# Quantum Information processes in atomic ensemble

In this note I will summarize the works by Polzik’s group. I will give the detials of the theoretical treatment in their works.

1. Introduction

Atoms are good at storing and processing quantum information because of its long coherent time. Light is suitable for long-distance quantum state transform. Hence, the coupling between atomic ensemble and light is the key factor for realizing quantum network.

As the number of ensemble is huge, we can treat it continuously, just as we treat light. Because of the collective coupling, the quantum information processes, such as teleportation, can be archieved without cavity, which is greatly decreased the difficulty of the experiments.

1.1. continuous discription of light and matter

Let’s firstly discuss the light/matter interaction. We will use continuous discription of the light and matter. We quantized electromagnetic field in one dimension and only care about on polarization mode (x-polerization)

$\displaystyle E= \sum_{\lambda} \sqrt{\frac{\hbar \omega_\lambda}{2 \epsilon_0 A L}} \big( a e^{ikz} + a^\dagger e^{-ikz} \big), \ \ \ \ \ (1)$

where {}{}

• A, cross seciton of the field mode;
• L, length of the mode;
• k, wave vector. Impose periodic boundary conditions, the ${k}$ space resolution ${\Delta k = 2\pi/L}$. When ${L \rightarrow \infty}$, ${\Delta k \rightarrow 0}$. ${k}$ space tends to continuous. We make changes

$\displaystyle \sum \Delta k \rightarrow \int \mathrm{d} k.$

We define a operator ${a(k) = \frac{a_\lambda}{\sqrt{\Delta k}}}$ with ${k \approx k_\lambda}$. The eletic field operator becomes

\displaystyle \begin{aligned} E = & \sum_\lambda \sqrt{\Delta k} \sqrt{\frac{\hbar \omega_\lambda} {2\epsilon_0 LA}} \big( a(k_\lambda) e^{ik_\lambda z} + a^\dagger (k_\lambda) e^{-ik_\lambda z} \big) \\ =& \sum_\lambda \Delta k \sqrt{\frac{\hbar \omega_\lambda}{4\pi \epsilon_0 A }} \big( a(k_\lambda) e^{ik_\lambda z} + a^\dagger (k_\lambda) e^{-i k_\lambda z} \big) \\ \rightarrow & \int \mathrm{d} k \sqrt{\frac{\hbar \omega_\lambda}{ 4\pi \epsilon_0 A}} \big( a(k) e^{ikz} + a^\dagger (k) e^{-ikz} \big). \end{aligned} \ \ \ \ \ (2)

${a(k), a^\dagger(k)}$ are continuous, have units of square root meters. The commutation is

$\displaystyle [ a(k), a^\dagger (k')] = \delta (k-k').$

The field Hamiltonian is

$\displaystyle H = \int \mathrm{d} k \hbar ck (a^\dagger (k) a(k) + \frac{1}{2} ).$

Now let’s transform the electric field into the spatial description.

$\displaystyle \left\{ \begin{array}{ll} a(z,t) =& \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} a(k,t) e^{ikz} \mathrm{d} k \\ a^\dagger (z,t) =& \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} a^\dagger (k,t) e^{-ikz} \mathrm{d} k \end{array} \right.$

where the commutation is ${[a(z,t), a^\dagger(z',t)] = \delta (z-z')}$. The dimension of ${a(z,t)}$ and ${a^\dagger(z,t)}$ is ${\frac{1}{\text{Length}}}$. ${a^\dagger(z,t) a(z,t) \mathrm{d}z}$ means the number of photons in the space between ${z}$ and ${z+\mathrm{d}z}$.

1.2. Maxwell-Bloch equation

The operator ${a(z,t)}$‘s motion follows Heisenberg equation. The Hamiltonian is ${H= H_{\mathrm{R}} + H_{\mathrm{atom}} + H_{\mathrm{int}}}$. ${H_{\mathrm{R}}}$ represents the Hamiltonian of pure radiation field. ${H_{\mathrm{atom}}}$ is the Hamiltonian of matter. ${H_{\mathrm{int}}}$ is the interaction Hamiltonian between light and matter. Therefore we get

\displaystyle \begin{aligned} \frac{\partial }{\partial {t}} a(z,t) =& \frac{1}{\sqrt{2\pi}} \int \frac{\partial}{\partial t} a(k,t) e^{ikz} \mathrm{d}k \\ =& \frac{1}{\sqrt{2\pi}} \int \frac{1}{i\hbar} \big[ a(k,t), H_{\mathrm{R}} + H_{\mathrm{int}} \big] e^{ikz} \mathrm{d}k. \\ \big[ a(k,t), H_{\mathrm{R}}] = &\int \mathrm{d}k' \hbar c k' \big[ a(k,t), a^\dagger (k',t) \big] a(k',t) \\ =& \hbar ck a(k,t). \end{aligned} \ \ \ \ \ (3)

