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An adventure in psychic communication:
Optical communication with invisible photons
M. Suhail Zubairy
Institute for Quantum Science and Engineering
and Department of Physics and Astronomy
Texas A&M University
The premise of this talk:
It always requires a “particle” (photon, electron …) to
communicate between two parties.
We propose a system such that there is no “particle” present in
the channel between the two parties!!!!
Optical communication with invisible photon!!
Possible implementation
S – Photon source emitting H photon
PBS – Polarizing beam splitter
SM – Switchable mirror
SPR – Switchable polarization rotator
MR – Mirror
OD -- Optical delay
PC – Pockel cell
H. Salih, Z. Li, M. Alamri, and M. S. Zubairy, Phys. Rev. Lett. 110, 170502 (2013)
Quantum communication
Quantum secure communication
Quantum key distribution protocols:
Bennett-Brassard protocol (BB-84)
Eckert-91 (E-91)
Bennett-92 (B-92)
.
.
Noh-09
Quantum key distribution protocols require
Real information carriers (photons)
Classical channel
Probabilistic outcome
This talk:
• Elements of quantum cryptography
– BB84
• Direct communication with invisible photons
– Mach-Zhender interferometer
– Interaction free measurement
– Counterfactual communication
Cryptography
Old fashioned cryptography
J BN IBQQZ UP CF IFSF UPEBZ
Key: A→Z, B→A, C→B …. Z→Y
I AM HAPPY TO BE HERE TODAY.
Problems:
• Sender and receiver should exchange key through secure channels
• It is possible for a clever eavesdropper to learn the key without the
knowledge of sender and receiver.
Public key algorithms: (RSA, …)
Sender and receiver exchange key on public channels
The ultimate security is not guaranteed.
Quantum cryptography:
Key is exchanged on public channel!
Eavesdropper can be traced immediately!
Magic or quantum mechanics!!
Quantum key distribution
BB-84
(Bennett-Brassard -84 protocol)
MESSAGE
ALICE
ENCRYPTION
EAVESDROPPER
MESSAGE
SCRAMBLED
MESSAGE
BOB
DECRYPTION
KEY
KEY
ALICE
Message
Add key
Scrambled text
1
0
1
0
1
1
1
1
0
1
0
1
0
1
1
1
0
1
0 1
0 1
0 0
TRANSMIT
BOB
Received message
1
1
0
1
1
1
0
0
Add key
0
1
1
0
1
0
0
1
Recovered message
1
0
1
1
0
1
0 1
Quantum cryptography
Bennett-Brassard 84 (BB’84) protocol
Single photons are carriers of information.
Two conjugate bases for polarization are chosen


0
,
0
45
0
90
1
,
135 
1
If polarization is measured in one basis (say  ), then the
possible outcome of polarization measurement in the conjugate
basis () is equally probable for the two directions.
Quantum key distribution
Alice:
Bob:
 


0 90




45 135




45 135
 


0 135


0
0
0
0
0
0
1

0
0

1
1
1






 





0
45
0
45
0



or  or  135 or 135 or
or 0
135 90
90
135 90
1
1
0
If there is an eavesdropper
50% he/she has same orientation of polarizer.
But 50% he/she has wrong orientation


0
0
0
0
Eve


Bob:


