Delayed choice experiment article on science alert

In summary: It's not so much a matter of interpretation as what the words mean. A particle is an object of zero or negligible size, a wave is an extended, diffuse thing. It is meaningless to talk of a object which is both diffuse and point-like.
  • #1
DaveC426913
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I'm familiar with the delayed choice experiment, and I'm trying to suss out this new setup.

http://www.sciencealert.com/reality-doesn-t-exist-until-we-measure-it-quantum-experiment-confirms

Reading this article is not very beneficial if there's a confounding typo. Paragraph 10:

When this second grating was added, it led to constructive or destructive interference, which is what you'd expect if the atom had traveled both paths, like a wave would. But when the second grating was added, no interference was observed, as if the atom chose only one path.

Should that last phrase be "...when the second grating was removed..." ?
 
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  • #2
It seems to be more subtle.
as if the atom chose only one path.
If the 2nd grating was removed, then there should remain only one grating which means there is only one path and no choice. So what choice is the author talking about here?

P.S.
Of course if the surface with the gratings is finite, there are an infinite number of path. So we should say "paths that go through (at least!) one of the gratings".
 
  • #3
OK, so it's not just me. The article misses some steps.

I have an urge to rewrite it so it makes sense. (I'm not sure why they needed to explain how they got a single atom of He, the whole BEC thing is a distraction to the main point.)
 
  • #5
Shyan said:
It seems to be more subtle.

If the 2nd grating was removed, then there should remain only one grating which means there is only one path and no choice. So what choice is the author talking about here?

P.S.
Of course if the surface with the gratings is finite, there are an infinite number of path. So we should say "paths that go through (at least!) one of the gratings".

The second grating is after the first, not next to it. The first scatters the atom's wavefunction into many wavelets - many paths. Without a second grating that is the end of the story and (pick your interpretation here) we only see the result of a single path each time an atom goes through. Over a large number we see a single lobe diffraction pattern, broad but with no interference bands or whatever the experiment gives. With the second grating in place, some of the wavelets are recombined making a more complicated pattern - one with interference. Recombining wavelets means that many (at least two!) paths are involved: the first grating might scatter the atom to the left and the right simultneously and the second recombines them. Presumably the authors decide to switch the second grating into position at the last moment, so if they (some randomizing contraption) don't put it in place, the atom is deemed to have traveled as one narrow wavelet all along so it chose to be on just one path. If they do put it in place, it is deemed to have been several wavelets - choosing several paths at the same time. Again it made that choice before the second grating was decided.

It all seems a bit contrived now that you can buy a Schrodinger Cat on Ebay.
 
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  • #6
Can someone explain once again why it can't be a particle with wavelike properties? Does our common sense interpretation require that they are exclusive?
 
  • #7
DaveC426913 said:
Can someone explain once again why it can't be a particle with wavelike properties? Does our common sense interpretation require that they are exclusive?
It's not so much a matter of interpretation as what the words mean. A particle is an object of zero or negligible size, a wave is an extended, diffuse thing. It is meaningless to talk of a object which is both diffuse and point-like.

However, assuming we are still talking about delayed choice scenarios, there is a huge amount of misleading nonsense repeated about electrons being waves if one thing is done but particles if another. This is incorrect, in both cases the pattern is that of a diffracted wave. The "two path" case is a broad lobe with interference bands across it. The "one path" case is a broad lobe without the banding. The broad lobe is a wave-like phenomenon. There is no suggestion of particle-like behaviour in either case.

The wave pattern reflects quantum mechanics, not the intrinsic nature of the electron.
 
  • #8
DaveC426913 said:
Can someone explain once again why it can't be a particle with wavelike properties? Does our common sense interpretation require that they are exclusive?

Tale a look at the hydrogen atom:
http://www.ciul.ul.pt/~ananunes/QM/Laguerres&Hydrogenatom.pdf [Broken]

What's the wavelike properties of the 1S state?

Thanks
Bill
 
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  • #9
bhobba said:
Tale a look at the hydrogen atom:
http://www.ciul.ul.pt/~ananunes/QM/Laguerres&Hydrogenatom.pdf [Broken]
What's the wavelike properties of the 1S state?
Thanks
Bill
It's wavelike because you can regard it as a standing wave. However you have explicitly rejected this idea elsewhere! Which is odd because

1) It is a solution of the Schrodinger wave equation and therefore a wave function
2) Wikipedia says it is a standing wave so it must be

The 1S state is a stationary state which is depicted as a spherically symmetrical orbital. However an orbital is a probability distribution, not the wavefunction. As the Wikipedia article says:
... this means that the particle has a constant probability distribution for its position, its velocity, its spin, etc.The wavefunction itself is not stationary: It continually changes its overall complex phase factor, so as to form a standing wave

Do you fundamentally disagree with this?
 
