# Time dependent perturbations and method of succesive approximations

1. May 11, 2005

### mattlorig

In chapter nine of Griffiths' Quatum Mechanics text, he talks about the method of succesive approximations as a method for solving a two level system in time dependent perturbation theory.

d(ca)/dt = f(t) cb --> ca_n = int[ f(t') * cb_n-1, dt', 0 t]
d(cb)/dt = g(t) ca --> cb_n = int[ g(t') * ca_n-1, dt', 0 t]

So, for the case were ca(0) = 1 and cb(0) = 0 one would get:

ca1 = int [f(t') * 0, dt', 0, t] = 0
cb1 = int [g(t') * 1, dt', 0, t]

but griffiths says ca1 = 1.

So, my question is the following. Is ca = ca_0 + ca_1 + ca_2 + ...
or, is ca ~ ca_n (with larger n being more precise)?

I hope my question was clear. I should really learn LATEX.

2. May 11, 2005

### mattlorig

Maybe I should clarify my question. Regarding the method of successive approxiamtions griffiths talks about in time dependent perturbation theory, is ca
= SUM (ca_n)
= LIM {ca_n}
?
Hopefully that's easier to understand than my first post.

3. May 11, 2005

### dextercioby

Nope,i don't really follow.Why would $c_{a,1}$ be 0...?

Daniel.

4. May 11, 2005

### mattlorig

because the integral of zero is zero.

5. May 11, 2005

### dextercioby

Okay,alright.What page of Griffiths ?

Daniel.

6. May 11, 2005

### mattlorig

p 302 of the (my book is black...I think it's the second newest edition). Griffiths does say "ca_2 includes the zeroth order term; the 2nd order correction would be the integral term alone". But, I think griffiths use of ca_n is ambiguous here. In any case, I'm pretty sure that if I just substitute cb_n-1 into the integral for ca_n, and let ca = SUM( ca_n), that would be correct.

7. May 11, 2005

### dextercioby

$$\left\{\begin{array}{c} \dot{c}_{a}(t)=\frac{1}{i\hbar}H'_{ab}(t)e^{-i\omega_{0}t}c_{b}(t)\\ \dot{c}_{b}(t)=\frac{1}{i\hbar}H'_{ba}(t)e^{i\omega_{0}t}c_{a}(t) \end{array}\right$$(1)

Initial conditions

$$\left\{\begin{array}{c}c_{a}(0)=1\\c_{b}(0)=0 \end{array}\right$$ (2)

Zero-th approximation

$$c_{a}^{(0)}(t)=1$$ (3)

$$c_{b}^{(0)}(t)=0$$ (4)

First order.Plug (3) & (4) in the equations (1) and integrate

$$\frac{dc_{a}^{(1)}(t)}{dt}=0\Rightarrow c_{a}^{(1)}(t)=\mbox{const}=c_{a}(0)=1$$ (5)

$$\frac{dc_{b}^{(1)}(t)}{dt}=\frac{1}{i\hbar}H'_{ba}(t)e^{i\omega_{0}t} \Rightarrow c_{b}^{(1)}(t)=\frac{1}{i\hbar}\int_{0}^{t} H'_{ba}(t')e^{i\omega t'} \ dt$$ (6)

Second order.Plug the first order approx given by (5) & (6) into the system (1).

$$\frac{dc_{a}^{(2)}(t)}{dt}=\frac{1}{i\hbar}H'_{ab}(t)e^{-i\omega_{0}t}\frac{1}{i\hbar} \int_{0}^{t} H'_{ba}(t')e^{i\omega t'} \ dt \Rightarrow c_{a}^{(2)}(t)=\mbox{const}-\frac{1}{\hbar^{2}}\int_{0}^{t} H'_{ab}(t')e^{-i\omega_{0}t'} \left[\int_{0}^{t'} H'_{ba}(t'')e^{i\omega t''} \ dt'' \right] \ dt'$$ (7)

Imposing the condition $c^{(2)}_{a}(0)=1$ (8),you get the formula (9.18) from Griffiths.

You try now for $c_{b}^{(2)}(t)$.See if you get what Griffiths says:it stays unchanged.

Go for the 3-rd order. Make sure you got it all clear.

Daniel.

Last edited: May 12, 2005
8. May 12, 2005

### mattlorig

Thanks Daniel for the help. I will find the 3rd order approximations to ca and cb today. I think my main mistake was just not realizing why ca_1 = 1. Now that that's clear, the rest should be fairly simple.

Also, after doing a bit more research, I found this method in my old Diff Eq book. Apparently it is known as Picard's Iteration Method.

Lastly, I wanted to thank you for always being kind and helping me through the (many) problems I've asked about. You've been extremeley kind and helpful on a number of occasions, and I appreciate it very much.

9. May 12, 2005

### dextercioby

Thank you.Well,it resembles in a way the method of iterations when solving the integral equation which gives birth to the Born series.

But it's different.In that case,u get the solution as an infinite sum of perturbative approximations.In this case,it's not a sum anymore.

$$c_{a}(t)\neq c_{a}^{(0)}(t)+c_{a}^{(1)}(t)+c_{a}^{(2)}(t)+...$$

,but $$c_{a}(t)\simeq c_{a}^{(0)}(t) \ \mbox{in the zero-th order}$$

$$c_{a}(t)\simeq c_{a}^{(1)}(t) \ \mbox{in the first order}$$

and so on.I hope u see the difference.U'll have to compare this case with the Born series (as i said before),with the Dyson series and with the series which appear in the stationary perturbative theory for the nondegerate energy levels and see where they look alike and where they are different.

Daniel.