B Does the Initial Emission in the Double Slit Experiment Have a Trajectory?

[bluecap]

I have 2 questions

1. I'm telling a friend that in the double slit experiment.. the electron has no trajectories between the emission and detection. But she commented the initial emission has trajectory. Is this correct? What is the term for this situation of the initial emission of the electron?

2. In the detection, how do you change the observable from position to let's say energy. Or what kind of detector do you have to put so it won't measure the position of the particle but some other observable (and what are they?)

Thank you.

 

[Simon Bridge]

I have 2 questions

1. I'm telling a friend that in the double slit experiment.. the electron has no trajectories between the emission and detection. But she commented the initial emission has trajectory. Is this correct? What is the term for this situation of the initial emission of the electron?

Not in the sense that the source emits an electron with a predetermined path to the slits. The whole setup is arranged so there is no way to tell what path the electron took to get from the source to the detector. Now: epistomologically this may be considered to be the same thing as saying there is no such path ... but that is not what is actually being said here.

The initial situation is described by the statistical geometry ... ie. it could be a (hemi-)spherical source: so the electron is equally likely to be emitted into any solid angle. We can, in principle, set up any profile we like. We could set up a profile so the source let's out a very narrow beam of electrons so tightly controlled that all of them will pass through a specific slit ... in which case, no diffraction pattern. There is also the issue that we cannot know exactly when the electron was emitted either... or the pattern vanishes.

2. In the detection, how do you change the observable from position to let's say energy. Or what kind of detector do you have to put so it won't measure the position of the particle but some other observable (and what are they?)

You use an "energy detector". note: energy and position can be measured simultaneously. You can work out the basics just by thinking of the electrons as bullets and the detectors are sand-bags.
Adjust for the understanding that electrons are charged.
How things get measured is a very important question so I want to answer in more detail if needed: what level do you need the answer at?

My goto guy for this stuff is Richard Feynman:
http://www.vega.org.uk/video/subseries/8
... it's 8 lectures aimed at a lay audience, see all 8 though.

 

[bluecap]

Not in the sense that the source emits an electron with a predetermined path to the slits. The whole setup is arranged so there is no way to tell what path the electron took to get from the source to the detector.
The initial situation is described by the statistical geometry ... ie. it could be a (hemi-)spherical source: so the electron is equally likely to be emitted into any solid angle. We can, in principle, set up any profile we like. We could set up a profile so the source let's out a very narrow beam of electrons so tightly controlled that all of them will pass through a specific slit ... in which case, no diffraction pattern. There is also the issue that we cannot know exactly when the electron was emitted either... or the pattern vanishes.

Usually what is the actual emission device.. is the electron in the atom pumped out and ejected outside?

You use an "energy detector". note: energy and position can be measured simultaneously. You can work out the basics just by thinking of the electrons as bullets and the detectors are sand-bags.
Adjust for the understanding that electrons are charged.
How things get measured is a very important question so I want to answer in more detail if needed: what level do you need the answer at?

What is the actual device that can detect only the energy? Does it mean no position of the electron could be detected? Actual experiments I need.. intermediate explanations.. Thank you!

My goto for this stuff is Richard Feynman:
http://www.vega.org.uk/video/subseries/8
... it's 8 lectures aimed at a lay audience, see all 8 though.[/QUOTE]

 

[Simon Bridge]

Usually what is the actual emission device.. is the electron in the atom pumped out and ejected outside?

Anything that fires electrons can be used as a source - ie a slug of Strontium 90. Traditionally a large van der Graaf generator is the source, and the beam is shaped using electromagnetic lenses and fired down an evacuated steel pipe.

 

[Simon Bridge]

What is the actual device that can detect only the energy? Does it mean no position of the electron could be detected? Actual experiments I need.. intermediate explanations.. Thank you!

... most any detector will do it. We use semiconductors (most often now) and crystals most of the time, though Geiger tubes are still in use.

Technically, the semiconductor (or whatever) is manufactured so that a charged particle passing through it's layers will trigger a flow of current (think - photodiode, but for charges rather than photons). What this means is that none of these things detect the position of the particle directly ... all you know is that the particle hit someplace on the material surface.

The detector itself is an earthed metal box with a hole in it - the semiconductor is mounted just behind the hole. It follows that only particles that pass through that hole will trigger the detector - which is how we have position information. If we just made a semiconductor sphere and put it around the source, we would not be able to tell the angular position of particles detected (we would know how far away the particle was when it was detected though). Since that is very expensive, we cannot help measuring position and energy at the same time. But you will notice that we do not measure exact positions. All this setup tells us is that some charged particle passed through the detector aperture somewhere.

 

[bluecap]

... most any detector will do it. We use semiconductors (most often now) and crystals most of the time, though Geiger tubes are still in use.

Technically, the semiconductor (or whatever) is manufactured so that a charged particle passing through it's layers will trigger a flow of current (think - photodiode, but for charges rather than photons). What this means is that none of these things detect the position of the particle directly ... all you know is that the particle hit someplace on the material surface.

But this information that the particle hit someplace on the material surface means there is still position.. but is it not when you chose the energy observable. There is literally no longer any position at all so the particle shouldn't even hit someplace on the material surface?

The detector itself is an earthed metal box with a hole in it - the semiconductor is mounted just behind the hole. It follows that only particles that pass through that hole will trigger the detector - which is how we have position information. If we just made a semiconductor sphere and put it around the source, we would not be able to tell the angular position of particles detected (we would know how far away the particle was when it was detected though). Since that is very expensive, we cannot help measuring position and energy at the same time. But you will notice that we do not measure exact positions. All this setup tells us is that some charged particle passed through the detector aperture somewhere.

"know how far away the particle" means there is still position. But there should 100% be no position at all.. is it not.
do you know of a site that has illustration of this pure energy detector.. why can't I find it in youtube.

 

[Simon Bridge]

But this information that the particle hit someplace on the material surface means there is still position.. but is it not when you chose the energy observable. There is literally no longer any position at all so the particle shouldn't even hit someplace on the material surface?

