electron went and the screen reveals that it behaves like a wave. In other words, what is the speed of the electron when it behaves like a wave?
2) Supposing that we manage to create a layout of the experiment where the slit is in the middle distance between the source and the screen, why does measuring the position of the electron at half way would determine the position at the end? Does the result changes if we move the slit closer to the source or to the screen? For all the electron “knows”, it was detected at a certain point, but it could be crossing many other slits after the one with the detector, thus start behaving like a wave, but it doesn’t. I seems like once it is detected, the rest of the experiment is already fully determined. So what if we put another barrier with two slits in front of the first one? Will the electron at that point begin to behave like a wave if no detectors determine which one of the second slits in went through? How does it even “knows” that it was crossing a second slit? Why would that behavior not happen if there were not this second barrier with slits?
3) (This question was actually in one of the other threads that I read but I didn’t catch a direct answer) If the detectors are OFF, does each single electron still leave only one mark in the screen? If that is the case, then many electrons need to be fired before the wave behavior can be appreciated on the screen, which can also be interpreted as each individual electron takes a different path after the slit. I don’t know if you understand where I’m going with this but thinking about it takes me back to the second question.
Thanks for reading and best regards,

 

[jfizzix]

1.) The uncertainty principle applies here in two ways. First, any measurement capable of telling which slit the electron went through (even using just timing) will destroy the subsequent interference pattern. Second, it isn't possible (as far as we know) to prepare an electron to be in a state with well defined timing and frequency (due to the energy time uncertainty principle). Because of this, if you prepared electrons in such a way that you could launch them at very precise times, and detect them on the screen, the uncertainty in frequency will be large enough that the electrons will be naturally incoherent with one another, and you won't be able to see interference anyway.

2.) The double slit experiment is something best taken as a big picture. The electron doesn't "know" it's been measured at one time and not another. The experiment is configured so that a given set of possible results is seen. Change the experiment, and change the results.

3.) Quantum mechanics is only really prepared to describe what you are likely to see if you measure something. If the detectors are off (assuming you mean also the screen), then you are not measuring anything. If you're asking what you see when the screen is "on", but there is no information that can tell you which path the electrons are going through, you should see a pattern of detector clicks, one at a time, until an interference pattern becomes visible.

 

[DrClaude]

You've hit upon an interesting point: if the electron doesn't "know" whether there will be other slits, how does it "know" whether to behave as a "wave" or not? It doesn't. Travelling through free space can be seen as passing a screen with an infinite number of infinitely small slits, and the resulting interference looks like a straight trajectory. This is the basic idea behind Feynman's path integral formulation of quantum mechanics.

If the detectors are OFF, does each single electron still leave only one mark in the screen? If that is the case, then many electrons need to be fired before the wave behavior can be appreciated on the screen, which can also be interpreted as each individual electron takes a different path after the slit.

Yes, each electron is always detected at a single point, and the interference pattern is built up from many electrons. This is one of the reason why wave-particle duality has been dropped. The electron is alwyas detect as a particle. It may seem that it goes through both slits at the same time, but if you try to detect it, you find that it either went through one slit or the other.

 

[tonyxon22]

The electron is alwyas detect as a particle. It may seem that it goes through both slits at the same time, but if you try to detect it, you find that it either went through one slit or the other.

But if you don't try to detect it, maybe you register its position as a particle as a single point, but it does not follow the feymann's path... Is like observation is affecting the path, making it straight instead of random...? But there is still the problem of the electron knowing or not if it is being observed, and at which point it can still change its mind and go in a different path...

 

[Derek Potter]

Hi. Rather than answer your questions directly it would be better to clear up some basic misunderstandings first.

1 The propagation of an electron is ALWAYS governed by a wavefunction. When you detect it, it ALWAYS displays particle properties. There is no question of it deciding to act like a wave or act like a particle as it travels.

2 It is true that the wave function can be derived as an integral of all possible paths but that's nothing to do with acting like a particle.

