1 Ph CS 219A Quantum Computation Lecture 6 Bell Inequalities Today we start a new topic We will explore more deeply how quantum correlations are different from classical ones Two parties in different laboratories who share quantum entanglement but are unable to communicate can perform ID: 912428
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Ph/CS 219A Quantum Computation Lecture 6. Bell Inequalities
1
Ph/CS 219AQuantum Computation
Lecture
6. Bell Inequalities
Today we start a new topic. We will explore more deeply how quantum correlations are different from classical ones.
Two parties in different laboratories who share quantum entanglement, but are unable to communicate, can perform tasks which would be impossible without the shared entanglement.
This discussion will also clarify the claim that probabilities are assigned to outcomes of measurements not because we are ignorant of a more complete description, but rather because the measurement process is intrinsically random.
Niels Bohr: “Anyone who is not shocked by quantum theory has not understood it.”
To introduce this topic, I will tell you a story about two physicists I know, Alice and Bob …
See Chapter
4
of the Lecture Notes.
Slide2Three coins on the table. Each is either heads or tails. You can uncover any one of the three coins, revealing whether it is heads of tails, but when you two the other two coins disappear --- you’ll never know whether those other two coins are heads or tails.
1
2
3
Slide3Three coins on the table. Each is either heads or tails. You can uncover any one of the three coins, revealing whether it is heads of tails, but when you two the other two coins disappear --- you’ll never know whether those other two coins are heads or tails.
1
2
3
Slide4Three coins on the table. Each is either heads or tails. You can uncover any one of the three coins, revealing whether it is heads of tails, but when you two the other two coins disappear --- you’ll never know whether those other two coins are heads or tails.
1
2
3
Slide5Three coins on the table. Each is either heads or tails. You can uncover any one of the three coins, revealing whether it is heads of tails, but when you two the other two coins disappear --- you’ll never know whether those other two coins are heads or tails.
1
2
3
Slide6Alice (Pasadena)
Bob (Waterloo)
Donald
(Denver)
There are many sets of coins, identically prepared by Donald.
For each of the three coins, in Pasadena or Waterloo, the probability is ½ that the coin is heads or tails.
1
2
3
1
2
3
Slide7Alice (Pasadena)
Bob (Waterloo)
There are many sets of coins, identically prepared by Donald.
For each of the three coins, in Pasadena or Waterloo, the probability is ½ that the coin is heads or tails.
But, if Alice and Bob both uncover the same coin, the outcomes are perfectly correlated.
Donald
(Denver)
1
2
3
1
2
3
Slide8Alice (Pasadena)
Bob (Waterloo)
There are many sets of coins, identically prepared by Donald.
For each of the three coins, in Pasadena or Waterloo, the probability is ½ that the coin is heads or tails.
But, if Alice and Bob both uncover the same coin, the outcomes are perfectly correlated.
Donald
(Denver)
1
2
3
1
2
3
Slide9Alice (Pasadena)
Bob (Waterloo)
There are many sets of coins, identically prepared by Donald.
For each of the three coins, in Pasadena or Waterloo, the probability is ½ that the coin is heads or tails.
But, if Alice and Bob both uncover the same coin, the outcomes are perfectly correlated.
Donald
(Denver)
1
2
3
1
2
3
Slide10Alice (Pasadena)
Bob (Waterloo)
There are many sets of coins, identically prepared by Donald.
For each of the three coins, in Pasadena or Waterloo, the probability is ½ that the coin is heads or tails.
But, if Alice and Bob both uncover the same coin, the outcomes are perfectly correlated.
Donald
(Denver)
1
2
3
1
2
3
Slide11Alice (Pasadena)
Bob (Waterloo)
Donald
(Denver)
1
2
3
1
2
3
There are many sets of coins, identically prepared by Donald.
For each of the three coins, in Pasadena or Waterloo, the probability is ½ that the coin is heads or tails.
But, if Alice and Bob both uncover the same coin, the outcomes are perfectly correlated.
We know it always works – we’ve checked it a million times.
Slide12Alice (Pasadena)
Bob (Waterloo)
There are many sets of coins, identically prepared by Donald.
For each of the three coins, in Pasadena or Waterloo, the probability is ½ that the coin is heads or tails.
But, if Alice and Bob both uncover the same coin, the outcomes are perfectly correlated.
We know it always works – we’ve checked it a million times.
