# Delayed choice experiment article on science alert

1. Jun 1, 2015

### DaveC426913

I'm familiar with the delayed choice experiment, and I'm trying to suss out this new setup.

Should that last phrase be "...when the second grating was removed..." ?

2. Jun 1, 2015

### ShayanJ

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".

3. Jun 1, 2015

### DaveC426913

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.)

4. Jun 1, 2015

5. Jun 2, 2015

### Derek Potter

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 travelled 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.

6. Jun 2, 2015

### DaveC426913

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. Jun 4, 2015

### Derek Potter

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. Jun 4, 2015

### Staff: Mentor

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

Last edited by a moderator: May 7, 2017
9. Jun 4, 2015

### Derek Potter

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?

Last edited by a moderator: May 7, 2017
10. Jun 4, 2015

### Staff: Mentor

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:

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

Thanks
Bill

Last edited: Jun 4, 2015
11. Jun 4, 2015

### Derek Potter

I'll take that as a "yes".
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.
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.
Of course.

12. Jun 4, 2015

### Staff: Mentor

So - there are none.

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

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. Jun 4, 2015

### vanhees71

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!

14. Jun 4, 2015

### Derek Potter

Wave-like observables? No, of course not.
You tell me! You had claimed
"The 1S state does not oscillate - its stationary". But the wavefunction does oscillate, stationary or otherwise.
Yes. But that does not mean you can't use the standing wave picture.

15. Jun 4, 2015

### Derek Potter

[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. Jun 4, 2015

### Staff: Mentor

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 in to (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

Last edited: Jun 4, 2015
17. Jun 4, 2015

### ShayanJ

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

18. Jun 4, 2015

### Staff: Mentor

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. Jun 4, 2015

### ShayanJ

I corrected that.

Yeah, phases are meaningless but only overall phases. Relative phases between different terms in a superposition do have meaning!

20. Jun 4, 2015

### Staff: Mentor

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