Back and forth nudging for quantum state estimation by continuous weak measurement

Abstract

We propose to apply the Back and Forth Nudging (BFN) method used for geophysical data assimilations [1] to estimate the initial state of a quantum system. We consider a cloud of atoms interacting with a magnetic field while a single observable is being continuously measured over time using homodyne detection. The BFN method relies on designing an observer forward and backwards in time. The state of the BFN observer is continuously updated by the measured data and tends to converge to the system's state. The proposed estimator seems to be globally asymptotically convergent when the system is observable. A detailed convergence proof and simulations are given in the 2-level case. An extension of the algorithm to the multilevel case is also presented.

Full text

Page 1

Back and forth nudging for quantum state estimation by continuous
weak measurement
Zaki Leghtas1, Mazyar Mirrahimi1, and Pierre Rouchon2
1INRIA Paris-Rocquencourt Domaine de Voluceau, BP105 78153 Le Chesnay Cedex, France
2Mines-ParisTech, Centre Automatique et Syst` emes, 60, boulevard Saint-Michel 75272 Paris Cedex, France
Abstract—We propose to apply the Back and Forth Nudging
(BFN) method used for geophysical data assimilations [1] to
estimate the initial state of a quantum system. We consider a
cloud of atoms interacting with a magnetic field while a single
observable is being continuously measured over time using
homodyne detection. The BFN method relies on designing an
observer forward and backwards in time. The state of the BFN
observer is continuously updated by the measured data and
tends to converge to the system’s state. The proposed estimator
seems to be globally asymptotically convergent when the system
is observable. A detailed convergence proof and simulations are
given in the 2-level case. An extension of the algorithm to the
multilevel case is also presented.
I. INTRODUCTION
Estimating the state of a quantum system is a fundamental
problem of great interest in quantum control. Amongst a
variety of applications, it is essential to verify the efficiency
of a quantum state preparation protocol [2], [3]. For this
reason, it is interesting to avoid the usual quantum state
tomography scheme which involves doing the experiment
many times and performing a strong projective measurement
of a new observable at each preparation. Indeed, since many
realizations of the preparation protocol are necessary to
obtain one state estimation, the fidelity of the preparation
protocol is averaged out over all these realizations. A new
approach overcomingthis problem was proposed and verified
experimentally in [4] where a controlled evolution is applied
to an ensemble average while an observable is continuously
measured. A Bayesian filter is then used to reconstruct the
quantum state from the measurement record. In this paper,
we consider a similar setting to the one in [4], [5], [6] (see
[7] for a recent review of these papers). We propose a new
approach inspired of the BFN method used in geophysical
