Quantum Optics I Rotating Frame Transformation
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.

Interaction Picture

Interaction Picture

Perturbation Hamiltonian HS=H0S+HintS(t)H^S = H^S_0 + H^S_\text{int} (t)

State vector ΨI(t)=eiH0StΨS(t)\ket{\Psi^I(t)} = \mathrm{e}^{\frac{\mathrm{i}}{\hbar} H_0^S t} \ket{\Psi^S(t)}

Operator OI(t)=eiH0StOS(t)eiH0StO^I(t) = \mathrm{e}^{\frac{\mathrm{i}}{\hbar} H_0^S t} O^S(t) \mathrm{e}^{-\frac{\mathrm{i}}{\hbar} H_0^S t}

For the operator H0H_0 itself, the interaction picture and Schrödinger picture coincide

H0I=eiH0StH0SeiH0St=H0SH^I_0 = \mathrm{e}^{\frac{\mathrm{i}}{\hbar} H_0^S t} H^S_0 \mathrm{e}^{-\frac{\mathrm{i}}{\hbar} H_0^S t} = H_0^S

This is easily seen through the fact that H0H_0 and eiH0St\mathrm{e}^{\frac{\mathrm{i}}{\hbar} H_0^S t} commute. From now on, we will omit the superscript representing pictures of H0H_0 .

For the density operator ρ(t)\rho(t),it has no difference from any other operators.

ρI(t)=npn(t)ψnI(t)ψnI(t)=npn(t)eiH0St/ψnS(t)ψnS(t)eiH0St/=eiH0St/ρS(t)eiH0St/\begin{aligned} \rho^I(t) & =\sum_n p_n(t)|\psi_{n}^I(t)\rangle\langle\psi_n^I(t)| \\ & =\sum_n p_n(t) \mathrm{e}^{\mathrm{i} H_{0}^S t / \hbar}|\psi_n^S(t)\rangle\langle\psi_n^S(t)| \mathrm{e}^{- \mathrm{i} H_{0}^S t / \hbar} \\ & =\mathrm{e}^{\mathrm{i} H_{0}^S t / \hbar} \rho^S(t) \mathrm{e}^{-\mathrm{i} H_{0}^S t / \hbar} \end{aligned}

Dynamics Transforming the Schrödinger equation into the interaction picture gives

itΨI(t)=HintI(t)ΨI(t)\mathrm{i} \hbar \frac{\partial}{\partial t} \ket{\Psi^I(t)} = H^I_\text{int}(t) \ket{\Psi^I(t)}

itOI(t)=[OI(t),H0]\mathrm{i} \hbar \frac{\partial}{\partial t} O^I(t) = [O^I(t),H_0]

A quantum state is evolved by the interaction Hamiltonian HintIH^I_\text{int} in the interaction picture, but a operator is evolved by the unperturbed Hamiltonian H0H_0 like in Heisenberg picture.

Expectation The expectation value of an operator in the interaction picture is a sandwich structure as well

ΨI(t)OI(t)ΨI(t)=ΨS(t)OSΨS(t)=ΨHOH(t)ΨH\braket{\Psi^I(t) | O^I(t) |\Psi^I(t)} = \braket{\Psi^S(t) | O^S |\Psi^S(t)} = \braket{\Psi^H | O^H (t) |\Psi^H}

Evolution Operator

The definition of evolution operator in the interaction picture is similar to Schrödinger picture

ΨI(t)=UI(t,t0)ΨI(t0)ΨS(t)=UI(t,t0)ΨS(t0)\ket{\Psi^I(t)} = U^I(t,t_0) \ket{\Psi^I(t_0)} \qquad \ket{\Psi^S(t)} = U^I(t,t_0) \ket{\Psi^S(t_0)}

Transforming the equation itUS(t,t0)=HS(t)US(t,t0)\mathrm{i} \hbar \frac{\partial}{\partial t} U^S(t,t_0) = H^S(t)U^S(t,t_0) in Schrödinger picture to interaction picture, we got

itUI(t,t0)=HintI(t)UI(t,t0)\mathrm{i} \hbar \frac{\partial}{\partial t} U^I(t,t_0) = H^I_\text{int}(t) U^I(t,t_0)

Compare it to the definition of ΨI(t)\ket{\Psi^I(t)} and the time-evolution equation of states, we can find

