Quantum Optics II Time dependent perturbation theory In time-dependent perturbation theory the main goal is to determine the time-evolution of a perturbed quantum system, with particular emphasis on calculating transition probabilities and modeling the irreversible decay of probability from a small quantum system coupled to a very large quantum system. The interaction picture will be used to discuss time-dependent perturbation.

## Time Dependent Perturbation Theory

### General Theory

Come back to Schrödinger picture. Suppose at $t=0$ we works out the eigenstates $\ket{n}$ of $H_0$ . We are interested in two of them, $\ket{f}$ and $\ket{i}$ (initial state and final state in Schrödinger picture). After time $t$ , the states in the Schrödinger picture become

$\ket{f,t} = \mathrm{e}^{-\frac{\mathrm{i}}{\hbar} E_f t} \ket{f} \qquad \ket{i,t} = \mathrm{e}^{-\frac{\mathrm{i}}{\hbar} E_i t} \ket{i}$

The questions is, what the transition probability $\mathcal{P}_{fi}(t)$ is for a system initially in the state $\ket{i}$ at time $t=t_0$ (could be zero) evolving into the state $\ket{f,t}$ at time $t$ ? Using the evolution operator, we can derive

$\mathcal{P}_{fi}(t) = \left| \langle f,t| U^S(t,t_0) |i,t_0\rangle\right|^2 = \left| \langle f| \mathrm{e}^{\frac{\mathrm{i}}{\hbar} E_f t} U^S(t,t_0) \mathrm{e}^{-\frac{\mathrm{i}}{\hbar} E_i t_0} |i\rangle\right|^2 = \left| \langle f| U^I(t,t_0) |i\rangle\right|^2$

According to the first order perturbation theory, that is, take the first order of the Dyson series, the transition probability is

\begin{aligned} \mathcal{P}_{fi}(t) &= \left| \langle f^S| U^I(t,t_0) |i^S\rangle\right|^2 \\ &= \left| \langle f^S| -\frac{\mathrm{i}}{\hbar} \int_{t_0}^t \mathrm{d} t_1\, H^I_\text{int}(t_1) |i^S\rangle\right|^2 \\ &= \left| \langle f^S| -\frac{\mathrm{i}}{\hbar} \int_{t_0}^t \mathrm{d} t_1\, \mathrm{e}^{\frac{\mathrm{i}}{\hbar} H_0^S t_1} H^S_\text{int}(t_1) \mathrm{e}^{-\frac{\mathrm{i}}{\hbar} H_0^S t_1} |i^S\rangle\right|^2 \\ &= \frac{1}{\hbar^2} \left| \int_{t_0}^t \mathrm{d} t_1\, \mathrm{e}^{\frac{\mathrm{i}}{\hbar} (E_f - E_i) t_1} \langle f| H^S_\text{int}(t_1) |i\rangle\right|^2 \end{aligned}

The superscripts $S$ indicating Schrödinger picture of kets $\ket{f}$ and $\ket{i}$ are omitted. Define $\omega_{fi} = \frac{E_f-E_i}{\hbar}$, we obtain the final formula of first-order perturbation theory.

Notice, however, this approach is valid only if $\mathcal{P}_{fi} \ll 1$.

### Sinusoidal Perturbation

Suppose $H^S_\text{int}(t) = W \sin \omega t$ , where $W$ is a time independent observation operator. Then

$\mathcal{P}_{fi}(t;\omega) = \frac{1}{\hbar^2} \left| \int_{t_0}^t \mathrm{d} t^\prime \, \mathrm{e}^{\mathrm{i} \omega_{fi} t^\prime} \sin \omega t^\prime \langle f| W |i\rangle\right|^2 = \frac{|\langle f| W |i\rangle|^2}{4 \hbar^2} \left|\frac{1-\mathrm{e}^{\mathrm{i}(\omega_{fi}+\omega)T}}{\omega_{fi}+\omega} + \frac{1-\mathrm{e}^{\mathrm{i}(\omega_{fi}-\omega)T}}{\omega_{fi}-\omega}\right|^2$

in which $T=t-t_0$ . Resonance requires $\omega \simeq \omega_{fi}$. We introduce rotating-wave approximation

$\left|\frac{1-\mathrm{e}^{\mathrm{i}(\omega_{fi}+\omega)T}}{\omega_{fi}+\omega}\right| = \left|\frac{\sin [(\omega_{fi}+\omega)T/2]}{(\omega_{fi}+\omega)/2}\right| \simeq0$

which means that anti-resonant term is negligible on average since it has large frequency $(\omega_{fi}+\omega \gg \omega_{fi}-\omega)$ and oscillates fast within a time period $T$. Therefore, the transition probability becomes

