Constant perturbations

Consider a constant perturbation which is suddenly switched on at time $t=0$:

$$H_1(t) = 0 \mbox{\hspace{2.1cm}for $t<0$}$$

$$H_1(t) = H_1\mbox{\hspace{1.75cm}for $t\geq 0$},$$ (792)

where $H_1$ is time-independent, 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 Eqs. (786)-(788) (with $t_0=0$),

$$c_n^{(0)}(t) = \delta_{in},$$ (793)

$$c_n^{(1)}(t) = -\frac{{\rm i}}{\hbar} H_{ni} \int_0^t \exp[ {\rm i} \omega_{ni} (t'-t)] dt'$$

$$= \frac{H_{ni}}{E_n - E_i} [1- \exp( {\rm i} \omega_{ni} t)],$$ (794)

giving

$$P_{i\rightarrow n}(t) \simeq \vert c_n^{(1)}\vert^2 = \frac{... ...vert^2} \sin^2\!\left[ \frac{(E_n-E_i) t}{2 \hbar}\right],$$ (795)

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

$$P_{i\rightarrow n}(t) = \frac{\vert H_{ni}\vert^2 t^2}{\hb... ...} {\rm sinc}^2\!\left[ \frac{(E_n-E_i) t}{2 \hbar}\right],$$ (796)

where

$${\rm sinc}(x)= \frac{\sin x}{x}.$$ (797)

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

$$\vert E_n - E_i\vert \stackrel {_{\normalsize <}}{_{\normalsize\sim}}\frac{2\pi \hbar}{t}.$$ (798)

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

$$\Delta E \Delta t \stackrel {_{\normalsize <}}{_{\normalsize\sim}}\hbar,$$ (799)

where $\Delta E = \vert E_n - E_i\vert$ is change in energy of the system associated with the transition, and $\Delta t = t$ is the time elapsed since the perturbation was switched on. Clearly, this result is just a manifestation of the well-known uncertainty relation for energy and time. This uncertainty relation is fundamentally different to the position-momentum uncertainty relation, since in non-relativistic quantum mechanics position and momentum are operators, whereas time is merely a parameter.

The probability of a transition which conserves energy (i.e., $E_n=E_i$) is

$$P_{i\rightarrow n} (t) = \frac{\vert H_{in}\vert^2 t^2}{\hbar^2},$$ (800)

where use has been made of ${\rm sinc}(0) = 1$. Note that this probability grows quadratically with time. This result is somewhat surprising, since 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 $i$ to any of the continuum of possible final states is

$$P_{i\rightarrow} (t) = \int P_{i\rightarrow n}(t) \rho(E_n) dE_n,$$ (801)

giving

$$P_{i\rightarrow} (t) = \frac{2 t}{\hbar} \int \vert H_{ni}\vert^2 \rho(E_n) {\rm sinc}^2(x) dx,$$ (802)

where

$$x=(E_n-E_i) t/2 \hbar,$$ (803)

and use has been made of Eq. (796). 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 centred on $E_n=E_i$. It follows that, in this limit, we can take $\rho(E_n)$ and $\vert H_{ni}\vert^2$ out of the integral in the above formula to obtain

$$P_{i\rightarrow[n]} (t) = \left.\frac{2\pi}{\hbar} \overli... ...t H_{ni}\vert^2} \rho(E_n) t \right\vert _{E_n\simeq E_i},$$ (804)

where $P_{i\rightarrow [n]}$ denotes the transition probability between the initial state $\vert i\rangle$ and all final states $\vert n\rangle$ which have approximately the same energy as the initial state. Here, $\overline{\vert H_{ni}\vert^2}$ is the average of $\vert H_{ni}\vert^2$ 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

$$\int_{-\infty}^{\infty} {\rm sinc}^2(x) dx = \pi.$$ (805)

Note that the transition probability, $P_{i\rightarrow [n]}$, 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,

$$w_{i\rightarrow [n]} = \frac{d P_{i\rightarrow [n]}}{dt},$$ (806)

giving

$$w_{i\rightarrow [n]} = \left.\frac{2\pi}{\hbar} \overline{\vert H_{ni}\vert^2} \rho(E_n) \right\vert _{E_n\simeq E_i}.$$ (807)

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

$$w_{i\rightarrow n} = \frac{2\pi}{\hbar} \vert H_{ni}\vert^2 \delta(E_n - E),$$ (808)

where it is understood that this formula must be integrated with $\int \rho(E_n) dE_n$ to obtain the actual transition rate.

Let us now calculate the second-order term in the Dyson series, using the constant perturbation (792). From Eq. (788) we find that

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

$$= \frac{\rm i}{\hbar} \sum_m \frac{H_{nm} H_{mi}}{E_m - E_i} \int... ...( {\rm i} \omega_{ni} t' ) - \exp( {\rm i} \omega_{nm} t'] \right) dt'$$

$$= \frac{{\rm i} t}{\hbar} \sum_m \frac{H_{nm} H_{mi}}{E_m - E_i} \left[ \exp( {\rm i} \omega_{ni} t/2) {\rm sinc}(\omega_{ni} t/2)\right.$$

$$\mbox{\hspace{2cm}}\left.- \exp( {\rm i} \omega_{nm} t/2) {\rm sinc}(\omega_{nm} t/2)\right].$$ (809)

Thus,

$$c_n(t) = c_n^{(1)}+ c_n^{(2)} = \frac{{\rm i} t}{\hbar} \exp( {\rm i} \omega_{ni} t/2) \lef... ... \frac{H_{nm} H_{mi}}{E_m - E_i}\right) {\rm sinc} (\omega_{ni} t/2)\right.$$

$$\left. - \sum_m\frac{H_{nm} H_{mi}}{E_m - E_i} \exp( {\rm i} \omega_{im} t/2) {\rm sinc}(\omega_{nm} t/2)\right],$$ (810)

where use has been made of Eq. (794). It follows, by analogy with the previous analysis, that

$$w_{i\rightarrow [n]} =\left. \frac{2\pi}{\hbar} \overline{... ... - E_i}\right\vert^2} \rho(E_n) \right\vert _{E_n \simeq E_i},$$ (811)

where the transition rate is calculated for all final states, $\vert n\rangle$, 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 last term on the right-hand side of Eq. (810) to average to zero (due to the oscillatory phase-factor) during the evaluation of the transition probability.

According to Eq. (811), 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 $\vert n\rangle$. The net transition, from $\vert i\rangle$ to $\vert n\rangle$, 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 above formula clearly breaks down if $H_{nm} H_{mi}\neq 0$ when $E_m = E_i$. This problem can be avoided by gradually turning on the perturbation: i.e., $H_1\rightarrow \exp(\eta t) H_1$ (where $\eta$ is very small). The net result is to change the energy denominator in Eq. (811) from $E_i-E_m$ to $E_i - E_m +{\rm i} \hbar \eta$.