Sudden Perturbations

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

$$H_1(t)=\left\{ \begin{array}{lll} 0 &\mbox{\hspace{0.5cm}}&\mbox{for $t<0$} \\ [0.5ex] H_1&&\mbox{for $t\geq 0$}\end{array}\right.,$$ (8.61)

where $H_1$ is independent of time, but is generally a function of the position, momentum, and spin operators. Suppose that the system is definitely in state $\vert i\rangle$ at time $t=0$ . According to Equations (8.58) and (8.59) (with $t_0=0$ ),

$$c_f^{\,(0)}(t) = \delta_{if},$$ (8.62)

$$c_f^{\,(1)}(t) = -\frac{{\rm i}}{\hbar}\, H_{fi} \int_0^t dt'\,\exp[\,{\rm i}\, ... ...{fi}\,(t'-t)]= \frac{H_{fi}}{E_f - E_i}\, [1- \exp(\,{\rm i}\,\omega_{fi}\,t)],$$ (8.63)

giving

$$P_{i\rightarrow f}(t) \simeq \left\vert c_f^{\,(1)}(t)\right\vert... ...vert E_f - E_i\vert^{\,2}}\, \sin^2\left[ \frac{(E_f-E_i)\,t}{2\,\hbar}\right],$$ (8.64)

for $i\neq f$ . The transition probability between states $\vert i\rangle$ and $\vert f\rangle$ can be written

$$P_{i\rightarrow f}(t) = \frac{\vert H_{fi}\vert^{\,2} \,t^{\,2}}{\hbar^{\,2}} \,{\rm sinc}^2\left[ \frac{(E_f-E_i)\,t}{2\,\hbar}\right],$$ (8.65)

where

$${\rm sinc}(x)\equiv \frac{\sin x}{x}.$$ (8.66)

The sinc function is highly oscillatory, and decays like $1/\vert x\vert$ at large $\vert x\vert$ . In fact, it is a good approximation to say that ${\rm sinc}(x)$ is small compared to unity except when $\vert x\vert \stackrel {_{\normalsize <}}{_{\normalsize\sim}}\pi$ . It follows that the transition probability, $P_{i\rightarrow f}$ , is negligibly small unless

$$\vert E_f - E_i\vert \stackrel {_{\normalsize <}}{_{\normalsize\sim}}\frac{2\pi\, \hbar}{t}.$$ (8.67)

Note that, in the limit $t\rightarrow \infty$ , only those transitions that conserve energy (i.e., $E_f=E_i$ ) have an appreciable probability of occurrence. At finite $t$ , is is possible to have transitions that do not exactly conserve energy, provided that

$${\mit\Delta} E \,{\mit\Delta} t \stackrel {_{\normalsize <}}{_{\normalsize\sim}}h,$$ (8.68)

where ${\mit\Delta} E = \vert E_f - E_i\vert$ is the change in energy of the system associated with the transition, and ${\mit\Delta} t = t$ is the time elapsed since the perturbation was switched on. This result is a manifestation of the well-known uncertainty relation for energy and time [53]. Incidentally, the energy-time uncertainty relation is fundamentally different from 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., $E_f=E_i$ ) is

$$P_{i\rightarrow f} (t) = \frac{\vert H_{fi}\vert^{\,2}\,t^{\,2}}{\hbar^{\,2}},$$ (8.69)

where use has been made of ${\rm sinc}(0) = 1$ . Note that this probability grows quadratically in 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 $H_1$ 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, $\vert i\rangle$ . It is helpful to define the density of states, $\rho(E)$ , where the number of final states lying in the energy range $E$ to $E+dE$ is given by $\rho(E)\,dE$ . Thus, the probability of a transition from the initial state $\vert i\rangle$ to one of the continuum of possible final states is

$$P_{i\rightarrow} (t) = \int_{-\infty}^\infty dE_f\,P_{i\rightarrow f}(t) \,\rho(E_f),$$ (8.70)

giving

$$P_{i\rightarrow} (t) = \frac{2\, t}{\hbar} \int_{-\infty}^\infty dx\, \vert H_{fi}\vert^{\,2}\, \rho(E_f) \,{\rm sinc}^2(x),$$ (8.71)

where

$$x=\frac{(E_f-E_i)\,t}{2\,\hbar},$$ (8.72)

and use has been made of Equation (8.68). We know that, in the limit $t\rightarrow \infty$ , the function ${\rm sinc}(x)$ is only non-zero in an infinitesimally narrow range of final energies centered on $E_f=E_i$ . It follows that, in this limit, we can take $\rho(E_f)$ and $\vert H_{fi}\vert^{\,2}$ out of the integral in the previous formula to obtain

$$P_{i\rightarrow[f]} (t) = \left.\frac{2\pi}{\hbar}\, \overline{\vert H_{fi}\vert^{\,2}} \,\rho(E_f)\,t\, \right\vert _{E_f\simeq E_i},$$ (8.73)

where $P_{i\rightarrow [f]}$ denotes the transition probability between the initial state, $\vert i\rangle$ , and all final states, $\vert f\rangle$ , that have approximately the same energy as the initial state. Here, $\overline{\vert H_{fi}\vert^{\,2}}$ is the average of $\vert H_{fi}\vert^{\,2}$ over all final states with approximately the same energy as the initial state. In deriving the previous formula, we have made use of the result [59]

$$\int_{-\infty}^{\infty} dx\,\,{\rm sinc}^2(x) = \pi.$$ (8.74)