Donald
(Denver)
1
2
3
1
2
3
Slide13Alice (Pasadena)
Bob (Waterloo)
Donald
(Denver)
1
2
3
1
2
3
There are many sets of coins, identically prepared by Donald.
For each of the three coins, in Pasadena or Waterloo, the probability is ½ that the coin is heads or tails.
But, if Alice and Bob both uncover the same coin, the outcomes are perfectly correlated.
We know it always works – we’ve checked it a million times.
Slide14Alice (Pasadena)
Bob (Waterloo)
There are many sets of coins, identically prepared by Donald.
For each of the three coins, in Pasadena or Waterloo, the probability is ½ that the coin is heads or tails.
But, if Alice and Bob both uncover the same coin, the outcomes are perfectly correlated.
We know it always works – we’ve checked it a million times.
Donald
(Denver)
1
2
3
1
2
3
Slide15Alice (Pasadena)
Bob (Waterloo)
Donald
(Denver)
1
2
3
1
2
3
There are many sets of coins, identically prepared by Donald.
For each of the three coins, in Pasadena or Waterloo, the probability is ½ that the coin is heads or tails.
But, if Alice and Bob both uncover the same coin, the outcomes are perfectly correlated.
We know it always works – we’ve checked it a million times.
Slide16Alice (Pasadena)
Bob (Waterloo)
Donald
(Denver)
1
2
3
1
2
3
Bob reasons:
-- We know the correlation is always perfect,
-- And surely what Alice does in Pasadena exerts no influence on what Bob finds when he uncovers a coin in Waterloo.
-- So, in effect, Alice and Bob, working together, can learn the outcome when any two of the coins are uncovered in Waterloo.
Slide17Alice (Pasadena)
Bob (Waterloo)
Donald
(Denver)
1
2
3
1
2
3
Bob reasons:
-- We know the correlation is always perfect,
-- And surely what Alice does in Pasadena exerts no influence on what Bob finds when he uncovers a coin in Waterloo.
-- So, in effect, Alice and Bob, working together, can learn the outcome when any two of the coins are uncovered in Waterloo.
Slide18Alice (Pasadena)
Bob (Waterloo)
Donald
(Denver)
1
2
3
1
2
3
Bob reasons:
-- We know the correlation is always perfect,
-- And surely what Alice does in Pasadena exerts no influence on what Bob finds when he uncovers a coin in Waterloo.
-- So, in effect, Alice and Bob, working together, can learn the outcome when any two of the coins are uncovered in Waterloo.
Slide19Alice (Pasadena)
Bob (Waterloo)
Donald
(Denver)
1
2
3
1
2
3
Bob reasons:
-- We know the correlation is always perfect,
-- And surely what Alice does in Pasadena exerts no influence on what Bob finds when he uncovers a coin in Waterloo.
-- So, in effect, Alice and Bob, working together, can learn the outcome when any two of the coins are uncovered in Waterloo.
Slide20Alice (Pasadena)
Bob (Waterloo)
Bell reasons:
Why? Because if you uncover all three coins, at least two have to be the same!
Slide21Alice (Pasadena)
Bob (Waterloo)
Alice and Bob did the experiment a million times, and found …
How could Bell’s prediction be wrong? Bell assumed the probability distribution describes our ignorance about the actually state of the coins under the black covers, and that there is no “action at a distance” between Pasadena and Waterloo. The lesson:
-- Don’t reason about “counterfactuals” (“I found H when I uncovered 1; I would have found either H or T if I had uncovered 2 instead, I just don’t know which.”) When the measurements are incompatible, then if we do measurement 1 we can’t speak about what would have happened if we had done measurement 2 instead.
-- Quantum randomness is not due to ignorance. Rather, it is intrinsic, occurring even when we have the most complete knowledge that Nature will allow.
-- Note that the quantum correlations do not allow A and B to send signals to one another.
Slide22Alice (Pasadena)
However, Alice and Bob did the experiment a million times, and found …
Bell inequality violations are seen in experiments with qubits encoded in photons, atoms,
electron spins, and
superconducting circuits.
There are “loopholes”:
Detection efficiency
Causality
“Free will”
Bell inequality violation has been verified experimentally since the 1980s, but the first “loophole free” experiments were achieved in 2015.
Alice and Bob shared a maximally entangled (Bell) pair of qubits, and each could perform a two-outcome measurement on her/his qubit in one of three possible ways. What did they measure?