UI(t,t0)=eiH0StUS(t,t0)eiH0St0U^I(t,t_0) = \mathrm{e}^{\frac{\mathrm{i}}{\hbar} H_0^S t} U^S(t,t_0) \mathrm{e}^{-\frac{\mathrm{i}}{\hbar} H_0^S t_0}

/* Warning: This is different from the definition of other operators in the interaction picutre! */

Dyson series

From itUI(t,t0)=HintI(t)UI(t,t0)\mathrm{i} \hbar \frac{\partial}{\partial t} U^I(t,t_0) = H^I_\text{int}(t) U^I(t,t_0) we can easily obtain

UI(t,t0)=1+(i)t0tdt1HintI(t1)+(i)2t0tdt1t0t1dt2HintI(t1)HintI(t2)+U^I(t,t_0) = 1+ \left( -\frac{\mathrm{i}}{\hbar}\right) \int_{t_0}^t \mathrm{d} t_1\, H^I_\text{int}(t_1) +\left( -\frac{\mathrm{i}}{\hbar}\right)^2 \int_{t_0}^t \mathrm{d} t_1 \int_{t_0}^{t_1} \mathrm{d} t_2 \, H^I_\text{int}(t_1) H^I_\text{int}(t_2) + \cdots

Rotating Frame

General transformation

A unitary transformation (or frame change) can be expressed in terms of a time-dependent Hamiltonian H(t)H(t) and unitary operator U(t)U(t). Under this change, the Hamiltonian transforms as

H(t)H~(t)=UHU+iU˙UH(t) \to \tilde{H}(t) = UHU^\dagger+\mathrm{i}\hbar \dot{U} U^\dagger

ψ(t)ψ~(t)=Uψ(t)\ket{\psi(t)} \to \ket{\tilde{\psi}(t)} = U \ket{\psi(t)}

Maintaining the form of Schrodinger equation

iddtψ~(t)=H~(t)ψ~(t)\mathrm{i} \hbar \frac{\mathrm{d}}{\mathrm{d} t} \ket{\tilde{\psi}(t)} = \tilde{H}(t) \ket{\tilde{\psi}(t)}

Warning: the expectation value of operators is no longer sandwich structure, the Hamiltonian, for example

ψ~(t)H~(t)ψ~(t)=ψ(t)UUHUUψ(t)+iψ(t)UU˙UUψ(t)=ψ(t)Hψ(t)+iψ(t)UU˙ψ(t)\begin{aligned} \braket{\tilde{\psi}(t) | \tilde{H}(t)|\tilde{\psi}(t)} &= \bra{\psi(t)}U^\dagger U H U^\dagger U \ket{\psi(t)}+\mathrm{i}\hbar \bra{\psi(t)}U^\dagger\dot{U} U^\dagger U \ket{\psi(t)} \\ &= \bra{\psi(t)} H \ket{\psi(t)}+\mathrm{i}\hbar \bra{\psi(t)}U^\dagger\dot{U} \ket{\psi(t)} \end{aligned}

an redundant term iψ(t)UU˙ψ(t)\mathrm{i}\hbar \bra{\psi(t)}U^\dagger\dot{U} \ket{\psi(t)} is attached.

Physical Meaning

There are two ways to comprehend the additional term iU˙U\mathrm{i}\hbar \dot{U} U^\dagger in the expression of transformed Hamiltonian H~\tilde{H}. The first way is to compare the situation with that in classical mechanics.

In classical mechanics, if an object undergoes circular motion with an angular frequency ω\omega , then its centrifugal potential energy is E=12mω2R2E = \frac{1}{2} m \omega^2 R^2. When switching to a reference frame rotating at the same angular frequency of ω\omega , an observer in this reference frame will find that the object is completely stationary. Then this observer will consider its energy as 00. The centrifugal potential energy disappears when performing rotating frame transformation.

In quantum mechanics, the energy is described by the Hamiltonian. iU˙U\mathrm{i}\hbar \dot{U} U^\dagger is that disappeared centrifugal potential energy. For example, in a two-level system H=ωggg+ωeeeH = \hbar \omega_g \ket{g}\bra{g} + \hbar \omega_e \ket{e}\bra{e} , for a unitary transformation like