3 Electrons (all objects actually) NEVER travel ballistically.

4 The idea that the behaviour of the wave depends on information is a derived result which requires a thorough understanding of quantum information in order to formulate a statement correctly. Otherwise just stick with interactions.

5 In a similar way, it all depends what you mean by ON and OFF. If OFF still interacts with the electrons it still destroys the interference - even if the information is not recorded. If OFF actually ceases to interact, then the interference pattern returns.

6 There is no particular reason why you shouldn't time an electron as it goes past. However you can't detect the wave itself, you can only take a positional measurement, which is a particle property. The wavefunction tells us the probability distribution of those measurements.

-DP

 

[bhobba]

This wave-particle duality stuff is one of those myths of QM that unfortuneately is hard to dispel:
http://arxiv.org/pdf/quant-ph/0609163.pdf

It a hangover of the semi-historical approach most texts tend to follow and De-Broglies hypothesis which it used as a justification for Schodinger's equation. It was consigned to the dustbin of history when Dirac came up with his transformation theory at the end of 1926:
http://www.lajpe.org/may08/09_Carlos_Madrid.pdf

Here is the double slit experiments explanation without using that myth:
http://arxiv.org/ftp/quant-ph/papers/0703/0703126.pdf

Whats going on is this. Each slit is a position measurement. After the slit it has a definite position so by the Heisenberg uncertainly relation it momentum is unknown. It's kinetic energy is still the same so the magnitude of its velocity doesn't change - its scattered in an unpredictable direction.

When you have two slits the wavefunction is a superposition of the the wavefunction at each slit and when you work through the math as detailed in the above link you get interference.

Thanks
Bill

 

[tonyxon22]

Just to clarify something:

If instead of two slits the experiment had only one, what would be the expected pattern on the screen? What I understand is that if the detector at the slit is ON then the expected pattern should be a line along the projection of the only slit. In the other hand, if the detector is OFF (no interaction with the electron) then the patter in the screen would be a cloud, something like points everywhere, without any interefence.

Is this correct?

 

[bhobba]

If instead of two slits the experiment had only one, what would be the expected pattern on the screen? What I understand is that if the detector at the slit is ON then the expected pattern should be a line along the projection of the only slit. In the other hand, if the detector is OFF (no interaction with the electron) then the patter in the screen would be a cloud, something like points everywhere, without any interefence.

The pattern for one slit is explained in the link I gave.

What detector are you talking about - a detector in the slits or at the screen?

Thanks
Bill

 

[Derek Potter]

What I understand is that if the detector at the slit is ON then the expected pattern should be a line along the projection of the only slit. ... Is this correct?

No it's not. Not even slightly

As I said, electrons NEVER travel ballistically.

The slit "prepares the wavefunction to a position state" i.e. limits it to a very narrow peak at the slit. After the slit, the wavefunction spreads out into a hemicylinder and thus the pattern is of more-or-less even illumination over the entire screen.

A detector can make no difference - the single slit already constrains the position fully so detecting it again cannot change anything.

edit - BTW, there seems to be a redface smilie attached to this message. I don't want it but I can't get rid of it.

 

[tonyxon22]

Then I find it very difficult to understand why with two slits and detectors you get the electrons stricking the area of the to lines of projection of the slits. More I think about this the less I understand it

 

[tonyxon22]

Is there any substantial difference in doing this experiment with electrons or with photons?

 

[Derek Potter]

(

Then I find it very difficult to understand why with two slits and detectors you get the electrons stricking the area of the to lines of projection of the slits. More I think about this the less I understand it

The reason you do not understand it is because it is not true! As I said, electrons NEVER travel ballistically.

Actually, what you describe is a very common misconception and is found all over poor-quality websites and videos so I'm not surprised you've picked it up. However, it is simply false. Electrons are always diffracted by the slits so that they cover the whole screen. You never get bar patterns like you suggest (unless, of course, the slits are so wide that diffraction is negligible and then interference is not visible anyway).

Yes, the whole thing can be done, in principle, with any particles using an appropriate set up. In fact I often forget whether we are talking about electrons or photons. I wouldn't recomment trying it with quarks or Higgs bosons though.

