Download Quantum Interference 3 Claude Cohen-Tannoudji Scott Lectures Cambridge, March 9

Quantum Interference 3
Claude Cohen-Tannoudji
Scott Lectures
Cambridge, March 9th 2011
Collège de France
1
Outline of lecture 3
1: Introduction
2: Entangled states
3: Examples of entangled states
4: Entangled states and non separability
5: Entangled states and which path information
6: Entangled states and the measurement process
7: Entangled states and two-photon interferences
2
Entangled states
Linear superpositions of product states of 2 systems 1 and 2
= 1 2 + 1 2
Cannot be written as a product of a state of system 1 by a
state of system 2.
Examples
- Two spins 1/2 in the singlet state
( 1+,2 )
1,2 + / 2
- Entangled internal and external degrees of freedom
g, p + e, p + k / 2
(
)
- Many particle entanglement
1+,2+,3 + 1,2,3 / 2
(
)
3
Reduced density operator for each subsystem
Even if the state of the whole system 1+2 is a pure state , the
state of each subsystem is not a pure state if is entangled
The predictions on system 1 can be entirely deduced from the
reduced density operator
(1) = Tr2 (1, 2)
where (1,2) is the density operator of 1+2. Idem for (2)
u1k (1) u1l = m u1k , v2m (1, 2) u1l , v2m
For example, for the singlet state of 2 spins , we have
1/ 2 0 (1) = 0 1 / 2 1/ 2 0 (2) = 0 1 / 2 Statistical mixtures with equal weights of 2 spins states
A useful identity
Tr = m u m u m = m u m u m = 4
Outline of lecture 3
1: Introduction
2: Entangled states
3: Examples of entangled states
4: Entangled states and non separability
5: Entangled states and which path information
6: Entangled states and the measurement process
7: Entangled states and two-photon interferences
5
Two-level atom e,g in a single mode cavity
Ee Eg = 0
Photon energy : At resonance (=0), the two states of the atom field system
e, N = 0 and g, N = 1
(N: number of photons)
are degenerate and far from all the other states of the sytem
These 2 states are coupled by the atom-field interaction VAF
e, 0 VAF g,1 = 0 / 2
0 = vacuum Rabi frequency
The system oscillates between the 2 states at the frequency 0
during a time t which can be adjusted by varying the transit time
of the atom through the cavity
Cavity Atom
Analogy with the RF pulses for a spin 1/2
6
Fictitious spin system associated
with the 2 states of the atom field system
+1 / 2
z
e, 0
1 / 2
z
g,1
An interaction lasting a time t produces a rotation for the
fictitious spin around the X axis with an angle = 0 t
= /2
= A /2 pulse transforms e,0 and g,1 into orthogonal linear
combinations of e,0 and g,1
A pulse transforms e,0 into g,1 and g,1 into e,0
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/ 2 pulse for the atom field system ( = / 2)
The atom in e enters the cavity which contains 0 photon
Initial state
e, 0
If = / 2, we have for the fictitious spin
+1 / 2
Z
1
+1 / 2
2
Z
i 1 / 2 Z Consequently, for the atom field system, we have
1
e, 0 e, 0 i g,1 2
The atom field interaction has transformed the initial state, which
is a product state of an atom state (g) by a field state (N=0), into
an atom-field entangled state.
In most cases, entanglement between two systems results from
an interaction between them
8
Entangled state of 2 atoms
One sends now 2 atoms through the cavity, one after the other
The first one, atom 1, enters the empty cavity (N=0) in e1.
If the interaction corresponds to a /2 pulse, the state of the AF
when atom 1 leaves the cavity is:
e1 , 0 i g1 ,1 / 2
The second atom, atom 2, enters the cavity in g2.
Just before it enters the cavity, the state of the whole system is:
g2 , e1 , 0 i g2 , g1 ,1 / 2
The interaction of atom 2 corresponds now to a pulse
g2 , g1 ,1 i e2 , g1 , 0
The atom 2 in g2 does not interact with the empty cavity
g2 , e1 , 0 g2 , e1 , 0
Finally,
g2 , e1 , 0 i g2 , g1 ,1 / 2 g2 , e1 , 0 e2 , g1 , 0 / 2
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Final state of the system
g2 , e1 , 0 e2 , g1 , 0 / 2 = 0 1
2
g2 , e1 e2 , g1 The cavity has returned to its empty state and the 2 atoms are in
an entangled state
The entanglement here is not due to a direct interaction between
the 2 atoms. Each atom interacts only with the cavity which
transmits information to atom 2 about its previous interaction
with atom 1
The cavity is a mediator between the 2 atoms. As its initial and
final states are the same (N=0) it acts like a catalyst
This experiment has been performed
Hagley, E., Maître, X., Nogues, G., Wunderlich, C., Brune, M.,
Raimond, J.-M. and Haroche, S. P.R.L.79, 1 (1997).