You are not making sense.
You do not know where I am, correct? But you agree that I do have a position, right?

Not knowing what the position is, is not the same as not having a position.

"know how far away the particle" means there is still position.

Yes - there is some information about position available to the experimenter - however vague.

But there should 100% be no position at all.. is it not.

Nobody is saying that you can have zero information about position of a particle that has been detected. Just the opposite - you cannot help but find some position information. For instance: once the particle has been detected, then we know it must be inside the observable Universe ... that is as vague as position information can get.

We can do that in theory - the wavefunction of a free particle is flat. We'd verify this by noticing it is equally likely to be detected anywhere and finding the probability of detceting one in as many places as we can think of. However there is no such thing as an absolutely free particle.

do you know of a site that has illustration of this pure energy detector.. why can't I find it in youtube.

... what you are looking for does not exist. Why would you suspect such a device could exist?

What has this got to do with quantum interference at slits?
Where are you going with this?

 

[bluecap]

You are not making sense.
You do not know where I am, correct? But you agree that I do have a position, right?

Not knowing what the position is, is not the same as not having a position.
Yes - there is some information about position available to the experimenter - however vague. Nobody is saying that you can have zero information about position of a particle that has been detected. Just the opposite - you cannot help but find some position information. For instance: once the particle has been detected, then we know it must be inside the observable Universe ... that is as vague as position information can get.

We can do that in theory - the wavefunction of a free particle is flat. We'd verify this by noticing it is equally likely to be detected anywhere and finding the probability of detceting one in as many places as we can think of. However there is no such thing as an absolutely free particle.

... what you are looking for does not exist. Why would you suspect such a device could exist?

What has this got to do with quantum interference at slits?
Where are you going with this?

Where I am going with is this. A particle when not detected has no intrinsic position or energy or spin, they are state vector or the wave function is unitary (or a ray in Hilbert space). I just want to be clear I understood this so I expected you to say the particle has no intrinsic position when you choose the energy observable. So you are saying it has vague position but is the vague position really intrinsic in the particle meaning it has really vague position and not just due to our apparatus focusing on energy observable and not position?

About quantum interference at slits. The detector either detects left or right hits. So if position is not an intrinsic property of particles. I'm imaging when the particle is emitted in the double slit experiment.. it can end up invisible. Do you know of other experimental setup where a particle exists first with position then it just vanishes into thin air?

 

[Simon Bridge]

Where I am going with is this. A particle when not detected has no intrinsic position or energy or spin, they are state vector or the wave function is unitary (or a ray in Hilbert space).

No. Before measurement, we cannot say anything about any of it's state variables.
That does not mean that it does not have a specific value for that variable. Just like you do not have not measured my position, body temperature, shoe size, whatever; but I do have a position, body temp, and shoe size.
Not knowing anything about something is just lack of knowledge.

It may be that I don't have feet - then the concept of "shoe size" does not apply to me... but you still do not know.
You seem to be saying that QM being silent about something means that the something does not exist. That is sloppy thinking.

I just want to be clear I understood this so I expected you to say the particle has no intrinsic position when you choose the energy observable. So you are saying it has vague position but is the vague position really intrinsic in the particle meaning it has really vague position and not just due to our apparatus focusing on energy observable and not position?

You were asking about what happens in real physical experiments, not what can be described in the mathematics. I replied accordingly. If you want to ask about idealized situations, sure - start a new thread.
What I am telling you is that knowing only that position has not been measured, we cannot say much about the position.
Not knowing the position of something is not the same as something not having a position, and it does not mean we can say nothing at all about the position.

About quantum interference at slits. The detector either detects left or right hits.

That does not sound like a standard setup. I don't know what you mean by "The detector either detects left or right hits.". A detector usually detects [a particle] or it does not. In the source+slits setup, you can put the detector anywhere you like. If you put it completely blocking the left slit, then it detects particles at the left slit or it does not. It tells you nothing about particles it does not detect.

So if position is not an intrinsic property of particles.

I don't know what you mean by an intrinsic position.
Maybe we are just using the words to mean different things. (See: "intrinsic spin" for example.)

I'm imaging when the particle is emitted in the double slit experiment.. it can end up invisible. Do you know of other experimental setup where a particle exists first with position then it just vanishes into thin air?

What you just described is nonsense. Quantum mechanics makes no such claims.
You are trying to make the theory say something it does not.
The QM description just says that we do not know how the particle got from emitter to detector. In fact, we can show the particle can be detected anywhere within the experiment by changing the detector position. But what path it took - QM is silent on that question.

Logically, and by experimental confirmation, any particle that makes it to the other side of the slits must have passed through one or the other of the slits.
What we find is that if we do any experiment where we can determine which slit an individual particle passes through, we must also disturb the setup in such a way that the diffraction pattern does not happen.

 

[PeterDonis]

Before measurement, we cannot say anything about any of it's state variables.
That does not mean that it does not have a specific value for that variable.

It also doesn't mean that it does. It depends on the state and the variable. If the state is an eigenstate of the operator associated with a particular variable ("observable" would be a better term), then it does have a specific value for that variable--more precisely, a measurement of that variable will yield a specific, definite value with certainty. If the state is not an eigenstate of the operator, however, then it doesn't; you can only make predictions about the probability of various values being measured.

So you are saying it has vague position but is the vague position really intrinsic in the particle meaning it has really vague position and not just due to our apparatus focusing on energy observable and not position?

If the particle is not in a position eigenstate (which it won't be given the apparatus you are describing), then not having a specific, definite value for position is intrinsic to the particle. But "vague position" is not a good way of describing this. See my description above.

 

[Simon Bridge]

It also doesn't mean that it does [have a value]. It depends on the state and the variable. If the state is an eigenstate of the operator associated with a particular variable ("observable" would be a better term), then it does have a specific value for that variable--more precisely, a measurement of that variable will yield a specific, definite value with certainty. If the state is not an eigenstate of the operator, however, then it doesn't; you can only make predictions about the probability of various values being measured.

I agree.

