Quantum Optics I Rotating Frame Transformation

The rotating frame transformation is a widely used technique in quantum optics, particularly for solving two-level and three-level systems. This article provides a detailed explanation of its formalism.

Interaction Picture

Interaction Picture

Perturbed Hamiltonian $H^S = H^S_0 + H^S_\text{int} (t)$

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

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

For the operator $H_0^S$ itself, the interaction picture and Schrödinger picture coincide

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

which follows directly from the fact that $H_0^S$ and $\mathrm{e}^{\frac{\mathrm{i}}{\hbar} H_0^S t}$ commute. From now on, we will omit the superscript indicating the picture of $H_0$ .

For the density operator $\rho(t)$, its transformation to the interaction picture adheres to the same rule as that for any other operator:

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

Dynamics Transforming the Schrödinger equation into the interaction picture gives

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

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

Therefore, a quantum state is evolved by the interaction Hamiltonian $H^I_\text{int}$ in the interaction picture, but an operator is evolved by the unperturbed Hamiltonian $H_0$ like in Heisenberg picture.

Expectation The expectation value of an operator in the interaction picture maintains the sandwich structure

$$\braket{\Psi^I(t) | O^I(t) |\Psi^I(t)} = \braket{\Psi^S(t) | O^S |\Psi^S(t)} = \braket{\Psi^H | O^H (t) |\Psi^H}$$

Evolution Operator

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

$$\ket{\Psi^I(t)} = U^I(t,t_0) \ket{\Psi^I(t_0)} \qquad \ket{\Psi^S(t)} = U^S(t,t_0) \ket{\Psi^S(t_0)}$$

Transforming the equation $\mathrm{i} \hbar \frac{\partial}{\partial t} U^S(t,t_0) = H^S(t)U^S(t,t_0)$ in Schrödinger picture to interaction picture, we get

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

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

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

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

Dyson series

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

$$U^I(t,t_0) = 1+ \left( -\frac{\mathrm{i}}{\hbar}\right) \int_{t_0}^t \mathrm{d} t_1\, H^I_\text{int}(t_1) +\left( -\frac{\mathrm{i}}{\hbar}\right)^2 \int_{t_0}^t \mathrm{d} t_1 \int_{t_0}^{t_1} \mathrm{d} t_2 \, H^I_\text{int}(t_1) H^I_\text{int}(t_2) + \cdots$$

Rotating Frame

Unitary transformation

A unitary transformation (or frame change) can be expressed using a unitary operator $U(t)$. Under this frame change, the Hamiltonian transforms as^[1]

$$H(t) \to \tilde{H}(t) = UHU^\dagger+\mathrm{i}\hbar \dot{U} U^\dagger$$

$$\ket{\psi(t)} \to \ket{\tilde{\psi}(t)} = U \ket{\psi(t)}$$

such that the form of Schrödinger equation is maintained

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

Warning: The expectation value of operators is no longer a sandwich structure. The Hamiltonian, as an example, is transformed as

$$\begin{aligned} \braket{\tilde{\psi}(t) | \tilde{H}(t)|\tilde{\psi}(t)} &= \bra{\psi(t)}U^\dagger U H U^\dagger U \ket{\psi(t)}+\mathrm{i}\hbar \bra{\psi(t)}U^\dagger\dot{U} U^\dagger U \ket{\psi(t)} \\ &= \bra{\psi(t)} H \ket{\psi(t)}+\mathrm{i}\hbar \bra{\psi(t)}U^\dagger\dot{U} \ket{\psi(t)} \end{aligned}$$

An redundant term $\mathrm{i}\hbar \bra{\psi(t)}U^\dagger\dot{U} \ket{\psi(t)}$ is attached. This is the difference between rotating frame and the interaction picture.

