Download Document

Nonlocal Boxes And All That
Daniel Rohrlich
Atom Chip Group, Ben Gurion University, Beersheba, Israel
21 January 2010
Axioms for special relativity
• The laws of physics are the same in all inertial reference
frames.
• There is a maximum signaling speed, and it is the speed of
light c.
Axioms for quantum mechanics
• Physical states are normalized vectors ψ(r), Ψ(r,t), ψ , (t) .
• Measurable physical quantities – “observables” – correspond
to Hermitian or (self-adjoint) operators on the state vectors.
• If a system is an eigenstate a with eigenvalue a of an
observable Â, then a measurement of  on a will yield a.
• Conversely, if a measurement of  on any state yields a, the
measurement leaves the system in an eigenstate a .
• The probability that a system in
a normalized state ψ can be
2
found in the state φ is φ ψ .
Axioms for quantum mechanics
• The time evolution of a quantum state (t ) is given by

i
 (t )  Hˆ  (t ) ,
t
where Ĥ is the Hamiltonian (kinetic energy + potential energy)
of the system in the state (t ) .
• The wave function of identical fermions (spin ½, ¾,…)
must
²
be antisymmetric under exchange of any pair of them; the wave
function of identical bosons (spin 0, 1,…) must be symmetric
under exchange of any pair of them.
“It’s like trying to derive special relativity from the wrong
axioms.” – Yakir Aharonov
• Fast objects contract in the direction of their motion.
• Moving clocks slow down.
• Twins cannot agree about who is the oldest.
• The observer determines the results of measurements.
•…
Question: What, indeed, is so “special” about special relativity?
Question: What, indeed, is so “special” about special relativity?
Answer: The two axioms so nearly contradict each other that
only a unique theory reconciles them.
Question: What, indeed, is so “special” about special relativity?
Answer: The two axioms so nearly contradict each other that
only a unique theory reconciles them.
Y. Aharonov and (independently) A. Shimony: Quantum
mechanics, as well, reconciles two things that nearly contradict
each other:
Question: What, indeed, is so “special” about special relativity?
Answer: The two axioms so nearly contradict each other that
only a unique theory reconciles them.
Y. Aharonov and (independently) A. Shimony: Quantum
mechanics, as well, reconciles two things that nearly contradict
each other:
Question: What, indeed, is so “special” about special relativity?
Answer: The two axioms so nearly contradict each other that
only a unique theory reconciles them.
Y. Aharonov and (independently) A. Shimony: Quantum
mechanics, as well, reconciles two things that nearly contradict
each other:
Two axioms for quantum mechanics
Axiom 1: There are nonlocal influences.
(What are they?)
Axiom 2: We cannot send superluminal
signals.
Can we derive quantum mechanics from these two axioms?
1. What is nonlocality?
Can we derive quantum mechanics from these two axioms?
1. What is nonlocality?
2. What does “no signalling” mean?
Can we derive quantum mechanics from these two axioms?
1. What is nonlocality?
Nonlocal correlations?
Aharonov-Bohm effect?
“Modular” dynamical variables?
2. What does “no signalling” mean?
Can we derive quantum mechanics from these two axioms?
1. What is nonlocality?
Nonlocal correlations?
Aharonov-Bohm effect?
“Modular” dynamical variables?
2. What does “no signalling” mean?
What is left of “no signalling” in
the limit c → ∞ of nonrelativistic
quantum mechanics?
Can we derive quantum mechanics from these two axioms?
1. What is nonlocality?
Nonlocal correlations
2. What does “no signalling” mean?
What is left of “no signalling” in
the limit c → ∞ of nonrelativistic
quantum mechanics?
Can we derive quantum mechanics from these two axioms?
1. What is nonlocality?
Nonlocal correlations
2. What does “no signalling” mean?
“No signalling” at any speed!
Quantum mechanics does not allow “signalling”
Introducing…Alice and Bob
Quantum mechanics does not allow “signalling”
Introducing…Alice and Bob
Alice
Bob
drawings by Tom Oreb
© Walt Disney Co.
Quantum mechanics does not allow “signalling”
Introducing…Alice and Bob, who share pairs in an
entangled state  and make spacelike separated
measurements on members of each pair.
Alice measures a fixed observable Aˆ ai  ai ai
on her member of each pair. Bob is free to choose
any observable Bˆ bi  bi bi to measure on his
member of each pair.
Alice
Question: Does prob(ai) depend on what
observableB̂ Bob chooses to measure?
Bob
drawings by Tom Oreb
© Walt Disney Co.
Quantum mechanics does not allow “signalling”
Introducing…Alice and Bob, who share pairs in an
entangled state  and make spacelike separated
measurements on members of each pair.
Alice measures a fixed observable Aˆ ai  ai ai
on her member of each pair. Bob is free to choose
any observable Bˆ bi  bi bi to measure on his
member of each pair.
Alice
Question: Does prob(ai) depend on what
observableB̂ Bob chooses to measure?
Answer: No, because prob(ai ) 