U=eiω1tgg+eiω2teeU = \mathrm{e}^{\mathrm{i} \omega_1 t} \ket{g}\bra{g} + \mathrm{e}^{\mathrm{i} \omega_2 t} \ket{e}\bra{e}

one can obtain with some calculation

iU˙U=ω1ggω2ee\mathrm{i}\hbar \dot{U} U^\dagger = -\hbar \omega_1 \ket{g}\bra{g} - \hbar \omega_2 \ket{e}\bra{e}

therefore, the stationary part is transformed into

H~=UHU+iU˙U=(ωgω1)gg+(ωeω2)ee\tilde{H} = UHU^\dagger+\mathrm{i}\hbar \dot{U} U^\dagger = \hbar(\omega_g -\omega_1) \ket{g}\bra{g}+ \hbar (\omega_e -\omega_2) \ket{e}\bra{e}

We could draw a conclusion that the effect of UU is to shift down the energy level of eigenstates by ω1\omega_1 and ω2\omega_2 .

The second way is to look back on Schrodinger equation. Still in our two-level system, supposing the initial state is

ψ(0)=cgg+cee\ket{\psi(0)} = c_g \ket{g} + c_e \ket{e}

The state at arbitrary time tt is then determined by iddtψ(t)=H(t)ψ(t)\mathrm{i} \hbar \frac{\mathrm{d}}{\mathrm{d} t} \ket{\psi(t)} = H(t) \ket{\psi(t)}, thus

ψ(t)=cgeiωgtg+ceeiωete\ket{\psi(t)} = c_g \,\mathrm{e}^{-\mathrm{i}\omega_g t} \ket{g} + c_e \, \mathrm{e}^{-\mathrm{i}\omega_e t} \ket{e}

where the energy of eigenstates, ωe\hbar \omega_e and ωg\hbar \omega_g, appear on the exponential of phase factor.

The unitary transformation ψ(t)ψ~(t)=Uψ(t)\ket{\psi(t)} \to \ket{\tilde{\psi}(t)} = U \ket{\psi(t)} actually shift the kets to

ψ~(0)=cgeiω1tg+ceeiω2teψ~(t)=cgei(ωgω1)tg+ceei(ωeω2)te\begin{gathered} \ket{\tilde{\psi}(0)} = c_g \,\mathrm{e}^{\mathrm{i}\omega_1 t} \ket{g} + c_e \, \mathrm{e}^{\mathrm{i}\omega_2 t} \ket{e} \\ \ket{\tilde{\psi}(t)} = c_g \,\mathrm{e}^{-\mathrm{i} (\omega_g-\omega_1) t} \ket{g} + c_e \, \mathrm{e}^{\mathrm{i}(\omega_e-\omega_2) t} \ket{e} \end{gathered}

it changes the oscillation frequency of the states.


We consider a three-level atom interacting with a bichromatic electric field. The Hamiltonian is

H=ωaaa+ωbbb+ωccc+[12Ω1eiω1tab+12Ω2eiω1tac+h.c.]H = \hbar \omega_a \ket{a}\bra{a} + \hbar \omega_b \ket{b}\bra{b} + \hbar \omega_c \ket{c}\bra{c} + \left[ \frac{1}{2}\hbar \Omega_1 \mathrm{e}^{-\mathrm{i} \omega_1 t} \ket{a}\bra{b} + \frac{1}{2}\hbar \Omega_2 \mathrm{e}^{-\mathrm{i} \omega_1 t} \ket{a}\bra{c} + \mathrm{h.c.} \right]

Performing such a unitary transformation

U=eiωataa+ei(ωaω1)tbb+ei(ωaω2)tccU = \mathrm{e}^{\mathrm{i} \omega_a t} \ket{a}\bra{a} + \mathrm{e}^{\mathrm{i} (\omega_a-\omega_1) t} \ket{b}\bra{b} + \mathrm{e}^{\mathrm{i} (\omega_a-\omega_2) t} \ket{c}\bra{c}

the Hamiltonian becomes

H~=ΔbbΔcc+[12Ω1ab+12Ω2ac+h.c.]\tilde{H} = -\hbar \Delta \ket{b}\bra{b} -\hbar \Delta \ket{c}\bra{c} + \left[ \frac{1}{2}\hbar \Omega_1 \ket{a}\bra{b} + \frac{1}{2}\hbar \Omega_2\ket{a}\bra{c} + \mathrm{h.c.} \right]

And to examine whether the interaction part of the Hamiltonian is now time-independent under your transformation, just substitute the transformed a~,b~,c~\ket{\tilde{a}},\ket{\tilde{b}},\ket{\tilde{c}} into the original expression and see if the oscillating factor of electric field is cancelled.