Quantum Optics I Bloch Sphere & Rotating Frame Transformation

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

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 $H_0$ 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 can be easily obtained from the fact that $H_0$ and $\mathrm{e}^{\frac{\mathrm{i}}{\hbar} H_0^S t}$ commute. From now on, we will omit the superscript representing pictures of $H_0$ .

For the density operator $\rho(t)$，it is no different from any other operators.

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

Dynamics Transforming the Schrödinger equation into the interaction picture gives

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

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

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

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

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

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 got

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

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

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

Bloch Sphere

Bloch Sphere

A pure state $\ket{\psi}$ is a representative of a equivalence class

$$\{ \mathrm{e}^{\mathrm{i}\chi} \ket{\psi}\mid\alpha\in\mathbb{R}\}$$

One can map the state space of a two-level system $\mathbb{C}^2$ to the unit 3D sphere $S^2$ by

$$\ket{\psi} = \cos\frac{\theta}{2} \ket{1}+\mathrm{e}^{\mathrm{i}\phi }\sin\frac{\theta}{2} \ket{0}$$

$$\boldsymbol{n}(\theta,\phi)=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$$

where $\theta \in [0,\pi]$ and $\phi \in [0,2\pi)$.

Bloch vector

We set the basis of the state space to $\{\ket{1},\ket{0}\}$, which means $\ket{1}=\begin{pmatrix}1\\ 0\end{pmatrix}$ and $\ket{0}=\begin{pmatrix}0 \\ 1\end{pmatrix}$.

/* Note: This definition is different from the order of kets that are usually chosen. The advantage of this kind of choice is that the Hamiltonian will have the form $H = \frac{1}{2} \hbar \omega_0 \sigma_z$ instead of $-\sigma_z$, since we set $\ket{0} \equiv \ket{g}$ and $\ket{1} \equiv \ket{e}$. */

State	Coordinates on $S^2$	$(\theta,\phi)$	Ket	Vector
$\ket{+}_z \equiv\ket{1}$	$(0,0,1)$	$(0,0)$	$\ket{1}$	$\begin{pmatrix}1\\ 0\end{pmatrix}$
$\ket{-}_z=\ket{0}$	$(0,0,-1)$	$(\pi,0)$	$\ket{0}$	$\begin{pmatrix}0 \\ 1\end{pmatrix}$
$\ket{+}_x$	$(1,0,0)$	$(\frac{\pi}{2},0)$	$\frac{\ket{1}+\ket{0}}{\sqrt{2}}$	$\frac{1}{\sqrt{2}}\begin{pmatrix}1\\ 1\end{pmatrix}$
$\ket{-}_x$	$(-1,0,0)$	$(\frac{\pi}{2},\pi)$	$\frac{\ket{1}-\ket{0}}{\sqrt{2}}$	$\frac{1}{\sqrt{2}}\begin{pmatrix}1\\ -1\end{pmatrix}$
$\ket{+}_y$	$(0,1,0)$	$(\frac{\pi}{2},\frac{\pi}{2})$	$\frac{\ket{1}+\mathrm{i}\ket{0}}{\sqrt{2}}$	$\frac{1}{\sqrt{2}}\begin{pmatrix}1\\ \mathrm{i}\end{pmatrix}$
$\ket{-}_y$	$(0,-1,0)$	$(\frac{\pi}{2},\frac{3\pi}{2})$	$\frac{\ket{1}-\mathrm{i}\ket{0}}{\sqrt{2}}$	$\frac{1}{\sqrt{2}}\begin{pmatrix}1\\ -\mathrm{i}\end{pmatrix}$

Theorem Antipodal points on the Bloch sphere corresponds to a pair of mutually orthogonal state vectors.

Rotation

A anticlockwise rotation around $\hat{n}$ by $\theta$ on the Bloch sphere is described by rotation operator

$$R(\theta) = \exp\left[-\frac{\mathrm{i}}{2} \theta (\hat{n}\cdot{\boldsymbol{\sigma}})\right]$$

Rotation around the x,y,z axis are

$$R_x= \exp\left(-\frac{\mathrm{i}}{2} \theta \sigma_x\right)= \begin{pmatrix} \cos\frac{\theta}{2} & -\mathrm{i}\sin\frac{\theta}{2} \\ -\mathrm{i}\sin\frac{\theta}{2} & \cos\frac{\theta}{2} \end{pmatrix}$$

$$R_y= \exp\left(-\frac{\mathrm{i}}{2} \theta \sigma_y\right)= \begin{pmatrix} \cos\frac{\theta}{2} & -\sin\frac{\theta}{2} \\ \sin\frac{\theta}{2} & \cos\frac{\theta}{2} \end{pmatrix}$$

$$R_z= \exp\left(-\frac{\mathrm{i}}{2} \theta \sigma_z\right)= \begin{pmatrix} \mathrm{e}^{-\mathrm{i}\theta/2} & 0 \\ 0 & \mathrm{e}^{\mathrm{i}\theta/2} \end{pmatrix}$$

Rotation around $(\cos\phi,\sin\phi,0)$ is defined as $R_\phi$, with

$$R_\phi(\theta)= \exp\left(-\frac{\mathrm{i}}{2} \theta \sigma_\phi\right)= \begin{pmatrix} \cos\frac{\theta}{2} & -\mathrm{i}\mathrm{e}^{-\mathrm{i}\phi}\sin\frac{\theta}{2} \\ -\mathrm{i}\mathrm{e}^{\mathrm{i}\phi}\sin\frac{\theta}{2} & \cos\frac{\theta}{2} \end{pmatrix}$$

Rotating Frame

General transformation

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

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

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

Maintaining the form of Schrodinger equation

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

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

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

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

Physical Meaning

There are two ways to comprehend the additional term $\mathrm{i}\hbar \dot{U} U^\dagger$ in the expression of transformed Hamiltonian $\tilde{H}$. 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 $\omega$ , then its centrifugal potential energy is $E = \frac{1}{2} m \omega^2 R^2$. When switching to a reference frame rotating at the same angular frequency of $\omega$ , an observer in this reference frame will find that the object is completely stationary. Then this observer will consider its energy as $0$. The centrifugal potential energy disappears when performing rotating frame transformation.

In quantum mechanics, the energy is described by the Hamiltonian. $\mathrm{i}\hbar \dot{U} U^\dagger$ is that disappeared centrifugal potential energy. 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 like

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

one can obtain with some calculation

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

therefore, the stationary part is transformed into

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

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

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

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

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

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

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

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

$$\begin{gathered} \ket{\tilde{\psi}(0)} = c_g \,\mathrm{e}^{\mathrm{i}\omega_1 t} \ket{g} + c_e \, \mathrm{e}^{\mathrm{i}\omega_2 t} \ket{e} \\ \ket{\tilde{\psi}(t)} = c_g \,\mathrm{e}^{-\mathrm{i} (\omega_g-\omega_1) t} \ket{g} + c_e \, \mathrm{e}^{\mathrm{i}(\omega_e-\omega_2) t} \ket{e} \end{gathered}$$

it changes the oscillation frequency of the states.

Example

Two-level system

Three-level system

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

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

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 Hamiltonian becomes

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

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