I think we can demonstrate that a particle has a position by measuring it - position exists, the particular value is a result of the measurement and what is being measured. Before the measurement, we don't know the position ... but we can know about the probability of the outcome of possible measurements. Do we have an example of a particle that exists and does not have a position?

I am being a bit pedantic about this because of the protest earlier that finding the particle on the dome detector (post #2/3?) still tells me something about it's position, and OP wanted to know a way to measure energy alone - with no knowledge of position. IRL this cannot be done - though I guess we could measure the energy of a particle confined to a well if we make the "detector aperture" from my earlier description bigger than the well. But even then, the resulting energy eigenstate would have a different position distribution... which would result in knowing more about position. I don't think the conditions for a measurement of energy not leading to some information about position is possible or indicated in the theory...

 

[Zafa Pi]

Not in the sense that the source emits an electron with a predetermined path to the slits. The whole setup is arranged so there is no way to tell what path the electron took to get from the source to the detector. Now: epistomologically this may be considered to be the same thing as saying there is no such path ... but that is not what is actually being said here.

Is there not an interpretations that says there is a path, but we don't know what it is, and an other that says there is no path? And, indeed they may be considered "epistomologically" the same.

There is also the issue that we cannot know exactly when the electron was emitted either... or the pattern vanishes.

Is this true? I thought that the time and position operators commute.

 

[bluecap]

No. Before measurement, we cannot say anything about any of it's state variables.
That does not mean that it does not have a specific value for that variable. Just like you do not have not measured my position, body temperature, shoe size, whatever; but I do have a position, body temp, and shoe size.
Not knowing anything about something is just lack of knowledge.

It may be that I don't have feet - then the concept of "shoe size" does not apply to me... but you still do not know.
You seem to be saying that QM being silent about something means that the something does not exist. That is sloppy thinking.

But without the Born Rule. Everything is just wave function. Lets' say born rule is overridden and suspended in an object. Then it won't collapse to any position or energy or spin.. so it would just literally vanish. A quantum object has no intrinsic position or energy or spin.. you make them appear by using an operator and Born rule. But if Born rule won't be deployed.. the properties would not even exist in our world or measurable. Come to think of it. Why do you deny this since this is what is taught in QM.

You were asking about what happens in real physical experiments, not what can be described in the mathematics. I replied accordingly. If you want to ask about idealized situations, sure - start a new thread.
What I am telling you is that knowing only that position has not been measured, we cannot say much about the position.
Not knowing the position of something is not the same as something not having a position, and it does not mean we can say nothing at all about the position.

That does not sound like a standard setup. I don't know what you mean by "The detector either detects left or right hits.". A detector usually detects [a particle] or it does not. In the source+slits setup, you can put the detector anywhere you like. If you put it completely blocking the left slit, then it detects particles at the left slit or it does not. It tells you nothing about particles it does not detect.

I don't know what you mean by an intrinsic position.
Maybe we are just using the words to mean different things. (See: "intrinsic spin" for example.)

What you just described is nonsense. Quantum mechanics makes no such claims.
You are trying to make the theory say something it does not.
The QM description just says that we do not know how the particle got from emitter to detector. In fact, we can show the particle can be detected anywhere within the experiment by changing the detector position. But what path it took - QM is silent on that question.

Logically, and by experimental confirmation, any particle that makes it to the other side of the slits must have passed through one or the other of the slits.
What we find is that if we do any experiment where we can determine which slit an individual particle passes through, we must also disturb the setup in such a way that the diffraction pattern does not happen.

 

[Zafa Pi]

I think we can demonstrate that a particle has a position by measuring it - position exists, the particular value is a result of the measurement and what is being measured. Before the measurement, we don't know the position ... but we can know about the probability of the outcome of possible measurements. Do we have an example of a particle that exists and does not have a position?

Are you saying that if we don't measure the position it still must exist (has some value x), we just don't know it? It seems that is the very logic, counterfactual definiteness, that leads to the Bell inequality, which we know is false.

 

[Simon Bridge]

Is there not an interpretations that says there is a path, but we don't know what it is, and an other that says there is no path? And, indeed they may be considered "epistomologically" the same.

Sure - there are interpretations that say all kinds of things. There's some that say the electron travels all possible paths at the same time. I think there is an insights article about it someplace...
What seems to have been suggested (bottom post #8) is that the particle ceases to exist when we are not looking at it.

Is this true? I thought that the time and position operators commute.

The commutation is irrelevant - I am talking about the conditions for a clear diffraction experiment.

The source has a low uncertainty on the energy of the electrons - so the uncertainty in the time they are emitted is large.
If we want to say precisely when they were emitted, then we have to accept a large uncertainty in the energy - and thus a wide range of overlapping diffraction patterns possibly washing the effect out. It's like giving the slits fuzzy edges.

ie. For slit separation $d\approx hc/E$, the interference maxima occur at $d\sin\theta_n = n(hc/E)$ right?
For $E\to E\pm\Delta E$, and propagate the uncertainties.

Are you saying that if we don't measure the position it still must exist (has some value x), we just don't know it?

No. I am saying that position exists without being measured, and our lack of knowledge of position does not mean the particle does not have a position.

It seems that is the very logic, counterfactual definiteness, that leads to the Bell inequality, which we know is false.

... wait, are you saying that the Bell inequality is known to be false?

 

[Simon Bridge]

But without the Born Rule. Everything is just wave function.

Nope. Everything is itself. You are confusing the map for the territory.

Lets' say born rule is overridden and suspended in an object. Then it won't collapse to any position or energy or spin.. so it would just literally vanish.

No - the mathematics we use to describe what can be known about it would no longer function in a useful way.
The thing itself would still behave the way it always does.
What has this got to do with interference at slits?

A quantum object has no intrinsic position or energy or spin..

What do you mean by "intrinsic" in this sense? Do you think that energy or spin do not exist unless they are measured? Have you not heard of the "intrinsic spin" of an electron?

you make them appear by using an operator and Born rule. But if Born rule won't be deployed.. the properties would not even exist in our world or measurable.