$\mathcal{P}_{fi}(t;\omega) = \frac{|\langle f| W |i\rangle|^2}{4 \hbar^2} \frac{\sin^2 [(\omega_{fi}-\omega)T/2]}{[(\omega_{fi}-\omega)/2]^2}$

Notice that the sinc-squared function sequence $\frac{N}{\pi} \frac{\sin^2 (Nx)}{(Nx)^2}$ approaches the behavior of delta function when $N \to \infty$. We naturally introduce a distribution $\delta_T(E)$ as follows

$\delta_T(E) = \frac{\sin^2\frac{ET}{2\hbar}}{\left(\frac{ET}{2\hbar}\right)^2} \frac{T}{2\pi\hbar} = \frac{2\hbar \sin^2(ET/2\hbar)}{\pi T E^2}$

which has the maximum value $\frac{T}{2\pi \hbar}$ at $E=0$ and its width of order is $\frac{2\pi\hbar}{T}$. It becomes delta function when $T \to \infty$. Therefore, the transition probability can be written as

$\mathcal{P}_{fi}(t;\omega) = |\langle f| W |i\rangle|^2 \frac{\pi T }{2\hbar} \delta_T(E_f-E_i-\hbar\omega)$ ### Continuum

For continuum, the states are labeled by the energy $E$ and some other parameters $\alpha$, and a significant concept in continuum is the density of state (DOS) $\rho(\alpha,E)$.

We shall calculate the transition probability of a system after time $T$ under a sinusoidal perturbation with frequency $\omega$ . From previous deductions, we can foresee that the system will be most likely to transit to those states with eigenenergy $E_f = E_i+\hbar \omega$. However, the eigenvalue $E_f$ might be degenerated, containing a variety of states $\ket{f,E_f}$ satisfying the condition, while their matrix elements $\braket{f,E_f | W|i}$ might be different even though they share the same energy. This makes the problem intractable.

Anyway, we start from the very beginning. The probability of transition is

$\sum_{f \neq i} \mathcal{P}_{fi}(t) = \int_{f \neq i} \mathcal{P}_{fi} (t) \rho(f,E_f) \, \mathrm{d} E_f \,\mathrm{d}f = \int_{f \neq i} \frac{|W_{fi}|^2}{4\hbar^2} \frac{\sin^2 [(\omega_{fi}-\omega)T/2]}{[(\omega_{fi}-\omega)/2]^2}\rho(f,E_f)\, \mathrm{d} E_f \, \mathrm{d} f$

Notice that, only when $E_f$ near $E_i + \hbar\omega$ does the sinc function reach its maximum, so $f \neq i$ actually means that the system will only transit into a finite range $\Delta E$ about $E_f$. So we obtain

$\mathcal{P}_{fi} = \frac{\pi T}{2\hbar} \int \left|\braket{f, E_f =E_i+\hbar \omega|W|i}\right|^2\rho(f,E_f=E_i+\hbar \omega) \, \mathrm{d}f$

The rate of transition $\Gamma$ is

$\Gamma = \frac{\mathrm{d}\mathcal{P}_{fi}}{\mathrm{d}T} = \frac{\pi}{2\hbar} \int \left|\braket{f, E_f =E_i+\hbar \omega|W|i}\right|^2\rho(f,E_f=E_i+\hbar \omega) \, \mathrm{d}f$

If $E_f$ has finite-fold degeneracy, the integration over $f$ will be replaced by summation

$\Gamma = \frac{\pi}{2\hbar} \sum_f \left|\braket{f, E_f|W|i}\right|^2 \rho(E_f)$

where $E_f = E_i + \hbar \omega$ . This is known as Fermi’s Golden Rule.

### Applicable Conditions

First, we use the first-order time dependent theory, which requires that $T$ is small enough without making transition probability $\mathcal{P}>1$. The characteristic time is given by

$\mathcal{P}_{fi} = \Gamma T < 1 \implies T < \frac{1}{\Gamma}$

Also, we approximate the sinc-squared function sequence as delta function, which requires that $T$ is large. But how to evaluate the order of magnitude of $T$ ? We hope $T$ is large enough to make $\frac{2\pi\hbar}{T}$, the width of $\delta_T(E)$, relatively small compared to the energy range of the system $\Delta E$ , so that $\delta_T(E)$ can pick out a narrow range around $E_f$ . The latter is given by the width of

$\Delta E = \text{width of }K(E)= \text{width of }\int \mathrm{d}f \,\left|\braket{f, E|W|i}\right|^2\rho(f,E)$

where $K(E)$ is a function that evaluates the interaction between initial states $\ket{i}$ and states with different energies $E$.

