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STOCHASTIC MODEL FOR THE NUMBER OF ATOMS IN A MAGNETO-OPTICAL TRAP

Published online by Cambridge University Press:  06 March 2006

Andrew L. Rukhin
Affiliation:
National Institute of Standards and Technology, and, University of Maryland at Baltimore County, Baltimore, MD 21250, E-mail: rukhin@math.umbc.edu
Ionut Bebu
Affiliation:
National Institute of Standards and Technology, and, University of Maryland at Baltimore County, Baltimore, MD 21250, E-mail: rukhin@math.umbc.edu
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Abstract

In this article a Markov chain for the distribution of single atoms is suggested and studied. We explore a recursive model for the number of atoms present in a magneto-optical trap under a feedback regime with a Poisson-distributed load. Formulas for the stationary distribution of this process are derived. They can be used to adjust the loading rate of atoms to maximize the proportion of time that a single atom spends in the trap. The (approximate) optimal regime for the Poisson loading and loss processes is found.

Type
Research Article
Copyright
© 2006 Cambridge University Press

1. INTRODUCTION

The ability to control individual atoms is crucial in nanotechnology and quantum information processing. Atom-on-demand technology will enable novel quantum computation schemes, unprecedented control over dopants in materials, and eventual realization of the ultimate goal of nanotechnology: atom-by-atom construction.

Here, we suggest a version of a recursive model for the number of atoms present in a magneto-optical trap (MOT) at any given time under a feedback regime with a Poissonian loss. Formulas for the stationary distribution of this process can be used to control the loading rate of atoms so as to maximize the proportion of time that a single atom spends in the trap. The (approximate) optimal regime for the Poisson loading and loss processes is found. More specifically, the values of loading and loss parameters maximizing the probability of exactly one atom in the trap are determined and confirmed through Monte Carlo simulation. These results make use of the Borel–Tanner distribution (Kingman [8]), which describes the total number of customers served before a queue vanishes with random arrival of customers and a constant time needed to serve each customer.

Optical quantum cryptography is built on the use of the single photon state [2, Sect. III.A]. Practical implementation mainly relies on faint laser pulses based on standard semiconductor lasers. The Electron and Optical Physics Division of the National Institute of Standards and Technology has built a prototype device that can produce and sustain one atom of a given species, upon demand, for a useful time interval. Hill and McClelland [6] described a new accurate method of trapping neutral atoms in a vacuum chamber that does not use faint pulses. To achieve the optimal stationary distribution for single atoms in our model, one needs a weak load regime and even weaker loss processes. However, under these conditions, the convergence to the stationary distribution is the slowest.

2. MARKOV CHAIN MODEL FOR THE NUMBER OF ATOMS IN A MAGNETO-OPTICAL TRAP

Suppose the door to the MOT could be opened the instant it became empty or closed at the instant it became occupied. Then the number of atoms in the trap can be modeled as a Markov chain. Because of physical limitations, the number of atoms in the trap cannot be monitored continuously: The trap can only be checked every T seconds. During that interval T, more than one atom can sneak inside the trap before the door is closed.

Let Rn be a Poisson random variable with parameter λ, representing the load (i.e., the random number of atoms entering the trap at time n) and denote by Xn the number of atoms in the MOT at time n,n = 1,2,…. For a fixed Xn = x, the random loss variable Yn has a Poisson distribution with parameter μx. A simple recursive model for the number Xn of atoms present in the MOT at step n under a feedback regime is

with Rn and (Xn−1,Yn−1) independent. As a simple example, suppose that at step n, there are three atoms in the MOT, so Xn = 3; according to the model, the trap is opened. If we lose two atoms, then at step n + 1, there is only one atom left in the trap (Xn+1 = 1), but if we lose all three atoms, Xn+1 is determined by a new load; notice that in the latter case, we only consider one step, although we have two time intervals. Our goal is to get a single atom with high probability; if we know that the trap empties, there is no point in stopping the process.

Clearly, Xn,n = 1,2,…, forms a countable state Markov chain in discrete time. If XnX, with the distribution of X being the steady-state (stationary) distribution for the chain, then the following equality of distributions holds:

Here, R is Poisson distributed with parameter λ, and for a fixed X = x, the random variable Y is Poisson distributed with parameter μx.

