Quantum Optics II Time dependent perturbation theory
fengxiaot Lv4

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=0t=0 we works out the eigenstates n\ket{n} of H0H_0 . We are interested in two of them, f\ket{f} and i\ket{i} (initial state and final state in Schrödinger picture). After time tt , the states in the Schrödinger picture become

f,t=eiEftfi,t=eiEiti\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 Pfi(t)\mathcal{P}_{fi}(t) is for a system initially in the state i\ket{i} at time t=t0t=t_0 (could be zero) evolving into the state f,t\ket{f,t} at time tt ? Using the evolution operator, we can derive

Pfi(t)=f,tUS(t,t0)i,t02=feiEftUS(t,t0)eiEit0i2=fUI(t,t0)i2\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

Pfi(t)=fSUI(t,t0)iS2=fSit0tdt1HintI(t1)iS2=fSit0tdt1eiH0St1HintS(t1)eiH0St1iS2=12t0tdt1ei(EfEi)t1fHintS(t1)i2\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 SS indicating Schrödinger picture of kets f\ket{f} and i\ket{i} are omitted. Define ωfi=EfEi\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 Pfi1\mathcal{P}_{fi} \ll 1.

Sinusoidal Perturbation

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

Pfi(t;ω)=12t0tdteiωfitsinωtfWi2=fWi2421ei(ωfi+ω)Tωfi+ω+1ei(ωfiω)Tωfiω2\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=tt0T=t-t_0 . Resonance requires ωωfi\omega \simeq \omega_{fi}. We introduce rotating-wave approximation

1ei(ωfi+ω)Tωfi+ω=sin[(ωfi+ω)T/2](ωfi+ω)/20\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 (ωfi+ωωfiω)(\omega_{fi}+\omega \gg \omega_{fi}-\omega) and oscillates fast within a time period TT. Therefore, the transition probability becomes

Pfi(t;ω)=fWi242sin2[(ωfiω)T/2][(ωfiω)/2]2\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 Nπsin2(Nx)(Nx)2\frac{N}{\pi} \frac{\sin^2 (Nx)}{(Nx)^2} approaches the behavior of delta function when NN \to \infty. We naturally introduce a distribution δT(E)\delta_T(E) as follows

δT(E)=sin2ET2(ET2)2T2π=2sin2(ET/2)πTE2\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 T2π\frac{T}{2\pi \hbar} at E=0E=0 and its width of order is 2πT\frac{2\pi\hbar}{T}. It becomes delta function when TT \to \infty. Therefore, the transition probability can be written as

Pfi(t;ω)=fWi2πT2δT(EfEiω)\mathcal{P}_{fi}(t;\omega) = |\langle f| W |i\rangle|^2 \frac{\pi T }{2\hbar} \delta_T(E_f-E_i-\hbar\omega)


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

We shall calculate the transition probability of a system after time TT under a sinusoidal perturbation with frequency ω\omega . From previous deductions, we can foresee that the system will be most likely to transit to those states with eigenenergy Ef=Ei+ωE_f = E_i+\hbar \omega. However, the eigenvalue EfE_f might be degenerated, containing a variety of states f,Ef\ket{f,E_f} satisfying the condition, while their matrix elements f,EfWi\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

fiPfi(t)=fiPfi(t)ρ(f,Ef)dEfdf=fiWfi242sin2[(ωfiω)T/2][(ωfiω)/2]2ρ(f,Ef)dEfdf\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 EfE_f near Ei+ωE_i + \hbar\omega does the sinc function reach its maximum, so fif \neq i actually means that the system will only transit into a finite range ΔE\Delta E about EfE_f. So we obtain

Pfi=πT2f,Ef=Ei+ωWi2ρ(f,Ef=Ei+ω)df\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 Γ\Gamma is

Γ=dPfidT=π2f,Ef=Ei+ωWi2ρ(f,Ef=Ei+ω)df\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 EfE_f has finite-fold degeneracy, the integration over ff will be replaced by summation

Γ=π2ff,EfWi2ρ(Ef)\Gamma = \frac{\pi}{2\hbar} \sum_f \left|\braket{f, E_f|W|i}\right|^2 \rho(E_f)

where Ef=Ei+ωE_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 TT is small enough without making transition probability P>1\mathcal{P}>1. The characteristic time is given by

Pfi=ΓT<1    T<1Γ\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 TT is large. But how to evaluate the order of magnitude of TT ? We hope TT is large enough to make 2πT\frac{2\pi\hbar}{T}, the width of δT(E)\delta_T(E), relatively small compared to the energy range of the system ΔE\Delta E , so that δT(E)\delta_T(E) can pick out a narrow range around EfE_f . The latter is given by the width of

ΔE=width of K(E)=width of dff,EWi2ρ(f,E)\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)K(E) is a function that evaluates the interaction between initial states i\ket{i} and states with different energies EE.