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  • #10
Derek Potter said:
It's wavelike because you can regard it as a standing wave.

A 1S state is nothing like a standing wave:
http://en.wikipedia.org/wiki/Standing_wave

Added later - where are the nodes and anti-nodes in the 1S state?

Wikipedia's other article not withstanding.

It says 'Therefore a stationary state is a standing wave that oscillates with an overall complex phase factor, and its oscillation angular frequency is equal to its energy divided by \hbar.'

The 1S state does not oscillate - its stationary.

Here is what's going on in stationary states:
https://www.physicsforums.com/threads/stationary-state.277647/

Its separable from the time dependant part which has no physical significance.

Thanks
Bill
 
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  • #11
I'll take that as a "yes".
bhobba said:
A 1S state is nothing like a standing wave:
http://en.wikipedia.org/wiki/Standing_wave
Added later - where are the nodes and anti-nodes in the 1S state?
Where indeed? The article considers only physical waves with real-valued properties. Complex values allow the standing wave to retain constant amplitude but for the phase to have a spatial dependency.
bhobba said:
Wikipedia's other article not withstanding.
It says 'Therefore a stationary state is a standing wave that oscillates with an overall complex phase factor, and its oscillation angular frequency is equal to its energy divided by \hbar.'
The 1S state does not oscillate - its stationary.
It doesn't say the state is oscillatory, it says the wave function is. The word "stationary" has a specific meaning which is nothing to do with any phase periodicity of the wave function.
bhobba said:
Here is what's going on in stationary states:
https://www.physicsforums.com/threads/stationary-state.277647/
Its separable from the time dependant part which has no physical significance.
Of course.
 
  • #12
Derek Potter said:
Where indeed? The article considers only physical waves with real-valued properties. Complex values allow the standing wave to retain constant amplitude but for the phase to have a spatial dependency.

So - there are none.

Derek Potter said:
It doesn't say the state is oscillatory, it says the wave function is.

And the difference between a state and wave-function is?

Derek Potter said:
The word "stationary" has a specific meaning which is nothing to do with any phase periodicity of the wave function..

Yes - its meaning is the time dependence is separable so its physically irrelevant. That's one of the reasons the wave-function as a wave is a crock - its physically unobservable. Of course there are interpretations where its physically real like BM but they are just that - interpretations.

Thanks
Bill
 
  • #13
This is an example for the bad habit to take the wave function as the state, but it's not the wave function which really represents the (pure) state but the corresponding ray in Hilbert space (i.e., the one-dimensional subspace). An energy eigenstate of a time-independent Hamiltonian has a wave function, which "separates" into a time and a space dependent part, i.e.,
$$u_E(t,\vec{x})=\exp(-\mathrm{i} E t) \Psi_E(\vec{x}).$$
So the time dependence is just a phase factor and thus irrelevant for the representation of the state. So there's nothing observable that's oscillating or somehow waving here. Instead the probability distribution for the position of the electron is given by
$$P(\vec{x}|\Psi_E)=|\Psi_E(\vec{x})|^2,$$
which is time-independent!

Wave-particle duality is an idea from a historically important precursor theory of quantum theory, usually dubbed "old quantum theory" but has no place in the modern QT anymore, except that in some respects the same mathematical techniques can be used to solve a problem as you use in classical field theories with wave-like solutions. That wave-like properties of quantum theory (in its wave-function formulation) are just a mathematical analogy but has nothing to do with what's observable in nature!
 
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  • #14
bhobba said:
So - there are none.
Wave-like observables? No, of course not.
bhobba said:
And the difference between a state and wave-function is?
You tell me! You had claimed
"The 1S state does not oscillate - its stationary". But the wavefunction does oscillate, stationary or otherwise.
bhobba said:
Yes - its meaning is the time dependence is separable so its physically irrelevant.
Yes. But that does not mean you can't use the standing wave picture.
bhobba said:
That's one of the reasons the wave-function as a wave is a crock - its physically unobservable. Of course there are interpretations where its physically real like BM but they are just that - interpretations.
This thread is about interpretation.
 
  • #15
[QUOTE="vanhees71, post: 5131067, member: 260864]
Wave-particle duality is an idea from a historically important precursor theory of quantum theory, usually dubbed "old quantum theory" but has no place in the modern QT anymore, except that in some respects the same mathematical techniques can be used to solve a problem as you use in classical field theories with wave-like solutions. That wave-like properties of quantum theory (in its wave-function formulation) are just a mathematical analogy but has nothing to do with what's observable in nature![/QUOTE]
I agree with what you're saying but the OP was confusing the localization of the wave function in different bases with wave-particle duality.
 