 

[bhobba]

Then I find it very difficult to understand why with two slits and detectors you get the electrons stricking the area of the to lines of projection of the slits. More I think about this the less I understand it

Did you read the following I posted before:
http://arxiv.org/ftp/quant-ph/papers/0703/0703126.pdf

Thanks
Bill

 

[bhobba]

Is there any substantial difference in doing this experiment with electrons or with photons?

Yes there is.

For photons position is not an observable.

Conceptually to start with, best to stick with electrons.

Thanks
Bill

 

[tonyxon22]

Did you read the following I posted before:
http://arxiv.org/ftp/quant-ph/papers/0703/0703126.pdf

Yes I did. I read all the three you sent. There was one about QM Myths and fact that for the moment I just read the slit experiment part but I’ll check it out fully later. Thanks for that

Actually, what you describe is a very common misconception and is found all over poor-quality websites and videos so I'm not surprised you've picked it up. However, it is simply false. Electrons are always diffracted by the slits so that they cover the whole screen. You never get bar patterns like you suggest (unless, of course, the slits are so wide that diffraction is negligible and then interference is not visible anyway).

If particles are always diffracted, what's the use of this experiment?

 

[Nugatory]

If particles are always diffracted, what's the use of this experiment?

You can still get interference or not, and that's the interesting part of the experiment. Diffraction is what makes the single-slit-open pattern illuminate the entire screen, not just a single bright spot with sharp edges behind the slit.

 

[bhobba]

If particles are always diffracted, what's the use of this experiment?

Its often used as a motivation for the QM formalism as an introduction to the wave-particle duality idea. Its a false idea but its often used.

In reality its explanation is a combination of two key ideas - that slits 'scatter' because they are a position measurement and we have the uncertainty relations plus when you have two of them its a superposition so you get interference.

Trouble is once the full quantum formalism is introduced they don't go back and show how it explains the experiment that motivated it.

Thanks
Bill

 

[Derek Potter]

For photons position is not an observable.

Why not?

 

[tonyxon22]

Because in order to detect a photon you need to absorve it, i think..

 

[Nugatory]

For photons position is not an observable.

Why not?

What would the position eigenstates of a photon be?

 

[Derek Potter]

If particles are always diffracted, what's the use of this experiment?

The experiments illustrate interference as well as diffraction, though the emphasis is usually on interference. "Which slit?" detector experiments - which are done though they don't do what you thought - illustrate loss of interference and thus loss of coherence.

 

[Derek Potter]

What would the position eigenstates of a photon be?

Same as for any continuous variable - an infinite set of delta functions for every possible value.

edited - made into an assertion, not a suggestion, mwah-ha-ha

 

[bhobba]

What are the dots on a photographic plate if not explicit records of where the photons were found when they were subjected to positional measurement by an emulsion of silver halide photon detectors?

It indicates where the photon interacted with the photographic plate - not its position.

That a photon has no position operator is a deep result of quantum field theory:
http://arnold-neumaier.at/physfaq/topics/position.html

Its usually not worried about in the double slit but we have some advanced posters on this forum so its best to be careful.

Thanks
Bill

 

[Derek Potter]

Sorry Bill, I deleted my comment, leaving yours hanging . Thanks for your answers.
Yes, I can see I need to mug up on QFT as I don't understand any of that paper. I think Nugatory may have been having a spot of fun.

 

[Derek Potter]

I stumbled across this paper: http://arxiv.org/abs/0711.0112 .
Obviously it's way over my head but the title and abstract sound promising.

Photon position eigenvectors lead to complete photon wave mechanics
"We have recently constructed a photon position operator with commuting components. This was long thought to be impossible, but our position eigenvectors have a vortex structure like twisted light. Thus they are not spherically symmetric and the position operator does not transform as a vector, so that previous non-existence arguments do not apply. ..."

So, does this remove the position operator problem after all ?

 

[bhobba]

So, does this remove the position operator problem after all ?