data assimilation [1] to reconstruct the state of the system
from the measured data. We design an observer, which is an
estimation of the quantum system’s state and feed in the data
continuously until the observer converges to the system’s
state. A similar proposal was outlined in [8]. However, the
method we propose has the advantage of extending naturally
to a multidimensional case and makes more use of the
specific dynamics which the system undergoes.We guess this
can strongly reduce the computation time of the estimation
and increase its robustness.
In section II we detail the problem settings and give the
dynamics equations of the BFN observer. Simulation plots
are then presented in section III to demonstrate the efficiency
of the state reconstruction protocol. A detailed convergence
proof of the observer is given in section IV. Finally, in
section V, we discuss the extension of this algorithm to the
multilevel case.
II. THE PROBLEM SETTING
We consider the experimental setting introduced in [4]. To
simplify the theoretical study, we suppose that the system
is a spin
2system (instead of a system of total angular
momentum equal to 3 or 4 as considered in [4]). It interacts
with a magnetic field in the x-y plane: the control, and a
probe. Homodyne detection of the probe, as explained in [5]
enables a weak continuous measurement of the spin system.
We suppose that all the parameters involved in the dynamics
are known. The dynamics of the spin ensemble is described
by the master equation:
1
d
dtρ(t)=
−i[Bx(t)σx+ By(t)σy,ρ(t)]
+
=
Γ(σzρ(t)σz− ρ(t))
Tr(σzρ(t)) .
(1)
y(t)
[.,.] is the Lie Bracket operator and Tr(.) is the trace
operator. We take ? = 1. ρ(t) is the density matrix of the
average ensemble at time t. It is a 2 × 2 positive Hermitian
matrix of trace 1. y is the measurement. σx,σy and σz are
the standard Pauli matrices. Γ > 0 gives the strength of the
coupling between the probe and the system. The aim is, from
a set of data {y(t)\t ∈ [0,T]}, to estimate the initial state of
the system ρ(0) which can be any pure or mixed state.
In order to do so we use Luenberger observers based on the
back and forth nudging (BFN) method [1], [8]. The designed
observer was first introduced, without BFN and for a non
dissipative system, in [9].
The idea is to design an observer on system (1) and another
on the same system but by changing t → T − t. And doing
this iteration n times. This is equivalent to supposing that
the system has a periodic dynamics of period 2T which is
symmetric with respect to t = T and that we are measuring
the system over a time interval [0,2nT]. This gives more
time for the observer to converge with a small gain and with
minimal amount of data. System (1) is referred to as the
”forward” system and the same system changing t to T − t
the ”backward” system. The index k introduced below refers
to the kthback and forth iteration of the algorithm. The letter
’f’ stands for ”forward” and ”b” for ”backward”. ˆ ρ is the
designed observer.
2011 American Control Conference
on O'Farrell Street, San Francisco, CA, USA
June 29 - July 01, 2011
978-1-4577-0079-8/11/$26.00 ©2011 AACC 4334