  • #16
Derek Potter said:
Wave-like observables? No, of course not..

It was a claim about nodes and anti nodes - why you bring wave like observables into it has me beat - and that's without even examining what such even mean.

Derek Potter said:
You tell me! You had claimed
"The 1S state does not oscillate - its stationary". But the wavefunction does oscillate, stationary or otherwise.

Its not hard.

## |u(t)> = ∫ |x><x|u(t)> ## (1)

By definition ## <x|u(t)> ## is the wave function. For a time independent Schroedinger equation the wave-function has the form from Vanhees post ie ## u_E(t,\vec{x})=\exp(-\mathrm{i} E t) \Psi_E(\vec{x}) ##

Substitute into (1) and you get a state of the form ## |u(t)> = \exp(-\mathrm{i} E t)|u> ##.

Now, in reality pure states are operators ## |u(t)><u(t)| ##. Substitution gives ## |u><u| ##. The time dependence has vanished. Its physically irrelevant. What this means would be an interesting discussion about the Scrodinger Equation, and if anyone wants to pursue it starting a new thread would be the way to go. But of relevance here is it disappears.

The point is, since its of no physical relevance, claiming its responsible for standing waves makes no sense.

Thanks
Bill
 
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  • #17
bhobba said:
It was a claim about nodes and anti nodes - why you bring wave like observables into it has me beat - and that's without even examining what such even mean.
Its not hard.

## |u(t)> = ∫ |x><x|u(t)> ## (1)

By definition ## <x|u(t)> ## is the wave function. For a time independent Schroedinger equation the wave-function has the form from Vanhees post ie ## u_E(t,\vec{x})=\exp(-\mathrm{i} E t) \Psi_E(\vec{x}) ##

Substitute into (1) and you get a state of the form ## |u(t)> = \exp(-\mathrm{i} E t)|u> ##.

Now, in reality pure states are operators ## |u(t)><u(t)| ##. Substitution gives ## |u><u| ##. The time dependence has vanished. Its physically irrelevant. What this means would be an interesting discussion about the Scrodinger Equation, and if anyone wants to pursue it starting a new thread would be the way to go. But of relevance here is it disappears.

The point is, since its of no physical relevance, claiming its responsible for standing waves makes no sense.

Thanks
Bill

But they seem to govern the evolution of the relative phases between different terms in a superposition. Doesn't that mean physical relevance?
 
  • #18
Shyan said:
But they seem to govern the evolution of the relative phases between different terms in a superposition. Doesn't that mean physical relevance?

First you picked up a version of my post with errors I corrected.

That said I don't know what you mean - remember phases are meaningless. That's another reason the wave picture is bad - I can multiply the supposed wave by a phase factor and it make no difference. Actual waves aren't like that.

Thanks
Bill
 
  • #19
bhobba said:
First you picked up a version of my post with errors I corrected.

That said I don't know what you mean - remember phases are meaningless.

Thanks
Bill
I corrected that.

Yeah, phases are meaningless but only overall phases. Relative phases between different terms in a superposition do have meaning!
 
  • #20
Shyan said:
Yeah, phases are meaningless but only overall phases. Relative phases between different terms in a superposition do have meaning!

Yes they do - but again how is that wave like?

Added Later - one can always take any function and do a Fourier transform on it and express it as a complex integral of supposed waves - does that make the function wave like?

Thanks
Bill
 
  • #21
bhobba said:
Yes they do - but again how is that wave like?

Thanks
Bill
That's not what I mean. I'm trying to say that you can't say the time dependence of stationary states don't have physical meaning. They surely have!
But well...yeah...they only have physical meaning in a superposition!
 
  • #22
Shyan said:
That's not what I mean. I'm trying to say that you can't say the time dependence of stationary states don't have physical meaning. They surely have! But well...yeah...they only have physical meaning in a superposition!

Take the 1S state - what's the superposition there? It is an eigenvalue of energy so its time dependence is a simple phasor - not a superposition.

Thanks
Bill
 
  • #23
bhobba said:
Take the 1S state - what's the superposition there? It is an eigenvalue of energy so its time dependence is a simple phasor - not a superposition.

Thanks
Bill
Yeah, I get it. My point was irrelevant to the discussion here!
 