Alice
0
0
50%
25% Wrong results
0 , 90
With equal Prob.
25% , 25%
Quantum key distribution protocols require:
Real information carriers (photons)
Classical channel
Probabilistic outcome
Question: Can we come up with a quantum communication protocol
that involves
Direct communication (no need for a prior quantum key
distribution)
No information carriers in the public channel
No classical channel needed for communication
Possible implementation
S – Photon source emitting H photon
PBS – Polarizing beam splitter
SM – Switchable mirror
SPR – Switchable polarization rotator
MR – Mirror
OD -- Optical delay
PC – Pockel cell
H. Salih, Z. Li, M. Alamri, and M. S. Zubairy, Phys. Rev. Lett. 110, 170502 (2013)
The simplest system in quantum optics:
Beam splitter:
|10〉 → cos 𝜃 10 + sin 𝜃|01〉
|01〉 → cos 𝜃 01 − sin 𝜃|10〉
Optical communication with NO photons in the transmission
channel!!
Mach-Zehnder Interferometer:
Beam-splitter transformation:
|10〉 → cos 𝜃 10 + sin 𝜃|01〉
|01〉 → cos 𝜃 01 − sin 𝜃|10〉
cos 𝜃 = 𝑅
Assume the input is |10〉
After the first beam splitter
|10〉 → cos 𝜃1 10 + sin 𝜃1 |01〉
After the second beam splitter, the output state is
cos 𝜃1 (cos 𝜃2 |10〉 + sin 𝜃2 |01〉) + sin 𝜃1 (cos 𝜃2 01 − sin 𝜃2 |10〉)
= cos(𝜃1 + 𝜃2 ) 10 + sin(𝜃1 + 𝜃2 ) |01〉
The photon state has an increasing amplitude for 01 for 𝜃 ≤ 𝜋/4
What happens when we block the path on RHS?
Assume the input is |10〉
After the first beam splitter
|10〉 → cos 𝜃 10 + sin 𝜃|01〉
Prob. of light lost on RHS = 𝑠𝑖𝑛2 𝜃
After the second beam splitter, the output state is
cos 𝜃(cos 𝜃 |10〉 + sin 𝜃|01〉)
Prob. of light at 𝐷1 = 𝑐𝑜𝑠 4 𝜃
Prob. of light at 𝐷2 = 𝑐𝑜𝑠 2 𝜃𝑠𝑖𝑛2 𝜃
cos 2𝜃 10 + sin 2𝜃|01〉
cos 𝜃(cos 𝜃 |10〉 + sin 𝜃|01〉)
When θ=π/4 (50/50 beam splitter), the output is
|01〉
(|10〉 + |01〉)/ 2
D2 clicks
Photon is absorbed: 50%
D1 and D2 click with 25% prob. each
D1 click means path blocked (but no photon) on RHS
Interaction-free measurement
A. C. Elitzur, and L. Vaidman, Found. Phys. 23, 987 (1993)
Object is not there: 50% (D2 clicks with certainty)
Object is there: 50%
(D1 and D2 click with equal prob.)
When D1 clicks (happens 25% times) we know the object is there
But the photon is on the LHS
Interaction-free measurement!!!!!
|10〉 → cos 𝜃 10 + sin 𝜃|01〉
|01〉 → cos 𝜃 01 − sin 𝜃|10〉
𝜃 = 𝜋/2𝑁
N beam-splitters with very small
transmittivity (t = sin(π/2N) ≪ 1)
Initial state 𝐢𝐬 |𝟏𝟎〉
After n cycles:
|10〉 → cos 𝑛𝜃 10 + sin 𝑛𝜃|01〉
At the end of the Nth cycle (𝑁𝜃 = 𝜋/2),
final state is |01〉
The detector 𝑫𝟐 clicks.
Photon is blocked at each step
After n cycles, the initial state 10 evolves to
|10〉 → 𝑐𝑜𝑠 𝑛−1 𝜃 (cos 𝜃 10 + sin 𝜃|01〉
𝜃 = 𝜋/2𝑁
Prob. for photon lost on RHS = (1 − 𝑐𝑜𝑠 2𝑁−2 𝜃)
Prob. for D1 clicking = 𝑐𝑜𝑠 2𝑁 𝜃
Prob. For D2 clicking = 𝑐𝑜𝑠 2𝑁−2 𝜃 𝑠𝑖𝑛2 𝜃
With N is large, 𝑐𝑜𝑠 2𝑁 𝜃 ≈ 1.
The photon is almost completely reflected
Quantum Zeno effect!!!
The detector 𝑫𝟏 clicks.
Protocol for counterfactual communication:
No photon in the public channel
Bob sends information to Alice!
Logic 0: Bob does not block at any stage
Logic 1: Bob blocks at every stage
Second chained version outside the first
chained structure.
The signal photon passes through M big
cycles composed by 𝐵𝑆𝑀 s with
𝜃𝑀 = 𝜋/2𝑀
For each mth cycle (𝑚 < 𝑀), there are
N small cycles composed by 𝐵𝑆𝑁 s with
𝜃𝑁 = 𝜋/2N
Initial state for the total system is |𝟏𝟎𝟎〉
Bob does not block the photon
at any stage (Logic 0)
After (M,N) cycles, 𝑫𝟏 clicks
H. Salih, Z. Li, M. Alamri, and M. S. Zubairy, Phys. Rev. Lett. 110, 170502 (2013)
Bob blocks the photon at all stages
(Logic 1)
After (M,N) cycles, 𝑫𝟐 clicks
H. Salih, Z. Li, M. Alamri, and M. S. Zubairy, Phys. Rev. Lett. 110, 170502 (2013)
Bob allows the photon to
pass through (logic 0).
The detector 𝑫𝟏 clicks.
Bob blocks the photon