Our goal is to find the distribution of X and to find the values of λ and μ that maximize the value π1 = P(X = 1). For this purpose, we determine the probability vector π = (π01,…) with πk = P(X = k), k = 0,1,2,….

Notice that the transition probabilities of the above Markov chain are

In particular,

with d0 = 1.

Observe that

which is maximized when μ = 0 (the value of λ being immaterial.)

3. STEADY-STATE DISTRIBUTION FOR THE NUMBER OF ATOMS

Theorem 1: For any λ,μ > 0, the Markov chain determined by transition probabilities given in Eq. (2.3) is ergodic. In fact, the chain is geometrically ergodic; that is, for some R(x) and ρ < 1,

n = 1,2,…. The vector π of stationary probabilities has the form given in Eq. (3.13) with the vector u satisfying Eq. (3.8).

Proof: We will prove that the Markov chain determined by Eq. (2.3) has a stationary distribution π by constructing it. For the stationary distribution to exist, we need

so π0 ≠ 0, and for k = 1,2,…, with vk = πk0,

Also,

. Introduce the lower triangular matrix B with the entries

For μ > 0, we have

, and for 0 < μ ≤ 1, the elements of B are in fact the probabilities of the Borel–Tanner distribution, so that, with eT = (1, 1,…), eTB = eT.

According to Eq. (2.3), the reduced transition probabilities matrix P formed by the elements {P(x|y),x,y = 1,2,…} can be written in the form

Here, D is the diagonal matrix with elements 1,2,…,dT = (d1,d2,…), and [ell ]T = (λ,λ2,…,λj/[(j − 1)!],…). Since P(0|y) = e−λλy/y!, one has for vT = (v1,v2,…),

On the other hand,

Combining these formulas, one obtains for u = Dv,

We will prove that the norm of B, as an operator in L2, is smaller than e−μ/2 (<1). Hence, IB is invertible in L2. To this end, we show that for any i = 1,2,…,

From definition (3.4),

so to prove Eq. (3.9), it suffices to demonstrate that

However, this inequality follows directly from comparison of the coefficients ikkik−1 ≤ (i − 1)k, k = 1,…,i − 1 . Since

, a theorem by Schur (Halmos [4, Problem 37], with pi ≡ 1, implies that the operator defined by the matrix B has norm in L2 less than e−μ/2 < 1. Note that b11 = e−μ, so that ∥B2e−μ.

Since [ell ] belongs to the space L2, the solution u of Eq. (3.8) is also in L2 and can be found from the formula

Now we can prove the existence and uniqueness of the stationary distribution π for μ > 0 and λ > 0. Start with u given by Eq. (3.12) and put v = D−1u. By the Cauchy–Schwartz inequality, vL1. The stationary distribution π is given by

The uniqueness of the stationary distribution π follows from the uniqueness of solution for Eq. (3.8). By Theorem 6.9 in Kemeny, Snell, and Knapp [7, p.135], the Markov chain with the transition probabilities given by Eq. (2.3) is ergodic.

Our argument shows that the operator ζIB is invertible if ζ > e−μ/2. It follows that for such values of ζ, one can find in L2 the solution u of the equation

for any right-hand side [ell ] in L2. This fact implies that one can solve the “drift” inequalities

with positive V and C = {0}. It follows from Nummelin [9, p.90] that our chain is geometrically ergodic, which concludes the proof of the theorem. █

The referee suggested a proof of geometric ergodicity by checking Eq. (3.15) with V(x) = βx for some β > 1:

Notice that as

, and for some 0 < ζ < 1, e−μ(1−1/β) < ζ. It follows that Eq. (3.15) holds for some finite xζ and positive β where C = {k : kxζ}.

To get a useful approximation to π1, we use the form of the matrix B and denote elements of

by αij = αij(μ), and put

.

Theorem 2: The probability π1 is given by

with an upper bound (3.24). An approximation (3.26) is valid, and a lower bound (3.31) for π1 holds, leading to approximation (3.36) for 0 < μ < 1. Small values of μ and λ are optimal.

