Rotating Frames

A rotating frame is a time-dependent unitary change of representation chosen to remove known rapid motion or expose slow dynamics. It is exact: no term is discarded merely by changing frames.

Fix the convention

$$\lvert\psi_R(t)\rangle = R(t)^* \lvert\psi(t)\rangle,$$

where $R(t)$ is unitary. Then the transformed Hamiltonian is

$$H_R(t) = R(t)^*H(t)R(t) - i\hbar R(t)^*\dot R(t).$$

The second term is the generator of the moving frame. Omitting it is the central rotating-frame error.

The Rotating-Wave Approximation is a later approximation that may become transparent in a rotating frame. It is not part of the exact transformation developed here.

Rotating-Frame Transformation

Let the laboratory-frame state satisfy

$$i\hbar \lvert\dot\psi\rangle = H(t)\lvert\psi\rangle.$$

Choose a differentiable unitary $R(t)$:

$$R(t)^*R(t)=I.$$

The transformed state

$$\lvert\psi_R\rangle = R^*\lvert\psi\rangle$$

has the same norm:

$$\langle\psi_R\vert\psi_R\rangle = \langle\psi\vert\psi\rangle.$$

The change is a representation choice, not a physical operation performed on the state. An active unitary operation would instead change the physical preparation while observables remained fixed. The same matrix can appear in both roles, so the interpretation and transformation rules must be stated.

Different authors choose $\lvert\psi_R\rangle=R\lvert\psi\rangle$ instead. That convention is valid but changes signs and adjoint placements. Formulas from different conventions should not be combined line by line.

Deriving the Transformed Hamiltonian

Differentiate the rotated state:

$$\lvert\dot\psi_R\rangle = \dot R^*\lvert\psi\rangle + R^*\lvert\dot\psi\rangle.$$

Insert $\lvert\psi\rangle=R\lvert\psi_R\rangle$ and the Schrödinger equation:

$$\begin{aligned} i\hbar \lvert\dot\psi_R\rangle &= i\hbar\dot R^*R \lvert\psi_R\rangle + R^*HR \lvert\psi_R\rangle. \end{aligned}$$

Differentiating $R^*R=I$ gives

$$\dot R^*R = -R^*\dot R.$$

Therefore

$$i\hbar \lvert\dot\psi_R\rangle = H_R(t)\lvert\psi_R\rangle,$$

with

$$H_R(t) = R^*HR - i\hbar R^*\dot R.$$

The transformation preserves self-adjointness. The operator $R^*\dot R$ is anti-self-adjoint:

$$\left( R^*\dot R \right)^* = -R^*\dot R,$$

so

$$i\hbar R^*\dot R$$

is self-adjoint.

The Frame Generator

Define the Hermitian frame generator

$$K_R(t) = i\hbar R(t)^*\dot R(t).$$

Then

$$H_R = R^*HR-K_R.$$

The two terms have distinct origins:

• $R^*HR$ is the laboratory Hamiltonian expressed in rotated coordinates;
• $-K_R$ is the inertial contribution caused by the moving frame.

If

$$R(t)=e^{-iGt/\hbar}$$

for a time-independent self-adjoint generator $G$, then

$$K_R=G,$$

and

$$H_R = e^{iGt/\hbar} H(t) e^{-iGt/\hbar} - G.$$

The subtraction of $G$ shifts energies in the rotating description. For a frame rotating at angular frequency $\omega$ about $z$, $G=\hbar\omega\sigma_z/2$ for a spin-half system.

For a more general time-dependent generator, one cannot automatically write $R(t)$ as an ordinary exponential of its instantaneous $K_R(t)$. Time ordering may be required.

