0
$\begingroup$

I can't seem to understand this. In Bohmian mechanics, particles have definite positions and are guided by a wave function governed deterministically by Schrodinger's equation. So if you knew a particle's initial position in a double slit experiment, you could predict exactly where it would end up. However, in standard quantum mechanics the final position of the particle is a question of probabilities.

This implies that in Bohmian mechanics we can never know exactly where the particle is, because then we could predict exactly where it would end up in a double slit experiment. Why is this? Or what have I misunderstood?

Thank you for any help.

$\endgroup$
0
$\begingroup$

The theory is deterministic and explicitly nonlocal: the velocity of any one particle depends on the value of the guiding equation, which depends on the configuration of the system given by its wave function; the latter depends on the boundary conditions of the system, which, in principle, may be the entire universe.

It is the guiding equation that is deterministic,i.e. exactly defined in space time, not the particle position which is still probabilistic.

It is analogous , as an example, to the statistical behavior of molecules in water waves. The track of the molecules themselves is deterministic but the behavior of the waves gives a statistical probability for the individual molecule's position.

That is why Bohmian mechanics are an interpretation of quantum mechanics, because they give the same physical predictions as the Copenhanded interpretation.

$\endgroup$
4
  • $\begingroup$ I see, thank you for your answer. Would this be accurate statement then: we are unable to accurately measure a particle's position because its position depends on all other particle positions in the universe? $\endgroup$
    – Amun
    Jan 12 at 13:53
  • $\begingroup$ @Amun for Bohmian mechanics, " the particle position, depends on calculations of the pilot wave which depends on all other boundary conditions" would be my statement. $\endgroup$
    – anna v
    Jan 12 at 14:58
  • $\begingroup$ And it is somehow in principle impossible for us to know the full boundary conditions? Could you say some more about what these boundary conditions consist of? $\endgroup$
    – Amun
    Jan 12 at 17:36
  • $\begingroup$ sorry , you would need somebody working in bohmian mechanics. $\endgroup$
    – anna v
    Jan 12 at 19:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.