Transcript Document
Quantum Optics II
Cozumel, Mexico, December 5-8, 2004
”Entanglement in Time and Space”
J.H. Eberly, Ting Yu, K.W. Chan, and M.V. Fedorov
University of Rochester / Prokhorov Institute
We consider entanglement as a dynamic property of quantum states and examine its
behavior in time and space. Some interesting findings: (1) adding more noise helps
fight phase-noise disentanglement, and (2) high entanglement induces spatial
localization, equivalent to a quantum memory force.
• Ting Yu & JHE, Phys. Rev. Lett. 93, 140404 (2004).
• K.W. Chan, C.K. Law and JHE, Phys. Rev. Lett. 88, 100402 (2002)
• JHE, K.W. Chan and C.K. Law, Phil. Trans. Roy. Soc. London A 361, 1519 (2003).
• M.V. Fedorov, et al., Phys. Rev. A 69, 052117 (2004).
1
Entanglement means a superposition of
conflicting information about two objects.
Superposition of
conflicting information,
but only one object.
Can you
handle the
conflicting
information
here?
Which face
is in the
back?
2
A pair of conflicts can be “entangled”
Try to see both at the same time.
Do they “flip” together?
3
Measurement cancels contradiction
A pair of boxes, but only one view of them
4
Bell States provide a simple example
Schrödinger-cat “Bell State”:
|
-
> =|C*>|N*> + |C>|N>
excited cat = C*, dead cat = C,
excited nucleus = N*, ground state = N
5
Bell States provide a simple example
Schrödinger-cat “Bell State”:
|
-
> =|C*>|N*> + |C>|N>
excited cat = C*, dead cat = C,
excited nucleus = N*, ground state = N
<N|
-
> = |C> (sorry, Cat)
6
Overview
Issue -- entanglement in time and space.
Illustration #1 -- two atoms are excited and
entangled but not communicating with each other.
Result #1 -- both atoms decay, diag e-2gt and offdiag e-gt , just as expected; but entanglement of
the atoms behaves qualitatively differently.
Illustration #2 -- two atoms fly apart in molecular
dissociation. Result #2 -- entanglement means
localization in space (a “quantum memory force”).
For detailed treatments:
Ting Yu & JHE, PRL 93 140404 (2004), and M.V. Fedorov, et al., PRA 69, 052117 (2004).
7
Illustration #1:
A
B
HAT = (1/2)wAsZA + (1/2) wBsZB , HCAV = kwkak†ak + knkbk†bk
HINT = k(gk*s-Aak† + gks+Aak) + k(fk*s-Bbk† + fks+Bbk)
At t=0 the joint initial state rAB is entangled and mixed (not pure). The atoms
only decay (no cavity feedback).
After t=0, what happens to entanglement?
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mixed initial states, C = 2/3
r
=
++
+-+
--
++
a
0
0
0
+- -+
0 0
1 1
1 1
0 0
-0
0
0
d
Initial state,
entangled and
mixed, where
d = 1-a.
C = concurrence
EOF
concurrence
1≥C≥0
9
Time Evolution Result:
r =
a
0
0
0
0 0
1 1
1 1
00
0
0
0
d
a(t) 0 0 0
0 b(t) z(t) 0
0 z*(t) c(t) 0
0 0 0 d(t)
Obvious point: the atoms go to their ground states, so d 3, and
the other elements decay to zero.
No surprise: the decay of r(t) is smooth, and exponential, measured
by usual natural lifetime:
s±A(t) = s±A(0) exp[-GAt/2 ± iwAt]
Reminder: the atoms decay independently.
10
Sol’n. in Kraus representation:
˜ (t) =
r
K (t) r(0) K†(t)
Kraus operators:
g 0 g 0
A
B
K1
0
1 0
1
0
K 3
w A
0 g B
0 0
0
1
g A
K 2
0
0 0
1 w B
0
0
0
K 4
w A
0 0
0 w B
0
0
gA = exp[-G t / 2 ] wA2 = 1- exp[-Gt ] , same for B.
for details, see Ting Yu and JHE, PRB 68, 165322 (2003)
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Kraus matrix evolution:
r
a
0
0
0
0 0
b z
z* c
0 0
0
0
0
d
Decays all depend on gA(t) = exp(-GAt/2) and gB(t) = exp(-GBt/2).
a(t) = gAgB, b(t) = gB2 +gA2 wB2,
c(t) = gA2 +gB2 wA2,
d(t) = wA2 +wA2 +wA2wA2, z(t) = gAgB , where wA2 = 1 - gA2, etc.
Usual Born-Markov solutions for the separate atoms:
s±A(t) = s±A(0) exp[-GAt/2 ± iwAt]
For detailed treatment: Ting Yu & JHE, PRL 93, 140404 (2004).
