Why does dbb need to assume pilot wave?

In summary, the De Broglie-Bohm interpretation conjunctures that:1) a wavefunction travels through both slits, but each particle has a well-defined trajectory and passes through exactly one of the slits2) there is a pilot wave3) the wavefunction satisfies the Schrodinger equation, which determines its travel velocity
  • #1
San K
911
1
The De Broglie–Bohm (DBB) interpretation conjunctures that:

a) the wavefunction travels through both slits, but each particle has a well-defined trajectory and passes through exactly one of the slits

b) there is a pilot waveNow, one can guess/understand why it needs to assume a) above.

However, what observation/phenomena, in a single particle interference experiments, induces the DBB to assume a pilot wave in addition to (a) above?
 
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  • #2
The wavefunction is the pilot wave.
 
  • #3
kith said:
The wavefunction is the pilot wave.

oops...thanks kith...at what velocity is it postulated to travel?
 
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  • #4
San K said:
oops...thanks kith...at what velocity is it postulated to travel?
The wave is postulated to satisfy the Schrodinger equation, which determines its travel velocity.
 
  • #5
Demystifier said:
The wave is postulated to satisfy the Schrodinger equation, which determines its travel velocity.

thanks Demystifier

is that velocity at, or above, the speed of light?
 
  • #6
Schrodinger equation describes nonrelativistic waves of massive particles, so the speed of wave is not bounded by the speed of light. Of course, it is only an approximation.
 
  • #7
Demystifier said:
Schrodinger equation describes nonrelativistic waves of massive particles, so the speed of wave is not bounded by the speed of light. Of course, it is only an approximation.

Thanks Demystifier, What does a massive particle mean? Is a photon/electron considered massive? or is it for all particles starting from the smallest ones till the big ones like Fullerene?

In the DBB - Why is there a need to assume that the speed of the wave is faster than light?

which observation/phenomena in the quantum (interference pattern) experiments is DBB trying to explain/rationalize by assuming waves travel almost instantaneously?

can not the -

the experimental observations, in various scenarios, of placing of an obstacle (and or detector) any time before the passage of the particle (as calculated by the speed of light),

be explained without having to assume that the wave is traveling instantaneously?
 
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  • #8
San K said:
In the DBB - Why is there a need to assume that the speed of the wave is faster than light?

I think I just got the answer, not sure if it's correct -

the photon is assumed to be entangled with some part of the experimental apparatus/environment at all points in time-space and that entanglement changes as the situation changes (due to movement of the particle/photon or change in the experimental setup etc).

this (change in) entanglement is assumed to happen (almost) instantaneously.
 
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  • #9
But since the pilot-wave is the same as the "normal" Schrödinger wave psi, both can "travel" faster than the speed of light, so it's not specific to dbb. In other words: QM is a non-relativistic theory, so why judge it by relativistic standards? But even still, dbb (for QM) seems to be compatible with relativity (although, as just claimed, there is no actual need for it to be so) since it can be proven that no information can be transmitted faster than the speed of light, which is all relativity requires anyway.

Also, "massive" simply means "mass [itex]m \neq 0[/itex]".

EDIT: I believe better two postulates for dBB (in the context of one-particle QM) are
1) There is a wavefunction [itex]\psi(\mathbf r, t) = R(\mathbf r, t) e^{i S(\mathbf r,t)}[/itex] governed by the Schrödinger equation.
2) There is a point-particle with position [itex]\mathbf X(t)[/itex] and velocity [itex]\mathbf v(t) = \frac{1}{m} \nabla S(\mathbf X(t),t)[/itex].
 
  • #10
mr. vodka said:
But since the pilot-wave is the same as the "normal" Schrödinger wave psi, both can "travel" faster than the speed of light, so it's not specific to dbb. In other words: QM is a non-relativistic theory, so why judge it by relativistic standards? But even still, dbb (for QM) seems to be compatible with relativity (although, as just claimed, there is no actual need for it to be so) since it can be proven that no information can be transmitted faster than the speed of light, which is all relativity requires anyway.