(logic 1).
The detector 𝑫𝟐 clicks.
Almost no photon in the public channel!!!!
Bob allows the photon to
pass through (logic 0).
The detector 𝑫𝟏 clicks.
Bob blocks the photon
(logic 1).
The detector 𝑫𝟐 clicks.
No photon in the public channel!!!!
ESSENTIAL PHYSICS IS QUANTUM ZENO EFFECT!!!!!
Probability of D1 clicking
𝜃𝑀 = 𝜋/2𝑀
Prob. for photon lost on RHS = (1 − 𝑐𝑜𝑠 2𝑀−2 𝜃)
Prob. of D1 clicking = 𝑐𝑜𝑠 2𝑀 𝜃
Prob. of D2 clicking = 𝑐𝑜𝑠 2𝑀−2 𝜃 𝑠𝑖𝑛2 𝜃
P1
100
0.00
80
0.75
0.90
60
M
0.95
1.00
40
20
1000
2000
N
3000
Probability of D2 clicking
𝜃𝑀 = 𝜋/2𝑀
P2
100
0.00
80
0.75
0.90
60
M
0.95
1.00
40
20
1000
2000
N
3000
P1
100
P2
100
0.00
80
0.00
80
0.75
0.75
0.90
0.90
60
60
1.00
40
20
0.95
M
M
0.95
1.00
40
20
2000
1000
3000
1000
N
P1 = 0.9051, P2 = 0.9055
2000
N
for
M = 25, N = 320
P1 = 0.9517, P2 = 0.9511 for
M = 50, N = 1250
P1 = 0.9837, P2 = 0.9837
M = 150, N = 10000
for
3000
Possible implementation
S – Photon source emitting H photon
PBS – Polarizing beam splitter
SM – Switchable mirror
SPR – Switchable polarization rotator
MR – Mirror
OD -- Optical delay
PC – Pockel cell
H. Salih, Z. Li, M. Alamri, and M. S. Zubairy, Phys. Rev. Lett. 110, 170502 (2013)
Experimental counterfactual communication with single photons
Jian-Wei Pan’s group from USTC, China
Experimental counterfactual communication with single photons
Security issue:
1. Since there is no signal photon in the public channel,
it is extremely difficult for Eve to obtain any information.
2. Eve can copy Alice’s setup (so her photon is also “invisible”)
and emits her photon into the channel to detect what Bob’s messages
are.
Fortunately, this attack can be avoided by a time delay setup
(This kind of attack is possible for any counterfactual scheme).
Summary
• Our quantum communication protocol does not rely on prior
exchange of secret key leading to direct communication.
• Under ideal conditions, no signal photon exists in the
communication channel!!!
• Security of communication can be guaranteed under conventional
attacks.
Questions concerning counterfactuality??
Is the probability of finding a signal
photon in the transmission channel
nearly zero?
According to Vaidman: It is zero
when Bob blocks but non-zero when
Bob does not Block
Vaidman: Given a click at D1, the
probability of finding the photon by
a non-demolition measurement of
the projection operator on the
transmission channel is one!
Forward (continuous line) and
backward (dashed line) evolving
wave functions of the photon – nonzero weak value
Weak Value
U L1
UL2
U L3
UL4
BS1
BS 2
BS 2
BS1
100  L1  L2  L3  L4
Pre-selected state at L2
 i  U L 2U L1 100
r A 
t
(B  C )
2
Post-selected state at L2 (D1 clicking)
 f  D1 U L 4U L 3
t
r A 
(B  C)
2
D1 :  f  i
Weak values:
AC 
 f C C i
 f i
t2
t2
 2 ; AB   2 ; AE  AF  0; 𝐴𝐴 = 1
2r
2r
2
 r4
“In the framework of these concepts we can state the following:
The photon did not enter the interferometer, the photon never
left the interferometer, but it was there. This is a new paradoxical
feature of a pre- and postselected quantum particle”
L. Vaidman, Phys. Rev. Lett 98 160403 (2007);
Phys. Rev. A 87, 052104 (2013)
We disagree!!!
1. In our analysis, the outcome
of clicks at D1, D2, and D3 are
identical if we keep the path E
open or block the path E
2. In Vaidman’s analysis, there is no
photon in path C (𝑨𝑪 = 𝟎) if we
block the path E but 𝑨𝑪 =
path E is open.
𝒕𝟐
𝟐𝒓𝟐
when
3. A weak measurement at C disturbs the interference.
A small photon amplitude leaks through to E.
Path through E to D1 becomes possible.
Example: Weak quantum non-demolition measurement:
𝐻 = ℏ𝜂 𝐶 𝐶 ⊗ |𝑏〉〈𝑏|
𝐶 𝐶 − System
𝑏 𝑏 − Meter
ℏ𝑔2
𝜂=
Δ
Weak measurement: 𝜂𝜏 ≪ 1
A  ( b  c )/ 2
Weak measurement
  f e iH  i  A
   f (1  iH )  i  A
b  c
b 
   f i 
 iAC