No - the model will just cease to function as a useful theory. It will stop being predictive.

Come to think of it. Why do you deny this [what?? be specific!] since this is what is taught in QM.

What you are saying does not make sense.

I expected you to say the particle has no intrinsic position when you choose the energy observable.

... but I did not say that so you appear to have gone into a tailspin. Is it your position that a article has no "intrinsic position" - please define that concept and present evidence to support it.
You were the one asking questions - you have received answers that you do not agree with... sorry about that.
Have you read @PeterDonis reply post #10?

Please answer the questions from my previous posts. If you will not answer questions or define your terms I cannot help you.

 

[bluecap]

Nope. Everything is itself. You are confusing the map for the territory.

No - the mathematics we use to describe what can be known about it would no longer function in a useful way.
The thing itself would still behave the way it always does.
What has this got to do with interference at slits?

What do you mean by "intrinsic" in this sense? Do you think that energy or spin do not exist unless they are measured? Have you not heard of the "intrinsic spin" of an electron?
No - the model will just cease to function as a useful theory. It will stop being predictive.What you are saying does not make sense.

... but I did not say that so you appear to have gone into a tailspin. Is it your position that a article has no "intrinsic position" - please define that concept and present evidence to support it.
You were the one asking questions - you have received answers that you do not agree with... sorry about that.
Have you read @PeterDonis reply post #10?

Please answer the questions from my previous posts. If you will not answer questions or define your terms I cannot help you.

By intrinsic. I mean existing even without measurement. But you really believe a particle has position even when not measured. Just wondering. What is your specialty in physics? Does it mean different physicists have different ideas? Is it not part of the mathematical formalism that when a state is not an eigenstate of the operator associated with a particular observable (as Peterdonis put it), it doesn't have the properties? Or is the properties like position being there or not there when unmeasured a part of interpretations that different physicists have different way of understanding it?

But then Simon point is that even when not measured, a particle has spin or energy for instance. Without this occurring, molecules would just collapse and objects cease to exist. This seems acceptable logic on his part.

However Peterdonis wrote that "It depends on the state and the variable. If the state is an eigenstate of the operator associated with a particular variable ("observable" would be a better term), then it does have a specific value for that variable--more precisely, a measurement of that variable will yield a specific, definite value with certainty. If the state is not an eigenstate of the operator, however, then it doesn't; you can only make predictions about the probability of various values being measured."

Peterdonis, is this true for all observables? For example.. position may not exist without the state being in an eigenstate of position.. but spin should definitely exist even without the state being in an eigenstate of spin or else atoms would cease to function. How do you resolve this?

 

[Simon Bridge]

I think we are getting bogged down in ontology... this can get very involved, see, for eg:
By intrinsic. I mean existing even without measurement.

By intrinsic. I mean existing even without measurement. But you really believe a particle has position even when not measured.

Show me a particle whose position has not been measured and I can answer the question.
What I have said repeatedly is that the lack of knowledge of the position of a particle is not the same as saying that the particle has no position.
The particle exists even when we are not looking at it.

In QM our possible knowledge of the position of a particle is given by the state vector.

Just wondering. What is your speciality in physics?

Quantum Mechanics :) Meaning nothing of course.

Does it mean different physicists have different ideas?

Of course different physicists have different ideas ... otherwise science would not work.

Is it not part of the mathematical formalism that when a state is not an eigenstate of the operator associated with a particular observable (as Peterdonis put it), it doesn't have the properties?

No.

Or is the properties like position being there or not there when unmeasured a part of interpretations that different physicists have different way of understanding it?

False dichotomy. Why can't it be neither?
How would you determine if a particle has energy?
Does the act of measuring the energy give the particle that energy or did the particle have energy all along?
I am maintaining that the particle had energy all along and the act of measuring determined that energy even if it was prepared in a superposition of energy eigenstates to begin with.

The quantum mechanics tells us the statistics of what is real - as for anything else, the theory is silent.
Cause and effect is tricky.

However Peterdonis wrote that "It depends on the state and the variable. If the state is an eigenstate of the operator associated with a particular variable ("observable" would be a better term), then it does have a specific value for that variable--more precisely, a measurement of that variable will yield a specific, definite value with certainty. If the state is not an eigenstate of the operator, however, then it doesn't; you can only make predictions about the probability of various values being measured."

... and I agreed with that statement.

If the particle state vector is described by a superposition of eigenstates for an observable, then the particle does not have a determined value for that observable. That is not the same as saying the particle does not exist for the purposes of that observable... which appears to be your assertion. If that is your idea then of course you think a particle in a superposition of position eigenstates must "vanish" - but all particles are in position superpositions all the time, even after position has been measured, are you saying all particles vanish? (probability of finding a particle in an eigenstate of position is zero.)

Remember how we get here - you asked about detectors for energy sans position ... when I gave an example where angular position was completely uncertain, you objected that radial position was measured to some uncertainty. Now you seem to be wanting to restrict position to an absolute value.

Where were you trying to go with that? What does this have to do with diffraction at slits?
What does this have to do with your original question about talking to someone ...

 

[bluecap]

I think we are getting bogged down in ontology... this can get very involved, see, for eg:

By intrinsic. I mean existing even without measurement.

Show me a particle whose position has not been measured and I can answer the question.

What I have said repeatedly is that the lack of knowledge of the position of a particle is not the same as saying that the particle has no position.

The particle exists even when we are not looking at it.

This is puzzling how a person that specializes in quantum mechanics could still diverged from the mainstream views. Unless you are talking about Bohmian trajectories? Because in conventional QM, the particle has no position when unmeasured although it has spin and energy. Only the "position" is unreal when unmeasured. I hope other experts here can share this to you and resolve all this. I read that many Ph.Ds in physics are not familiar with Bell's Inequity. It shows that a particle has no properties displayed when unmeasured.. because if it has.. you can simultaneously measured it's position and momentum for two entangled particles that is sent away. So it has really no position when unmeasured.

In QM our possible knowledge of the position of a particle is given by the state vector.