Similar papers have been discussed here before - and no it doesn't.

Like I said - its a deep result of QFT - there is really no escaping it.

Thanks
Bill

 

[Derek Potter]

OK, let's try this then. There is no position operator. But we can localise a photon with a pinhole. How come I, with my negligible grasp of QFT, can say exactly where the photon is but QFT tells me that you can't have a valid probability calculation even in principle? I can even use a series of slits, very closely spaced. As long as they are aligned some photons will struggle through - the "position" is "measured" several times in quick succession and it does not change. If the slits are misaligned nothing gets through - the photon doesn't jump around between measurements.

So I'm wondering whether the "no position operator" result applies to the position vector - all three coordinates at once - but doesn't rule out an operator for y alone, leaving x out of it (x is not implied by the position of the screens if the y measurement means the photon is not located in x!)
edit - added: Delocation in x would not affect the interference picture.

 

[bhobba]

But we can localise a photon with a pinhole.

You have localised the field - not the photon.

To understand the difference you need to study QFT:
https://www.amazon.com/dp/019969933X/?tag=pfamazon01-20

Thanks
Bill

 

[Nugatory]

Same as for any continuous variable - an infinite set of delta functions for every possible value.

we can localise a photon with a pinhole.

I'm going to try a non-QFT response, but please don't forget that there is some serious hand-waving in this answer - I'm just trying to convince you that cannot always trust first-quantization stuff when applied to photons.

Localizing something is not the same thing as saying that it is in an eigenstate of a position operator. When a particle passes through a pinhole, that narrows down the region of space in which it may be found, but the position is still uncertain because of the non-zero size of the pinhole itself. The only way to really nail the position down to that delta function that you're thinking of is if the pinhole is of size zero (that is, model the transmissability through the screen with a delta function). You can make this work with a point particle such as an electron - but you will get zero electromagnetic field passing through a size-zero hole, so no photon at all instead of a photon of precisely known position.

Another problem you'll find is that the entire formalism of position delta functions found by superimposing all possible momentum states and vice versa that you're suggesting we use comes from solving Schrodinger's $H\Psi=E\Psi$. However, that's not a relativistic equation, so there's no particular reason to expect it to work for photons - and if you do try, you'll find yourself looking at a $p^2/2m$ term, a fairly strong hint that you shouldn't be applying this when $m$ is equal to zero.

As I said, there is some serious hand-waving going on here. But it should be enough to reinforce the point that Bhobba made above, that "Conceptually to start with, best to stick with electrons" so as to avoid the complications of a proper relativistic treatment of position.

 

[Derek Potter]

Well, this is the problem, Bill referred me to an arcane paper which, if I have interpreted the jargon correctly, showed that a position operator cannot be defined for massless spin-1 particles. Which effectively limits its applicability to photons. Neutrinos, even back in the good old days when they were massless, would have had a position operator as they have s<=1/2.
"Theorem. An irreducible representations of the full Poincare group with mass m>=0 and finite spin has a position operator transforming like a 3-vector and satisfying the canonical commutation relations if and only if either m>0 or m=0 and s<=1/2 (but s=0 if only the connected poincare group is considered)."
On a more concrete note (concrete note - wave-particle ) I'm not sure that the finite size of the pinhole is a real issue. Certainly the pinhole is a superposition of deltas but that just means that a more physically realistic model would use a basis of "pinhole" functions. Neither position nor momentum states but a combination. As a bonus, this does away with the infinite momentum implied by a position delta. And why is it possible to get an electron through a delta pinhole but not a photon? I would have thought that the argument applies equally to the electron i.e. no electrons squeeze through a zero-diameter hole. Plus the infinite momentum problem. Surely we just take the scenario as a limiting case?

I am not trrying to nit-pick at what was intended to be hand-waving, but I'm having serious difficulty in understanding exactly what is meant by not having a position operator when it is quite clear that we just made a position measurement but QFT apparently tells us that we didn't.

Hope that made sense, I'm more than half asleep right now. Cheers.