Bob
(Waterloo)
Slide23Ph/CS 219A Quantum Computation Lecture 6. Bell Inequalities
23
Okay, what’s
really
going on?
Donald
prepares a maximally entangled state of two qubits:
For Alice, the three coins that
can be uncovered are three possible single-qubit observables that she may choose to measure, and same for Bob. These three operators are
noncommuting
, so measuring one disturbs a measurement of the others. That is why when Alice uncovers a coin (measures one of the observables), the other coins disappear (she can’t know what would have happened if she had measured these observables instead).
Heads and tails are two possible outcomes of measuring an observable which has eigenvalues +1 and -1. Because the state prepared by Donald is
entangled,
Alice’s measurement outcomes are correlated with Bob’s. The observables they measure have the form.
You can check:
Therefore:
(repeated indices summed)
Slide24Ph/CS 219A Quantum Computation Lecture 6. Bell Inequalities
24
Okay, what’s
really
going on?
When Alice and Bob “uncover
the same coin” their measurement axes are
anti-aligned
. Then they always get the same measurement outcome.
When Alice and Bob “uncover different coins” the angle between their measurement axes is 60 degrees. In that case they get the same measurement outcome with probability ¼.
Bell was wrong, because he assumed incorrectly that he could assign probabilities of outcomes to measurements that were not performed.
The randomness of the outcomes is not due to ignorance --- it is
intrinsic
.
Slide25Ph/CS 219A Quantum Computation Lecture 6. Bell Inequalities
25
Clauser
-Horne-
Shimony
-Holt
(CHSH) Inequality
Suppose the randomness
of the outcomes is due to ignorance. If we had complete information we would know with certainty the values of:
A
B
bits
a
,a
’
1
1
b,b
’
correlated
Notice that:
Consider:
This is the CHSH Inequality.
(
Clauser
was a Caltech physics BA, 1964.)
For a joint probability distribution governing these four binary
variables:
Slide26Ph/CS 219A Quantum Computation Lecture 6. Bell Inequalities
26
Clauser
-Horne-
Shimony
-Holt
(CHSH) Inequality
A
B
bits
a
,a
’
1
1
b,b
’
correlated
The CHSH inequality
is violated.
Slide27Ph/CS 219A Quantum Computation Lecture 6. Bell Inequalities
27
Clauser
-Horne-
Shimony
-Holt
(CHSH) Inequality
This the choice
of binary observables that
maximally
violates the CHSH inequality.
(
Cirel’son
Inequality)
Slide28Ph/CS 219A Quantum Computation Lecture 6. Bell Inequalities
28
Clauser
-Horne-
Shimony
-Holt
(CHSH) Inequality
Bell inequality violations are seen in experiments with qubits encoded in photons, atoms,
electron spins, and
superconducting circuits.
There are “loopholes”:
Detection
efficiency. Alice and Bob don’t have a fair sample because detection sometimes fails.
Causality. E.g., A measures first, and by the time B measures, a signal could have traveled A
B.
“Free will
”. Hidden variables govern not only the outcomes, but also what A and B “choose” to measure.
Bell inequality violation has been verified experimentally since the 1980s, but the first “loophole free” experiments were achieved in 2015
. These simultaneously close the detection and causality loopholes.
Bell inequality violations have been seen in experiments where A and B measurement choices of what to measure are governed by brightness fluctuations in stars hundreds of light years apart.
Slide29Ph/CS 219A Quantum Computation Lecture 6. Bell Inequalities
29
CHSH Game
CHSH inequality:
A
B
bits
a
y
x
b
Goal:
correlated
Averaged uniformly over inputs, no “classical strategy” can win the game with success probability better than .75
For “quantum strategies”, the highest possible success probability improves to
Slide30Ph/CS 219A Quantum Computation Lecture 6. Bell Inequalities
30
Bell inequalities more
g
enerally
Now
a, b
are possible measurement “settings” for Alice and Bob, taking
m
possible values.
And
x, y
are possible measurement “outcomes” taking
v
possible values.
A
B
bits
a
y
x
b
correlated
In a “local configuration” there is a definite
x
for each
a
(
which does not depend on
b
)
, and a definite
y
for each
b
(which does not depend on
a
).
A “local model” also known as a “local hidden variable theory” (LHVT) assigns a probability to each local configuration.
Bell inequalities
are constraints on
p(
xy|ab
)
that apply to any local model.