Quantum Mechanics :) Meaning nothing of course.

Of course different physicists have different ideas ... otherwise science would not work.

No.

False dichotomy. Why can't it be neither?

How would you determine if a particle has energy?

Does the act of measuring the energy give the particle that energy or did the particle have energy all along?

I am maintaining that the particle had energy all along and the act of measuring determined that energy even if it was prepared in a superposition of energy eigenstates to begin with.

The quantum mechanics tells us the statistics of what is real - as for anything else, the theory is silent.

Cause and effect is tricky.

... and I agreed with that statement.

If the particle state vector is described by a superposition of eigenstates for an observable, then the particle does not have a determined value for that observable. That is not the same as saying the particle does not exist for the purposes of that observable... which appears to be your assertion. If that is your idea then of course you think a particle in a superposition of position eigenstates must "vanish" - but all particles are in position superpositions all the time, even after position has been measured, are you saying all particles vanish? (probability of finding a particle in an eigenstate of position is zero.)

Remember how we get here - you asked about detectors for energy sans position ... when I gave an example where angular position was completely uncertain, you objected that radial position was measured to some uncertainty. Now you seem to be wanting to restrict position to an absolute value.

Where were you trying to go with that? What does this have to do with diffraction at slits?

What does this have to do with your original question about talking to someone ...

Double slit is the most convenient to imagine the state vector and its evolution. I'm sharing QM to someone about this article https://www.wired.com/2014/06/the-new-quantum-reality/ which starts:

"For nearly a century, "reality" has been a murky concept. The laws of quantum physics seem to suggest that particles spend much of their time in a ghostly state, lacking even basic properties such as a definite location and instead existing everywhere and nowhere at once. Only when a particle is measured does it suddenly materialize, appearing to pick its position as if by a roll of the dice.

This idea that nature is inherently probabilistic — that particles have no hard properties, only likelihoods, until they are observed — is directly implied by the standard equations of quantum mechanics."

The article was written by Natalie Wolchover whose serious articles are permitted in physicsforums.

 

[Nugatory]

This thread seems to have drifted away from the oroiginal questions for a while now...

I have 2 questions
1. I'm telling a friend that in the double slit experiment.. the electron has no trajectories between the emission and detection. But she commented the initial emission has trajectory. Is this correct? What is the term for this situation of the initial emission of the electron?

At the moment of emission, the electron is very localized. There is a high probability that we we will find it very near a particular point in space, namely the point where the emitter is located. However, as time passes the wave function evolves as described by Schrodinger's equation; it spreads out and the probability of finding at any particular location becomes smaller. Eventually, however, the electron interacts with something (a detector, a droplet of vapor in a cloud chamber, a random dust particle, ...) and then the wave function collapses. Immediately after that interaction, the electron is highly localized again.

So there's an initial position (the interaction with the emitter leaves the electron is a localized state) and there's a final position (the interaction with the detector leaves the electron in a localized state) but it doesn't follow that there is a trajectory in between. We often refer to the first interaction as "preparation" and the second as "measurement".

2. In the detection, how do you change the observable from position to let's say energy. Or what kind of detector do you have to put so it won't measure the position of the particle but some other observable (and what are they?)

You measure the energy of a charged particle by passing it through a magnetic field or between the plates of a parallel plate capacitor before you detect it.

 

[vanhees71]

This thread seems to have drifted away from the oroiginal questions for a while now...At the moment of emission, the electron is very localized. There is a high probability that we we will find it very near a particular point in space, namely the point where the emitter is located. However, as time passes the wave function evolves as described by Schrodinger's equation; it spreads out and the probability of finding at any particular location becomes smaller. Eventually, however, the electron interacts with something (a detector, a droplet of vapor in a cloud chamber, a random dust particle, ...) and then the wave function collapses. Immediately after that interaction, the electron is highly localized again.

That's not correct. In order to have interference you need a state with pretty well determined momentum, which means the electron should not be too well localized. In transverse direction the wave packet of the Schrödinger wave should be broad enough such that it covers both slits with not too much variation in intensity. Otherwise you don't see the quantum interference effect the double-slit experiment is supposed to demonstrate. In other words, the position of the electron must be sufficiently indetermined at the position of the slits such that it is not possible to distinguish through which slit it's gone in order to get interference fringes at observation on the screen. The interference is of the partial amplitudes for going through either slit. That's most intuitive in the description in terms of the Feynman path integral, and it's described in great detail in the famous book by Feynman and Hibbs.

At detection on the screen the electron interacts however locally with the detector material and thus leaves a well-localized track on the screen. You don't need to envoke the highly problematic idea of a collapse here.

 

[PeterDonis]

I think we can demonstrate that a particle has a position by measuring it

We can demonstrate that the particle is in a position eigenstate immediately after measurement. (But it won't stay in that state, since such a state is not an eigenstate of the Hamiltonian and so will not remain the same under time evolution.) But that does not show that the particle was in a position eigenstate before the measurement. Measurement can change the particle's state.

Do we have an example of a particle that exists and does not have a position?

Sure, any particle that is not in a position eigenstate.

even then, the resulting energy eigenstate would have a different position distribution... which would result in knowing more about position.

Having a different position distribution is not the same as "knowing more" about position. You are still talking about an electron (or other quantum particle) as though it were a classical object. It isn't.

You asked the OP earlier if him not knowing your position means you don't have one. You are a classical object--more precisely, setting up an experiment in which quantum interference effects between you and something else would be practically impossible. So it works fine to say that you have a position even if nobody knows its exact value. (Even you might not know it if, for example, you were taken somewhere blindfolded.) But that does not work for quantum particles, because we can run practical experiments where interference effects are observed, and where the probabilities that arise when you square quantum amplitudes cannot be given a simple ignorance interpretation.

 

[PeterDonis]

I thought that the time and position operators commute.

There is no time operator in QM. Time is a parameter, not an operator.

 

[PeterDonis]

This idea that nature is inherently probabilistic — that particles have no hard properties, only likelihoods, until they are observed — is directly implied by the standard equations of quantum mechanics.