2
2




   f  i e iC b  c / 2
Unitary transformation
b  ( b  i c ) / 2 ; c  (i b  c ) / 2
A  ( b  c )/ 2
Observable quantitities:
2
1
Pb   f  i [1  sin( AC )]
2
2
1
Pc   f  i [1  sin( AC )]
2
 f i
2
 Pb  Pc  r 4

AC 
1

arcsin
Pc  Pb
Pb  Pc
Weak measurement
2
1
Pb   f  i [1  sin( AC )]
2
2
1
Pc   f  i [1  sin( AC )]
2
 f i

2
 Pb  Pc  r 4
Pc  Pb
AC 
arcsin

Pb  Pc
1
A  ( b  c )/ 2
Tempting claims:
Weak measurement has no influence on the final outcome
Photon exists in path C but not in E
Does weak measurement disturb the system?
2
1
Pb   f  i [1  sin( AC )]
2
2
1
Pc   f  i [1  sin( AC )]
2
 f i

2
 Pb  Pc  r 4
Pc  Pb
AC 
arcsin

Pb  Pc
1
The probability of finding a photon at E:
PE  t 2 2 2 / 8
(Second order in 𝜼𝝉)
How can a second-order 𝑃𝐸 yield linear values for 𝑃𝑏 and 𝑃𝑐 ?
How can a second-order 𝑃𝐸 yield linear values for 𝑃𝑏 and 𝑃𝑐 ?
2
1
Pb   f  i [1  sin(AC )]
2
2
1
Pc   f  i [1  sin(AC )]
2
 f i