As ${\frac{\partial}{\partial z} a(z,t) = \frac{1}{ \sqrt{2\pi}} \int ik a(k,t) e^{ikz} \mathrm{d}k}$, we get

$\displaystyle \frac{1}{\sqrt{2\pi}} \int \big[ a(k,t), H_{\mathrm{R}} \big] e^{ikz} \mathrm{d}k = -c \frac{\partial}{\partial z} a(z,t). \ \ \ \ \ (4)$

Combining Eq. (3) and (4) we get

$\displaystyle \big( \frac{\partial}{\partial t} + c \frac{\partial}{ \partial z} \big) a (z,t) = \frac{1}{i\hbar} \big[ a(z,t),H_{\mathrm{int}} \big]. \ \ \ \ \ (5)$

This is what to say Maxwell-Bloch equation, which describes the light field affected by atoms through the interaction ${H_{\mathrm{int}}}$.

If the radiation field is restricted to the narrow band, that is ${\omega \approx \omega_0}$, we get

\displaystyle \begin{aligned} E=& \int \mathrm{d}k \sqrt{\frac{\hbar \omega_0}{4\pi \epsilon_0 A}} \big( a(k,t) e^{ikz} + a^\dagger (k,t) e^{-ikz} \big) \\ =& \sqrt{\frac{\hbar \omega_0}{2\epsilon_0 A}} \big( a(z,t) + a^\dagger (z,t) \big). \end{aligned} \ \ \ \ \ (6)

In free space, we get time and space dependence ${a(z,t)= a(0,t-\frac{z}{c})}$. Let’s define ${a(t)= \sqrt{c} a(z,t)}$, ${a^\dagger(t) a(t)}$ is the flux of photons at time ${t}$. For the light in the vacuum state we get

$\displaystyle \big[ a(t), a(t') \big] = \delta (t-t').$

• 1.3. Continuous Matter Operators

For simplicity, we only consider the collection of two-level atoms, which are coupled to the light by dipole transition. ${|g\rangle}$ and ${|e \rangle}$ are ground and excited states. The interaction Hamiltonian is

$\displaystyle H_{\mathrm{int}} = - \sum_j \vec{d}_j \cdot \vec{E} (\vec{R}_j), \ \ \ \ \ (7)$

where ${\vec{d}_j = -e \vec{r}_j}$ is the dipole operator for the ${j}$‘th atom, ${\vec{R}_j}$ is the location of the ${j}$‘th atom.

If we consider the light is linearly polarized along ${x}$, we get Hamiltonian

$\displaystyle H_{\mathrm{int}} = \sqrt{\frac{\hbar \omega_0 }{2\epsilon_0 A}} \sum_j \big( d^* \sigma_{eg}^{(j)} a(z_j,t) + d \sigma_{ge}^{(j)} a^\dagger (z_j,t) \big). \ \ \ \ \ (8)$

It is convienient to define continuous spin operator

$\displaystyle \sigma_{ge} (z,t)= \frac{1}{\rho A \mathrm{d}z} \sum_{z \in [z,z+\mathrm{d}z ]} \sigma_{ge}^{(j)} (t), \ \ \ \ \ (9)$

where ${\rho A \mathrm{d}z}$ is the number of atoms in the slice ${[z,z+\mathrm{d} z]}$, ${\sigma_{ge}(z,t)}$ is dimensionless. Rewriting Hamiltonian via continuous spin operator, we get

$\displaystyle H_{\mathrm{int}} = \sqrt{\frac{\hbar \omega_0}{2\epsilon_0 A}} \int_0^L \big( d^* \sigma_{eg} (z,t) a(z,t) + d \sigma_{ge} (z,t) a^\dagger (z,t) \rho A \mathrm{d}z. \ \ \ \ \ (10)$

Combining Maxwell-Bloch equation, we get

$\displaystyle \big( \frac{\partial}{\partial t} + c \frac{\partial}{\partial z} \big) a(z,t) = -i g \rho A \sigma_{ge} (z,t), \ \ \ \ \ (11)$

where ${g= \sqrt{\omega_0/2\omega_0\hbar A} d}$. Together with the Heisenberg equation of motion for ${\sigma_{ge}(z,t)}$, we have the coupled quantum Maxwell-Bloch equations describing light/matter interaction.