First of all, saying that particles "have no hard properties" until they are observed, is not the same as saying that particles do not exist until they are observed.

Second, it's also not quite correct. If a particle is in an eigenstate of an observable, then it has a definite value for the property corresponding to that observable, even if it is not measured. At least, that's what "the standard equations of quantum mechanics" say.

Third, these "hard properties" are artifacts of our classical intuitions. According to "the standard equations of quantum mechanics", a quantum system always has a perfectly definite state. It's just that most of these states don't look like the kinds of states we're used to from classical physics. But that's a problem with our intuitions, not nature.

 

[PeterDonis]

in conventional QM, the particle has no position when unmeasured although it has spin and energy.

This is not correct. The particle can have a definition position when unmeasured if it is in a position eigenstate. The particle does not have a definite energy when unmeasured unless it is in an energy eigenstate. And although the magnitude of a particle's spin is definite when unmeasured (as long as we're only looking at single particles--it isn't definite when unmeasured if we are looking at multi-particle systems), its direction is not.

 

[zonde]

1. I'm telling a friend that in the double slit experiment.. the electron has no trajectories between the emission and detection. But she commented the initial emission has trajectory. Is this correct? What is the term for this situation of the initial emission of the electron?

Whether particle has a trajectory or not between emission and detection can be answered within particular interpretation. If you do not specify interpretation discussion will end nowhere.

2. In the detection, how do you change the observable from position to let's say energy. Or what kind of detector do you have to put so it won't measure the position of the particle but some other observable (and what are they?)

Detector always has classical position. There is no detector that won't measure position of particle to some extent.
You measure other observables by making a setup where particle (or scattered particles) ends up at different detectors based on that observable (see Nugatory's example in post #20).

 

[vanhees71]

This is not correct. The particle can have a definition position when unmeasured if it is in a position eigenstate. The particle does not have a definite energy when unmeasured unless it is in an energy eigenstate. And although the magnitude of a particle's spin is definite when unmeasured (as long as we're only looking at single particles--it isn't definite when unmeasured if we are looking at multi-particle systems), its direction is not.

There is no position eigenstate. That's most simple to see in the position representation. Then the "eigenvector" of position with the eigenvalue $\vec{x}_0$ would be represented by
$$u_{\vec{x}_0}(\vec{x})=\delta^{(3)}(\vec{x}-\vec{x}_0).$$
That's clearly not a square-integrable function and thus does not represent a state.

In fact, no value in the continuous part of the spectrum of a self-adjoint operator is a true eigenvalue and thus also has no true eigenvector. You have generalized eigenvectors, which are distributions. Formally they belong to a larger space than the Hilbert space, namely the dual space of the domain (and thus also the co-domain) of the self-consistent operator (keyword: "Rigged Hilbert Space", see Ballentine, Quantum Mechanics for a non-formal introduction).

Another way to see this is the uncertainty relation for position and momentum,
$$\Delta p_x \Delta x \geq \hbar/2.$$
This means that there exists no state for which either $x$ or $p_x$ are determined accurately. You can make either of the standard deviations of these quantities as small as you like, but you can never make it exactly 0.

 

[DrChinese]

A quantum object has no intrinsic position or energy or spin.. you make them appear by using an operator and Born rule. But if Born rule won't be deployed.. the properties would not even exist in our world or measurable. Come to think of it. Why do you deny this since this is what is taught in QM.

You might have read what Simon said earlier, which if read one way would lead to this confusion. Of course the Heisenberg Uncertainty Principle describes the limits on simultaneous knowledge of quantum properties, he is not denying that at all. Go back to the original question, as Nugatory says. Obtaining a sharp position when the particle is finally detected does not tell you anything about the trajectory it traverses. Ditto with obtaining sharp energy measurements.

The question in the double slit is always about why the interference arises (or not) in various scenarios. So I might ask you to rephrase/review your original question in those terms. You asked about whether there is an initial trajectory. The answer is that often the initial trajectory is well defined, but must be "blurred" prior to arriving at the double slit. Otherwise, there can be no interference. With light, for example, there may be a single slit prior to the double slit - this has the effect of making the light essentially coherent. It then had a definite trajectory upon emission, but that stops at some point if there is to be interference.

 

[PeterDonis]

There is no position eigenstate.

I am oversimplifying, yes. This is a "B" level thread.

 

[PeterDonis]

Detector always has classical position.

This is interpretation dependent.

 

[zonde]

This is interpretation dependent.

Not quite. Any interpretation has to reproduce our macroscopic observations.

 

[Zafa Pi]

That's not correct. In order to have interference you need a state with pretty well determined momentum, which means the electron should not be too well localized. In transverse direction the wave packet of the Schrödinger wave should be broad enough such that it covers both slits with not too much variation in intensity. Otherwise you don't see the quantum interference effect the double-slit experiment is supposed to demonstrate. In other words, the position of the electron must be sufficiently indetermined at the position of the slits such that it is not possible to distinguish through which slit it's gone in order to get interference fringes at observation on the screen. The interference is of the partial amplitudes for going through either slit. That's most intuitive in the description in terms of the Feynman path integral, and it's described in great detail in the famous book by Feynman and Hibbs.

At detection on the screen the electron interacts however locally with the detector material and thus leaves a well-localized track on the screen. You don't need to envoke the highly problematic idea of a collapse here.

After "That is not correct." What you say seems correct and informative. However the "That is not correct." refers to Nugatory's statement:

"This thread seems to have drifted away from the oroiginal questions for a while now...At the moment of emission, the electron is very localized. There is a high probability that we we will find it very near a particular point in space, namely the point where the emitter is located. However, as time passes the wave function evolves as described by Schrodinger's equation; it spreads out and the probability of finding at any particular location becomes smaller. Eventually, however, the electron interacts with something (a detector, a droplet of vapor in a cloud chamber, a random dust particle, ...) and then the wave function collapses. Immediately after that interaction, the electron is highly localized again."

What is incorrect about it?