Also, "massive" simply means "mass [itex]m \neq 0[/itex]".

EDIT: I believe better two postulates for dBB (in the context of one-particle QM) are
1) There is a wavefunction [itex]\psi(\mathbf r, t) = R(\mathbf r, t) e^{i S(\mathbf r,t)}[/itex] governed by the Schrödinger equation.
2) There is a point-particle with position [itex]\mathbf X(t)[/itex] and velocity [itex]\mathbf v(t) = \frac{1}{m} \nabla S(\mathbf X(t),t)[/itex].

thanks mr. vodka...cheers

however why is the (Schrödinger) function assumed/postulated to "travel/effect" faster than light?...in the first place
 
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  • #11
It does not necessarily travel faster than light, there is just nothing forbidding it in a non-relativistic theory. Think of Newtonian mechanics: particles can travel faster than light; they're not postulated to, it's just that since Newtonian mechanics is non-relativistic, there's no upper bound to velocity.

That being said, it seems like QM is more generally applicable than Newtonian mechanics even on the matter of relativity. For example, non-locality could have been a wrong prediction of QM, due to necessary relativistic corrections in those conditions, however it seems non-locality is a fact, so in a sense although QM is a non-relativistic theory, it is more widely valid than simply in "non-relativistic cases".
 
  • #12
Even in relativistic QM, if you start with a plane wave:

$$Ae^{i(kx-\omega t)} = Ae^{i(px-Et)/\hbar}$$

and calculate the phase velocity ##v_p = \omega / k## using the relativistic formulas for momentum and energy, you find that ##v_p > c##! This is a common exercise in intro modern physics textbooks.

What matters of course is the group velocity ##v_g = d\omega / dk## which does always turn out to be < c.
 
  • #13
jtbell said:
Even in relativistic QM, if you start with a plane wave:

$$Ae^{i(kx-\omega t)} = Ae^{i(px-Et)/\hbar}$$

and calculate the phase velocity ##v_p = \omega / k## using the relativistic formulas for momentum and energy, you find that ##v_p > c##! This is a common exercise in intro modern physics textbooks.

What matters of course is the group velocity ##v_g = d\omega / dk## which does always turn out to be < c.

thanks jtbell.

thus...do some "massless effects/cause" have to travel faster than light in order to sustain certain laws of physics...(for example - law of conservation of momentum/energy)?
 

Related to Why does dbb need to assume pilot wave?

1. Why is the pilot wave theory necessary for understanding dbb?

The de Broglie-Bohm (dbb) theory, also known as the pilot wave theory, is necessary because it provides a framework for understanding the behavior of quantum particles. It offers a different perspective on the wave-particle duality of quantum mechanics and helps to explain phenomena that cannot be fully understood through other interpretations of quantum mechanics.

2. How does the pilot wave theory explain quantum entanglement?

The pilot wave theory explains quantum entanglement as a result of the entangled particles being connected by a pilot wave, which guides their movements and ensures that they remain correlated. This theory offers a more intuitive explanation for entanglement compared to other interpretations of quantum mechanics.

3. Can the pilot wave theory be tested or proven?

While the pilot wave theory is a valid interpretation of quantum mechanics, it is difficult to test or prove. This is because it is based on non-local hidden variables, which cannot be directly observed. However, there have been experiments that support certain aspects of the pilot wave theory, such as the double-slit experiment.

4. Is the pilot wave theory widely accepted by scientists?

The pilot wave theory is a valid and well-studied interpretation of quantum mechanics, but it is not widely accepted by all scientists. Many physicists still prefer the Copenhagen interpretation, which does not involve hidden variables. However, the pilot wave theory has gained more attention and acceptance in recent years.

5. What are the implications of accepting the pilot wave theory?

If the pilot wave theory is accepted, it would mean that quantum particles have definite positions and trajectories, even though they may appear to behave probabilistically. This could potentially have implications for our understanding of reality, as it challenges the idea of an observer-dependent reality in quantum mechanics.

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