# Iteration method for density equations

1. Jun 27, 2014

### Robert_G

hi there; I need some help with the following formulas

In the interaction picture.

$\frac{d}{dt}\rho(t)=\frac{1}{i\hbar}[V(t),\rho(t)]$ (1)

Then

$\rho(t+\Delta t) = \rho(t)+\frac{1}{i\hbar}\int_t^{t+\Delta t} dt' [V(t'),\rho(t')]$ (2)

This equation can be iterated. and it is

$\Delta \rho(t)=\frac{1}{i\hbar}\int_t^{t+\Delta t} dt' [V(t'),\rho(t')]+(\frac{1}{i\hbar})^2 \int_t^{t+\Delta t} dt' \int_t^{t'}dt''[\underbrace{V(t'),[V(t'')}_{Note\;t'\;and\;t''},\rho(t'')]]$ (3)

$\Delta \rho(t) = \rho(t+\Delta t) - \rho(t)$

I can understand the eq.(2), but not the eq. (3).
Is anybody know how to get the equation (3). and why do we want to do such calculation?

2. Jun 27, 2014

### maajdl

Formula (2) is an exact formula.
By substituting formula (2) into formula (2), you get formula (3).
Formula (3) is therefore also an exact formula.

These formulas are typically used for perturbation calculations.
These are approximations where ρ is constant without the interaction V: ρ(t) ~ρ(0) for V=0.
Replacing ρ(t') = ρ(0) in formula (2) is then a first order approximation.
Replacing ρ(t") = ρ(0) in formula (3) is then a second order approximation.
The second order approximation should be better than the first order ... if this sequence is converging!

3. Jun 27, 2014

### Robert_G

If formula (2) is already the exact solution. It should contain all the information, why we still iterate it to get the perturbation. It does not make sence to me.

4. Jun 27, 2014

### maajdl

It's an exact formula, but not really a solution.
To calculate ρ(t) from equation (2), you need to know ρ(t). (!)
Equation (2) replaces the differential equation (1), by an integral equation (2).
The equation (2) is not easier to solve for ρ(t) than equation (1).

However, equation (2) can be the starting point for an approximate solution.
If ρ(t)=ρ(0) is a good zero order approximation, then equation (2) gives you the first order correction.
Similarly, equation (3) is a second order correction.

Of course, going from these formal equations to a practical solution might not be a piece of cake.
These formal solutions are only a starting point for more developments.

Try these ideas on some exercices.