Time Dependent Perturbations (2)

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).


valentin68

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