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8.6: Sudden Perturbations

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Consider, for example, a constant perturbation that is suddenly switched on at time t=0 :

H1(t) =0 \( \mbox{\hspace{2.0cm}for \) H1(t) =H1\( \mbox{\hspace{1.75cm}for \), ???

where H1 is time-independent, but is generally a function of the position, momentum, and spin operators. Suppose that the system is definitely in state |i at time t=0 . According to Equations ???-??? (with t0=0 ),

c???n(t) =δin, ??? c???n(t) $ = -\frac{{\rm i}}{\hbar}\, H_{ni} \int_0^t dt'\,\exp[\,{\rm i}\, ...
...{ni}\,(t'-t)]= \frac{H_{ni}}{E_n - E_i}\, [1- \exp(\,{\rm i}\,\omega_{ni}\,t)],$ ???

giving

$ P_{i\rightarrow n}(t) \simeq \vert c_n^{\ref{1}}\vert^{\,2} = \frac{4...
...vert E_n - E_i\vert^{\,2}}\, \sin^2\left[ \frac{(E_n-E_i)\,t}{2\,\hbar}\right],$ ???

for in . The transition probability between states |i and |n can be written

Pin(t)=|Hni|2t22sinc2[(EnEi)t2], ???

where

sinc(x)sinxx. ???

The sinc function is highly oscillatory, and decays like 1/|x| at large |x| . It is a good approximation to say that sinc(x) is small except when |x|<π . It follows that the transition probability, Pin , is small except when

|EnEi|<2πt. ???

Note that in the limit t only those transitions that conserve energy (i.e., En=Ei ) have an appreciable probability of occurrence. At finite t , is is possible to have transitions which do not exactly conserve energy, provided that

ΔEΔt<h, ???

where ΔE=|EnEi| is the change in energy of the system associated with the transition, and Δt=t is the time elapsed since the perturbation was switched on. This result is just a manifestation of the well-known uncertainty relation for energy and time. Incidentally, the energy-time uncertainty relation is fundamentally different to the position-momentum uncertainty relation, because (in non-relativistic quantum mechanics) position and momentum are operators, whereas time is merely a parameter.

The probability of a transition that conserves energy (i.e., En=Ei ) is

Pin(t)=|Hin|2t22, ???

where use has been made of sinc???=1 . Note that this probability grows quadratically with time. This result is somewhat surprising, because it implies that the probability of a transition occurring in a fixed time interval, t to t+dt , grows linearly with t , despite the fact that H1 is constant for t>0 . In practice, there is usually a group of final states, all possessing nearly the same energy as the energy of the initial state |i . It is helpful to define the density of states, ρ(E) , where the number of final states lying in the energy range E to E+dE is given by ρ(E)dE . Thus, the probability of a transition from the initial state i to any of the continuum of possible final states is

Pi(t)=dEnPin(t)ρ(En), ???

giving

Pi(t)=2tdx|Hni|2ρ(En)sinc2(x), ???

where

x=(EnEi)t/2, ???

and use has been made of Equation ???. We know that in the limit t the function sinc(x) is only non-zero in an infinitesimally narrow range of final energies centered on En=Ei . It follows that, in this limit, we can take ρ(En) and |Hni|2 out of the integral in the above formula to obtain

Pi[n](t)=2π¯|Hni|2ρ(En)t|EnEi, ???

where Pi[n] denotes the transition probability between the initial state |i and all final states |n that have approximately the same energy as the initial state. Here, ¯|Hni|2 is the average of |Hni|2 over all final states with approximately the same energy as the initial state. In deriving the above formula, we have made use of the result

dxsinc2(x)=π. ???

Note that the transition probability, Pi[n] , is now proportional to t , instead of t2 .

It is convenient to define the transition rate, which is simply the transition probability per unit time. Thus,

wi[n]=dPi[n]dt, ???

giving

wi[n]=2π¯|Hni|2ρ(En)|EnEi. ???

This appealingly simple result is known as Fermi's golden rule. Note that the transition rate is constant in time (for t>0 ): i.e., the probability of a transition occurring in the time interval t to t+dt is independent of t for fixed dt . Fermi's golden rule is sometimes written

win=2π|Hni|2δ(EnE), ???

where it is understood that this formula must be integrated with dEnρ(En) to obtain the actual transition rate.

Let us now calculate the second-order term in the Dyson series, using the constant perturbation ???. From Equation ??? we find that

c???n(t) $ = \left(\frac{-{\rm i}}{\hbar}\right)^2 \sum_m H_{nm} H_{mi} \int...
...i}\,\omega_{nm}\,t'\,) \int_0^{t'} \, dt'' \,\exp(\,{\rm i} \,\omega_{mi}\,t\,)$ $ =\frac{\rm i}{\hbar} \sum_m \frac{H_{nm} \,H_{mi}}{E_m - E_i} \in...
...[\exp(\,{\rm i}\,\omega_{ni}\,t'\,) - \exp(\,{\rm i}\,\omega_{nm}\,t']\,\right)$ $ = \frac{{\rm i}\,t}{\hbar} \sum_m \frac{H_{nm} H_{mi}}{E_m - E_i}...
...t/2)- \exp(\,{\rm i}\,\omega_{nm} \,t/2) \,{\rm sinc}(\omega_{nm}\,t/2)\right].$ ???

Thus,

cn(t)=c???n+c???n $ = \frac{{\rm i}\,t}{\hbar} \exp(\,{\rm i}\,\omega_{ni}\,t/2)\, \l...
... \frac{H_{nm}\,H_{mi}}{E_m - E_i}\right)\, {\rm sinc} (\omega_{ni}\,t/2)\right.$ mHnmHmiEmEiexp(iωimt/2)sinc(ωnmt/2)], ???

where use has been made of Equation ???. It follows, by analogy with the previous analysis, that

$ w_{i\rightarrow [n]} =\left. \frac{2\pi}{\hbar}\, \overline{ \lef...
...\,H_{mi}}{E_m - E_i}\right\vert^{\,2}} \rho(E_n) \right\vert _{E_n \simeq E_i},$ ???

where the transition rate is calculated for all final states, |n , with approximately the same energy as the initial state, |i , and for intermediate states, |m whose energies differ from that of the initial state. The fact that EmEi causes the last term on the right-hand side of Equation ??? to average to zero (due to the oscillatory phase-factor) during the evaluation of the transition probability.

According to Equation ???, a second-order transition takes place in two steps. First, the system makes a non-energy-conserving transition to some intermediate state |m . Subsequently, the system makes another non-energy-conserving transition to the final state |n . The net transition, from |i to |n , conserves energy. The non-energy-conserving transitions are generally termed virtual transitions, whereas the energy conserving first-order transition is termed a real transition. The above formula clearly breaks down if HnmHmi0 when Em=Ei . This problem can be avoided by gradually turning on the perturbation: i.e., $ H_1\rightarrow \exp(\eta\,t)\,
H_1$ (where η is very small). The net result is to change the energy denominator in Equation ??? from EiEm to $ E_i -
E_m +{\rm i}\,\hbar\,\eta$ .

Contributors

  • Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)


This page titled 8.6: Sudden Perturbations is shared under a not declared license and was authored, remixed, and/or curated by Richard Fitzpatrick.

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