Bohmian Mechanics: Do photons travel faster than c in double slit experiment?

[Demystifier]

I'll let Demystifier speak for Bohemian Mechanics since this is his area.

Actually, my area is Bohmian (not Bohemian) mechanics.
But you may be surprised that I made a contribution to the "Bohemian" as well:
http://xxx.lanl.gov/pdf/physics/0702069 [Am.J.Phys.76:143-146,2008]
(See the lyrics after the Abstract.)

 

[Hans de Vries]

Actually, my area is Bohmian (not Bohemian) mechanics.

Oops

But you may be surprised that I made a contribution to the "Bohemian" as well:
http://xxx.lanl.gov/pdf/physics/0702069 [Am.J.Phys.76:143-146,2008]
(See the lyrics after the Abstract.)

 

[ytuab]

I will take a look and comment later.

Edit: I've just did it. They don't really measure any rotation of the neutrons. They just measure interference, which is theoretically explained by rotations of wave functions.

The experiment (Phys Lett. 54A 1975 425) measured spatial rotation of the neutrons.

Did you really read this paper or something about this experiment?

They actually "spatially" rotated the neutrons by an angle of 4pai using "rotation by precession" in this real world . (The result is just consistent with 4 pai. It is not accidential.)
Spatiall rotation is consistent with the rotaions of spinor wavefunction ( you say spatial rotation is different from the rotation of the spinor wavefunction.)

Of course, we can not look at the spinning neutrons directly.
So they measured interference. (you say only about this last step. you didn't say the method of rotaion (this is most important.))

In my personal view, I DON'T believe "the spin of the electron".
But you believe in the quantum mechanics. So you misunderstand this experiment.

 

[Hans de Vries]

The experiment (Phys Lett. 54A 1975 425) measured spatial rotation of the neutrons.

They actually "spatially" rotated the neutrons by an angle of 4pai using "rotation by precession" in this real world . (The result is just consistent with 4 pai. It is not accidential.)

The 4pi rule is valid in all rotational directions. If the rotation is around
the spin pointer then you can measure only the interference. (The spin
pointer doesn't change direction) However, a rotation around any other
axis will rotate the spin pointer itself and change the direction of the
magnetization which is measurable.

Regards, Hans

 

[ytuab]

The 4pi rule is valid in all rotational directions. If the rotation is around
the spin pointer then you can measure only the interference. (The spin
pointer doesn't change direction) However, a rotation around any other
axis will rotate the spin pointer itself and change the direction of the
magnetization which is measurable.

Regards, Hans

I see.
What you say is indeed interesting.

But I didn't know about the other direction.

 

[Demystifier]

In my personal view, I DON'T believe "the spin of the electron".

What exactly do you mean by that?

 

[ytuab]

What exactly do you mean by that?

I don't believe "the spin of the electron".

Because as I have stated before, "spin" has some serious problems.

For example, the electron is too small. So by equating the spin angular momentum to 1/2 hbar, spinning sphere speed leads to 100 times the speed of light. (If the electron size is classical radius.)

The classical radius size is as big as nuclear size of the hydrogen .

Spinning electoron has magnetic moment and spin angular momentum.
And one electron has the charge and mass.
Magnetic moment is basically caused by the rotaion of the charge, isn't it?
Angular momentum is basically caused by the rotaion of the mass, isn't it?

Probably Nobody believes "spin" is a actual spinning of the electron.
In spite of that, why do we call it "spin"?
It is inconsistent.

 

[Demystifier]

I don't believe "the spin of the electron".

Because as I have stated before, "spin" has some serious problems.

For example, the electron is too small. So by equating the spin angular momentum to 1/2 hbar, spinning sphere speed leads to 100 times the speed of light. (If the electron size is classical radius.)

The classical radius size is as big as nuclear size of the hydrogen .

Spinning electoron has magnetic moment and spin angular momentum.
And one electron has the charge and mass.
Magnetic moment is basically caused by the rotaion of the charge, isn't it?
Angular momentum is basically caused by the rotaion of the mass, isn't it?