Example

Two-level system

Consider a two-level atom interacting with a bichromatic electric field. The Hamiltonian is

$$H =\frac{1}{2}\hbar\omega_0 \sigma_z +\frac{1}{2} \hbar\Omega \left[\mathrm{e}^{\mathrm{i}(\omega t+\phi)} +\mathrm{e}^{-\mathrm{i}(\omega t+\phi)}\right] \sigma_x$$

Perform the following unitary transformation

$$U=\exp \left( \frac{\mathrm{i}}{2} \omega t \sigma_z \right)$$

/* Note: The rotatation frequency is $\omega$ instead of $\omega_0$! */ 

the rotated Hamiltonian will become

$$\tilde{H} = UHU^\dagger+\mathrm{i}\hbar \dot{U} U^\dagger = -\frac{1}{2}\hbar\delta \sigma_z + \frac{1}{2} \hbar \Omega (\sigma^\dagger\mathrm{e}^{-\mathrm{i}\phi} + \sigma\mathrm{e}^{\mathrm{i}\phi})$$

The above derivations can still be interpreted as a version of the [interaction picture](#Interaction Picture). Instead of selecting the entire time-independent part, $\frac{1}{2}\hbar\omega_0 \sigma_z$, as $H_0$, we choose $\frac{1}{2}\hbar\omega \sigma_z$ as $H_0$. The time-dependent interaction part then becomes

$$H_\text{int}(t) = -\frac{1}{2}\hbar\delta \sigma_z + \frac{1}{2} \hbar\Omega \left[\mathrm{e}^{\mathrm{i}(\omega t+\phi)} +\mathrm{e}^{-\mathrm{i}(\omega t+\phi)}\right] \sigma_x$$

In this context, the unitary transformation aligns precisely with the interaction picture transformation $U = \mathrm{e}^{\frac{\mathrm{i}}{\hbar} H_0 t}$.

Two-level system

If the Hamiltonian is initially written as

$$H = \hbar \omega_0 \ket{e}\bra{e} + \frac{1}{2} \hbar\Omega \left[\mathrm{e}^{\mathrm{i}(\omega t+\phi)} +\mathrm{e}^{-\mathrm{i}(\omega t+\phi)}\right] (\ket{e}\bra{g} + \ket{g}\bra{e})$$

The unitary transformation should be chosen as

$$\begin{aligned} U &= \mathrm{e}^{\mathrm{i}\omega t |e\rangle\langle e|} = I+\sum_{n=1}^\infty \frac{(\mathrm{i}\omega t)^n}{n!} (|e\rangle\langle e|)^n = I+\sum_{n=1}^\infty \frac{(\mathrm{i}\omega t)^n}{n!} |e\rangle\langle e| \\ &=I+(\mathrm{e}^{\mathrm{i}\omega t}-1) |e\rangle\langle e| = |g\rangle\langle g| + \mathrm{e}^{\mathrm{i}\omega t}|e\rangle\langle e| \end{aligned}$$

the rotated Hamiltonian will become

$$\tilde{H} = UHU^\dagger+\mathrm{i}\hbar \dot{U} U^\dagger = -\hbar\delta \ket{e}\bra{e} + \frac{1}{2} \hbar \Omega (\sigma^\dagger\mathrm{e}^{-\mathrm{i}\phi} + \sigma\mathrm{e}^{\mathrm{i}\phi})$$

Three-level system

Consider a three-level atom interacting with a bichromatic electric field. The Hamiltonian is^[2]

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

Performing such a unitary transformation

$$U = \mathrm{e}^{\mathrm{i} \omega_a t} \ket{a}\bra{a} + \mathrm{e}^{\mathrm{i} (\omega_a-\omega_1) t} \ket{b}\bra{b} + \mathrm{e}^{\mathrm{i} (\omega_a-\omega_2) t} \ket{c}\bra{c}$$

the rotated Hamiltonian will become

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

To check if the interaction part of the Hamiltonian will become time-independent under certain transformation, just substitute the transformed $\ket{\tilde{a}},\ket{\tilde{b}},\ket{\tilde{c}}$ into the original expression and see if the oscillating factor of electric field is cancelled.

