Jaarboek no. 89. 2010/2011 - Koninklijke Maatschappij voor ...

qubit can now be described as follows: with prob-
ability |α0| 2 , bit 0 is observed and the second qubit
collapses to |β0/0|0>+β1/0|1>, and with probability
|α1| 2 , bit 0 is observed and the second qubit col-
lapses to |β0/1|0>+β1/1|1>. The corresponding holds
when the measurement is done in another basis.
An important example is an EPR pair (named after
Einstein, Podolsky and Rosen), whose state is given
by (|0>⊗|0>+|1>⊗|1>)/√2. Measuring one of the two
qubits in the computational basis has the effect
that a random bit x is observed and the other qubit
collapses to |x>. Since (|0>⊗|0>+|1>⊗|1>)/√2=(H0>⊗H|
0>+H|1>⊗H|1>)/√2 (again, an easy computation), the
corresponding also holds when measuring one of
the two qubits in the Hadamard basis: a random
bit x is observed and the other qubit collapses to
H|x>. It is important to realize that the collapse of
the other qubit happens instantaneously, as soon as
the first qubit is measured, even if the two qubits
are far apart. This is an important feature of quantum
mechanics, called nonlocality. We see in the
upcoming section some surprising consequence
of nonlocality. We stress that nonlocality does not
contradict relativity, as it does not enable for fasterthan-speed-of-light
propagation of matter, energy
or information. Beyond measuring qubits, one can
also transform the state of qubits (or any quantum
system). This is done by means of an arbitrary unitary
transformation U, acting on the corresponding
state space (C2⊗C2 in case of two qubits). If the state
is given by the vector |ψ, then U transforms it into
the state U|ψ>. We do not discuss this any further, we
merely point out that in general such a transformation
is not the composition of qubit-transformations
that act individually on the qubits, but, instead, U
acts coherently on all the qubits.
Nonlocal Games
In a nonlocal game, two parties, usually called Alice
and Bob, play against a referee. In general, a nonlocal
game works as follows. The referee chooses two
΄questions΄ x and y according to some fixed probability
distribution П, and sends x to Alice and y
to Bob. Upon receipt of x and y, respectively, Alice
and Bob need to send back respective ΄answers΄ a
Natuurkundige voordrachten I Nieuwe reeks 89
Quantum Information
and b to the referee. Depending on his questions
and the received answers, the referee then decides
whether Alice and Bob jointly win this execution of
the game or whether they jointly lose. Formally, they
win if and only if a, b, x, y satisfies some (fixed) predicate
V(a, b|x, y). What makes the game interesting is
that Alice and Bob are not allowed to communicate
during the execution of the game. Thus, Alice only
knows x but not y when she provides her answer a,
and Bob only knows y but not x when he provides
his answer b. One way to ensure that Alice and Bob
do not communicate is to give them their questions
at exactly the same time, and to give them very little
time to provide their answers, less time than a signal
would require to travel from Alice to Bob. Under the
widely accepted assumption that information cannot
travel faster than with the speed of light, this
obviously prevents Alice and Bob from communicating.
What we are interested in is the maximal probability
with which Alice and Bob can win a given
nonlocal game. We stress that Alice and Bob know
the game, specified by П and V, and they are allowed
to communicate before the game starts and share
information and decide on some common strategy.
But once the game starts and Alice and Bob receive
their questions, they are on their own. A simple
yet important example of a nonlocal game is the
so-called CHSH game (named after Clauser, Horne,
Shimony and Holt). In the CHSH game, x and y are
chosen as two random and independent bits, and
a and b (also bits) need to satisfy a⊕b=x^y (where
⊕ is the logical ΄xor΄, i.e. addition mod 2, and ^ is
the logical ΄and΄). A convincing line of reasoning
for the analysis of the CHSH game goes as follows.
Because they cannot communicate, Alice and Bob
must compute their respective answers as deterministic
functions of the corresponding questions:
a = f(x) and b = g(y). (Alice and Bob might use ‘common
and/or independent’ randomness; but every
possible choice of the randomness fixes deterministic
functions f and g). Given this observation, it is a
straightforward computation (e.g. by enumerating
all possible functions f: {0,1}→{0,1} and g: {0,1}→{0,1})
that Alice and Bob cannot win the CHSH game with
57