Quantum Computing
MAS 725
Hartmut Klauck
NTU
12.3.2012
Topics today
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Superdense coding
Distinguishing quantum states
Bell inequalities
Superdense Coding
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Alice has two bits of classical information she wants
to send to Bob
We allow quantum communication
Can we achieve this with sending one qubit (and
using entanglement)?
Superdense Coding
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Communicate two bits without error:
 Even when Alice and Bob share entanglement,
sending 1 classical bit is not enough
 Without entanglement Alice needs to send 2
qubits [Holevo 73]
 Superdense Coding allows to send 2 bits with 1
qubit of communication and the consumption of
1 EPR pair
Note the two bits to be communicated are not yet
known when Alice and Bob exchange the EPR pair
Superdense Coding
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[Bennett and Wiesner 92]
Alice applies a unitary transformation to her qubit
xy=00 ) I (Identity)
xy=01 ) Z (Phase Flip)
xy=10 ) X (Bit Flip)
xy=11 ) i¢Y
Bit and Phase Flip
Superdense Coding
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The result is one of the four Bell states!
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xy=00:
xy=01:
xy=10:
xy=11:
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1/21/2 (|00i+|11i)
1/21/2 (|00i-|11i)
1/21/2 (|01i+|10i)
1/21/2 (|01i-|10i)
Bob receives Alice’s qubit
Since the states are mutually orthogonal for all 4 inputs, we
can measure them and perfectly distinguish them
Measure in the Bell basis
Superdense Coding
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Alternative: replace measuring in the Bell basis by:
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Apply the unitary operation that maps the Bell basis
onto the standard basis
For this simply reverse the circuit to generate Bell
states
Then measure in the standard basis
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A remark
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The superdense coding protocol is optimal:
 Suppose Alice and Bob share an arbitrary
entangled state (independent of the classical
message they want to send)
 They still have to send n/2 qubits to
communicate n classical bits
Distinguishing quantum states
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We have a known set of quantum states
S={ |1 i ,…, |ki }
Input: a state from S
Desired output: index i of the given state
Identification Problem
Suppose all states in S are pairwise orthogonal
 We can solve the problem perfectly: construct
an observable with subspaces Si=span(|ii), and
projectors |iihi|
Distinguishing states that are not
orthogonal
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Let S={ |i , |i } with h  | i  0
Can we solve the indentification problem?
|i =  |i +  |i with = h|i
A projective measurement is linear, i.e.
”-proportion” of |i behaves just like |i
With probability |h|i|2 the output on |i is 
(if the output on |i is always )
Hence the identification problem cannot be solved
without error!
Which features of quantum
mechanics are „nonclassical“?
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Classical physics is deterministic
Example: Rolling a die
 The reason we model this with probabilities is that we
don‘t know the initial state of the die, and that computing
the exact motions is hard
Quantum mechanics is intrinsically probabilistic
God does not play dice
Which features of quantum
mechanics are „nonclassical“?
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[Einstein Podolsky Rosen 35] claimed that quantum mechanics
gives an incomplete picture of reality
What does that mean?
There might be a deeper „classical“ theory that allows to
eliminate the probabilistic predictions of quantum mechanics
by referring to „hidden parameters“
The state 1/21/2 (|00i+|11i) on two spatially separated qubits
exhibits „spooky actions at a distance“: when measured it
behaves like two perfectly correlated coin tosses, but (a large
enough) distance would not allow communication
Quantum mechanics vs. local
and realistic theories
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A theory is
 local, when spatially separated systems do not
influence each other
 realistic, when properties are determined by the
system that is measured already before a
measurement
• Properties can be predicted with certainty in principle
• Properties are objective
Bell inequalities
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Charlie prepares two particles (in some state) and
gives one to Alice and Bob each
Alice and Bob are separated
Alice‘s particle has properties for Alice PQ and PR
with values Q and R
Bob‘s particle has properties PS and PT with values S
and T
Values of the properties are „objective“, i.e., are
determined before the measurement takes place
Q,R,S,T 2 {-1,1}
Bell inequalities
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Rewrite QS+RS+RT-QT = Q( S-T ) + R( S+T )
S-T=0 or S+T=0, and QS+RS+RT-QT = § 2
We bound the expectation E[QS+RS+RT-QT] · 2
The random experiment here: Charlie‘s preparation of the
particles
There is no randomness after Charlie has prepared them
Now we have probabilities P(q,r,s,t) of the event that the
particles satisfy Q=q, R=r, S=s, T=t
Expectation E[QS+RS+RT-QT] · 2
By linearity:
E[QS]+E[RS]+E[RT]-E[QT] · 2
Bell inequalities
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Assumptions behind the inequality: There are
probabilities P(q,r,s,t) [realism], and the
measurements are independent [locality]
Measuring EPR pairs
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State of the two particles: 1/21/2 (|01i-|10i)
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We show that there are observables such that
E[QS]+E[RS]+E[RT]-E[QT] ¸ 2¢ 21/2
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Experimental test decides in favor of quantum
mechanics [Aspect 82]
Hidden parameters
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We give a slightly different view of this result via games
A classical model with hidden parameters:
Alice and Bob receive particles prepared by Charlie
 This means:
 There is shared classical information, which determines
the value for all measurements performed by Alice and
Bob
 Remark: given a realistic theory with hidden parameters
such a list exists
For the rest of the protocol/game Alice and Bob are separated
(spatially and causally)
Hidden parameters
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The task of Alice and Bob:
 Alice and Bob can initially agree on a strategy, then no further
communication is allowed
 Alice and Bob then receive a random bit x resp. Y
 They may use more shared randomness
 They produce outputs a,b
 The protocol succeeds, if xÆy = a©b
x,y describe the property that has to be measured
What is the winning probability of the best protocol?
