Rotating Wave (Physics 325)

1. Consider the Rotating wave approximation (cf. Griffiths Problem 9.7) in units with $\hbar = 1$,

$H′_{ba} = (V_{ba}/2)e^{−i\omega t}$; 

$H′_{ab} = (V_{ab}/2)e^{i\omega t}$;

(a) Show that $\dot{c}a = (−iV_{ab}/2)e^{i(\omega-\omega_0)t}*c_b$; 

$\dot{c}b = (−iVba/2)e^{−i(\omega-\omega_0)t}*c_a$

(b) Evaluate $\ddot{c_b}$ and derive a homogeneous 2nd order differential equation for $c_b$.

(c) Assuming independent solutions of the form $exp(iλt)$, find the solution for $c_b(t)$ such that $c_b(0) = 0$ and the transition rate $R = (d/dt)|cb(t)|^2$. Compare with Griffiths Equation 9.28.

Rotating wave approximation is

$H_{ba}’=V_{ba}/2*exp⁡(-iωt)$   and $H_{ab}’=V_{ab}/2*exp⁡(iωt)$


$(\dot{c_a}) ̇=-i/ℏ H_{ab}’*exp⁡(-iω_0 t)*c_b$     and $(c_b ) ̇=-(i/ℏ) H_{ba}’*exp⁡(iω_0 t)*c_a$
Therefore $(ℏ=1)$:

$(\dot{c_a}) ̇=-i*V_ab/2*exp⁡[i(ω-ω_0 )t]*c_b$    and $(c_b ) ̇=-i*V_ba/2*exp⁡[-i(ω-ω_0 )t]*c_a$

$\ddot{c_b} ̈=-(iV_ba)/2 [-i(ω-ω_0 ) ]*exp[-i(ω-ω_0 )t]*c_a-i*V_ba/2*exp⁡[-i(ω-ω_0 )t]*(\ddot{c_a}) ̇$

In the first approximation $c_a=1$  so that

$\ddot{c_b} ̈=-(V_ba (ω-ω_0))/2*e^[-i(ω-ω_0 )t] -(V_{ab} V_{ba})/4*c_b$

$\ddot{c_b }+(V_{ab} V_{ba})/4 c_b=-(V_{ba} (ω-ω_0))/2*e^[-i(ω-ω_0 )t]$

This is second order ODE with free term.


For solution


One has

$[-λ^2+(V_ab V_ba)/4] e^iλt=-V_ba/2*(ω-ω_0 )*exp⁡(-i(ω-ω_0 )t)$

So that

$λ=-(ω-ω_0 )   and-λ^2+(V_ab V_ba)/4+λ*V_ba/2=0$
From equation 9.25 one has

$V_ba=V_ab^*    so that-λ^2+λ*V_ba/2+|V_ba |^2/4=0$
$λ=V_ba/4 (1±√(5/4))$

Now I am lost because for solutions like $c_b (t)=exp⁡(iλt)$    never fulfull the condition $c_b (0)=0$
I suppose


So that

$R_{ab}=d/dt |c_b (t) |^2=d/dt  exp⁡(2*|λ|*t)$  with $|λ|>0$


Equation 9.28 from Griffits is referring to the probability of transition not to its rate.

$P_{ab}=|c_b |^2=|V_{ab} |^2/ℏ*\sin^2⁡(λt/2)/λ^2$     with $ℏ=1$
So that

$R_{ab}=(dP_ab)/dt=|V_{ab} |^2*\sin⁡(λt)/λ^2 *λ/2=(|V_{ab} |^2)/2λ*\sin⁡(λ*t)≈V_{ab}/2*\sin⁡(λ*t)$