# Probability fluid velocity and quantum entanglement question

1. Jan 27, 2013

### JPaquim

Hey everyone, I'm a second year undergraduate student in aerospace engineering and I've been learning a bit of quantum mechanics for the past few weeks, for recreational purposes.
I've been following this textbook: http://www-thphys.physics.ox.ac.uk/people/JamesBinney/qb.pdf

On pages 40-41, the author makes an analogy between the probability density function associated with the wave function, and regular incompressible fluid flow/flow of electrons in a conductive wire. In the process, he associates a probability current density $J$ and a probability fluid velocity $v$. Then he mentions that in a state of well-defined momentum the probability fluid velocity reduces to classical particle velocity. First of all, how exactly does the computation $\nabla(\vec{x}\cdot\vec{p}) = \vec{p}$ work? What sort of product rule should I use and how will it all work out? And does this way of interpreting the wave function as being somehow represented by a fluid have any real utility? Is it something I should pursue, or is it useful only for this particular situation? What interpretation should I have of the fact that in a situation of definite momentum, the fluid velocity coincides with the particle velocity?

Another perhaps more general question regarding functional analysis. The author defines the product of two operators as simply function composition. But when he differentiates a product of operators, he uses the regular product rule, instead of the chain rule. How come?

And a very simple question, just to see if I've got the basics of quantum entanglement down. In the EPR experiment, the fact that one of the electron's z spin is measured to be +1/2 and subsequently the other one's x spin is measured to be +1/2 places the first one back in a state of indefinite z spin, right? So subsequent measurements of the z spin of the first could yield either +1/2 or -1/2, right? And supposedly that information can travel faster than light, being a so called "spooky action at a distance".

Cheers