Operators change form for density matrix equations?

1. May 19, 2015

Agrippa

Imagine applying an operator to a wave-function:

$\psi_t(x_1, x_2, ..., x_n) \rightarrow \frac{L_n(x)\psi_t(x_1, x_2, ..., x_n)}{||\psi_t(x_1, x_2, ..., x_n)||}$

Where $\psi _t(x_1, x_2, ..., x_n)$ is initial system state vector, denominator is normalization factor, and Ln(x) is a linear operator equal to:

$L_n(x) = \frac{1}{(\pi r^2_c)^{3/4}}e^{-(q_n - x)^2 / 2r^2_c}$

So, position wave-function is multiplied by a Gaussian (with width $r_c$; $q_n$ is position operator for nth particle, and x is random spatial variable where Gaussian multiplication is centred).

When this exact same equation is presented in a master equation for density matrix we get:

$\frac{d}{dt}\rho(t) = -\frac{i}{\hbar}[H, \rho(t)] - T[\rho(t)]$

where H is standard quantum Hamiltonian and T[] represents effect of the operator. In position representation:

$<x|T[\rho(t)]|y> = [1 - e^{-(x - y)^2 / 4r^2_c}]<x|T[\rho(t)]|y>$

Clearly, form of Gaussian function has changed, but why? Standard presentations (e.g. pp.30-33) never explain the change.
Is there anyone out there who knows the math well enough to be able to explain why all the changes occur e.g. why do we replace the initial fraction with "1 - "? And why replace $2r^2_c$ with $4r^2_c$?

What would go wrong if we simply replaced $1 - e^{-(x - y)^2 / 4r^2_c}$ with $\frac{1}{(\pi r^2_c)^{3/4}}e^{-(q_n - x)^2 / 2r^2_c}$ ?

2. May 24, 2015

Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?