Scheme initially proposed for trapped ions by I. Cirac and
P.Zoller
10
Sources of entangled photons
Cascade J=0 J=1 J=0 (case of Ca)
0
1 / 3
px
x
1 / 3
pz
x
1
y
z
1 / 3
py
z
1
1
y
1
2
x
2
y
x
1
z
y
= 1 y, 2 y + 1 x, 2 x / 2
0
Parametric down conversion of type 2 in a BBO crystal
1
2
One photon is absorbed. Two photons
1 and 2 are emitted, the atom returning
to the ground state
Conservation of energy
= 1 + 2
Because of the phase matching conditions in the BBO crystal, the 2
photons have orthogonal linear polarizations and the directions of
emission of these 2 photons lie on 2 cones.
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A recent improvement
If 1=2=/2, a photon emitted
along one of the intersections of
the 2 cones can have one of the
2 polarizations, the other one
having the orthogonal polarization
Entangled state
=
+
Interest of such a method: no solid angle limitation. Possibility to
inject the pair of photons in 2 fibers where they can propagate over
kilometers.
P. Kwiatt, K. Mattle, W. Weinfurter, A. Zeilinger, A. Sergienko, Y. Shih,
Phys. Rev. Lett. 75, 4337 (1995)
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Outline of lecture 3
1: Introduction
2: Entangled states
3: Examples of entangled states
4: Entangled states and non separability
5: Entangled states and which path information
6: Entangled states and the measurement process
7: Entangled states and two-photon interferences
13
Conceptual importance of entangled states
1 – They clearly reveal the existence of non intuitive
quantum correlations.
Consider 2 spins 1/2 which are in the singlet state and which are
moving apart from each other.
If one measures Sz on the first spin and if one finds +1 (in units of
/2), one is sure that Sz is equal to -1 for the second spin.
Idem if one measures Sx or Sy (Isotropy of the singlet state).
Einstein, Podolsky et Rosen (1935) conclude that the quantum
description of phenomena is incomplete.
Their argument: measuring Sz or Sx on the first spin does not
influence the second spin which is very far. If the value of Sz or Sx
is sure for the second, this means that this value for the second
spin was pre-existing before the measurement on the first spin.
“Local realism”.
How can one reconcile this conclusion with the fact that
Sz et Sx do not commute and cannot have simultaneously
well defined values
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2 – An important advance: Bell’s inequalities
Suppose that there are additional variables not included in
the usual quantum description. They characterize the state of
the system when it is created and they are described by a
probability density P() positive and normalized.
If one admits that the results of the measurements on 1 only
depend on and on the measurement apparatus of 1 (and
not on the apparatus measuring 2 – locality assumption), one
can derive (Bell 1964) inequalities obeyed by certain
combinations of correlation signals between results of
measurements performed on 1 and 2.
But, the important point is that the predictions of usual
quantum mechanics violate these inequalities.
One must then confront experimentally the predictions of
quantum mechanics with those, which are different, of a more
complete and more local description of physical phenomena in
terms of additional parameters.
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3 – Experimental tests give a clear evidence for a violation
of Bell’s inequalities and for a confirmation of the
quantum predictions.
Several generations of experiments with increasing accuracy:
• Freedman, Clauser 1972-76
• Fry, Thomson 1976
• Aspect, Grangier, Roger, Dalibard 1981-82
• Kwiatt, Mattle, Weinfurter, Zeilinger 1995
4 – Conclusion
As far they can be from each other, the 2 systems appearing in
an entangled state cannot be considered as separate entities.