Optimal strategy: both always output 0
They win with probability 3/4
If a locally realistic theory with hidden parameters can „explain“
quantum theory, then both theories need to predict the same
outcome. I.e., every quantum strategy needs to have success · ¾
A quantum protocol
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Alice and Bob share the state
1/21/2 (|01i-|10i)
x=0, Alice measures in the standard basis and outputs the result
x=1, Alice applies the
rotation
measures, outputs the result
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y=0, Bob measures, outputs the complement of the result
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y=1, Bob rotates with
measures, outputs complement
Rotation gate
Qubit |0i is mapped to
cos(/8) |0i + sin(/8) |1i
Probability of measuring 0 is then
cos2(/8)
Analysis
Case 1: x=0, y=0; both measure in the standard
basis, output a,b with a=b, hence a©b=0,
result is always correct
 Case 2: x=0, y=1;
A) Alice measures a=0 (with prob. ½)
Remaining state is |01i
Bob now has |1i, i.e., b=0 with prob. cos2(/8),
error is 1-cos2(/8)= sin2(/8)
B) Alice measures a=1, like case 2 a)
 Case 3: x=1, y=0; analogous , error sin2(/8)
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Analysis
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Case 4: x=y=1, now the desired output is a©b=1
Alice and Bob both rotate before measuring
Alice rotates by -/8, Bob rotates by +/8
Now the two measurements are independent
(angle is ¼/4), and the error is ½
 A tedious calculation shows that the state after
the rotations is ½ ( |00i + |01i - |10i +|00i )
Total error:
¼ (0 + 2¢sin2(/8) + 1/2) ¼ 0.2
Hence the success probability is larger than in any
classical theory
Back to Bell inequality
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R: result of measuring in the standard basis (§ 1)
Q: result of measuring after Alice’s rotation (§ 1)
S: result of measuring in the standard basis (§ 1)
T: result measuring after Bob’s rotation
(§ 1)
E[RS]=1, E[QS]=E[RT]=cos2(/8)
E[QT]=1/2
Hence E[RS]+E[QS]+E[RT]-E[QT]=
1 + 2 cos2(/8) - 1/2 ¼ 2.2 > 2
Was we know
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Quantum effects are fundamentally different from classical
physics
In toy problems (Deutsch Josza) quantum computers can beat
classical computers
Do we get speedups for interesting problems?
Is it possible to solve NP-complete problems efficiently on
quantum computers?
How about uncomputable functions??
Goals of quantum computing
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(Building a quantum computer)
Design efficient algorithms
Design and analyze protocols for quantum
communication/entanglement assisted
communication
Quantum cryptography
Fault tolerance
Simon’s Problem
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Given: Black box function f:{0,1}n!{0,1}n
Promise: there is an s2{0,1}n with s0n
 For all x: f(x)=f(x©s)
 For all x,y: x  y©s ) f(x)  f(y)
Find s !
Example: f(x)=2bx/2c
Then for all k: f(2k)=f(2k+1); s=00 ... 01
Simon’s algorithm solve the problem in time/queries poly(n)
Every classical randomized algorithm (even with errors) needs
(2n/2) queries to the black box