Thus the second condition is

$\frac{2\pi\hbar}{T} \ll \Delta E \implies T \ll \frac{\hbar}{\Delta E}$

Therefore we have $\Delta E \gg \Gamma$.

## Long-time behavior

### Exponential Decay

In order to study the evolution of the system on a timescale that could be long compared to $\Gamma^{-1}$ one must use a non-perturbative method. The means of solving this type of problem was introduced in 1930 by Weisskopf and Wigner.

The Schrodinger equation version of time-dependent perturbation theory actually does the following things: Assume $\ket{\psi(t)} = \sum b_n \mathrm{e}^{-\mathrm{i} E_n t/ \hbar} \ket{\varphi_n}$, and then obtain

$\mathrm{i} \hbar \frac{\mathrm{d}}{\mathrm{d} t} b_n(t)=\sum_k \mathrm{e}^{\mathrm{i} \omega_{n k} t} W_{n k}(t) b_k(t)$

And do Taylor expansion, compare terms with the same order:

$b_n(t)=b_n^{(0)}(t)+ b_n^{(1)}(t)+ b_n^{(2)}(t)+\cdots$

Take the first order, assume $b_n(0) = \delta_{ni}$, we get

$\mathrm{i} \hbar \frac{\mathrm{d}}{\mathrm{d} t} b_n(t)= \sum_k \mathrm{e}^{\mathrm{i} \omega_{n k} t} W_{n k}(t) b_k(t) = \sum_{k \neq n} \mathrm{e}^{\mathrm{i} \omega_{n k} t} W_{n k}(t) b_k(t)$

The second equals sign comes from $W$ only have non-diagonal matrix element. Notice that, all calculations are strictly accurate by now. If we separately consider the evolution of initial state $\ket{i}$ and the other states $\ket{k}$, the ODE with be

$\begin{cases} \mathrm{i} \hbar \frac{\mathrm{d}}{\mathrm{d} t} b_i(t)= \sum_k \mathrm{e}^{\mathrm{i} \omega_{ik} t} W_{i k}(t) b_k(t) \\ \mathrm{i} \hbar \frac{\mathrm{d}}{\mathrm{d} t} b_k(t)= \sum_{k^\prime} \mathrm{e}^{\mathrm{i} \omega_{k k^\prime} t} W_{k k^\prime}(t) b_{k^\prime}(t) \approx \mathrm{e}^{\mathrm{i} \omega_{k i} t} W_{k i}(t) b_i(t) \end{cases}$

In the second equation, we do an approximation that only the interaction between the initial state and other states are considered. The interactions within other state themselves are neglected. We can then substitute the second equation into the first one:

$\mathrm{i} \hbar \frac{\mathrm{d}}{\mathrm{d} t} b_i(t) = \sum_k \mathrm{e}^{\mathrm{i} \omega_{ik} t} W_{i k}(t)\int_0^t \frac{1}{\mathrm{i}\hbar} \mathrm{e}^{\mathrm{i} \omega_{k i} t^\prime} W_{k i}(t^\prime) b_i(t^\prime) \, \mathrm{d}t^\prime$

\begin{aligned} \dot{b}_i(t) &= -\frac{1}{\hbar^2} \int \mathrm{d} f \int\mathrm{d} E_f \int_0^t \mathrm{d} t^\prime \rho(f,E_f) \mathrm{e}^{\mathrm{i} \omega_{fi} (t^\prime-t)} |W_{fi}|^2 \sin\omega t \sin \omega t^\prime b_i(t^\prime) \\ &\sim -\frac{1}{4\hbar^2} \int \mathrm{d} f \int\mathrm{d} E_f \int_0^t \mathrm{d} t^\prime\rho(f,E_f) \mathrm{e}^{\mathrm{i} (\omega_{fi}-\omega) (t^\prime-t)} |W_{fi}|^2b_i(t^\prime) \\ &= -\frac{1}{4\hbar^2} \int\mathrm{d} E_f \int_0^t \mathrm{d} t^\prime K(E_f) \, \mathrm{e}^{\mathrm{i} (\omega_{fi}-\omega) (t^\prime-t)} b_i(t^\prime) \end{aligned}

Assume that the derivative $\dot{b}_i(t)$ has only a very short memory of the previous values of $b(t^\prime)$ between $0$ and $t$ . Actually, it depends only on the values of at times immediately before. So we can substitute $b(t^\prime)$ with $b(t)$. And since $t$ is not the integration variable, it can be extracted out.