We have following commutation

$\displaystyle [\sigma_{\mu\nu}(z,t), \sigma_{\mu'\nu'}(z',t)] = \frac{1}{\rho A} ( \sigma_{\mu\nu'} \delta_{\nu\mu'} - \sigma_{\mu'\nu} \delta_{\nu'\mu} ) \delta(z-z').$

The dimension of ${\delta(z-z')}$ is ${\frac{1}{\text{Length}}}$.

2. QND measurement induced entanglement

In this section we will discuss the scheme to realize QND measurement on matter and using the measurement to generate entanglement between atomic ensembles. The results we get in the last section will be widely used in this section.

The level struction of the atoms and schematic setup is shown in Lu-Ming Duan et al. PRL 85, 5643(2000).

The atomic ensemble is of a pencil shape with Fresnel number ${F= A/\lambda_0 L \approx 1}$. The input laser pulse is linearly polarized with to orthogonal mode.

$\displaystyle E^+ = (z,t)= \sqrt{\frac{\hbar \omega}{4\pi \epsilon_0 A}} \sum_{i=1,2} a_i (z,t) e^{i(k_0 z - \omega_0 t)}, \ \ \ \ \ (12)$

where ${\omega_0 = \frac{2\pi c}{\lambda_0}}$ is the frequency of the laser. ${a^\dagger_i (z,t) a_i (z,t) \mathrm{d}z}$ is the photon number of ${i}$th mode in the area ${[z, z+\mathrm{d}z]}$. The dimension of ${a(z,t)}$ is ${\frac{1}{\sqrt{Length}}}$. Suppose that ${\langle a_i(0,t) \rangle = \alpha_t}$. The number of photons is ${2N_p = 2c \int_0^T | \alpha_t |^2 \mathrm{d}t \gg 1}$.

Define Stokes operators

\displaystyle \begin{aligned} S_x^p=& \frac{c}{2} \int_0^L (a_1^\dagger a_2 - a_2^\dagger a_1) d \tau, \\ S_y^p = & \frac{c}{2i} \int_0^L ( a_1^\dagger a_2 - a_2^\dagger a_1) d\tau, \\ S_z^p = & \frac{c}{2} \int_0^L (a_1^\dagger a_1- a_2^\dagger a_2) d \tau. \end{aligned} \ \ \ \ \ (13)

The commutation of them is

\displaystyle \begin{aligned} {S_x^P,S_y^p} =& \frac{ic^2}{2} \int_0^L \int_0^L \big[ a_1^\dagger a_1 \delta(\tau-\tau') - a_2^\dagger a_2 \delta(\tau-\tau') \big] d\tau d\tau' \\ =& \frac{ic^2}{2} \int_0^T (a_1^\dagger a_1- a_2^\dagger a_2) d\tau \\ =& i S_z^p. \end{aligned} \ \ \ \ \ (14)

${[S^p_y, S^p_z] = iS_x^p}$; ${[S_z^p, S_x^p]= iS_y^p}$.

For ouer coherent input, ${\langle S_x^p \rangle = N_p}$, ${\langle S_y^p \rangle = \langle S_z^p \rangle = 0}$. Because ${N_p \gg 1}$, we can treat ${S_x^p}$ classically. We define ${X^p = \frac{S_y^p}{\sqrt{\langle S_x^p \rangle}}}$, ${P^p = \frac{S_z^p}{\sqrt{\langle S_x^p \rangle}}}$. We get ${[X^p, Y^p]=i}$.

We use continuous atomic operators,

$\displaystyle \sigma_{\nu\mu} (z,t) = \lim_{\delta z \rightarrow \infty} \frac{1}{\rho A \delta z} \sum_{z \in (z,z+\delta z)} \sigma^{(j)}_{\nu\mu} (t), \ \ \ \ \ (15)$

where ${\nu,\mu = 1,2,3,4}$. The Hamiltonian is

$\displaystyle H_I = \hbar \sum_{i=1,2} \int_0^L \big[ g \sigma_{i,i+2} a_i (z,t) + \mathrm{H.C.} \big] \rho A \mathrm{d} z, \ \ \ \ \ (16)$

where ${g= \sqrt{\frac{\omega_0}{2\omega_0 A \hbar}} d}$. In the large detuing limit, we can adiabatically eliminated the upper state population from the equation of motion.