2
 Pb  Pc  r 4
Pc  Pb
AC 
arcsin

Pb  Pc
1
The probability of finding a photon at E: 𝑃𝐸 = 𝑡 2 𝜂 2 𝜏 2 /8
2
2
1 i 2 t
1 4 2 t2
Pb 
r  i   r  r

2
4
2
4
1 4
t2
1
 r [1  sin( 2  )]   f  i
2
2r
2
2
[1  sin(AC )]
RESOLUTION:
The probability of finding a photon at E: 𝑃𝐸 = 𝑡 2 𝜂 2 𝜏 2 /8
1 i 2 t
Pb 
r  i 
2
4
2
1
  f i
2
2
2
1 4 2 t2
 r r

2
4
[1  sin(AC )]
In a weak measurement at C the
probability of click at D1 is the
modulus square of the sum of two
amplitudes:
(i) Reflected amplitude along path A
(ii)Transmitted amplitude along path FCE
Joint measurement:
 f U L 4 E E U L 3 C C U L 2U L1  i
 f U L 4U L 3U L 2U L1  i
t2
 2
2r
Joint measurement at C
and E at the same time
and in the same setup
Summary
• Our quantum communication protocol does not rely on prior
exchange of secret key leading to direct communication.
• Under ideal conditions, no signal photon exists in the
communication channel!!!
• Security of communication can be guaranteed under conventional
attacks.
Experimental counterfactual communication with single photons
Jian-Wei Pan’s group from USTC, China
Experimental Results
•M=4, N=2
•Heralded single photon source: from a PDC source with the
brightness of 1.4×107 pairs/s
•The deviation between experimental and theoretical results is due
to finite visibility of the interferometers
MESSAGE
ALICE
ENCRYPTION
EAVESDROPPER
MESSAGE
SCRAMBLED
MESSAGE
BOB
DECRYPTION
KEY
KEY
ALICE
Message
Add key
Scrambled text
1
0
1
0
1
1
1
1
0
1
0
1
0
1
1
1
0
1
0 1
0 1
0 0
TRANSMIT
BOB
Received message
1
1
0
1
1
1
0
0
Add key
0
1
1
0
1
0
0
1
Recovered message
1
0
1
1
0
1
0 1
Complementarity (Niels Bohr)
16th September 1927 at the
International Physics Congress, Como, Italy
Two observables are complementary if precise knowledge of
one of them implies that all possible outcomes of measuring the
other one are equally probable
Position-momentum
Spin components
Polarization
Time delay system (spatial relativity)
Determining who has right of reading Bob’s messages
Control signal
Alice
Public classical channel
Bob
Transmission channel
Measurement photon
OD2
Quantum channel
SD
(one arm of the interferometer)
The time the photon spending in OD2 :
The period SD is switched off :
t


T
Alice
Transmission channel
Bob

OD2
SD
No Authorization
Eve
Suppose it takes the time T for Eve’s photon to get to Bob’s devise.
Security condition:
Signal 4
Signal 3
The minimum time interval (
    t
Signal 2
) is determined by
Signal 1
t
RSA encryption: (1978)
Public key:
N = p q (p, q are primes and must remain secret )
e ─ encryption key, e is relatively prime to (p-1)(q-1)
Private key:
d = e-1(mod (p-1)(q-1))
13 mod 6 = 1 etc.
Decryption key
Message
m
Encrypted message
c = me (mod N)
Decrypted message
cd (mod N) = m ☺
Message recovered
RSA encryption: (1978)
Public key:
N = p q (p, q are primes and must remain secret )
e ─ encryption key, e is relatively prime to (p-1)(q-1)
p = 47
Private key:
q = 71
d = e-1(mod (p-1)(q-1))
N = p q = 3337
Decryption key
(p-1)(q-1)=46x70=3220
Message
m
e=79 ( randomly chosen )
Encrypted message
-1 (mod 3220)=1019
d=79
c = me (mod N)
Decrypted message
m = 688
cd (mod N) = m ☺
c=68879 (mod 3337 )=1570
15701019 (mod 3337)=688