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Entanglement evolution:
Entanglement has its
own rules, and follows
the atom decay law
only exceptionally.
Entanglement can be
completely lost in a
finite time!
art by Curtis Broadbent
Ting Yu & JHE, Phys. Rev. Lett. 93, 140404 (2004).
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Noise + Entanglement
1
1
r (1 F) I + (4F 1) | ( ) ( ) | Werner state density matrix
3
3
W
Werner qubits undergo only
off-diagonal relaxation under
the influence of phase noise.
All entanglement of a Werner
state is destroyed in a finite
time by pure phase noise.
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Two Noises + Entanglement
Pure off-diagonal relaxation of qubits
Add some diagonal relaxation, for
example via vacuum fluctuations.
Add diag. to off-diag. relaxation of qubits
Werner entanglement gets some
protection from added noise!
Ting Yu & JHE (in preparation).
15
Spatial localization and entanglement
With just two objects, high
entanglement can be reached by
allowing each object a wide variety of
different states.
The idealized two-particle wave
function used by Einstein, Podolsky
and Rosen in their famous 1935 EPR
paper used continuous variables
(infinite number of states) to get the
maximum degree of entanglement.
Experiments following the original
EPR “breakup” scenario have
not been done yet.
16
Physical examples of breakup
Down conversion (K ≈ 4.5 — 1000’s)
Huang and Eberly, JMO 40, 915 (1993)
Law, Walmsley and Eberly, PRL 84, 5304 (2000)
Law and Eberly, PRL 92, 127903 (2004)
Raman scattering (K > 100)
Chan, Law, and Eberly, PRA 68, 022110 (2003)
Chan, et al., JMO 51, 1779 (2004).
Spontaneous emission (K ≈ 1)
Chan, Law and Eberly, PRL 88, 100402 (2002)
Fedorov, et al. (in preparation, 2004)
Ionization/Dissociation (K > 10)
Fedorov, et al., PRA 69, 052117 (2004).
Chan and Eberly, quant-ph 0404093
17
EPR system is created by break-up
The variables entangled are positions (x1 and x2), or
momenta (k1 and k2). Perfect correlation is implied in the
EPR wavefunction:
(x , x ) (x x )
0
1
2
1
2
How much correlation is realistic? How to measure it?
Original paper: Einstein, Podolsky and Rosen, Phys. Rev. 47, 777 (1935).
18
Localization - Entanglement
•
All information is in . We can plot the
two-particle density ||2 vs. x1 and x2.
•
Knowledge of one particle gives
information about the other particle.
•
Joint localization information is packet
entanglement.
Fedorov ratios
for particle
localization:
x1
x2
F1
F2
cond
x1
x2cond
||2
We can calculate
these for a simple
dissociation model.
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Time-dependent EPR example
Given a dissociation rate gd, a post-breakup diatomic is:
R2
r
r
(r1,r2;t) ~ 2 (vt r)expg d t exp
2
r
v
4R
0
relative part
rel
CM part
2
(r vt)g d v
CM
2
R R0
M.V. Fedorov, et al., PRA 69, 052117 (2004) / quant-ph/0312119.
20
Joint Probability Density |total|2
rel
2
CM
2
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Dynamics of localization
What do we know and when do we know it?
• Massive particles spreading wavepackets:
x x(t) and X X(t)
• EPR pairs: [x, P] = 0 and [X, p] = 0
nonlocality
• Spreading is governed by the free-particle Hamiltonian.
• Time evolution is merely via phase in the momentum picture:
k12
˜ (k ;0)
exp
1
2
4k
1
k12
ht 2
˜
(k1;t)
exp
i
k1
2
4m1
4k
1
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Dynamics of localization - F ratios
Plots of |(t)|2
vs. x1 and x2 :
Experiments track localization via packet spreading (i.e.,spatial
variances). The two-particle ratio h(t) = ∆x/2∆X is a convenient
parameter [*] connected with dynamical evolution.
Calculation
x1
F1
x1cond
Inferred dependence
B
F(t) ~ A h(t) +
h(t)
* Chan, Law and Eberly, PRL 88, 100402 (2002).
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Universal man-in-street theory
Model the breakup state as double-Gaussian [*].
1
x
X * K.W. Chan and JHE,
(x , x )
exp
exp
quant-ph 0404093
ab
4x
4X
2
0
1
2
2
2
2
0
0
This makes it easy to calculate the Fedorov ratios (F1 ~ F2 )
at t=0 and for later times.