Probably Nobody believes "spin" is a actual spinning of the electron.
In spite of that, why do we call it "spin"?
It is inconsistent.

I completely agree that spin as a kind of rotational motion does not exist. Still, the effect that we call "spin" exists. We should probably use a different word for this effect, but it is just a matter of terminology, not of physics.

Nevertheless, there are reasons for calling it "spin": The sum of this and the ordinary angular momentum is conserved.

 

[ArjenDijksman]

2. If the guiding wave affect the particle shouldn't the particle affect the wave as well? If the answer is yes how is this happening?

The answer is - no. (Otherwise, it would be in contradiction with standard QM.)

I wasn't aware of this detail of Bohmian quantum mechanics and I don't understand why a particle couldn't affect its pilot wave. Couder and Fort have shown experimentally that the particle and its steering wave must be considered as a whole. If the particle ceases to exist, the pilot wave disappears. And reversely, if the pilot wave is disturbed, the particle is unable to survive: http://www.physorg.com/news78650511.html.

 

[Demystifier]

I don't understand why a particle couldn't affect its pilot wave.

If it did, then pilot wave would not satisfy the Schrodinger equation. On the other hand, all experiments confirm that the wave satisfies the Schrodinger equation.

 

[ArjenDijksman]

Hello Demystifier. Thanks for your answer.

If it did, then pilot wave would not satisfy the Schrodinger equation. On the other hand, all experiments confirm that the wave satisfies the Schrodinger equation.

Could you be more precise? The pilot wave and the particle satisfy the same Schrödinger equation, the wave as a whole and the particle as a quantum entity. So if the particle affects the wave, the wave still satisfies the Schrödinger equation. For example, if the particle is a little bit disturbed before the two slits, the pilot wave will readjust its phase to the new state of the particle, satisfying of course the perturbed Schrödinger equation.

Regards,
Arjen

 

[zenith8]

I don't understand why a particle couldn't affect its pilot wave.

In classical physics there is an interplay between particle and field - each generates the dynamics of the other. In de Broglie-Bohm pilot wave theory \Psi acts on the positions of particles but, evolving as it does autonomously via Schroedinger's equation, it is not acted upon by the particles.

One may think this is unaesthetic, but while it may be reasonable to require reciprocity of actions in classical theory, this cannot be regarded as a logical requirement of all theories that employ the particle and field concepts, especially one involving a nonclassical field.

However, as you imply, there is in fact a kind of back-action. This arises from the standard notion that the shape of the quantum field of a particle is determined by the shape of the environment (which consists of many particles, and is part of the boundary conditions put into the Schroedinger equation before solving it, even in conventional QM).

Normally in QM this 'back-action' is not taken into account. The wave guides the particles but back-action of the particle onto the wave is not systematically calculated. Of course, the back-action is physically real since the particle movement determines the initial conditions for the next round of Schroedinger calculation, but there is no systematic way to characterize such feedback. The reason this works in practice is that the back-action may not exert any systematic effect.

There is a fair of amount of interesting speculation lurking in dark corners of the internet that there is actually a systematic effect in systems which are self-organizing. That is - 'life' is what happens when a physical system uses its own nonlocality in its organization (Note to moderators: don't ban me - I'm just repeating what I heard). In this case a feedback loop is created, as follows: system configures itself so as to set up its own pilot wave, which in turn directly affects its physical configuration, which then affects its non-local pilot wave, which affects the configuration etc..

This sort of thing has never been systematically developed in the pilot-wave literature, largely because the people who talk about it on the internet are in the main well-known wackos (I won't name names, because it's probably against the rules). However, there is something interesting in this idea of 'back-action'. Bohm and Hiley even mention it in their Undivided Universe testbook - though in a very different context. (p. 345-346 if you're interested).

