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Parallelism between quantum density matrix and a probability distribution in classical mechanics

Physics Asked on June 1, 2021

In quantum statistical mechanics we often define a density matrix as

$rho = sum_{i} p_{i} | Psi_{i} rangle langle Psi_{i} | $.

Its time volution is determined by the equation:

$rho left(tright) = U left(tright) rho U^{dagger} left( t right) $ and it allows us to determine the expectation value of any observable $O$ of the system at any time as

$langle O (t) rangle = mathrm{Tr} left[ rho(t) O right] $. Defining Heisenberg’s time evolution as $O(t)=U^{dagger} left( t right) O U left( t right) $ it is easy to see the following two things:

  1. $langle O rho(t) rangle = mathrm{Tr}left[ rho O(t) right] $

  2. $ mathrm{Tr} left[ rho(t) O(t) right] = mathrm{Tr}left[ rho O right] $

Number two is particularly interesting as it means that time evolution of both the probability distribution and the observable leaves measurements unchanged.

I tried to take this as a starting point for a statistical description of classical mechanics. I therefore defined a "classical state" as a probability distribution over the phase space $(textbf{Q}, textbf{P} )$: $phi ( textbf{Q} , textbf{P})$. I know how observables evolve in classical mechanics: $frac{partial}{partial t} f_{t} left( textbf{Q}, textbf{P} right) = left{f_{t} , H right} $. Using relation 2) (which I expect to be true in classical physics as well) I define the classical state at time $t$ such that:

$constant = int dtextbf{Q} dtextbf{P} phi_{t} ( textbf{Q} , textbf{P}) f_{t} left( textbf{Q}, textbf{P} right) $. By deriving both sides of the equation in $t$ and substituting the classical equations of motion for the function $f_{t} $ I get:

$0 = int dtextbf{Q} dtextbf{P} left[ dot{phi}_{t} f_{t} + phi_{t} right{f_{t} , H left} right] $. By expanding Poisson’s brackets, integrating by parts in N dimensions, and equating the integrand to zero I get to the equation:

$dot{phi}_{t} = left{ phi_{t}, H right} $, which kind of looks like Liouville’s theorem but it’s not because it’s missing a $-$ sign in front of the brackets. I expected to get exactly Liouville’s theorem and I don’t understand what went wrong. Maybe relation 2) doesn’t work in classical mechanics, or it needs a little fixing, for example if I claim
$constant = int dtextbf{Q} dtextbf{P} phi_{-t} ( textbf{Q} , textbf{P}) f_{t} left( textbf{Q}, textbf{P} right) $ instead, I should get Liouville’s theorem. Maybe I have to reverse time?

One Answer

  1. $ mathrm{Tr} left[ rho(t) O(t) right] = mathrm{Tr}left[ rho O right] $

Number two is particularly interesting as it means that time evolution of both the probability distribution and the observable leaves measurements unchanged.

This is a false premise. In quantum mechanics the evolution operator affects either the density matrix (in the Schrödinger picture) or the operators (in the Heisenberg picture), but not both.

Answered by Emilio Pisanty on June 1, 2021

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