\displaystyle \begin{aligned} H_{eff} =& \sum_{i=1,2} \hbar \int_0^L \big[ \frac{g^2}{\Delta} \sigma_{ii} a^\dagger_i (z,t) a_i (z,t) \rho A \mathrm{d}z \big]\\ =& \int_0^L \frac{g^2}{\Delta} \rho A \big[ (\sigma_{11} - \sigma_{22}) (a_1^\dagger a_1 - a_2^\dagger a_2) + \sigma_{11} a_2^\dagger a_2 + \sigma_{22} a_1^\dagger a_1 \big] \mathrm{d} z \\ =& \frac{g^2 \rho A}{\Delta} \int_0^L \big[ \frac{1}{2} (\sigma_{11} - \sigma_{22} )( a^\dagger_1 a_1 - a_2^\dagger a_2) + \frac{1}{2} N_a N_p \big] \mathrm{d} z, \end{aligned} \ \ \ \ \ (17)

where ${N_a}$ is the number of atoms. Neglect the constant energy part, which only induces the global phase, The effective Hamiltonian is

$\displaystyle H_{eff} = \hbar \frac{2g^2}{\Delta L} \sqrt{N_aN_p} P^a P^p. \ \ \ \ \ (18)$

The Heisenberg equation is

\displaystyle \begin{aligned} \dot{X}^p = &\frac{1}{i\hbar} \big[X^p, H_{eff}] \\ =& \frac{2g^2}{\Delta L} \sqrt{N_a N_p} P^p. \end{aligned} \ \ \ \ \ (19)

We can eayily get the input-output relation as

\displaystyle \begin{aligned} {X^P}' =& X^P + \frac{2g^2}{\Delta L} \sqrt{N_aN_p} P^a \frac{L}{c} \\ =& X^P + \frac{2g^2}{\Delta c} \sqrt{N_aN_p} P^a, \\ {X^a}' =& X^a + \frac{2g^2}{\Delta c} \sqrt{N_aN_p} P^p, \\ {P^\beta}' =& P^\beta, \quad \beta=a,p. \end{aligned} \ \ \ \ \ (20)

Here we neglect the loss terms, which is small.

Using input-output relation (20), we get

\displaystyle \begin{aligned} {X^P_2}' =& X_1^P + \kappa \big( P^a_1 + P^a_2 \big), \\ {X_1^a}' + {X_2^a}' = & X_1^a + X_2^a + 2\kappa P^p. \end{aligned} \ \ \ \ \ (21)

where ${\kappa = \frac{2g^2}{\Delta c} \sqrt{N_aN_p}}$. Is is easy to get the viarance of the operators

\displaystyle \begin{aligned} \delta({X_2^P}') =& \frac{1}{2} + \kappa^2 \\ \delta({X_1^a}' + {X_2^a}') = &\delta(X_1^a) + \delta(X_2^a) + \delta(P^p) (2\kappa)^2 \\ =& \frac{1}{2} + \frac{1}{2} + 4\kappa^2 \frac{1}{2} \\ =& 1+ 2\kappa^2. \end{aligned} \ \ \ \ \ (22)

Here we use the fluctuation ${\delta(X_1^a) = \delta(X_2^a) = \delta(P^p) =\frac{1}{2}}$.

Suppose that ${\langle S_x^a \rangle_1 = \langle S_x^a \rangle_2 = N_a}$, we get

$\displaystyle \big[ X_1^a + X_2^a, P_1^a + P_2^a ] = 2i. \ \ \ \ \ (23)$

$\displaystyle \Rightarrow \delta(X_1^a + X_2^a) \delta(P_1^a + P_2^a) =1.$

After measurement on ${{X^p}'}$, the pile up noise in the variable ${X_1^a + X_2^a}$ is ${(1+2\kappa^2)}$. The uncertainty relation must be hold on. Therefore, ${\delta(P_1^a + P_2^a) = \frac{1}{1+2\kappa^2}}$. We rotate the collective atomic spins, ${X_1^p \rightarrow -P_1^a, P_1^a \rightarrow X_1^a , X_2^a \rightarrow P_2^a, P_2^a \rightarrow -X_2^a}$. Then we measure again. The QND measurement will be on ${X_1^a - X_2^a}$ and makes it squeezed, ${\delta(X_1^a - X_2^a) = \frac{1}{1 + 2\kappa^2}}$. We have prepared a two-mode squeezed state. The squeezing parameter ${r}$ is given by

$\displaystyle r = \frac{1}{2} \ln (1+ 2\kappa^2)$

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