They form a single non separable entity
Quantum non separability
16
Outline of lecture 3
1: Introduction
2: Entangled states
3: Examples of entangled states
4: Entangled states and non separability
5: Entangled states and which path information
6: Entangled states and the measurement process
7: Entangled states and two-photon interferences
17
Young’s double slit experiment
Can one determine through which slit the particle passes in a
Young’ double slit experiment
A
p
S
P
B
- p
If the particle follows the path SAP (SBP), the screen receives a
momentum kick p (- p). By measuring the variation of the
momentum of the screen, one could hope to determine if the
particle has followed the path SAP or SBP
But the screen is also a quantum object with a momentum
spread p. The path followed by the particle can be determined
only if p << p. But because of Heisenberg relations the
position of the screen has a spread x so large that the
interference pattern is washed out
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Entanglement and which path information
A
E+
+
S
B
E
P
r
The final state of the particle + screen system is an entangled state
fin + E+ + E
E+ ( E ) : Initial state of the screen displaced by p (- p)
+ ( ) : state of the particle going from A (B) to P
The probability to find the particle at the point P is equal to
(r ) = fin r r fin
2
2
= + ( r ) E+ E+ + ( r ) E E
* + 2 E E+ Re + ( r ) ( r )
(
)
19
Entanglement and which path information (2)
The interference fringes come only from the last term
This term is multiplied by the scalar product E E+
If the state of the screen determines unambiguously the path
of the particle, the 2 states E+ and E- must be clearly distinct
without any overlap. Their scalar product must be equal to 0
so that the fringes vanish
This result can be extended to any quantum device which
could be introduced for determining the path of the atom. If the
device is efficient, i.e. if its two final states are different, the
interference fringes disappear. One cannot observe fringes
and simultaneously know the path of the atom
Illustration in precise terms of the principle of complementarity
introduced by Niels Bohr stating that the wave and particle
aspects are only revealed with specific arrangements which
are not the same for the wave and particle aspects
20
Outline of lecture 3
1: Introduction
2: Entangled states
3: Examples of entangled states
4: Entangled states and non separability
5: Entangled states and which path information
6: Entangled states and the measurement process
7: Entangled states and two-photon interferences
21
Entanglement and quantum measurement theory
Ideal measurement process (Von Neumann model)
S : Microscopic system to be measured
Eigenstates i
with eigenvalues ai
Initial state : 0
M : Measuring apparatus
S in i interacts with M in 0
i 0 i i
To each state i of S corresponds a well defined state i of M
2 states i and j of S corresponding to 2 different states
i and j of M are clearly distinguishable : i j = ij
S : can be considered as a “needle”
Linearity of quantum mechanics
ci i 0 i
c
i
i
i i
Entangled state
22
Difficulties associated with macroscopic coherences
At the end of the measurement process, linear superpositions
of states where M is in macroscopically different states i and
j with i j appear
Macroscopic coherences
Not usual for our perception of the macroscopic world
A well known example
Schrödinger cat
Radioactive atom in the excited state A* in the presence of a
cat which is alive. Emission of a ray (A*A) which triggers
the release of a poison which kills the cat
If the atom is in c1 A* + c2 A , the total system is in
c1 A* , cat alive + c2 A , cat dead
The cat may be alive and dead at the same time!
23
A possible solution
Coupling of M with the environment E
We focus on 2 states 1 and 2 coupled to the 2 states 1 and 2
Assumptions on the coupling of M to the environment E
The coupling M-E does not change the occupations of 1 and 2
If E starts from the state 0
1 0 1 1
2 0 2 2
The scalar product 1 2 tends to 0 at a much shorter time
scale than the one governing the evolution of 1 1 and 2 2
Linearity of quantum mechanics
S ME =
1
1 + 2 0 0
2
1
1 1 1 + 2 2 2 2
The correlations 1 1 and 2 2 are not modified
24
Final reduced density operator of S +M
S M = TrE S ME = TrE S ME S ME
S M
1
= i i i i
2 i
1
+ i i j j
2 i j
TrE i i
= i i =1
TrE i j
= j i
It is the second line of this equation which gives rise to
macroscopic coherences.
The coupling with E introduces a multiplicative factor j i
which tends rapidly to 0.This coupling suppresses macroscopic
coherences.
“Decoherence” due to the coupling with the environment
25
A simple example of decoherence
The measuring apparatus M consists of a big particle coupled
to the system S to be measured and displaced by an amount
proportional to ai if the system is in the state i. The position
of this big particle appears as a needle which measures ai
The environment E in which M is immersed is a bath of light
particles colliding with M. These collisions give rise to a
brownian motion of M
If S is in a linear superposition of 1 and 2 , macroscopic
coherences appear for the big particle M which can be in a linear
superposition of 2 wave packets located at 2 different positions
separated by a macroscopic distance
How fast are these spatial coherences of M destroyed by the
collisions with the bath E of light particles colliding with M?
26
State of the big particle M
Linear superposition of 2 wave packets of width separated by a
distance 2a large compared to .
(x + a)
(x a)
(x) : wave packet
centered in x=0
-a
0
( )
F.T. of (x) : ( p) = (1 / 2 ) e
+a
x
(x) = 1 / 2 (x a) (x + a) Momentum distribution
P(p) p = / a
i p a/
e+ i p a/ ( p)
2
(p)=F.T. of (x)
(
P( p) = 2 ( p) sin 2 p a / )
The coherence between the 2
wave packets separated by 2a
gives rise to fringes appearing in
p the momentum distribution with
a fringe spacing p = /a .