Here are some Mathematica code to play with: [Link].

Physical Intuition

As the name suggests, the rotating frame transformation aims to extract the rotational component from the dynamics of quantum states.

We can make an analogy to classical mechanics. In classical mechanics, consider an object undergoing circular motion with angular frequency $\omega$ in the lab frame. By switching to a rotating frame with the same angular frequency $\omega$, the dynamics are significantly simplified because the circular motion effectively vanishes. However, this frame transition also introduces an fictitious centrifugal force $\boldsymbol{F}_c = m\omega^2 \boldsymbol{r}$ to balance the dynamics.

Similarly, in quantum mechanics, the term $\mathrm{i}\hbar \dot{U} U^\dagger$ reflects the adjustment required in the Hamiltonian when transitioning to a rotating reference frame. This reflects how the Hamiltonian governs the dynamics of quantum states, analogous to Newton’s second law $\boldsymbol{F} = m \boldsymbol{a}$, governing the dynamics of objects in classical mechanics. Just as a fictitious force like $\boldsymbol{F}_c$ is introduced in the classical case, the adjustment term $\mathrm{i}\hbar \dot{U} U^\dagger$ reflects the modifications to the Hamiltonian required when moving to the rotating frame in quantum mechanics.

Moreover, in quantum mechanics, the situation is slightly different. Atomic states actually rotate on the Bloch sphere at an angular velocity of $\omega_0$, while the chosen rotating frame has a frequency $\omega$, synchronized with the frequency of the optical field rather than the atom’s intrinsic frequency. This choice is made to eliminate the time-dependent phase in the interaction Hamiltonian $H_\text{int}(t)$, transforming it into a static Hamiltonian that can then be solved through diagonalization.

We can quantitatively evaluates the effect of such unitary transformations. For example, in a two-level system $H = \hbar \omega_g \ket{g}\bra{g} + \hbar \omega_e \ket{e}\bra{e}$ , for a unitary transformation such that

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

the additional term will be

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

Therefore, the stationary part is transformed to

$$H_0 \to H_0 +\mathrm{i}\hbar \dot{U} U^\dagger = \hbar(\omega_g -\omega_1) \ket{g}\bra{g}+ \hbar (\omega_e -\omega_2) \ket{e}\bra{e}$$

We could draw a conclusion that the effect of $U$ is shifting down the energy level of eigenstates by $\omega_1$ and $\omega_2$ .

If we look back on the Schrödinger equation, for a system with an initial state

$$\ket{\psi(0)} = c_g \ket{g} + c_e \ket{e}$$

the state at arbitrary time $t$ will be evolved by $\mathrm{i} \hbar \frac{\mathrm{d}}{\mathrm{d} t} \ket{\psi(t)} = H(t) \ket{\psi(t)}$, a.k.a.

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

The unitary transformation $\ket{\psi(t)} \to \ket{\tilde{\psi}(t)} = U \ket{\psi(t)}$, where $U = \mathrm{e}^{\mathrm{i} \omega_1 t} \ket{g}\bra{g} + \mathrm{e}^{\mathrm{i} \omega_2 t} \ket{e}\bra{e}$, will shift the angular frequency of the states to

$$\ket{\tilde{\psi}(t)} = c_g \,\mathrm{e}^{-\mathrm{i} (\omega_g-\omega_1) t} \ket{g} + c_e \, \mathrm{e}^{\mathrm{i}(\omega_e-\omega_2) t} \ket{e}$$

These tricks help find proper unitary transformations in three-level or multi-level systems.

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1. Wikipedia. Unitary transformation (quantum mechanics)[EB/OL]. [2024-12-27]. https://en.wikipedia.org/wiki/Unitary_transformation_(quantum_mechanics). ↩︎

2. lh1962. 相干布居囚禁与绝热布居转移[EB/OL]. 2017-12-21[2024-12-27]. https://chaoli.club/index.php/4037. ↩︎