Quantum Mechanics – How Does De Broglie-Bohm Theory Explain the Stern-Gerlach Experiment?

measurement-problemquantum mechanicsquantum-interpretationsquantum-spin

The Copehagen interpretation of QM explains the Stern–Gerlach experiment by asserting that a particle is in a superposition of states and doesn't have a definite spin until measured. However, the de Broglie–Bohm interpretation suggests that there is always a definite position of the particle, with a well-defined initial state (which may not be measurable with enough precision), including its spin. This initial state defines the path the particle will take through the apparatus.

How does one explain the results that the spin of a particle is always measured with its full magnitude along the plane of measurement? I would guess it has something to do with the pilot-wave's effects on the particle as its angular momentum is changed by the magnetic field.

Best Answer

How does one explain the results that the spin of a particle is always measured with its full magnitude along the plane of measurement?

In Bohmian mechanics, the spinor valued wave function (packet), under the influence of the apparatus, splits into two discrete packets only one of which guides the Bohmian particle; the other packet is 'empty'.

Which packet the particle ends up in is determined by the initial configuration.

On this view, spin is not a property of the particle but wavefunction

From the article "Bohmian Mechanics" at SEP

Bohmian mechanics makes sense for particles with spin, i.e., for particles whose wave functions are spinor-valued. This occurs because the Stern-Gerlach magnets are so designed and oriented that a wave packet (a localized wave function with reasonably well defined velocity) directed towards the magnet will, by virtue of the Schrödinger evolution, separate into distinct packets — corresponding to the spin components of the wave function and moving in the discrete set of directions. The particle itself, depending upon its initial position, ends up in one of the packets moving in one of the directions.

The probability distribution for the result of such a Stern-Gerlach experiment can be conveniently expressed in terms of the quantum mechanical spin operators — for a spin-1/2 particle given by certain 2 by 2 matrices called the Pauli spin matrices — in the manner alluded to above. From a Bohmian perspective there is no hint of paradox in any of this — unless we assume that the spin operators correspond to genuine properties of the particles.


Very interesting link. So spin is a property of the wave function, and not of the particle?

From the paper "What you always wanted to know about Bohmian mechanics but were afraid to ask"

Thus, in general, 'measurements' do not measure anything in the closer meaning of the term. The only exception being of course position measurements, and, in some sense momentum-measurements. The latter do indeed measure the asymptotic (Bohmian) velocities. Hence, the only properties of a Bohmian particle are its position and its velocity.

Just as $\psi$ is no classical field, the Bohmian particles are no classical particles, i.e. they are no bearers of properties other than position. Therefore a physical object like e.g. an electron should not be confused with the Bohmian particle at position $Q_i$ . It is represented by the pair $(\psi, Q_i )$.

Related Question