As it is normally explained, it’s definitely fake. There is no reason to believe particles turn into waves when you’re not looking and turn back into particles when you look, and believing this demonstrably leads to irreconcilable paradoxes. Dmitry Blokhintsev was correct that the particles are just particles, and the “wave” is a property of its stochastic dynamics over an ensemble of systems. The wave is part of the nomology: it tells you how the particles stochastically behave in the aggregate, but the particles are still particles at all times. Ontologically, they are particles. Nomologically, their stochastic dynamics in an ensemble of systems converges to wave-like behavior.
The fact that you can make predictions without one or the other doesn’t mean that either, or both, don’t conribute to equally powerful predictive models.
Well in de Broglie-Bohm you have a particle and a wave as separate but connected entities. That can’t be a duality (which requires two equally valid but mutually exclusive descriptions).
It mostly sounds fake, because it’s never explained very well.
Just the fact that something can be correlated more strongly than anything classical but also not allow signaling might be. I don’t know, I find it hella confusing, even if you can show that pretty straightforwardly without any actual quantum mechanics.
QM is a lot easier to understand when we stop pretending a theory that only gives you statistical results somehow has no relevance to statistics. Every “paradox” can always be understood and resolved by applying a statistical analysis. If you apply such a statistical analysis to entangled systems, let’s say you have two qubits with their own bit values b1 and b2, you find that if you apply a unitary operator to just b1, there are cases where the way in which this stochastically perturbs b1 has a dependence upon the value of b2.
You could not send a signal to b2 by perturbing b1 because perturbing b1 has no effect on b2, rather, the way in which b1 stochastically changes merely depends upon the current state of b2. You might think maybe you could send a signal the other way. If b1 depends upon b2, then you could perturb b2 to alter b1. But the dependence is always symmetrical, such that if you apply a stochastic perturbation to b2 the way in which it will change will depend upon the value of b1, and so it becomes a vicious circle.
It is non-local in the sense that the way in which one changes depends upon the value of the other far away, but not in the sense that perturbing one locally alters the value of the one far away, and the dependence is always symmetrically mutual, so there is no way to signal between them.
the wave-particle duality in quantum mechanics
As it is normally explained, it’s definitely fake. There is no reason to believe particles turn into waves when you’re not looking and turn back into particles when you look, and believing this demonstrably leads to irreconcilable paradoxes. Dmitry Blokhintsev was correct that the particles are just particles, and the “wave” is a property of its stochastic dynamics over an ensemble of systems. The wave is part of the nomology: it tells you how the particles stochastically behave in the aggregate, but the particles are still particles at all times. Ontologically, they are particles. Nomologically, their stochastic dynamics in an ensemble of systems converges to wave-like behavior.
Not a fact. If it was, the de Broglie-Bohm theory wouldn’t work.
The fact that you can make predictions without one or the other doesn’t mean that either, or both, don’t conribute to equally powerful predictive models.
But it does mean the duality is not logically necessary.
Honestly, I’d’ve thought it meant the exact opposite.
Well in de Broglie-Bohm you have a particle and a wave as separate but connected entities. That can’t be a duality (which requires two equally valid but mutually exclusive descriptions).
It mostly sounds fake, because it’s never explained very well.
Just the fact that something can be correlated more strongly than anything classical but also not allow signaling might be. I don’t know, I find it hella confusing, even if you can show that pretty straightforwardly without any actual quantum mechanics.
QM is a lot easier to understand when we stop pretending a theory that only gives you statistical results somehow has no relevance to statistics. Every “paradox” can always be understood and resolved by applying a statistical analysis. If you apply such a statistical analysis to entangled systems, let’s say you have two qubits with their own bit values b1 and b2, you find that if you apply a unitary operator to just b1, there are cases where the way in which this stochastically perturbs b1 has a dependence upon the value of b2.
You could not send a signal to b2 by perturbing b1 because perturbing b1 has no effect on b2, rather, the way in which b1 stochastically changes merely depends upon the current state of b2. You might think maybe you could send a signal the other way. If b1 depends upon b2, then you could perturb b2 to alter b1. But the dependence is always symmetrical, such that if you apply a stochastic perturbation to b2 the way in which it will change will depend upon the value of b1, and so it becomes a vicious circle.
It is non-local in the sense that the way in which one changes depends upon the value of the other far away, but not in the sense that perturbing one locally alters the value of the one far away, and the dependence is always symmetrically mutual, so there is no way to signal between them.