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

[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 Schrödinger'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.