Page 2

For the forward system consider:
d
dtˆ ρf
k(t)=
−i[Bx(t)σx+ By(t)σy, ˆ ρf
k(t)]
+Γ(σzˆ ρf
(Γ + γ)σz(ˆ yf
?
k(t)σz− ˆ ρf
k(t))
−
k(t) − y(t))
?
(2)
ˆ yf
k(t)=
Tr
σzˆ ρf
k(t)
.
For the backward system consider:
d
dtˆ ρb
k(t)=
i[Bx(T − t)σx+ By(T − t)σy, ˆ ρb
k(t)]
−
Γ(σzˆ ρb
(Γ − γ)σz(ˆ yb
?
k(0) = ˆ ρf
k(t)σz− ˆ ρb
k(t))
+
k(t) − y(T − t))
?
k(T) and ˆ ρf
(3)
ˆ yb
k(t)=
Tr
σzˆ ρb
k(t)
.
Noting that ˆ ρb
γ > 0. In (2), γ′= Γ+γ is the gain of the forward observer.
It is written in this form for the convenience of the proof.
We initialize the observer ˆ ρf
and of trace 1. We typically take ˆ ρ(0) =1
identity matrix. This way we make no a priori assumption
on the initial state.
Remark 1: Notice that the observer introduced above is
trace preserving and stays Hermitian for all time. However
it does not preserve positivity.
We define for all t ∈ R+(k is defined as: k = E(t
where E represents the integer part)
k(0) = ˆ ρb
k−1(T) and
0(0) = ˆ ρ(0) to be Hermitian
2I where I is the
2T)
˜ ρ(t)=ˆ ρf
k(t − 2kT) − ρ(t − 2kT)
if t ∈ [2kT,(2k + 1)T[
ˆ ρb
k(t − (2k + 1)T) − ρ(2(k + 1)T − t)
if t ∈ [(2k + 1)T,2(k + 1)T[
(4)
˜
Bx(t)=
Bx(t − 2kT)
if t ∈ [2kT,(2k + 1)T[
Bx(2(k + 1)T − t)
if t ∈ [(2k + 1)T,2(k + 1)T[
˜
By(t)=
By(t − 2kT)
if t ∈ [2kT,(2k + 1)T[
By(2(k + 1)T − t)
if t ∈ [(2k + 1)T,2(k + 1)T[
Tr(σz˜ ρ(t)) .Z(t)=
(5)
Let
V : A (hermitian) → Tr?A2?
V is definite positive. For all k ∈ N we have:
,
d
dtV (˜ ρ(t))=
−4ΓV (˜ ρ(t)) − 2γ(Z(t))2
if t ∈ [2kT,(2k + 1)T[
4ΓV (˜ ρ(t)) − 2γ(Z(t))2
if t ∈ [(2k + 1)T,2(k + 1)T[ .
(6)
=
(7)
We note, for all k ∈ N, Vk= V (˜ ρ(2kT)). We define the
function g such that for all t ∈ R+
02468 10
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
Iterations (k)
Tr(˜ ρ(2kT ))2
Fig. 1.
to zero when k goes to infinity.
Notice that the upper envelope (i.e Vk= Tr?(˜ ρ(2kT))2?) goes
g(t)=
e4Γ(t−2kT)if t ∈ [2kT,(2k + 1)T[
e−4Γ(t−2(k+1)T)if t ∈ [(2k + 1)T,2(k + 1)T[ ,
Vk+1− Vk
=
−2γ(
?(2k+1)T
2kT
g(t)Z2(t)dt
+
?2(k+1)T
0 .
(2k+1)T
g(t)Z2(t)dt)
(8)
≤
(Vk)k is a decreasing sequence which is studied in more
detail in Section IV. Before looking into the convergence
proof, we present some simulations which show the robust-
ness of the convergence of ˆ ρf
k(0) towards ρ(0) when k goes
to infinity.
III. SIMULATIONS
For the simulations of figures 1, 2, 3 and 4, we take:
Γ = 0.25 kHzγ = 0.25 kHz
∀t
?Bx(t)2+ By(t)2= 10 kHz
We take Bx(t) = B0cos(θ(t)) and By= B0sin(θ(t)). B0=
10 kHz. For θ(t), at 10 equally spaced times between 0 and
T, we take a random value between 0 and 2π. Using a cubic
spline interpolation, we build θ(t) over [0,T].
10 iterations are simulated: k = 0,..,10. 10% Gaussian
noise was added to the measurement, Bxand By.
We initialize the estimator in the completely mixed state
ˆ ρf
0(0) = ˆ ρ(0) =
Bloch sphere by taking a random Hermitian positive matrix
satisfying: Tr?ρ(0)2?= Tr(ρ(0)) = 1.
4335
T = 1 ms
1
2I. We randomly initialize ρ on the