$\dot{b}_i(t) = -\frac{b_i(t)}{4\hbar^2} \int\mathrm{d} E_f \int_0^t \mathrm{d}t^\prime K(E_f) \mathrm{e}^{\mathrm{i} (\omega_{fi}-\omega) (t^\prime-t)}$

For the long-time behavior, $t$ is relatively large, so that we can set $t \to \infty$ and use the Fourier transform of Heaviside function Proof

$F(\omega)=\int_0^{\infty} e^{\mathrm{i} \omega t}\, \mathrm{d} t=\pi \delta(\omega)+\frac{1}{\mathrm{i} \omega}$

we obtain

\begin{aligned} \dot{b}_i(t) &= -\frac{b_i(t)}{4\hbar^2} \int\mathrm{d} E_f \int_0^t \mathrm{d}t^\prime K(E_f) \mathrm{e}^{\mathrm{i} (\omega_{fi}-\omega) (t^\prime-t)}\\ &= -\frac{b_i(t)}{4\hbar^2} \int\mathrm{d} E_f \int_{-t}^0 \mathrm{d} (t^\prime-t) K(E_f) \mathrm{e}^{\mathrm{i} (\omega_{fi}-\omega) (t^\prime-t)}\\ &\sim -\frac{b_i(t)}{4\hbar^2} \int\mathrm{d} E_f \int_{-\infty}^0 \mathrm{d} \tau K(E_f) \mathrm{e}^{\mathrm{i} (\omega_{fi}-\omega) \tau}\\ &= -\frac{b_i(t)}{4\hbar^2} \int\mathrm{d} E_f \int_0^{\infty} \mathrm{d} \tau K(E_f) \mathrm{e}^{-\mathrm{i} (\omega_{fi}-\omega) \tau}\\ &= -\frac{b_i(t)}{4\hbar^2} \int \mathrm{d}E_f \left[ \pi \delta(\omega_{fi}-\omega) + \mathrm{i} \frac{1}{\omega_{fi}-\omega} \right] K(E_f) \end{aligned}

taking advantage of $\delta(ax) = \frac{1}{|a|} \delta(x)$, we can transfer $\delta(\omega_{fi}-\omega)$ to $\delta(E_{fi}-E)$, and therefore

$\dot{b}_i(t) = -\frac{b_i(t)}{4\hbar} \left[ \pi K(E_i + \hbar \omega) + \mathrm{i} \int \frac{K(E_f)\, \mathrm{d} E_f }{E_f-(E_i + \hbar \omega)} \right]$

For the first term, with $\Gamma = \frac{\pi}{2\hbar} K(E_i +\hbar\omega)$, that is actually $\frac{\Gamma}{2}$! For the second term, the factor $\frac{1}{E_f - (E_i + \hbar\omega)}$ actually picks out the effect of $K(E)$ around $E_i + \hbar \omega$. Notice that $K(E)$ has a unit of energy, we define the integral as $\mathcal{E}$. The evolution of $b_i(t)$ is thus described with

$\dot{b}_i(t) = -b_i(t) \left(\frac{\Gamma}{2} + \mathrm{i}\frac{\mathcal{E}}{\hbar}\right)$

which implies the exponential decay

$b_i(t) = \mathrm{e}^{-\Gamma t/2} \mathrm{e}^{-\mathrm{i} \mathcal{E}t/\hbar}$

### Distribution of Final State

Substitute $b_i(t) = \mathrm{e}^{-\Gamma t/2} \mathrm{e}^{-\mathrm{i} \mathcal{E}t/\hbar}$ into $\mathrm{i} \hbar \frac{\mathrm{d}}{\mathrm{d} t} b_k(t) \approx \mathrm{e}^{\mathrm{i} \omega_{k i} t} W_{k i}(t) b_i(t)$, we find:

$b_k (t) = \frac{W_{ki}}{2} \frac{1-\mathrm{e}^{-\Gamma t/2}\mathrm{e}^{\mathrm{i}(E_k - E_i -\mathcal{E})t/\hbar}}{E_k - E_i - \mathcal{E} + \mathrm{i}\hbar\Gamma/2}$

For long time behavior, the numerator approaches $1$. The energy distribution of the final state will be

$\mathcal{P}_k(t) = \frac{|W_{ki}|^2}{4} \frac{1}{(E_k-E_i - \mathcal{E})^2 + \hbar^2 \Gamma^2/4}$