 ai , b j
2
j
Bob
drawings by Tom Oreb
© Walt Disney Co.
Quantum mechanics does not allow “signalling”
Introducing…Alice and Bob, who share pairs in an
entangled state  and make spacelike separated
measurements on members of each pair.
Alice measures a fixed observable Aˆ ai  ai ai
on her member of each pair. Bob is free to choose
any observable Bˆ bi  bi bi to measure on his
member of each pair.
Alice
Question: Does prob(ai) depend on what
observableB̂ Bob chooses to measure?
Bob
Answer: No, because prob(ai ) 


  ai 
b j b j  ai 


j




 ai , b j
2
j
drawings by Tom Oreb
© Walt Disney Co.
Quantum mechanics does not allow “signalling”
Introducing…Alice and Bob, who share pairs in an
entangled state  and make spacelike separated
measurements on members of each pair.
Alice measures a fixed observable Aˆ ai  ai ai
on her member of each pair. Bob is free to choose
any observable Bˆ bi  bi bi to measure on his
member of each pair.
Alice
Question: Does prob(ai) depend on what
observableB̂ Bob chooses to measure?
Answer: No, because prob(ai ) 
Bob
  ai
2

 ai , b j
2
j
.
drawings by Tom Oreb
© Walt Disney Co.
“Superquantum” correlations
Back to the question:
Can we derive quantum mechanics from the two axioms of
nonlocal correlations and no “signalling” at any speed?
This is a grandiose ambition.
“Superquantum” correlations
Back to the question:
Can we derive quantum mechanics from the two axioms of
nonlocal correlations and no “signalling” at any speed?
This is a grandiose ambition.
We could, however, entertain a more modest ambition:
Could we derive a part of quantum mechanics from these two
axioms?
“Superquantum” correlations
Back to the question:
Can we derive quantum mechanics from the two axioms of
nonlocal correlations and no “signalling” at any speed?
This is a grandiose ambition.
We could, however, entertain a more modest ambition:
Namely, could we derive Tsirelson’s bound from these two
axioms?
Alice’s black box can nonlocally influence Bob’s.
Alice
Bob
Alice’s black box can nonlocally influence Bob’s.
A, A′, B, and B′ are physical quantities having
values ±1.
Alice
measures
A or A′
Bob
measures
B or B′
Alice’s black box can nonlocally influence Bob’s.
A, A′, B, and B′ are physical quantities having
values ±1.
Alice
measures
A or A′
CHSH inequality
|CL(A,B)+CL(A,B′)+CL(A′,B)−CL(A′,B′)| ≤ 2
Bob
measures
B or B′
Alice’s black box can nonlocally influence Bob’s.
A, A′, B, and B′ are physical quantities having
values ±1.
Alice
measures
A or A′
CHSH inequality
|CL(A,B)+CL(A,B′)+CL(A′,B)−CL(A′,B′)| ≤ 2
Bob
measures
B or B′
Tsirelson’s bound
|CQ(A,B)+CQ(A,B′)+CQ(A′,B)−CQ(A′,B′)| ≤ 2√2
Alice’s black box can nonlocally influence Bob’s.
A, A′, B, and B′ are physical quantities having
values ±1.
Alice
measures
A or A′
CHSH inequality (locality)
|CL(A,B)+CL(A,B′)+CL(A′,B)−CL(A′,B′)| ≤ 2
Bob
measures
B or B′
Tsirelson's bound
|CQ(A,B)+CQ(A,B′)+CQ(A′,B)−CQ(A′,B′)| ≤ 2√2
Alice’s black box can nonlocally influence Bob’s.
A, A′, B, and B′ are physical quantities having
values ±1.
Alice
measures
A or A′
CHSH inequality (locality)
|CL(A,B)+CL(A,B′)+CL(A′,B)−CL(A′,B′)| ≤ 2
Bob
measures
B or B′
Tsirelson's bound (“no signalling”?)
|CQ(A,B)+CQ(A,B′)+CQ(A′,B)−CQ(A′,B′)| ≤ 2√2
How to beat Tsirelson’s bound in two easy steps:
1. For any measurement of A, A′, B, and B′, let the outcomes 1
and −1 be equally likely.
2. Let CSQ(A,B) = CSQ(A,B′) = CSQ(A′,B) = 1 = −CSQ(A′,B′).
B
A′
45º
A
45º
45º
B′
Then CSQ(A,B)+CSQ(A,B′)+CSQ(A′,B)−CSQ(A′,B′) = 4.
“Superquantum” correlations
Example of CSQ(φ):
