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 $H_{ab}’=V_{ab}/2*exp(iωt)$
a)
$(\dot{c_a}) ̇=-i/ℏ H_{ab}’*exp(-iω_0 t)*c_b$ 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 $(c_b ) ̇=-i*V_ba/2*exp[-i(ω-ω_0 )t]*c_a$
b)
$\ddot{c_b} ̈=-(iV_ba)/2 [-i(ω-ω_0 ) ]*exp[-i(ω-ω_0 )t]*c_a-i*V_ba/2*exp[-i(ω-ω_0 )t]*(\ddot{c_a}) ̇$
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.
c)
For solution
$c_b=exp(iλt)$
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
$0>λ=V_ba/4*(1-√(5/4))$
So that
$R_{ab}=d/dt |c_b (t) |^2=d/dt exp(2*|λ|*t)$ with $|λ|>0$
$R_{ab}=2|λ|*exp(2|λ|t)=V_ba/2*exp(2*|λ|*t)$
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
$R_{ab}=(dP_ab)/dt=|V_{ab} |^2*\sin(λt)/λ^2 *λ/2=(|V_{ab} |^2)/2λ*\sin(λ*t)≈V_{ab}/2*\sin(λ*t)$