There is also quite a lot of speculation that the two-way traffic between pilot-wave and particle configurations provides a possible mechanism for consciousness. See Paavo Pylkkanen's Mind, Matter, and the Implicate Order book (2007) or the Cambridge Pilot-wave http://www.tcm.phy.cam.ac.uk/~mdt26/pilot_waves.html" (lectures 7 and 8).

Peter Holland has explored some deeper ideas related to this question in his work on a possible Hamiltonian formulation of pilot-wave theory. They're a bit technical, but see the following papers:

Hamiltonian theory of wave and particle in quantum mechanics I: Liouville's theorem and the interpretation of the de Broglie-Bohm theory P. Holland (2001)

Hamiltonian theory of wave and particle in quantum mechanics II: Hamilton-Jacobi theory and particle back-reaction P. Holland (2001)

 

[Demystifier]

Could you be more precise? The pilot wave and the particle satisfy the same Schrödinger equation, the wave as a whole and the particle as a quantum entity. So if the particle affects the wave, the wave still satisfies the Schrödinger equation. For example, if the particle is a little bit disturbed before the two slits, the pilot wave will readjust its phase to the new state of the particle, satisfying of course the perturbed Schrödinger equation.

Can you write down the equations that support your claims above? More precisely, can you write down the equation that describes how particle affects the wave? Without the equations, any further discussion of that would be pointless.

 

[ArjenDijksman]

However, there is something interesting in this idea of 'back-action'. Bohm and Hiley even mention it in their Undivided Universe testbook - though in a very different context. (p. 345-346 if you're interested).

Thanks for your complete answer. Because mutual back action between wave and particle is common good in classical physics, it's surprising that neither De Broglie nor Bohm developed the "two way relationship between wave and particle" (p.346) apart from the short section you mentioned in The Undivided Universe. All the more because that would have given the wealth of "experimental clues" they were longing for, through parametric adjustment of macroscopic pilot-wave experiments like those of Couder and Fort mentioned earlier.

Can you write down the equations that support your claims above? More precisely, can you write down the equation that describes how particle affects the wave? Without the equations, any further discussion of that would be pointless.

You're right, it's better to be precise. The process I visualized was the following:

• State vector |\Psi> of (any) quantum particle satisfies i.hbar.d|\Psi>/dt = H.|\Psi>, where H is the hamiltonian matrix.
• At the same time wave-function \Psi(x,t) of the pilot wave satisfies i.hbar.d\Psi(x,t)/dt = H.\Psi(x,t).

Suppose the particle is affected by the interaction with another particle (Compton scattering, collision...):

• New state vector |\Psi'> of particle satisfies i.hbar.d|\Psi'>/dt = (H+H').|\Psi'>, where H' is the part of the hamiltonian caused by interaction with other particle.
• The old pilot wave being completely out of phase with the particle, the new pilot-wave will have as wave-function \Psi'(x,t) satisfying i.hbar.d\Psi'(x,t)/dt = (H+H').\Psi'(x,t).

 

[Demystifier]

You're right, it's better to be precise. The process I visualized was the following:

• State vector |\Psi> of (any) quantum particle satisfies i.hbar.d|\Psi>/dt = H.|\Psi>, where H is the hamiltonian matrix.
• At the same time wave-function \Psi(x,t) of the pilot wave satisfies i.hbar.d\Psi(x,t)/dt = H.\Psi(x,t).

Suppose the particle is affected by the interaction with another particle (Compton scattering, collision...):

• New state vector |\Psi'> of particle satisfies i.hbar.d|\Psi'>/dt = (H+H').|\Psi'>, where H' is the part of the hamiltonian caused by interaction with other particle.
• The old pilot wave being completely out of phase with the particle, the new pilot-wave will have as wave-function \Psi'(x,t) satisfying i.hbar.d\Psi'(x,t)/dt = (H+H').\Psi'(x,t).

OK, now I see what you mean.
My answer is the following. There is no "old" and "new" wave function. There is only one wave function that corresponds to your Psi'. At early times the influence of H' on Psi' may be negligible so that Psi' can be approximated by Psi at early times. Still, the wave function is only one.