Note that the transition probability, $P_{i\rightarrow [f]}$ , is now proportional to $t$ , instead of $t^{\,2}$ .

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

$$w_{i\rightarrow [f]} = \frac{d P_{i\rightarrow [f]}}{dt},$$ (8.75)

which implies that

$$w_{i\rightarrow [f]} = \left.\frac{2\pi}{\hbar}\, \overline{\vert H_{fi}\vert^{\,2}} \,\rho(E_f) \right\vert _{E_f\simeq E_i}.$$ (8.76)

This appealingly simple result is known as Fermi's golden rule [29,44]. Note that the transition rate is constant in time (for $t>0$ ): that is, 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 more usually written

$$w_{i\rightarrow f} = \frac{2\pi}{\hbar} \,\vert H_{fi}\vert^{\,2}\, \delta(E_f - E_i),$$ (8.77)

where it is understood that this formula must be integrated with $\int dE_f\,(\cdots)\,\rho(E_f)$ in order to obtain the actual transition rate.

Let us now calculate the second-order term in the Dyson series, using the constant perturbation (8.64). From Equation (8.60), we find that (with $t_0=0$ )

$$c_f^{\,(2)}(t) = \left(\frac{-{\rm i}}{\hbar}\right)^2 \sum_m H_{fm} \,H_{mi} \i... ... i}\,\omega_{fm}\,t'\,) \int_0^{t'} dt'' \,\exp(\,{\rm i} \,\omega_{mi}\,t''\,)$$

$$=\frac{\rm i}{\hbar} \sum_m \frac{H_{fm} \,H_{mi}}{E_m - E_i} \in... ...[\exp(\,{\rm i}\,\omega_{fi}\,t'\,) - \exp(\,{\rm i}\,\omega_{fm}\,t')\,\right]$$ (8.78)

$$= \frac{{\rm i}\,t}{\hbar} \sum_m \frac{H_{fm} \,H_{mi}}{E_m - E_... ...a_{fm} \,t}{2}\right) \,{\rm sinc}\left(\frac{\omega_{fm}\,t}{2}\right)\right].$$

Thus,

$$c_f(t) = c_f^{\,(1)}+ c_f^{\,(2)} = \frac{{\rm i}\,t}{\hbar} \exp\left(\frac{\,{\rm i}\,\omega_{fi}... ...i}}{E_m - E_i}\right)\, {\rm sinc} \left(\frac{\omega_{fi}\,t}{2}\right)\right.$$

$$\left. - \sum_m\frac{H_{fm}\,H_{mi}}{E_m - E_i} \exp\left(\frac{\... ...ega_{im}\,t}{2}\right)\,{\rm sinc}\left(\frac{\omega_{fm}\,t}{2}\right)\right],$$ (8.79)

where use has been made of Equation (8.66). It follows, by analogy with the previous analysis, that

$$w_{i\rightarrow [f]} =\left. \frac{2\pi}{\hbar}\, \overline{ \lef... ...\,H_{mi}}{E_m - E_i}\right\vert^{\,2}} \rho(E_f) \right\vert _{E_f \simeq E_i},$$ (8.80)

where the transition rate is calculated for all final states, $\vert f\rangle$ , with approximately the same energy as the initial state, $\vert i\rangle$ , and for intermediate states, $\vert m\rangle$ whose energies differ from that of the initial state. The fact that $E_m\neq E_i$ causes the final term on the right-hand side of Equation (8.82) to average to zero during the evaluation of the transition probability. (See Exercise 4.)

According to Equation (8.83), a second-order transition takes place in two steps. First, the system makes a non-energy-conserving transition to some intermediate state $\vert m\rangle$ . Subsequently, the system makes another non-energy-conserving transition to the final state $\vert f\rangle$ . The net transition from state $\vert i\rangle$ to state $\vert f\rangle$ 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 previous formula clearly breaks down if $H_{fm}\,H_{mi}\neq 0$ when $E_m = E_i$ . This problem can be avoided by gradually turning on the perturbation: that is, $H_1\rightarrow \exp(\eta\,t)\, H_1$ (where $\eta$ is very small). The net result is to change the energy denominator in Equation (8.83) from $E_i-E_m$ to $E_i - E_m +{\rm i}\,\hbar\,\eta$ .