Thus the second condition is

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

Therefore we have ΔEΓ\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 Γ1\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 ψ(t)=bneiEnt/φn\ket{\psi(t)} = \sum b_n \mathrm{e}^{-\mathrm{i} E_n t/ \hbar} \ket{\varphi_n}, and then obtain

iddtbn(t)=keiωnktWnk(t)bk(t)\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:

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

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

iddtbn(t)=keiωnktWnk(t)bk(t)=kneiωnktWnk(t)bk(t)\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 WW only have non-diagonal matrix element. Notice that, all calculations are strictly accurate by now. If we separately consider the evolution of initial state i\ket{i} and the other states k\ket{k}, the ODE with be

{iddtbi(t)=keiωiktWik(t)bk(t)iddtbk(t)=keiωkktWkk(t)bk(t)eiωkitWki(t)bi(t)\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:

iddtbi(t)=keiωiktWik(t)0t1ieiωkitWki(t)bi(t)dt\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

b˙i(t)=12dfdEf0tdtρ(f,Ef)eiωfi(tt)Wfi2sinωtsinωtbi(t)142dfdEf0tdtρ(f,Ef)ei(ωfiω)(tt)Wfi2bi(t)=142dEf0tdtK(Ef)ei(ωfiω)(tt)bi(t)\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 b˙i(t)\dot{b}_i(t) has only a very short memory of the previous values of b(t)b(t^\prime) between 00 and tt . Actually, it depends only on the values of at times immediately before. So we can substitute b(t)b(t^\prime) with b(t)b(t). And since tt is not the integration variable, it can be extracted out.

b˙i(t)=bi(t)42dEf0tdtK(Ef)ei(ωfiω)(tt)\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, tt is relatively large, so that we can set tt \to \infty and use the Fourier transform of Heaviside function Proof

F(ω)=0eiωtdt=πδ(ω)+1iωF(\omega)=\int_0^{\infty} e^{\mathrm{i} \omega t}\, \mathrm{d} t=\pi \delta(\omega)+\frac{1}{\mathrm{i} \omega}

we obtain

b˙i(t)=bi(t)42dEf0tdtK(Ef)ei(ωfiω)(tt)=bi(t)42dEft0d(tt)K(Ef)ei(ωfiω)(tt)bi(t)42dEf0dτK(Ef)ei(ωfiω)τ=bi(t)42dEf0dτK(Ef)ei(ωfiω)τ=bi(t)42dEf[πδ(ωfiω)+i1ωfiω]K(Ef)\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 δ(ax)=1aδ(x)\delta(ax) = \frac{1}{|a|} \delta(x), we can transfer δ(ωfiω)\delta(\omega_{fi}-\omega) to δ(EfiE)\delta(E_{fi}-E), and therefore

b˙i(t)=bi(t)4[πK(Ei+ω)+iK(Ef)dEfEf(Ei+ω)]\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 Γ=π2K(Ei+ω)\Gamma = \frac{\pi}{2\hbar} K(E_i +\hbar\omega), that is actually Γ2\frac{\Gamma}{2}! For the second term, the factor 1Ef(Ei+ω)\frac{1}{E_f - (E_i + \hbar\omega)} actually picks out the effect of K(E)K(E) around Ei+ωE_i + \hbar \omega. Notice that K(E)K(E) has a unit of energy, we define the integral as E\mathcal{E}. The evolution of bi(t)b_i(t) is thus described with

b˙i(t)=bi(t)(Γ2+iE)\dot{b}_i(t) = -b_i(t) \left(\frac{\Gamma}{2} + \mathrm{i}\frac{\mathcal{E}}{\hbar}\right)

which implies the exponential decay

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

Distribution of Final State

Substitute bi(t)=eΓt/2eiEt/b_i(t) = \mathrm{e}^{-\Gamma t/2} \mathrm{e}^{-\mathrm{i} \mathcal{E}t/\hbar} into iddtbk(t)eiωkitWki(t)bi(t)\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:

bk(t)=Wki21eΓt/2ei(EkEiE)t/EkEiE+iΓ/2b_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 11. The energy distribution of the final state will be

Pk(t)=Wki241(EkEiE)2+2Γ2/4\mathcal{P}_k(t) = \frac{|W_{ki}|^2}{4} \frac{1}{(E_k-E_i - \mathcal{E})^2 + \hbar^2 \Gamma^2/4}