Page 3

00.20.40.60.81
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
Time(ms)


Measurement
Estimated measurement
Fig. 2.
Gaussian noise was added to the data. The estimated measurement is
obtained by simulating (1) with ρ(0) = ρf
Measurement and estimated measurement versus time. 10%
k(0) and k = 10.
1
2
1
2
0
0.5
1
Re(ρ(0))
1
2
1
2
0
0.5
1
Re(ˆ ρf
10(0))
1
2
1
2
−0.5
0
0.5
Im(ρ(0))
1
2
1
2
−0.5
0
0.5
Im(ˆ ρf
10(0))
Fig. 3. The density matrix of the system and its estimator at time t = 0.
0 0.20.4
Time (ms)
0.6 0.81
−1.5
−1
−0.5
0
0.5
1
1.5


Bx/B0
By/B0
Fig. 4. The magnetic fields Bx(t) and By(t).
IV. CONVERGENCE PROOF
Theorem 1: Consider system (1) and the observers (2),
(3). Take T > 0 (which may be arbitrarily small). For
all Bx,By
∈ C2([0,T],R) such that Bx(0)d
By(0)d
dtBx(0) ?= 0 and ∀ˆ ρ(0) which is Hermitian and of
dtBy(0) −
trace 1, we have
lim
t→∞Tr?˜ ρ2(t)?= 0 ,
where ˜ ρ is defined in (4).
Remark 2: The convergence outlined in theorem 1 is in
two steps:
1) limt→∞Z(t) = 0
2) limt→∞Tr?˜ ρ2(t)?= 0
The observer (2), (3) is designed such that we always have
convergence of the estimated measurement to the measure-
ment: limt→∞Z(t) = 0. Since the system is observable:
(σz,σx,σy) and its commutators span the space of all trace-
less 2 × 2 Hermitian matrices, we can find fields (Bx,By)
such that limt→∞Z(t) = 0 implies limt→∞Tr?˜ ρ2(t)?= 0.
the simulations above, we chose T ∼ 1/Γ. If T is too small,
we have little information on the system and therefore the
state estimation will lose robustness to noise. On the other
hand, if T is too large, the system’s state ρ(t) will converge
(because of the damping Γ) and the numerical integration of
the backwards observer (3) becomes unstable when noise is
added.
Proof:
For any piecewise continuous function f we define:
f(2kT+) = limt→2kT,t>2kTf(t).
From (8), we know that (Vk)k is a decreasing sequence.
Besides, for all k ∈ N, Vk ≥ 0, hence (Vk)k converges to
a limit that we note by V∞. Summing (8) between 0 and
N ∈ N∗:
Remark 3: Theorem 1 holds for all T > 0. However, in
VN− V0
=−2γ
?(2N+2)T
0
g(t)Z2(t)dt
∀t ∈ R+g(t) ≥ 1, hence
Since ∀t ∈ R+Z2(t) ≥ 0,?∞
From (6) and (7) we have ∀u ∈ [0,2T] and ∀k ∈ N:
V (˜ ρ(2kT + u)) ≤ V (˜ ρ(2kT)) ,
hence, for all t ∈ R+we have V (˜ ρ(t)) ≤ V0. ˜ ρ is therefore
bounded and belongs to the ball centered around 0 and of
radius V0.
We are now going to prove that (Vk)k converges to zero
when k goes to infinity, and from (9) we will conclude that
V (˜ ρ(t)) converges to zero when t goes to infinity. In order
to prove the convergence of (Vk)k, we are going to prove
that Z(2kT),d
when k goes to infinity.
Weconsider
Bx,By
Z
is
C3
˜
Bx,˜ By,d
Bx,d
By,d2
Bx,d2
over S, therefore
dt3˜ ρ are bounded over S and
therefore
dt3Z are bounded over S. Since Z
is continuous over R+and
is uniformly continuous over R+, so Z2is uniformly
?(2N+2)T
0
Z2(t)dt ≤
V0−VN
2γ
.
0Z2(t)dt exists and is finite.
(9)
dtZ(2kT+),d2
dt2Z(2kT+) all converge to zero
C2([0,T],R)
?
functions.
+ 1)T[.
are bounded
over
dt˜
S=
dt2˜ By
k∈N]kT,(k
and
˜ ρ
dt˜
dt2˜
dt˜ ρ,d2
dt2Z,d3
d
dt2˜ ρ,d3
d
dtZ,d2
d
dtZ is bounded over S, Z
4336

