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My questions are about de Broglie-Bohm "pilot wave" interpretation of quantum mechanics (a.k.a. Bohmian mechanics).

  1. Do quasiparticles have any meaning in Bohmian mechanics, or not? Specifically, is it possible to trace the motion of a quasiparticle (e.g. a phonon, or a hole) by watching Bohmian trajectories?

  2. Bohmian mechanics provides some explanation of difficulties related to quantum measurement process. But imagine that, in some kind of "theory of everything", all known elementary particles (leptons, quarks, gluons etc.) are actually quasiparticles (real particles are always confined). Would the explanation, provided by Bohmian mechanics, survive in this case?

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I think that your inherent suggestion that the Bohmian mechanics suffers from an inconsistency because of quasiparticles is right. QM shows that it doesn't matter whether one says that something is a particle or quasiparticle: the Hilbert spaces and dynamics and probabilistic predictions are isomorphic. In the Bohmian mechanics, it always matters because particles' degrees of freedom are "beable" while other e.g. quasiparticles' degrees of freedom are not. For the same reason, quantum computers can't simulate the reality in Bohmian mechanics. It just doesn't work. – Luboš Motl Oct 6 '11 at 12:03
@Luboš Motl: Electron (a particle) in Hydrogen atom is always in a mixed state - it does not have an electron wave function $\psi (\vec{r}_e)$. Instead, it is a quasi-particle that may be in a pure state $\psi (\vec{r})$ where $\vec{r}=\vec{r}_e-\vec{r}_p$. As well, proton is also in a mixed state but the center of mass may be in a pure state like $exp\left ( i(\vec{P}\vec{R}-\frac{P^2}{2M_A\hbar}t)\right)$.There are no such real particles with $\mu$ and $M_A=m_e+m_p$; the corresponding variables describe quasi-particles. It is namely quasi-particle energy in atom that is quantized. – Vladimir Kalitvianski Oct 6 '11 at 13:13
The answer to your question, @Vladimir, is obviously No: every quantum system (whether it is a particle or quasiparticle) may always be found in a pure state. That's why any quantum system may exhibit 100% destructive interference etc., something that would indeed be impossible if an object were "forced" to be partly mixed, and this is one reason why any picture of the quantum phenomena that tries to be more classical than QM (including Bohmian mechanics) is inevitable inconsistent with the tests of interference and other empirically established features of the quantum world. – Luboš Motl Oct 6 '11 at 15:57
What do you mean by "meaning"? How can you say that quasiparticles have any meaning in standard quantum mechanics? They're just a very good way of describing an emergent phenomenon; they don't really exist in the usual sense of the word. – Peter Shor Oct 8 '11 at 21:27
Can I also say that after thinking it over, I suspect there may be a very good question lurking in here. Namely: do the trajectories of quasiparticles in Bohmian mechanics obey the same rules as the trajectories of particles in Bohmian mechanics? And if not, what rules do they obey? – Peter Shor Oct 10 '11 at 22:21

Nobody wrote an answer, so I'll give it a try

  1. Indeed in Bohmian QM quasiparticles are not on the same ground as ordinary particles. For example consider phonons in a crystal lattice. From a Bohmian POV, the atom position observables have fundamental significance. But observables natural from the phonon POV, such as phonon number are not functions of the atom positions. Of course these observables are still functions of atom positions + momenta, so in principle they can be assigned values along a Bohmian trajectory. However, this approach has serious issues. For one thing the "phononic" observables are still not going to play a symmetric role with the atomic positions. For another, the phonon number will not be integer*! This shows it hardly makes sense to talk about phonon trajectories
  2. Actually, the elementary particles are "quasiparticles". They are excitations of the quantum fields. This indeed means the Bohmian approach runs into trouble, since in Bohmian QFT the fields become the fundamental observables, making it inconsistent with nonrelativistic Bohmian QM. Actually it's not the only problem of Bohmian QFT: it also fails to be Lorentz invariant

*In the simplest harmonic model, the occupation number of each mode is a linear function of its energy i.e. something quadratic in both positions and momenta

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  1. Given that Bohmian mechanics (better named de Broglie-Bohm or dBB theory) contains the Schrödinger equation, it contains quantum theory completely, and therefore all things which make sense in quantum theory remain meaningful. Of course, for quasiparticles nobody plans to construct trajectories.

And it is also not possible to simply "watch" Bohmian trajectories.

  1. dBB theory is usually presented as a theory for particles, but this is far too restrictive. The better way is to present it as a theory for the configuration space: It defines a trajectory for the configuration $q(t)$.

So, the configuration may be, as well, a field $f(x)$ on the space. Then the "trajectory" is $f(x,t)$.

The mathematics of dBB theory works if the Hamiltonian has the form $H=p^2 + V(q)$. This works nicely with relativistic field theories $L = (d_tf)^2-(d_xf)^2$ + interaction terms gives momentum $p(x)=d_t$ $f(x)$ and gives a Hamiltonian quadratic in $p(x)$.

All the advantages provided by dBB theory do not depend on the question what is the particular configuration space. Thus, they would survive whatever the choice of Q is. All one has to care for is a way to obtain a choice of a configuration space so that $H=p^2 + V(q)$. For bosonic field theories no problem, for fermionic field theories proposals exist too. In the worst case, one can consider partial realizations (like defining trajectories only for bosons). This is less beautiful but preserves the major advantages (realism, causal explanation, no measurement problem).

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It's just an elementary mistake to say that a "theory A containing the key equation of theory B" may do everything that B does. Theories are only equivalent if their objects may be mapped in a one-to-one way, the observable evolution laws are isomorphic in both, and therefore if A isn't "missing" anything from B, but it also has nothing extra on top of B. Bohmian mechanics has extra "beables" - classical particle positions etc. (which are sometimes said to be measured instead of the collapsed wave functions) - so it is clearly not equivalent to QM. – Luboš Motl May 16 at 14:11

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