27
Understanding the decoherence
As shown in lecture 2, the momentum distribution function P(p)
is the Fourier transform of the spatial correlation function G(a).
3
G( a) = d p P( p) exp i p. a / (
)
The oscillations of P( p) with a period p = / a reveal the
existence of a spatial coherence length extending over a
distance scaling as a
Understanding the decoherence, i.e. the destruction of spatial
coherences, is thus equivalent to understand how the brownian
motion of P induced by the collisions with the light particles of
E destroy the oscillations of P(p)
28
Momentum diffusion
The particle M diffuses in the gas by undergoing collisions. As
in Brownian motion, there is a momentum diffusion. Each state
with a well defined momentum acquires after a time t a
momentum dispersion p given by the equation:
( p )
2
= 2 D t
where D is the momentum diffusion coefficient.
After a time t, the momentum distribution of the particle M,
starting from the superposition of 2 wave packets described
above, is the convolution product of the initial momentum by a
curve of width p=(2Dt)1/2.
When t increases, the fringe contrast of P(p) diminishes and
the fringes progressively disappear. There is no coherence left
between the 2 wave packets and the state of M has become a
statistical mixture of 2 wave packets.
29
Relaxation time TR of the spatial coherence
This time is called « decoherence » time if the distance
between the 2 wave packets is of a macroscopic scale.
The fringes disappear when the broadening due to momentum
diffusion becomes equal to the spacing of the fringes of P(p)
which are a signature of the coherence between the 2 wave
packets. TR is thus given by the equation:
( )
p
2
( )
= 2DTR = p
2
= 22 / a2
1
2D 2
= 2 2a
TR It in interesting now to compare the decoherence rate 1/TR to
the damping rate of the average momentum of the particle M
given by the equation:
d p / dt = p
: friction coefficient
30
Comparison of the decoherence rate 1/TR to the damping
rate of the mean momentum
According to the fluctuation-dissipation theorem, D and are
related (Einstein 1905):
D / = 3m k BT
It follows that:
2
6 m k BT 2
1
2D 2
3 a
= 2 2 a =
a =
2 2
TR 2 T The decoherence rate 1 / TR is thus larger than the damping
rate of the mean momentum by a factor equal to the square
of the distance a between the 2 wave packets divided by the
thermal de Broglie wavelength T.
Since T is very small (on the order of 10-11 m at T=300K), TR is
much larger than , which clearly shows that superpositions of
macroscopically different states are rapidly destroyed.
The spatial diffusion coefficient decreases when the density of
light particles increases: M is embedded in E
1 1 and 2 2 vary slowly while 1 2 0 rapidly
31
Outline of lecture 3
1: Introduction
2: Entangled states
3: Examples of entangled states
4: Entangled states and non separability
5: Entangled states and which path information
6: Entangled states and the measurement process
7: Entangled states and two-photon interferences
32
Photo detection signals for a quantum field
Field operator
+
(+ ) ( ) ( ) (+ ) Ê( r ,t) = Ê ( r ,t) + Ê ( r ,t) with E ( r ,t) = Ê ( r ,t) (+ ) Ê ( r ,t) = Ei âi exp i ki .r i t modes i
( )
(
)
Ê (+ ) Ê ( ) : positive (negative) frequency component of the field operator Ê
âi ( âi+ ):annihilation (creation) operator of a photon of the mode i
Ei :normalization coefficient
Single counting rate wI
wI ( r ,t):probability to detect one photon at point r and at time t
wI ( r ,t) = Ê ( ) ( r ,t) Ê (+ ) ( r ,t) :state of the field
Double counting rate wII
wII ( r ,t; r , t ):probability to detect one photon at r ,t and another one at r , t ( ) ( ) (+ ) (+ ) wII ( r ,t; r , t ) = Ê ( r ,t) Ê ( r , t ) Ê ( r , t ) Ê ( r ,t) 33
Two-photon quantum state
= 1A 1B
One photon in mode A, one photon in mode B
Can one observe interference fringes on wI for such a state?
Using
1A â +A = 0 A
â A 1A = 0 A
â A 0 A = 0
0 A â +A = 0
one gets : wI ( r ,t) = E A2 + EB2
and idem with A B
and idem with A B
1A 1B = 0
No interference term (no coherence between the 2 modes)
Can one observe interference fringes on wII?