Question: can we guess what happens to localization?
x2
X2
(x, X;t) ~ exp
exp
2
2
4(x 0 + i t /2)
4(X 0 + i t /2M)
24
Entanglement migration to phase
x2
X2
(x, X;t) ~ exp
exp
2
2
4(x 0 + i t /2) 4(X 0 + i t /2M)
Therefore P(x, X; t) = (x, X; t)2 has two similar real exponents.
x x
2
1
1
2
2
2 x 0 + ( t /2 ) 2 x 02
When these exponents are added, the
nonseparable x1x2 term 0 at a specific
time t0:
t 2 M x X
0
0
0
2
m
m1
X + 2X
1 M 1 M 2
2 X 02 + ( t /2M) 2 X 02
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Quantum memory force (QMF)
Atom
Photon
The dissociation example has a close analog
in spontaneous emission. These atom-photon
space functions show a “force” arising from
shared quantum information, a “quantum
memory force” (QMF). The first four
bound states are shown for Schmidt number
K = 3.5, which is slightly “beyond-Bell.,” i.e.,
K > 2.
M.V.Fedorov, et al. (in preparation).
Chan-Law-Eberly, PRL 88, 100402 (2002)
26
Summary / dynamics of entanglement
•
Entanglement dynamics are largely unknown (time or space)
•
Noisy environment kills entanglement but not intuitively
•
Individual atom decay is not a guide for entanglement
•
Diag. + off-diag. noise has a cancelling effect
•
EPR-type breakup is ubiquitous / creates two-party correlation
•
Conditional localization vs. entanglement ?
•
Packet dynamics, Fedorov ratio and control parameter h
•
Man-in-street theory and phase entanglement
•
Memory effects enforce spatial configurations (QMF)
27
Acknowledgement
Research supported by NSF grant PHY-00-72359, MURI Grant DAAD19-99-10215, NEC Res. Inst. grant, and a Messersmith Fellowship to K.W. Chan.
References
Ting Yu & J.H. Eberly, PRL 93, 140404 (2004) and in preparation.
M.V. Fedorov, et al., PRA 69, 052117 (2004).
C.K. Law and J.H. Eberly, PRL 92, 127903 (2004).
M.V. Fedorov, et al., PRA (in preparation).
K.W. Chan, C.K. Law and J.H. Eberly, PRL 88, 100402 (2002).
K.W. Chan and J.H. Eberly, quant-ph/0404093.
A. Einstein, B. Podolsky and N. Rosen, Phys. Rev. 47, 777 (1935).
28
More sophisticated Schmidt analysis
Any bipartite pure state can be
written as a single discrete sum:
•
•
0 (x1, x 2 ) n n (x1 ) n (x 2 )
n 0
Continuous basis discrete basis
Unique association of system 1 to system 2
e.g., (x1 x 2 ) n (x1) n (x 2 )
n
Size of ent.
K
n
2
n
1
29
Discretization of continuum information,
the Schmidt advantage
Unique
mode pairs
Continuous-mode basis
Pure-state non-entropic
measure of entanglement:
Schmidt-mode basis
Schmidt number counts experimental
modes, provides practical metric
30
Interpreting K, the Schmidt number
K = 1, no entanglement.
K = 2, perfect Bell
states.
K = 5, beyond Bell,
more information.
K = 10, still more info.
Quantum info is always
discrete and countable.
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Estimation of K for photodissociation
Comparing the photodissociation process with the doubleGaussian model, we identify x0 = v ⁄ gd and X0 = R0.
If we take R0 = 10 nm, v 2E ~ c
1 eV
~ 103 m s ,
10 GeV
and define d = gd-1, then with d in sec,
h
x 0
v
~ ~ 1011 d
2X 0 g d R0
1 1
K h + 100
2 h
K
d (s)
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Retreat of entanglement into phase
The Fedorov ratios for double-Gaussian 0 (x1 ,x 2 ) :
Position:
Momentum:
where
x1
F1
c o cosh r K C(t)
x1
k1
G1
c o cosh r K
k1
e
2 r
e 2r + (t /t 0 ) 2
, so C(t) 1 .
2r
2
1+ e (t /t0 )
t0
2mx 0 X 0
Note non-equivalence of k-space and x-space
for these experimentally measurable quantities.
33
34
Double-Gaussian Schmidt analysis
For the man-in-street double-Gaussian model (with m1 = m2)
0 (x1, x 2 ) n 1n (x1 ) n2 (x 2 )
n 0
i
The Schmidt modes are the number states n (x i ) x i n
1 x 0 2X 0
K n n
+
2 2X0 x0
while from the actual wave function we had inferred
B
cond
F1 x1 / x1 Ah(t) +
h(t)
and one finds:
2
1
with h(t) = ∆x(t)/2∆X(t). These are the same, except for spreading!
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