 

[bhobba]

Well, this is the problem, Bill referred me to an arcane paper which, if I have interpreted the jargon correctly, showed that a position operator cannot be defined for massless spin-1 particles.

Nugatory admitted it was hand-wavey and at that level I think he did a reasonable job. Where you argument breaks down for photons is they aren't point particles like electrons - but purely excitations of the field. Of course elections are also excitations of the electron field in QFT - that why it hand-wavey.

If you want the real deal study QFT. The text I linked to, while challenging (QFT is challenging doesn't matter how you look at it) can be done after a first exposure to proper QM - not hand-wavey popularisations - but the real deal such as Susskinds book:
https://www.amazon.com/dp/0465036678/?tag=pfamazon01-20

If the issue really interests you start a new thread. Be warned - this quickly gets into some really deep waters.

Thanks
Bill

 

[Derek Potter]

Thanks for the book recommendations, Bill. I shall try to get hold of a copy of Quantum-Field-Theory-Gifted-Amateur and see whether my physics degree gave me an adequate, if rusty, "first exposure to proper QM".

 

[bhobba]

Thanks for the book recommendations, Bill. I shall try to get hold of a copy of Quantum-Field-Theory-Gifted-Amateur and see whether my physics degree gave me an adequate, if rusty, "first exposure to proper QM".

It should.

Thanks
Bill

 

[TrickyDicky]

Where you argument breaks down for photons is they aren't point particles like electrons - but purely excitations of the field. Of course elections are also excitations of the electron field in QFT - that why it hand-wavey.

I'm scratching my head at this. Where do you get photons are less of a point particle(in the QFT, particle physics current sense of elementary particle) than electrons?

 

[TrickyDicky]

You can still get interference or not, and that's the interesting part of the experiment. Diffraction is what makes the single-slit-open pattern illuminate the entire screen, not just a single bright spot with sharp edges behind the slit.

I know what you mean but it is important to remark that there is no difference between difffraction and interference, they are the same physical phenomenon. One can then make distinctions in intensity or the specific form it manifests but it is important not to perpetuate the misleading picture of the double slit that almost all sources portrait, as commented also in #12.

 

[TrickyDicky]

I'm going to try a non-QFT response, but please don't forget that there is some serious hand-waving in this answer - I'm just trying to convince you that cannot always trust first-quantization stuff when applied to photons.

Localizing something is not the same thing as saying that it is in an eigenstate of a position operator. When a particle passes through a pinhole, that narrows down the region of space in which it may be found, but the position is still uncertain because of the non-zero size of the pinhole itself. The only way to really nail the position down to that delta function that you're thinking of is if the pinhole is of size zero (that is, model the transmissability through the screen with a delta function). You can make this work with a point particle such as an electron - but you will get zero electromagnetic field passing through a size-zero hole, so no photon at all instead of a photon of precisely known position.

Another problem you'll find is that the entire formalism of position delta functions found by superimposing all possible momentum states and vice versa that you're suggesting we use comes from solving Schrodinger's $H\Psi=E\Psi$. However, that's not a relativistic equation, so there's no particular reason to expect it to work for photons - and if you do try, you'll find yourself looking at a $p^2/2m$ term, a fairly strong hint that you shouldn't be applying this when $m$ is equal to zero.

Not entering into the photon position issue, your two examples are a hand-wavey form of saying the position delta function formalism is only possible in a time-independent setting like the tunneling zero size pinhole first example or the TISE $H\Psi=E\Psi$ second example. But again, the double slit experiment cannot be simplified in that form even in hand-waving: the wave function is perturbed in a time-dependent way in its time evolution by the slits. This makes it collapse to the measured dot at the screen. Ignoring this is keeping with the misleading picture of the experiment that so many sources perpetuate.

 

[bhobba]

I'm scratching my head at this. Where do you get photons are less of a point particle(in the QFT, particle physics current sense of elementary particle) than electrons?

They do not have position like you expect of a point particle - in fact that pretty much the definition of a point particle.