 

[bluecap]

This is not correct. The particle can have a definition position when unmeasured if it is in a position eigenstate. The particle does not have a definite energy when unmeasured unless it is in an energy eigenstate. And although the magnitude of a particle's spin is definite when unmeasured (as long as we're only looking at single particles--it isn't definite when unmeasured if we are looking at multi-particle systems), its direction is not.

I was in general discussion with a friend about double slit. I told her there is no trajectory in between emission and detection. She said the emission and detection is both particle so she assumes there is trajectory. I want to tell her it is possible for the detection to be detecting purely the energy eigenstate without any position. Is it possible for the detector not to be in any position eigenstate (meaning no position) but only energy (or spin or other variable) eigenstate? Any experimental setup you can mention with your context (and not Simon's). Can you please give other examples where a particle is sent with position and received without any position so I can convince her that position is not a priori?

Simon earlier replies kept saying there was still position even if it is not in position eigenstate. I don't know what's going on with him, hope he can clarify it because I want to know just how many physicists into quantum mechanics still want to keep classical picture of quantum particle and thought there was really position even if not in any position eigenstate.

 

[PeterDonis]

Any interpretation has to reproduce our macroscopic observations.

Sure, but if you are claiming that "detectors always have classical positions" is a macroscopic observation, then you're going to end up with a definition of "having a classical position" that is perfectly consistent with the detector being a quantum mechanical system in which none of the individual quantum constituents (atoms, electrons, whatever) have classical positions, and in which the detector registering a position measurement does not mean the measured quantum object has a classical position. (And anyway, that would be taking this topic well beyond what can be discussed in a "B" level thread.)

 

[PeterDonis]

I told her there is no trajectory in between emission and detection. She said the emission and detection is both particle so she assumes there is trajectory.

And this is precisely what you cannot "assume". If we were talking about classical objects, you could; but we aren't, we're talking about quantum objects. And with quantum objects you cannot assume classical trajectories; you have to devise an experimental setup that actually measures a definite trajectory. And if you do that, you will find that the interference goes away.

Is it possible for the detector not to be in any position eigenstate (meaning no position) but only energy (or spin or other variable) eigenstate?

The eigenstates I have been talking about are states that the measured quantum object could be in. They are not states of the detector. In the usual simple analysis of an experiment like the double slit, the detector is not assigned a state at all; it's just a label for "whatever it is that produces the measurement result". Immediately after a measurement, the measured quantum object is in an eigenstate of the measured observable whose eigenvalue is the measurement result.

Can you please give other examples where a particle is sent with position and received without any position so I can convince her that position is not a priori?

It's not a matter of what is sent/received. It's what happens in between. In the double slit experiment, if there is no apparatus set up to measure which slit the particle goes through, then you cannot assume it has a definite trajectory during the experiment. At the end of the experiment, the particle will be detected at some position on the detector, so it has a definite position then. (Actually, even then it's not in a position eigenstate, since as vanhees71 pointed out there is no such thing; instead it's in a state with a very narrow spread of the wave function in the position basis. But that's getting beyond a "B" level discussion.) But you cannot assume that the particle took a particular trajectory through space to get to that ending position.

If we modify the double slit experiment to detect which slit the particle went through, then we can say it has a definite position as it passes through the slit, since we are measuring it. (Again, it's actually a state with a narrow spread in the position basis.) Since this is also sufficient to eliminate the interference pattern at the final detector, it is usually assumed that the particle takes a definite trajectory throughout the experiment in this case. But strictly speaking, we can't assume that, because we only measure the position at the slit; we don't measure it in between the source and the slit or in between the slit and the detector. It's just that whether or not we assume a definite trajectory during these unmeasured portions of the experiment makes no difference to our analysis of the results.

Simon earlier replies kept saying there was still position even if it is not in position eigenstate.

I can't speak for him, but I would interpret what he said as follows: position is always a well-defined observable, whether or not a particular particle is in an eigenstate of it. The same would be true for any other observable. It's just emphasizing the distinction between observables and states of the system. Observables are associated with measuring devices--detectors; which observable a given detector measures depends on how it is constructed. Position is a well-defined observable because we know how to construct detectors that measure it (and how to model those detectors mathematically).

It's also important to realize that ordinary language is a very poor tool to use when trying to understand this subject. The proper tool is math. Which also means it's very difficult to properly treat this subject at a "B" level. It really helps to take the time to get a solid background in the underlying math.

 

[bluecap]

And this is precisely what you cannot "assume". If we were talking about classical objects, you could; but we aren't, we're talking about quantum objects. And with quantum objects you cannot assume classical trajectories; you have to devise an experimental setup that actually measures a definite trajectory. And if you do that, you will find that the interference goes away.
The eigenstates I have been talking about are states that the measured quantum object could be in. They are not states of the detector. In the usual simple analysis of an experiment like the double slit, the detector is not assigned a state at all; it's just a label for "whatever it is that produces the measurement result". Immediately after a measurement, the measured quantum object is in an eigenstate of the measured observable whose eigenvalue is the measurement result.
It's not a matter of what is sent/received. It's what happens in between. In the double slit experiment, if there is no apparatus set up to measure which slit the particle goes through, then you cannot assume it has a definite trajectory during the experiment. At the end of the experiment, the particle will be detected at some position on the detector, so it has a definite position then. (Actually, even then it's not in a position eigenstate, since as vanhees71 pointed out there is no such thing; instead it's in a state with a very narrow spread of the wave function in the position basis. But that's getting beyond a "B" level discussion.) But you cannot assume that the particle took a particular trajectory through space to get to that ending position.

If we modify the double slit experiment to detect which slit the particle went through, then we can say it has a definite position as it passes through the slit, since we are measuring it. (Again, it's actually a state with a narrow spread in the position basis.) Since this is also sufficient to eliminate the interference pattern at the final detector, it is usually assumed that the particle takes a definite trajectory throughout the experiment in this case. But strictly speaking, we can't assume that, because we only measure the position at the slit; we don't measure it in between the source and the slit or in between the slit and the detector. It's just that whether or not we assume a definite trajectory during these unmeasured portions of the experiment makes no difference to our analysis of the results.