Proof: Using the identity (IB)−1 = I + B(IB)−1, the αij can be found from the recurrent formula

for ij. Note that

One has

so that using Eq. (3.13), we obtain Eq. (3.17). Note that

It is easy to see by induction that

. If for k = 1,…,j − 1, αk1 ≤ α11 /k, then for j ≥ 2,

Also, as iαijjαjj,

From Eq. (3.22), we obtain

which shows that for this probability to be close to unity, one must have μ ≈ 0 or λ must be large. For

it follows from Eqs. (3.22) and (3.23) that limn→∞ π1(n) = π1 and

where

is the incomplete gamma function.

An argument similar to that in Eq. (3.23) can be used to demonstrate that for any fixed j, iαij is a nonincreasing sequence in i = j,j + 1,…, so that the limit, δj = limi→∞ iαij, exists. Since ∥(IBT)−12 ≤ [1 − ∥B2]−1 and, for a fixed l, blp is a decreasing sequence in pl, we get, for ij + 2 from Eq. (3.18),

so that

For 0 < μ < 1, limi→∞ i2bij = 0, and, therefore, δj = δj+1 for any j, demonstrating that δj ≡ δ. It follows that

Using Eq. (3.23), a lower bound for π1 is obtained,

To get a formula for δ when 0 < μ < 1, recall that for any positive λ,

As neither αkj nor dk depend on λ, it follows that for any

, which implies that

Indeed,

since, as in the derivation of the expected value of the Borel–Tanner distribution (e.g., formula (16) in Haight and Breuer [3]),

These results lead to an approximation for π1; if one replaces u1 by λα11 + λ2α21 + δ(eλ − 1 − λ − λ2/2) and sets u2 ≈ λ2α22 + δ(eλ − 1 − λ − λ2/2), uj ≈ δ(eλ − 1 − λ − λ2/2),j ≥ 3. With

denoting the integral exponential function and C denoting Euler's constant, the approximate formula for π1 when 0 < μ < 1 takes the form

For small values of μ,

This ratio is a decreasing function of λ; for small λ,π1 ∼ λ/(λ + μ), so that π1 → 1 as λ ↓ 0,μ/λ → 0. Therefore small values of λ and μ = o(λ) are optimal. █

Theorem 2 gives a method for approximating the probability of interest π1. Given ε, from Eq. (3.26) we determine the level of truncation n. From Eq. (3.25), using the recurrence formulas (3.18) and (3.19), we obtain the approximate value π1n of π1 such that |π1 − π1n| ≤ ε. Figure 1 depicts the probability π1 for λ,μ ∈ (0,2.5) using ε = 10−6.

The plot of the probability π1 as a function of λ and μ.

The referee has suggested using the geometric distribution as the loss distribution. Thus, assume that Yn = min(Xn,G), where G is a random variable with geometric distribution over nonnegative integers with parameter p. Then the inverse of IB can be obtained explicitly. For any load distribution R with the finite first moment, the same technique as in Theorem 1 shows that the stationary distribution is obtained from Eq. (3.13):

In particular,

which is an increasing function in p for any load distribution such that 1 − P(R = 0) ≥ P(R = 1)E(R). The latter condition holds for the Poisson load distribution. Therefore, values of p close to 1 and λ ≈ 0 are optimal for this distribution.

4. DISCUSSION

The bound in Eq. (3.26) suggests that the convergence to the stationary distribution (although geometric) is the slowest for small μ. For this reason, it takes more time to attain the optimal regime than for the stationary distribution associated with other parametric values. Theorem 2 gives a method for computing an approximation π1n as defined in Eq. (3.25) of the probability of interest π1 with any given accuracy ε.

Gibson and Seneta [1] and Heyman [5] gave a method for computing a numerical approximation of the steady-state distribution π of a Markov chain with states 0,1,…, and a transition probabilities matrix P. The idea is to approximate π by πn, the steady-state distribution of the chain obtained by truncating P, keeping the first n states and normalizing it in a convenient way (e.g., by linear normalization or, by row normalization). The conditions (A1)–(A4) in Heyman [5] hold for our case; conditions (A2)–(A4) are easy to check and (A1) follows from Theorem 1, if its implementation is practical.

Acknowledgments

The work of the first author was supported in part by the Defense Research Projects Agency under the DARPA Quantum Information Science and Technology (QuIST) program. The authors are grateful to the referee for penetrating comments.

References

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Figure 0

The plot of the probability π1 as a function of λ and μ.