Observables and Density Operators

To preserve expectation values, transform an observable as

$$A_R(t) = R(t)^*A(t)R(t).$$

Then

$$\langle\psi_R\vert A_R \vert\psi_R\rangle = \langle\psi\vert A \vert\psi\rangle.$$

A density operator transforms in the same way:

$$\rho_R = R^*\rho R.$$

The trace rule is invariant:

$$\operatorname{Tr}(\rho_RA_R) = \operatorname{Tr}(\rho A).$$

If $A$ has explicit laboratory-frame time dependence, its rotated derivative is

$$\frac{dA_R}{dt} = R^* \frac{\partial A}{\partial t} R + \frac{i}{\hbar} [K_R,A_R].$$

This formula is useful for distinguishing explicit time dependence of the physical operator from apparent time dependence caused by the moving coordinates.

Measurement projectors, effects, and coupling operators must be transformed consistently. Rotating the Hamiltonian alone while interpreting unrotated measurement axes can produce correct algebra and incorrect laboratory predictions.

Propagators in a Rotating Frame

Let $U(t,t_0)$ be the laboratory propagator. The rotating-frame propagator is

$$U_R(t,t_0) = R(t)^* U(t,t_0) R(t_0).$$

It maps the rotated initial state to the rotated final state:

$$\lvert\psi_R(t)\rangle = U_R(t,t_0) \lvert\psi_R(t_0)\rangle.$$

Differentiating $U_R$ reproduces the transformed Hamiltonian $H_R$. The formula also shows why both endpoint frames matter. Replacing it by $R^*(t)U(t,t_0)$ silently assumes $R(t_0)=I$.

Composition is preserved:

$$U_R(t_2,t_1) U_R(t_1,t_0) = U_R(t_2,t_0).$$

Thus a rotating frame changes representation without changing unitarity or the consistency of propagation.

Interaction Picture as a Special Frame

Suppose

$$H(t)=H_0+V(t)$$

and let $U_0(t,t_0)$ be generated by $H_0$. Choose

$$R(t)=U_0(t,t_0).$$

Then

$$i\hbar\dot R=H_0R,$$

so the frame term cancels the transformed $H_0$:

$$\begin{aligned} H_R &= U_0^*(H_0+V)U_0 - i\hbar U_0^*\dot U_0\\ &= U_0^*VU_0. \end{aligned}$$

This is the interaction Hamiltonian. The Interaction Picture is therefore a rotating frame selected by a Hamiltonian split.

Not every useful rotating frame is an interaction picture in the same practical sense. A drive-frequency frame may be chosen to make resonant structure visible even when its generator is not the full solvable part of the Hamiltonian.

Spin in a Rotating Field

Consider a spin-half system in a circularly rotating transverse field:

$$\begin{aligned} H(t) &= \frac{\hbar\omega_0}{2}\sigma_z\\ &\quad+ \frac{\hbar\Omega}{2} \left[ \cos(\omega t)\sigma_x + \sin(\omega t)\sigma_y \right]. \end{aligned}$$

Choose

$$R(t) = e^{-i\omega t\sigma_z/2}.$$

The operator rotation gives

$$R^* \left[ \cos(\omega t)\sigma_x + \sin(\omega t)\sigma_y \right] R = \sigma_x.$$

The frame generator is

$$K_R = i\hbar R^*\dot R = \frac{\hbar\omega}{2}\sigma_z.$$

Therefore the exact rotating-frame Hamiltonian is time independent:

$$H_R = \frac{\hbar\Delta}{2}\sigma_z + \frac{\hbar\Omega}{2}\sigma_x, \qquad \Delta=\omega_0-\omega.$$

No rotating-wave approximation was used. A circularly rotating field is exactly stationary in the co-rotating frame.

The effective field in the rotating frame points along

$$(\Omega,0,\Delta).$$

The spin precesses about this tilted axis. Resonance means $\Delta=0$, so the effective axis is transverse.