I know what happens in between where there is no position when it's not eigenstate of position, but for totally new laymen, the best way to convince them is if the emission starts with position and detection ends up with no position. Can you think of any present or proposed experiments where this can be demonstrated or done? Best is if the double slit detector can be put in state to measure the spin or energy of the particle instead of position as the double slit is the best example we have now. I want an example because I'll write an article about it for laymen in my school. Thanks.

I can't speak for him, but I would interpret what he said as follows: position is always a well-defined observable, whether or not a particular particle is in an eigenstate of it. The same would be true for any other observable. It's just emphasizing the distinction between observables and states of the system. Observables are associated with measuring devices--detectors; which observable a given detector measures depends on how it is constructed. Position is a well-defined observable because we know how to construct detectors that measure it (and how to model those detectors mathematically).

It's also important to realize that ordinary language is a very poor tool to use when trying to understand this subject. The proper tool is math. Which also means it's very difficult to properly treat this subject at a "B" level. It really helps to take the time to get a solid background in the underlying math.

 

[PeterDonis]

for totally new laymen, the best way to convince them is if the emission starts with position and detection ends up with no position

I'm not sure that is the best strategy, because it implies that an experiment where position is measured at both ends does mean there is a definite trajectory in between.

My personal opinion about QM pedagogy is that half measures only make it worse. It's better to attack classical intuitions, like the intuition that there must be a definite trajectory, right from the start. My experience is that any attempt to water it down just means it takes longer for people to fully understand how different quantum behavior is from what they're used to.

Best is if the double slit detector can be put in state to measure the spin or energy of the particle instead of position

This can't be done. The whole point of the double slit is the interference pattern on the detector; the intensity varies with position on the detector.

 

[Zafa Pi]

In Simon Bridge's post #15 we see the following exchange:
Zafa: Are you saying that if we don't measure the position it still must exist (has some value x), we just don't know it?
Simon: No. I am saying that position exists without being measured, and our lack of knowledge of position does not mean the particle does not have a position.

I don't understand why Simon said "No." It is clear to me we are saying the same thing, which both bluecap and I believe to be false.

We also have:
Zafa: It seems that is the very logic, counterfactual definiteness, that leads to the Bell inequality, which we know is false.
Simon: ... wait, are you saying that the Bell inequality is known to be false?

Bell's Theorem is valid and in it there is a proof of Bell's inequality. But the hypothesis of the theorem assumes the existence of (unknown) values of unmeasured entities, i.e. counterfactual definiteness (or realism, or hidden variables, or. determinism). QM experiments show the inequality is not valid, thus the hypothesis is not valid, contrary to your statement at the top.
I think this is what bluecap was referring to in post #19.

 

[Simon Bridge]

We can demonstrate that the particle is in a position eigenstate immediately after measurement. (But it won't stay in that state, since such a state is not an eigenstate of the Hamiltonian and so will not remain the same under time evolution.) But that does not show that the particle was in a position eigenstate before the measurement. Measurement can change the particle's state.

I agree - I think this may be a semantic thing.

Sure, any particle that is not in a position eigenstate.

Can we demonstrate that it does not have a position at all in the way OP is saying?
Can you show me a particle in a position eigenstate immediately after measurement?
iirc the probability of finding one should be zero right?

Remember I am responding to a claim here.

Initially it was that not knowing the position of a particle means it does not have a position.
Next that not having a position is the same as the particle vanishing.

When we say we have located a particle at a particular position - it is always in a superposition of position eigenstates about that position. That is how we describe it in maths right? In the example of diffraction at slits, we say we detect the particle at position of the detector ... what we are saying is that a detector with a particular aperture width detected the particle.

I am taking the position that it is sensible to talk about the position of a particle when it is not in a position eigenstate.

The spin-up particle may be in box A or box B ... we look and see it is in box A.
The question arises: was it always in box A before we looked?
The answer is: it depends ... in some situations the theory is silent on that answer.
I am making a distinction between that and saying that the particle did not exist until we looked.

Having a different position distribution is not the same as "knowing more" about position. You are still talking about an electron (or other quantum particle) as though it were a classical object. It isn't.

OK, perhaps I chose poor wording: we still have information about the position (OP wanted to have no information about the position). Our knowledge of the position is different from before - ie. it would be a stationary state. It may be that the new distribution is narrower than the previous one - in which case we have a better localized particle.

You asked the OP earlier if him not knowing your position means you don't have one. You are a classical object--more precisely, setting up an experiment in which quantum interference effects between you and something else would be practically impossible. So it works fine to say that you have a position even if nobody knows its exact value. (Even you might not know it if, for example, you were taken somewhere blindfolded.) But that does not work for quantum particles, because we can run practical experiments where interference effects are observed, and where the probabilities that arise when you square quantum amplitudes cannot be given a simple ignorance interpretation.

I agree there - but that is not what I was trying to say.

As a matter of semantics I can agree that we can say any particle that is not in a position eigenstate does not have a position ... isn't it more useful to say that the classical concept of absolute position does not apply to quantum particles, though there exist some situations where the concept of position gets used for these things in a sensible way and explain. Otherwise we are stuck in the "the particle does not have a position therefore it does not exist" that I was responding to... especially since no particle detected is in a position eigenstate, that would suggest that no particle detected exists. I don't think that is sensible... but maybe I misread?

In QM positions exist - the theory tells us the statistics, which is sometimes counter-intuitive as with the nonlocal stuff and the interference, but it does not tell us how those statistics arose. A QM position involves a probability distribution, it is a statistical thingy. I'm sure there is a way to express this clearly.

 

[PeterDonis]

I think this may be a semantic thing.

I think it's two problems: (1) we're trying to use ordinary language when we should be using math; (2) we're in a "B" level thread and this isn't really a "B" level issue.

 

[PeterDonis]

Thread closed for moderation.

Edit: This topic cannot be discussed further at the "B" level and the thread will remain closed.