Thanks
Bill

 

[TrickyDicky]

They do not have position like you expect of a point particle - in fact that pretty much the definition of a point particle.

Thanks
Bill

That might be a definition of a classical point-particle, but since you are referring to QFT and position is not an observable there, that cannot be a difference between electrons and photons in QFT.
All it's happenning here is that photons are not easily(if at all, but some people have tried) describable in terms of NRQM wave functions, something not particularly surprising.

 

[bhobba]

That might be a definition of a classical point-particle, but since you are referring to QFT and position is not an observable there, that cannot be a difference between electrons and photons in QFT.

Come again.

I gave a link that showed position is an observable for electrons but not for photons. Obviously that is the difference.

Nugatory gave a hand-wavey argument - but that's all it is.

Thanks
Bill

 

[TrickyDicky]

Come again.

I gave a link that showed position is an observable for electrons but not for photons. Obviously that is the difference.

You are not even making the distinction between NRQM and QFT. There are not photons in NRQM, EM interaction is treated semiclassically there.

 

[bhobba]

You are not even making the distinction between NRQM and QFT. There are not photons in NRQM, EM interaction is treated semiclassically there.

Exactly how does that change anything?

Thanks
Bill

 

[TrickyDicky]

Exactly how does that change anything?

You need to specify in which setting you frame your assertions, you seem to be switching from non-relativistic QM to QFT as it fits you. Now in nrqm there are no photons to begin with in a stric sense for the mathematical treatment, so the problem doesn't arise in the terms you refer to. It so happens that phenomenologically, in terms of interference in the double slit experiment doesn't make much difference to do the experiment with photons or electrons at a first approximation(Davisson-Germer and Young experiments equivalence) .

If your assertions where framed in QFT terms(as you hinted at several times by alluding to QFT texts) they don't make much sense, since position is not an observable regardless if one talks about electrons or photons. Now can yo decide if you are referring to the nrqm or the QFT setting when speaking about position?

 

[bhobba]

You need to specify in which setting you frame your assertions, you seem to be switching from non-relativistic QM to QFT as it fits you

The link I gave was crystal clear - QFT.

But since position is not an observable for photons an ordinary wave-function does not exist.

Thanks
Bill

 

[TrickyDicky]

The link I gave was crystal clear - QFT.

But since position is not an observable for photons an ordinary wave-function does not exist.

Thanks
Bill

Nevermind, I simply figured you were aware that position was not an observable for electrons or any other particle in QFT either.

 

[bhobba]

Nevermind, I simply figured you were aware that position was not an observable for electrons or any other particle in QFT.

Did you read the link I gave?

I know its quite advanced but the conclusion was clear - for massive particles:
'Since the physical irreducible representations of the Poincare group are uniquely determined by mass and spin, we see that in the massive case, a position operator must always exist'

For photons:
'Therefore, the concept of a photon position is necessarily subjective, since it depends on the POVM used, hence on the way the measurement is performed. It does not describe something objective.'

Its not a bog deal. Most don't worry about it in the double slit experiment. Its simply, like I said, we have some advanced posters on this forum so I like to be exact.

Thanks
Bill

 

[TrickyDicky]

Did you read the link I gave?

I just did, it is obvious it is not exactly about QFT but about relativistic QM of particles. In the RQM formalism you have position eigenstates, in QFT position is a label on operators.

Its not a bog deal. Most don't worry about it in the double slit experiment. Its simply, like I said, we have some advanced posters on this forum so I like to be exact.

That is commendable, and while I agree it is no big deal in the context of this thread, it is advisable to get it right when choosing a reference, I recommend pages 25-6 of Srednicki.

 

[Howard]

Not an answer, just some basics. An electron is a field permeating space. We perceive it as a particle or wave. It's best not to think of fundamental entities as particles or waves, but as fields. Vibration in a field is perceived as a wave or particle depending on the experimental conditions. Fields react to create other fields or properties, eg, the Higgs field reacts with the electron field to give the electron its mass.
Regards, Howard.