The only non zero matrix elements are
â +A âB+ âB â A , âB+ â +A â A âB , âB+ â +A âB â A , â +A âB+ â A âB They are all equal to 1 and we get :
2 2
wII ( r ,t; r , t ) = 2E A EB 1+ Re exp i( k A k B ).( r r ) i( A B )(t t ) We have now interference fringes: once a photon is detected in r ,t, the
probability to detect the second one in r , t is an oscillating function
of r r and t t {
}
34
Physical discussion
What are the “objects” which interfere in wII?
Another equivalent expression for wII
wII ( r ,t; r , t ) = Ê ( ) ( r ,t) Ê ( ) ( r , t ) Ê (+ ) ( r , t ) Ê (+ ) ( r ,t) = Ê ( ) ( r ,t) Ê ( ) ( r , t ) f f Ê (+ ) ( r , t ) Ê (+ ) ( r ,t) f
where we have introduced the closure relation over a complete set of states f
In fact, the only state f which gives a non zero contribution is the
vacuum state because we have 2 annihilation operators acting on a
2-photon state. We can thus write:
2
(+ ) (+ ) wII ( r ,t; r , t ) = 0 Ê ( r , t ) Ê ( r ,t) = 0 Ê
(+ )
A
( r , t ) Ê B(+ ) ( r ,t) + Ê B(+ ) ( r , t ) Ê A(+ ) ( r ,t) 2
We have also used the fact that, since we start from a state with one
photon , one photon , one of the 2 annihilation operators must
destroy the photon , the other the photon .
35
Two interfering paths
It thus appears that wII is the modulus squared of an amplitude which
is itself the sum of 2 amplitudes
2
wII ( r ,t; r , t ) = A1 + A2
(+ ) (+ ) (+ ) (+ ) A1 = 0 Ê A ( r , t ) Ê B ( r ,t) A2 = 0 Ê B ( r , t ) Ê A ( r ,t) (
)
There are 2 paths leading from the initial 2-photon state to the final
state where they have been both detected:
First path : photon A detected in r ,t photon B detected in r , t Second path : photon B detected in r ,t photon A detected in r , t Photon r ,t
Photon r , t Photon r ,t
Photon r , t 36
Conclusion
The objects which interfere in this experiment are not light
waves. They are transition amplitudes which describe
different possible paths leading the system from an initial
state to a final state.
One understands in this way that interference phenomena
can appear in physical processes involving not one but
several photons.
Multiparticle interferometry
U. Fano, Am. J. Phys., 29, 539 (1961) who applies these ideas to
the interpretation of the Hanbury Brown and Twiss effect
For more details, see Photons and Atoms, Complement AIII, p. 204
C. C-T, J. Dupont-Roc and G.Grynberg, Wiley (1989)
37
Connection with entangled states
Another equivalent expression for wII
wII ( r ,t; r , t ) = 1A1B Ê ( ) ( r ,t) Ê ( ) ( r , t ) Ê (+ ) ( r , t ) Ê (+ ) ( r ,t) 1A1B
= r ,t Ê ( ) ( r , t ) Ê (+ ) ( r , t ) r ,t
( )
( )
r ,t
( )
where
(+ ) = Ê ( r ,t) 1A1B
i
k
.
r
t
i
k
.r t
(
)
(
= E A â A e A A + E B âB e B B ) 1A1B
i k .r t
i k .r t
= E A e ( A A ) 0 A1B + E B e ( B B ) 1A0 B
The first detection of a photon at r, t transforms the initial
uncorrelated state 1A1B into an entangled state, linear
superposition of 0 A1B and 1A 0 B with coefficients depend
depending on the coordinates r, t of the first detection
The first detection at r, t establishes quantumcorrelations
between the 2 modes A and B
depending on r, t which explain
why the second detection at r , t depend on r r and t - t 38
Conclusion
Atomic physics is an ideal playground for illustrating and testing
quantum concepts
Most of the devices that we are using in our daily life (labtops, mobile
phones, MRI,..) are based on quantum physics? Quantum concepts are
not always easy to grasp, in particular the concept of quantum
interference, but they are essential and we must learn them.
Entangled states were considered a few decades ago as a topic for
philosophical discussions about reality. They are now very useful for
practical applications (quantum cryptography) and look very promising
for future developments (quantum computers?)
One can hope that the present theoretical and experimental activity will
lead to a better understanding of the foundations of quantum
mechanics
- How to reduce decoherence?
- Frontier between the classical and quantum worlds?
- How to reconcile the evolution described by Schrödinger equation
and the reduction of the wave packet?
39