Page 4

continuous over R+. What’s more?∞
that limt→∞Z2(t) = 0 and hence
0Z2(t)dt exists and is
finite. We can conclude by applying Barbalat’s lemma [10]
lim
t→∞Z(t) = 0 ,
(10)
and in particular
lim
k→∞Z(2kT) = 0 .
(11)
Since the derivatives of Z are not continuous over R+
but only over S, we cannot directly apply Barbalat’s lemma
to
dtZ(t) and
d
d2
dt2Z(t).
Suppose that
goes to infinity.
There exists ǫ > 0 and a sequence (tn)n such that
limn→∞tn= ∞ and
can be treated in exactly the same way).
Since
dt2Z(t) is bounded over S we have: ∃η ∈]0,T/2[
such that for all n ∈ N and t ∈ [−tmin
t) −
and tmax
n
= min((E(tn/T)+1)T −tn,η). E represents the
integer part.
Hence, for all t ∈ [−tmin
d
dtZ(tn)−(d
t) −
tmin
n
+ tmax
n
≥ η.
|Z(tn+ tmax
|
tn−tmin
n
≥ηǫ
2
> 0 .
d
dtZ(t) does not converge to zero when t
d
dtZ(tn) > ǫ (or
d
dtZ(tn) < −ǫ which
d2
n
,tmax
n
] |d
dtZ(tn+
d
dtZ(tn)| <ǫ
2. Where tmin
n
= min(tn− E(tn/T)T,η)
n
dtZ(tn+t)) ≥
ǫ
2=
,tmax
n
] we have:
d
dtZ(tn+ t) =
dtZ(tn)−|d
2. Also, notice that T ≥
dtZ(tn)−dd
dtZ(tn+
d
dtZ(tn)| ≥ ǫ −
ǫ
n
) − Z(tn− tmin
?tn+tmax
n
)| =
n
d
dtZ(t)dt|
This is in contradiction with (10), we therefore conclude
that
d
dtZ(t)
and in particular:
lim
t→∞
=0 ,
(12)
lim
k→∞
d
dtZ(2kT+)=0 .
(13)
Suppose that
k goes to infinity, there exists ǫ > 0 and a sequence (kn)n
such that limn→∞kn = ∞ and
d3
dt3Z is bounded over S, there exists 0 < η < T such that for
all n ∈ N and 0 < t < η |d2
ǫ
2. Hence:
d2
dt2Z(2kT+) does not converge to zero when
d2
dt2Z(2knT+) > ǫ, since
dt2Z(2knT+t)−d2
dt2Z(2knT+)| <
|d
dtZ(2knT + η) −d
?2knT+η
≥ηǫ
2
> 0 ,
dtZ(2knT+)| =
|
2knT
d2
dt2Z(t)dt|
which contradicts (12). Hence:
lim
k→∞
d2
dt2Z(2kT+)=0 .
(14)
We note
X(t)
Y (t)
=
=
Tr(σx˜ ρ(t))
Tr(σy˜ ρ(t)) .
We recall that from (11)(13)(14):











lim
k→∞Z(2kT) = 0
d
dtZ(2kT+) = 0
d2
dt2Z(2kT+) = 0 .
lim
k→∞
lim
k→∞
(15)
Using (1), we find that (15) implies:









lim
k→∞Z(2kT) = 0
lim
k→∞
d
Bx(2kT)Y (2kT) −d
˜
Bx(2kT)Y (2kT) −˜
By(2kT)X(2kT) = 0
lim
k→∞
dt
˜
dt
˜
By(2kT)X(2kT) = 0 .
(16)
Notice that˜
same holds for their derivatives. We take Bx,By such that
Bx(0)d
dtBx(0) ?= 0. (16) implies:


This is equivalent to limk→∞V (˜ ρ(2kT)) = 0. (9) enables
us to conclude that:
Bx(2kT) = Bx(0) and˜
By(2kT) = By(0), the
dtBy(0) − By(0)d





lim
k→∞Z(2kT) = 0
lim
k→∞X(2kT) = 0
lim
k→∞Y (2kT) = 0 .
lim
t→∞Tr?˜ ρ2(t)?= 0 .
V. EXTENSION TO THE MULTILEVEL CASE
We now consider a system of total angular momentum F.
The dimension of the system is d = 2F +1, and the density
matrix ρ(0) belongs to the set of positive d × d Hermitian
matrices of trace 1. There are therefore d2−1 parameters to
identify. In order to extend the proof of the two level case
(d = 2) to the multilevel case, we would need to prove that
Z(t),d
dtd2−2Z(t) all converge to zero when t goes
to infinity. If the system is observable, we would be able to
conclude that we can find a control such that ˜ ρ(t) converges
to zero. Two complications arise from considering a system
of higher dimension:
First, the need to extract information from further derivatives
of the measurement record systematically reduces the robust-
ness of the state estimation. One direction of improvement
would be to design a nonlinear observer which preserves
positivity. Such an observer is given in [11] for a non
dissipative system (Γ = 0). The difficulty is to build an ob-
server which stays positive even in the backwards dynamics
which is unstable due to the dissipation term in Γ. Such an
dtZ(t),..,dd2−2
4337