CSQ(φ)
1
φ
0
45º
90º
135º
180º
−1
0º
S. Popescu and DR, Found. Phys. 24 (1994) 379
“Superquantum” correlations
Q. Superquantum correlations do not imply quantum
mechanics. Do they imply uncertainty?
A. Yes! Let Bob measure σz(B). If σz(B) = 1, then σz(A) = 1. Then
Δσx(A) = 1 = Δσy(A).
This uncertainty is irreducible because, since CSQ(φ) violates
the Tsirelson bound, it violates also the CHSH inequality.
More recent results:
• Superquantum correlations render all communication
complexity problems trivial.
W. van Dam, Thesis, U. Oxford (1999); W. van Dam,
preprint quant-ph/0501159 (2005)
• All correlations stronger than quantum correlations may
render communication complexity problems trivial.
G. Brassard et al., Phys. Rev. Lett. 96 (2006) 250401
From “superquantum correlations” to the “nonlocal box”
“Mathematicians are like Frenchmen: whatever you say to
them they translate into their own language and forthwith it is
something entirely different.”
― Goethe
From “superquantum correlations” to the “nonlocal box”
{
Nonlocal box: PNL(a,b|x,y) =
Translation:
½
0
a + b ↔ (a + b) mod 2
if a + b = xy
otherwise
From “superquantum correlations” to the “nonlocal box”
{
Nonlocal box: PNL(a,b|x,y) =
Translation:
½
0
if a + b = xy
otherwise
↔
↔
↔
↔
x=0
x=1
y=0
y=1
Alice’s result is 1 ↔
Alice’s result is –1 ↔
Bob’s result is 1 ↔
Bob’s result is –1 ↔
a=1
a=0
b=1
b=0
Alice measures A
Alice measures A′
Bob measures B
Bob measures B′
Axioms for quantum mechanics
1. What is nonlocality?
We’ve tried nonlocal correlations. What
about nonlocal equations of motion
(nonlocal cause and effect)?
2. What does “no signalling” mean?
“No signalling” at superluminal speeds
Axioms for quantum mechanics
Introducing…Jim “the Jammer”
J. Grunhaus, S. Popescu and DR, Phys. Rev. A53 (1996) 3781
drawings by Tom Oreb
© Walt Disney Co.
Axioms for quantum mechanics
Introducing…“jamming”
Alice
measures
A or A′
Jim “the Jammer”
Bob
measures
B or B′
drawings by Tom Oreb
© Walt Disney Co.
Q. Can “Jim the Jammer” jam the nonlocal correlations of
Alice and Bob?
A. Yes, if “jamming” satisfies two conditions:
1. The unary condition – For any measurement of A, A′, B,
and B′, the outcomes 1 and −1 are equally likely, with or
without jamming.
2. The binary condition – The intersection of the forward light
cones of Alice’s and Bob’s measurements lies in the
forward light cone of Jim’s action.
A configuration that violates the binary condition:
b
a
j
A configuration that obeys the binary condition:
b
a
j
This configuration, too, obeys the binary condition!
j
b
a
Back to the “nonlocal box”
{
Nonlocal box: PNL(a,b|x,y) =
Translation:
½
0
if a + b = xy
otherwise
↔
↔
↔
↔
x=0
x=1
y=0
y=1
Alice’s result is 1 ↔
Alice’s result is –1 ↔
Bob’s result is 1 ↔
Bob’s result is –1 ↔
a=1
a=0
b=1
b=0
Alice measures A
Alice measures A′
Bob measures B
Bob measures B′
Back to the “nonlocal box”
{
Nonlocal box: PNL(a,b|x,y) =
½
0
if a + b = xy
otherwise
Does this nonlocal box violate “information causality”?
Information causality: If Alice sends Bob m (classical) bits,
Bob has access to a range of at most m bits of Alice’s data.
Case m = 0: “no signalling”.
M. Pawłowski, T. Paterek, D. Kaszlikowski, V. Scarani, A.
Winter and M. Żukowski, Nature 461 (2009) 1101
Back to the “nonlocal box”
{
Nonlocal box: PNL(a,b|x,y) =
½
0
if a + b = xy
otherwise
Does this nonlocal box violate “information causality”?
Information causality: If Alice sends Bob m (classical) bits,
Bob has access to a range of at most m bits of Alice’s data.