Linear Drive and RWA Preview

Now consider a linearly oscillating drive:

$$H(t) = \frac{\hbar\omega_0}{2}\sigma_z + \hbar\Omega \cos(\omega t)\sigma_x.$$

Use the same $R(t)$. The transformed drive is

$$\begin{aligned} R^* \left[ \hbar\Omega \cos(\omega t)\sigma_x \right] R &= \frac{\hbar\Omega}{2}\sigma_x\\ &\quad+ \frac{\hbar\Omega}{2} \left[ \cos(2\omega t)\sigma_x - \sin(2\omega t)\sigma_y \right]. \end{aligned}$$

The rotating frame separates a static co-rotating term from a counter-rotating term at $2\omega$:

$$\begin{aligned} H_R(t) &= \frac{\hbar\Delta}{2}\sigma_z + \frac{\hbar\Omega}{2}\sigma_x\\ &\quad+ \frac{\hbar\Omega}{2} \left[ \cos(2\omega t)\sigma_x - \sin(2\omega t)\sigma_y \right]. \end{aligned}$$

Dropping the last line is the rotating-wave approximation. The frame change is exact; the deletion is controlled only when the fast term has a sufficiently small net effect on the observables and time scales of interest.

The detailed scale conditions, generalized Rabi frequency, and Bloch–Siegert correction belong to Rotating-Wave Approximation.

Relation to Periodic and Floquet Dynamics

For a $T$-periodic laboratory Hamiltonian, a periodic frame

$$R(t+T)=R(t)$$

produces a $T$-periodic transformed Hamiltonian. Its one-period operator is

$$U_{F,R}(t_0) = R(t_0)^* U_F(t_0) R(t_0).$$

The Floquet eigenphases and quasienergies are unchanged because the operators are unitarily conjugate.

If $R(t)$ is not periodic over $T$, the endpoint frames differ and the relation is not a simple conjugacy. Quasienergy representatives can appear shifted, and the chosen period or enlarged period must be tracked.

Rotating frames can expose a useful effective Hamiltonian, but an exact Floquet operator still includes all laboratory-frame evolution. Approximate frame Hamiltonians should not be substituted into Floquet claims without an error estimate.

AMO and Quantum-Control Uses

Rotating frames are routine in:

• nuclear and electron spin resonance;
• atomic and molecular spectroscopy;
• laser-driven transitions;
• trapped-ion and neutral-atom control;
• superconducting-qubit gates;
• cavity and circuit QED;
• dynamical decoupling and average-Hamiltonian theory;
• Floquet engineering.

They make detuning, phase, and slowly varying envelopes explicit. A pulse phase often becomes the azimuthal direction of a transverse control axis in the rotating frame.

Links to Quantum Control explains how the exact frame transformation becomes a pulse-design tool. Protocol design, calibration, composite pulses, robustness, and open-system errors then continue in Rabi and Ramsey Control, Pulse Sequences, and Driven Open Systems.

Common Mistakes

• Omitting the frame term $-i\hbar R^*\dot R$.
• Combining formulas from $R^*\lvert\psi\rangle$ and $R\lvert\psi\rangle$ conventions.
• Treating a passive frame change as an active physical pulse.
• Transforming the Hamiltonian but not states, observables, or measurement operators consistently.
• Assuming a rotating frame is automatically an approximation.
• Assuming every rotating-frame Hamiltonian is time independent.
• Dropping counter-rotating terms before identifying their frequencies and amplitudes.
• Applying the circular-drive exact result to a linearly polarized drive.
• Forgetting the endpoint factor $R(t_0)$ in the transformed propagator.
• Claiming Floquet quasienergies are unchanged under a nonperiodic frame without tracking endpoint phases.

References

• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Volume Two, Wiley, 1977.
• C. P. Slichter, Principles of Magnetic Resonance, 3rd ed., Springer, 1990.
• L. Allen and J. H. Eberly, Optical Resonance and Two-Level Atoms, Dover, 1987.
• C. Cohen-Tannoudji, J. Dupont-Roc, and G. Grynberg, Atom–Photon Interactions: Basic Processes and Applications, Wiley, 1992.
• D. A. Lidar and T. A. Brun, eds., Quantum Error Correction, Cambridge University Press, 2013.

