Consider the time-dependent perturbation given by

$H’ (x,t)=V’ (x)*\cos(ωt)$ for $0<t<T$. In this case the equation that gives the derivatives of the first order (time dependent) coefficients is

$(dc_n^1)/dt=-i/ℏ*∑_{(m=1)}^∞ c_m^0 V_{nm}’*\cos(ωt)*exp(iω_{nm} t)$

with $ω_{nm}=ω_n-ω_m$ the frequencies of the levels and $ω$ the frequency of the perturbation.

Integrate to find the first order coefficients.

We integrate

$c_n^1 (t)=-i/h_bar ∑_{(m=1)}^∞ c_m^0 V_nm’ ∫_0^t \cos (ωt)*\exp(iω_{nm} t)dt$

All coefficients $c_m^0$ are zero except the $l$-th coefficient for initial level which is $=1$

$c_n^1 (t)=-(i/ℏ) V_{nl}’ ∫_0^t\cos (ωt)*\exp (iω_{nl} t)dt=$

$=-(iV_{nl}’/2ℏ) ∫_0^t [\exp[i(ω_nl+ω)t]+\exp[i(ω_nl-ω)t]]dt$

$c_n^1 (t)=-(V_{nl}’/2ℏ)*[\frac{\exp [i(ω_{nl}+ω)t]-1}{(ω_{nl}+ω)}+\frac{\exp [i(ω_{nl}-ω)t]-1}{(ω_nl-ω)}]$

Which of the two terms in the expression will have a larger contribution to $c_n^1 (t)$. Explain.

Suppose the perturbation is light so that $ω(photon)\sim ω_{nl}$. In this case the second term denominator is very small $ω_{nl}-ω→0$ (so that the second term is very big) and the first term does not count.

Assuming you can neglect the smaller of the two terms, determine the transition probability, $P_n$ for this system. Explain how you arrived at the answer.

The transition probability is just (as in question 2 solved first)

$P_n (t)=|c_n (t) |^2$

$c_n (t)=-(V_{nl}’/2ℏ)*\frac{\exp [i(ω_{nl}-ω)t]-1}{(ω_{nl}-ω)}=$

$=-(V_{nl}’/2ℏ)*\frac{exp[i(ω_{nl}-ω)t/2]}{(ω_{nl}-ω)]}*{\exp[i(ω_{nl}-ω)t/2]-\exp[-i(ω_{nl}-ω)t/2] }$

$c_n (t)=-(V_{nl}’/2ℏ)*\frac{\sin[(ω_{nl}-ω)t/2]}{(ω_{nl}-ω)}*\exp[i(ω_{nl}-ω)t/2]$

$P_n (t)=[|V_{nl}’ |^2/(4ℏ^2 )]*\frac{\sin^2[(ω_{nl}-ω)t/2]}{(ω_{nl}-ω)^2}$

This is called the Fermi Golden Rule (for the probability of transition).