Page 5

observer would reduce the size of the admissible ρ(0)s given
a noisy measurement record, the robustness will therefore
be increased. However, no dramatic improvement should be
expected since the information on some matrix elements
of ρ(0) are hidden in high derivatives of the measurement
record.
Second, although the observability criteria insures the ex-
istence of a control such that limt→∞Z(t) = 0 implies
limt→∞˜ ρ(t) = 0, we don’t have any well known method to
find such a control. The higher the dimension, the harder it is
to find a control which makes the data y(t) informationally
complete about the initial state ρ(0).
We now give some simulations which show that our BFN
protocol still works well for a system of total angular
momentum F = 1 (d = 3). Consider the system:
d
dtρ(t)
y(t)
=−i[H(t),ρ(t)] + ΓD[O]ρ(t)
Tr(Oρ(t)) .
(17)
=
Where H is the system’s Hamiltonian and D[O] the Lind-
blad superoperator. We have: H(t) = gFµB(Bx(t)Fx+
By(t)Fy) + βΓFx2. gF,µB,Γ and β are positive constants,
Bx,By are the controls and Fx,Fy,Fz are the angular
momentum operators. O is the observable, and we take
O =√ΓFz. D[O]ρ(t) = Oρ(t)O†−1
The superscript†stands for conjugate transpose. The term
βΓFx2is necessary to insure the observability of the system
[12]. We now consider the following observers:
2(OO†ρ(t)+ρ(t)O†O).
d
dtˆ ρf
k(t)=−i[H(t), ˆ ρf
γO(ˆ yf
k(t) − y(t))
Tr
?
k(t)] + ΓD[O]ˆ ρf
k(t)
−
=
(18)
ˆ yf
k(t)Oˆ ρf
k(t)
?
,
d
dtˆ ρb
k(t)=i[H(T − t), ˆ ρb
γO(ˆ yb
k(t) − y(T − t))
Tr?Oˆ ρb
k(0) =
k(t)] − ΓD[O]ˆ ρb
k(t)
−
=
(19)
ˆ yb
k(t)
k(t)?
,
with the conditions: ˆ ρb
k−1(T). We initialize the observer at ˆ ρf
Id is the 3 × 3 identity matrix.
For the numerical simulations in figures 5, 6 and 7 we
take:
gF= 1µBB0= 30
ˆ ρf
k(T) and ˆ ρf
0(0) =1
k(0) =
3Id where
ˆ ρb
Γ = 1
γ = 1β = 10T = 1
We take Bx(t) = B0cos(θ(t)) and By(t) = B0sin(θ(t)).
θ(t) is found using a numerical search routine aiming to max-
imize a certain criteria (entropy), as explained in [13]. 10%
noise is added to the controls Bx,Byand 10% noise is added
to the measurement y(t). We take ρ(0) =1
2


1
0
1
0
0
0
1
0
1


Notice that the estimated measurement is almost identical
to the measurement y(t) (figure 6). Also, the sequence
01020
Iterations (k)
3040 50
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
Tr (˜ ρ(2kT ))2
Fig. 5.
decreases and seems to converge to zero.
The upper envelope is Vk = Tr?(˜ ρ(2kT))2?. Notice that it
00.2 0.4 0.60.81
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
Time(ms)


Measurement y(t)
Estimated measurement
Fig. 6.
Gaussian noise was added to the data. The estimated measurement is
obtained by simulating (17) with ρ(0) = ρf
Measurement and estimated measurement versus time. 10%
k(0) and k = 50.
1
2
3
1
2
3
0
0.5
1
1
2
3
1
2
3
0
0.5
1
ρ50(0)
ˆ
f
||
ρ(0)
||
Fig. 7.
We plot the modulus of each matrix element.
The density matrix of the system and its estimator at time t = 0.
(Vk)k decreases and seems to converge to zero (figure
5). This enables us to reconstruct the initial state with a
96% fidelity where the fidelity F is computed as follows
F = Tr
are needed than in the 2 level case (50 as opposed to 10)
to achieve a similar fidelity. Each back and forth iteration
takes about 0.1 seconds so the presented simulation takes
???
ˆ ρf
50(0)ρ(0)
?
ˆ ρf
50(0)
?
[14]. More iterations
4338