Delayed Choice with both slits closed, What happens?

[Rohan de Silva]

I am reading about this most famous experiment in physics, the double-slit experiment.

My question is very simple: in the "Delayed Choice" version, have they done the test where they close BOTH SLITS (at once) after the photon has passed the slits??

If so, what have been the result?

Can you point me to a paper or some article on this?

 

[Nugatory]

in the "Delayed Choice" version, have they done the test where they close BOTH SLITS (at once) after the photon has passed the slits??

You'll have to tell us more about exactly which experiment you're asking about - "the 'Delayed Choice' version" isn't very specific. Kim's 1999 delayed choice experiment is a good example of a delayed choice experiment, and it's not clear what your question means in this context.

 

[Rohan de Silva]

Thanks. Yes, the quantum eraser is the most sophisticated version of the experiment. I am interested in the less complex delayed-choice only experiment, where instead of closing one slit after the photon has passed the slit, you close BOTH slits after the photon has passed the slits.

 

[PeterDonis]

instead of closing one slit after the photon has passed the slit, you close BOTH slits after the photon has passed the slits.

This description is not really correct. You're not measuring the photon at the slits, so you can't say the photon passes the slits at any particular time. Photons have a nonzero amplitude to travel at speeds other than the speed of light.

 

[Rohan de Silva]

This description is not really correct. You're not measuring the photon at the slits, so you can't say the photon passes the slits at any particular time. Photons have a nonzero amplitude to travel at speeds other than the speed of light.

So, then how can they close one slit AFTER the photon has passed the slit??
This experiment, with one slit closed after the photon has passed slit has been done, right??

 

[PeterDonis]

So, then how can they close one slit AFTER the photon has passed the slit??

That's not what they are doing. It's unfortunate that that ordinary language is used to describe the experiment, since it is prone to misunderstandings such as yours.

What they are doing is this: they measure the time at which the photon leaves the source; they know the distance from the source to the slits; they take that distance divided by the speed of light to calculate the time when, notionally, the photon has "passed" the slits; then they do whatever they are going to do to alter the conditions--such as close one slit--later than that calculated time. But to actually think of that calculated time as "the time the photons pass the slits"--instead of as just a notional calculation--requires you to assume that the photons are classical particles moving at the speed of light. They aren't.

 

[Rohan de Silva]

That's not what they are doing. It's unfortunate that that ordinary language is used to describe the experiment, since it is prone to misunderstandings such as yours.

What they are doing is this: they measure the time at which the photon leaves the source; they know the distance from the source to the slits; they take that distance divided by the speed of light to calculate the time when, notionally, the photon has "passed" the slits; then they do whatever they are going to do to alter the conditions--such as close one slit--later than that calculated time. But to actually think of that calculated time as "the time the photons pass the slits"--instead of as just a notional calculation--requires you to assume that the photons are classical particles moving at the speed of light. They aren't.

Thanks. This is a good explanation.

So, I know that photon is not a "classical" particle and it's position adhere to Schrodinger equation.

Do they close the slit when the entire Schrodinger equation probability distribution passes the slit??

 

[vanhees71]

Photons, in fact, have no position at all. It's just not possible to define, what the position of a photon might be. That is, because they are massless particles of spin 1. You cannot define a position observable for any massless particle with a spin ##\geq 1##. This follows from the mathematics underlying relativistic quantum field theory, and the photon cannot be described by the Schrödinger equation, because it's massless. The Schrödinger equation only applies to massive particles and refers to the non-relativistic limit. From another point of view you can also argue that non-relativistic quantum mechanics with massless particles does not lead to a quantum theory which can be physically interpreted in any way to make sense. The mathematics you'd need to study for all these fascinating insights is the theory of Lie groups and its representations and its relation to quantum theory. You find a very good treatment of the non-relativistic case in Ballentine, Quantum Mechanics. For the relativistic case the best treatment is found in Weinberg, Quantum Theory of Fields, vol. I.

 

[Rohan de Silva]

What is the "non-relativistic limit"?

Is it a speed close to speed of light?

I have read that the double-slit experiment has been done with massive atoms as well. So, don't they obey the Schrodinger equation?

 

[PeterDonis]

I know that photon is not a "classical" particle and it's position adhere to Schrodinger equation.

No, this is not correct. The Schrodinger equation, as @vanhees71 mentioned, only works for non-relativistic particles, which means, roughly, particles moving much slower than the speed of light--or more precisely, particles that do not have any amplitude to move at or near the speed of light. Photons obviously do not meet this restriction.

Do they close the slit when the entire Schrodinger equation probability distribution passes the slit??

They close the slit based on the calculation I described earlier, which has nothing to do with the Schrodinger equation at all. It's a calculation based on the (wrong) assumption that a photon is a classical particle that moves at the speed of light.

There is no way to close the slit only when the entire wave function is past the slit, if you want to close the slit during the experiment at all. Until the particle is observed hitting the detector at the other end, the wave function will have some nonzero amplitude to not yet have passed the slit.

I have read that the double-slit experiment has been done with massive atoms as well. So, don't they obey the Schrodinger equation?

They do--at least, they do as long as the atoms are moving much slower than the speed of light (or more precisely, that whatever state the atoms are in, it has no amplitude to move at or near the speed of light). I don't know if delayed choice experiments have been done with such atoms, though. If they were, the time at which the delayed choice is made would have to be calculated based on some reasonable estimate of the speed of the particle, such as its expectation value for velocity. As I noted above, there is no way to wait until the particle's wave function is entirely past the slit before closing it, if you want to do a delayed choice experiment at all.

 

[vanhees71]

Of course, entanglement and the possibility of delayed-choice experiments based on it, is not restricted to photons. It's only technically easy to create entangled photon pairs and make high-precision experiments in quantum optics with photons. I'm not aware of a concrete delayed-choice experiment with massive particles. If there is one, I guess it's most likely to involve either neutrons or ultracold trapped atomic gases.