  • #24
Shyan said:
That's not what I mean. I'm trying to say that you can't say the time dependence of stationary states don't have physical meaning. They surely have!
But well...yeah...they only have physical meaning in a superposition!
Well, the superposition of two energy-eigen states with different eigenvalues is no longer a stationary state. Then you have written a time-dependent state as superposition of energy eigenstates. Since the Hamiltonian is self-adjoint, it provides a complete set of (generalized) eigenvectors, which implies that you can write any state as (generalized) superposition of energy eigenstates.

An energy eigenstate itself is a stationary state, because it's time dependence is only in one overall phase factor. In the superposition of energy eigenstates with different energy eigenvalues you have a physically relevant time dependence, because the time dependence is not in one overall phase factor. That's all. It's very simple!
 
  • #25
bhobba said:
Yes they do - but again how is that wave like?

Added Later - one can always take any function and do a Fourier transform on it and express it as a complex integral of supposed waves - does that make the function wave like?

Thanks
Bill
Sure does. *If* the components show independent wave-like evolution. We do this all the time in engineering, switching from the time domain to the frequency domain whenever it suits us.
 
  • #26
Derek Potter said:
Sure does. *If* the components show independent wave-like evolution.

Sorry - the Fourier transform theorem is clear on this. Any function - even weird ones like the Dirac delta Function can be decomposed into complex phasors. That does not make the function wave-like. Is a spike of infinite height wavelike - you are really stretching it if you think so - but I can't stop you.

Then there are other considerations. Take the wavefunction of two entangled particles - it propagates in a six dimensional complex space. How is that wavelike?

Thanks
Bill
 
  • #27
You'd have a job stopping me as that's exactly what we do. It's commonplace to consider the propagation of a delta function in a linear system - it's called the impulse response!

I don't personally believe there is an invisible physical space of a googol dimensions. The wavefunction as defined in conventional QM cannot be physical. Not sure whether this is resolved in MWI, I suspect it is.
 
  • #28
Derek Potter said:
The wavefunction as defined in conventional QM cannot be physical

Then why you argue it's a wave or wave like has me beat.

BTW it is physical in BM.

Thanks
Bill
 
  • #29
bhobba said:
Then why you argue it's a wave or wave like has me beat.
Thanks
Bill
Because we are attempting - well I am, I can't speak for you - to dispel the myth that the entity displays particle-like behaviour if you move a detector, take a measurement, blow your nose or whatever the latest YouTube bangs on about. It doesn't. The diffraction pattern remains exactly that, a diffraction pattern, a different pattern in the different cases but a diffraction pattern all the same.
 
  • #30
Derek Potter said:
Because we are attempting - well I am, I can't speak for you - to dispel the myth that the entity displays particle-like behaviour if you move a detector, take a measurement, blow your nose or whatever the latest YouTube bangs on about. It doesn't. The diffraction pattern remains exactly that, a diffraction pattern, a different pattern in the different cases but a diffraction pattern all the same.

That's got nothing to do with anything - certainly not the wave particle duality.

I don't know any science advisor on this forum that adheres to it except as a heuristic and maybe useful at the beginner level.

Thanks
Bill
 
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  • #31
bhobba said:
That's got nothing to o with anything - certainly not the wave particle duality.
I don't know any science advisor on this forum that adheres to it except as a heuristic and maybe useful at the beginner level.
Thanks
Bill
I rest my case.
 

1. What is a delayed choice experiment?

A delayed choice experiment is a type of thought experiment in quantum mechanics that explores the concept of wave-particle duality. It involves observing the behavior of a quantum system after it has already been measured, which can potentially change the outcome of the initial measurement.

2. How does a delayed choice experiment work?

In a delayed choice experiment, a quantum system is first sent through a series of detectors that measure its wave-like properties. Then, the experimenter can choose to either observe the system's behavior or not. If they choose not to observe, the system will continue to behave as a wave. However, if they do observe, the system's behavior will collapse into a particle-like state.

3. What is the significance of the delayed choice experiment?

The delayed choice experiment challenges our understanding of causality and the idea that an event can only be influenced by events that occurred before it. It suggests that the act of observation can have a retroactive effect on the behavior of a quantum system, which goes against our classical understanding of cause and effect.

4. What is the purpose of the "Delayed choice experiment" article on Science Alert?

The purpose of the article is to discuss a recent study published in the journal Physical Review Letters, which used a delayed choice experiment to demonstrate the retroactive effect of observation on quantum systems. The article aims to explain the experiment and its implications in a way that is accessible to a general audience.

5. How does the delayed choice experiment relate to other quantum mechanics concepts?

The delayed choice experiment is closely related to other concepts in quantum mechanics, such as the observer effect and the uncertainty principle. It also raises questions about the nature of reality and the role of consciousness in shaping our understanding of the universe.

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