We will concentrate on the case m = 1.
M. Pawłowski, T. Paterek, D. Kaszlikowski, V. Scarani, A.
Winter and M. Żukowski, Nature 461 (2009) 1101
Back to the “nonlocal box”
{
Nonlocal box: PNL(a,b|x,y) =
½
0
if a + b = xy
otherwise
Case m = 1: Alice has two fresh, independent bits a0 and a1.
Bob receives a number, either 0 or 1, and his task is to obtain
the corresponding bit from Alice, i.e. either a0 or a1. Alice
sends him one classical bit. There is no other communication
between them.
Using classical physics, Bob cannot succeed.
Back to the “nonlocal box”
{
Nonlocal box: PNL(a,b|x,y) =
½
0
if a + b = xy
otherwise
Case m = 1: Alice has two fresh, independent bits a0 and a1.
Bob receives a number, either 0 or 1, and his task is to obtain
the corresponding bit from Alice, i.e. either a0 or a1. Alice
sends him one classical bit. There is no other communication
between them.
Using classical physics, Bob cannot succeed.
Using quantum correlations, Bob cannot succeed.
Back to the “nonlocal box”
{
Nonlocal box: PNL(a,b|x,y) =
½
0
if a + b = xy
otherwise
Case m = 1: Alice has two fresh, independent bits a0 and a1.
Bob receives a number, either 0 or 1, and his task is to obtain
the corresponding bit from Alice, i.e. either a0 or a1. Alice
sends him one classical bit. There is no other communication
between them.
Using classical physics, Bob cannot succeed.
Using quantum correlations, Bob cannot succeed.
Using a nonlocal box, Bob can succeed.
Back to the “nonlocal box”
{
Nonlocal box: PNL(a,b|x,y) =
½
0
if a + b = xy
otherwise
Protocol for Alice and Bob:
1. Alice chooses, as input to the nonlocal box, x  a0  a1.
2. Alice adds (mod 2) the output a from the nonlocal box to a0.
She sends the result, i.e. a  a0 , to Bob.
3. Bob, for his input to the nonlocal box, chooses y = 0 if he
wants a0 and y = 1 if he wants a1. To the output b he adds the
one bit from Alice, and this sum, b  a  a0, is his response.
Back to the “nonlocal box”
{
Nonlocal box: PNL(a,b|x,y) =
½
0
if a + b = xy
otherwise
Analysis:
If Bob chooses y = 0, then a  b  xy  0. Thus Bob’s response
reduces from b  a  a0 to a0 , as required.
If Bob chooses y = 1, then a  b  xy  x  a0  a1. Thus
b  a  a0 reduces to a0  a1  a0  a1, as required.
Back to the “nonlocal box”
{
Nonlocal box: PNL(a,b|x,y) =
½
0
if a + b = xy
otherwise
Analysis:
If Bob chooses y = 0, then a  b  xy  0. Thus Bob’s response
reduces from b  a  a0 to a0 , as required.
If Bob chooses y = 1, then a  b  xy  x  a0  a1. Thus
b  a  a0 reduces to a0  a1  a0  a1, as required.
Back to the “nonlocal box”
{
Nonlocal box: PNL(a,b|x,y) =
½
0
if a + b = xy
otherwise
Analysis:
If Bob chooses y = 0, then a  b  xy  0. Thus Bob’s response
reduces from b  a  a0 to a0 , as required.
If Bob chooses y = 1, then a  b  xy  x  a0  a1. Thus
b  a  a0 reduces to a0  a1  a0  a1, as required.
Discussion
• The (simplest) example involves the most extreme nonlocal
box, the PR box. But any nonlocal box violating Tsirelson’s
bound violates information causality.
• So are we done? No, because the correlations of some
nonlocal boxes (“noisy” nonlocal boxes) do not exceed
Tsirelson’s bound yet are not quantum correlations.
• In any case, “information causality” is a lovely name, but
what does it mean? What is the connection with causality?
Discussion
To quote from Pawłowski et al.: “Note that the extreme
violation of Information Causality observed here means that
Bob can learn perfectly either bit, not that he can learn both
Alice's bits simultaneously – the latter would imply signaling.”
Is that so?