Exercises

1. Derive the rotating-frame Hamiltonian for the convention $\lvert\psi_R\rangle=R^*\lvert\psi\rangle$.

Solution

Differentiate:

$$\lvert\dot\psi_R\rangle = \dot R^*\lvert\psi\rangle + R^*\lvert\dot\psi\rangle.$$

Using $\lvert\psi\rangle=R\lvert\psi_R\rangle$ and

$$i\hbar\lvert\dot\psi\rangle=H\lvert\psi\rangle$$

gives

$$i\hbar\lvert\dot\psi_R\rangle = \left( i\hbar\dot R^*R + R^*HR \right) \lvert\psi_R\rangle.$$

Because $\dot R^*R=-R^*\dot R$,

$$H_R = R^*HR - i\hbar R^*\dot R.$$

2. Show that density-operator expectation values are frame invariant.

Solution

Use

$$\rho_R=R^*\rho R, \qquad A_R=R^*AR.$$

Then

$$\begin{aligned} \operatorname{Tr}(\rho_RA_R) &= \operatorname{Tr} \left( R^*\rho RR^*AR \right)\\ &= \operatorname{Tr} \left( R^*\rho AR \right)\\ &= \operatorname{Tr}(\rho A), \end{aligned}$$

where unitarity and cyclicity of the trace were used.

3. Derive the exact rotating-frame Hamiltonian for the circularly rotating spin field.

Solution

For

$$R=e^{-i\omega t\sigma_z/2},$$

one has

$$K_R = i\hbar R^*\dot R = \frac{\hbar\omega}{2}\sigma_z.$$

The transverse combination transforms as

$$R^* \left[ \cos(\omega t)\sigma_x + \sin(\omega t)\sigma_y \right] R = \sigma_x.$$

Therefore

$$\begin{aligned} H_R &= \frac{\hbar\omega_0}{2}\sigma_z + \frac{\hbar\Omega}{2}\sigma_x - \frac{\hbar\omega}{2}\sigma_z\\ &= \frac{\hbar(\omega_0-\omega)}{2}\sigma_z + \frac{\hbar\Omega}{2}\sigma_x. \end{aligned}$$

This result is exact.

4. Identify the approximation in the linearly driven two-level model.

Solution

After the exact frame transformation,

$$\begin{aligned} H_R(t) &= \frac{\hbar\Delta}{2}\sigma_z + \frac{\hbar\Omega}{2}\sigma_x\\ &\quad+ \frac{\hbar\Omega}{2} \left[ \cos(2\omega t)\sigma_x - \sin(2\omega t)\sigma_y \right]. \end{aligned}$$

Nothing has yet been approximated. The rotating-wave approximation drops the final, rapidly oscillating line and keeps

$$H_{\rm RWA} = \frac{\hbar\Delta}{2}\sigma_z + \frac{\hbar\Omega}{2}\sigma_x.$$

Its validity requires the counter-rotating term to have a small accumulated effect, typically under weak near-resonant driving and appropriate observation times.

5. Derive the rotating-frame propagator and its periodic-frame Floquet relation.

Solution

Because

$$\lvert\psi(t)\rangle = U(t,t_0)\lvert\psi(t_0)\rangle,$$

one has

$$\begin{aligned} \lvert\psi_R(t)\rangle &= R(t)^* U(t,t_0) R(t_0) \lvert\psi_R(t_0)\rangle. \end{aligned}$$

Thus

$$U_R(t,t_0) = R(t)^*U(t,t_0)R(t_0).$$

If $R(t+T)=R(t)$, then

$$\begin{aligned} U_{F,R}(t_0) &= U_R(t_0+T,t_0)\\ &= R(t_0)^* U_F(t_0) R(t_0). \end{aligned}$$

The two Floquet operators are unitarily conjugate and have the same eigenphases.