### Lecture 3

```Quantum Computing
MAS 725
Hartmut Klauck
NTU
5.3.2012
Some linear algebra:
 Vector space V (dimension d)
 Subspaces: U µ V and U is also a vector space (dim e < d)
 There is an orthonormal basis:
v1,...,ve,...,vd, span(v1,...,ve)=U
 Projection onto U:
 PU = i=1...e | v iih vi |
 Example: Projector P=|v1ih v1|
P |v1i = |v1ih v1| v1i = |v1i
P ( |v1i+ |v2i)= |v1i + |v1ih v1| v2i = |v1i
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Measurements (II)
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A? B iff v?w for all v2 A, w2 B
A © B=V iff A? B and
for all v2 V: v =  u +  w,
u2A, w2B, kuk,kvk = 1, ||2 + ||2=kvk2
Measurements (III)
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Hilbert spaces V, dim k
Observable:
System l · k of subspaces S0,...,Sl-1, pairwise
orthogonal
Probability of measuring i is
k Proj(Si) |i k2
The state |i collapses to
Proj(Si) |i / kProj(Si) |ik
(renormalized)
Observables correspond to a measurement device
We only consider projections measurements
Measurements (IV)
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A matrix A is Hermitian, if A=Ay
Hermitian matrices have only real eigenvalues and
are diagonalizable:
 Uy A U is diagonal for some unitary U
The eigenspaces of A decompose Cn
Let ¸(i) denote an eigenvalue of A, and Pi the
projection onto its eigenspace
 Then A=i ¸(i) Pi
Measurements (V)
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A Hermitian matrix is a concise representation of an
observable
 Eigenvalues correspond to measurement
outcomes
 Eigenspaces decompose Cn
Measurements: example
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Two qubits, living in C4
Observable:
S0=span(|00i,|01i)
S1=span(|10i,|11i)
S0 ? S1
This observable corresponds to measuring the first
qubit: S0 indicates that it is 0, S1 indicates 1
Measuring an EPR-Pair
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The state is 1/21/2 ¢ (|00i+|11i)
We measure the first qubit
Result:
If we measure 0, then the state collapses to |00i
If we measure 1 we get |11i
Each happens with probability ½
Qubit 2 collapses right after measuring qubit 1
The qubits act like a shared public coin toss. This is even true if
the qubits are spatially separated
Summary
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Hilbert space: register holding a quantum state
Vectors: states
Unitary transformation: evolution of states
(computation)
Observables: for measuring the computation‘s
output
The probability distribution on results: output of the
computation
Projected and normalized vector: the remaining
quantum state
Measurement is the only way to extract information
from a quantum state
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No-Cloning Theorem
Bell states
Quantum Teleportation
No Cloning
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Suppose we are given a quantum state |i
Can we make a copy?
Copying classical information is trivial (ask the music
I.e., there is a unitary transformation
Un: Un|xi|0i = |xi|xi for all x2{0,1}n
But then by linearity U1 1/2.5(|0i+|1i) |0i=
1/21/2 (|00i+|11i)
 1/21/2(|0i+|1i) times 1/2.5(|0i+|1i)
No Cloning Theorem
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Theorem: There is no unitary U, such that for all
quantum states |i on n qubits:
U |i |0mi = |i|i|()i
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for some m und |()i (which is garbage)
I.e. there is no universal way to copy unknown
quantum states without error!
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[Dieks, Wootters/Zurek 82]
Proof: No Cloning
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Let a linear U be given (this also fixes m). Then
U |0ni |0mi =|0ni |0ni |0i
U |1ni |0mi =|1ni |1ni |1i
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Then also (due to linearity)
U 1/21/2 ( |0ni+|1ni ) |0mi
= 1/21/2 ( U |0ni|0mi + U |1ni|0mi )
= 1/21/2 ( |02ni |0i + |12ni |1i )
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But we wanted (for some |2i)
1/2 (|0ni+|1ni) (|0ni+|1ni) |2i
Proof: No Cloning (II)
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We have 1/21/2 ( |02ni |0i + |12ni |1i )
We want 1/2 (|0ni+|1ni) (|0ni+|1ni) |2i
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Claim: This is not the same!
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Proof of the claim: we compute the inner product.
Same state ) inner product = 1
h  |  i = h |i ¢ h|i
h  +  | i=h|i + h|i
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Proof No Cloning (III)
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We get 1/21/2 ( |02ni |0i + |12ni |1i )
We expect 1/2 (|0ni+|1ni) (|0ni+|1ni) |2i
Inner product:
1/23/2 ¢ (
( h0n|0ni+h0n|1ni ) ¢
( h0n|0ni+h0n|1ni ) ¢ h0|2i )
+( h1n|0ni+h1n|1ni) ¢
( h1n|0ni+h1n|1ni) ¢ h1|2i ) )
= 1/23/2 ¢ (1 ¢ 1 ¢ a0,2 + 1 ¢ 1 ¢ a1,2) .
Then |1/23/2 ¢ ( a0,2+a1,2)| · 2/23/2 < 1
No Cloning
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No unitary can map both a state and another state
that are not orthogonal in a way that wakes a copy
It is possible to clone quantum states with small
success probability
Bell States
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Consider the following basis (John Bell)
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This is orthonormal in C4
Generating Bell states
x
H
|x,yi
y
CNOT Gate: CNOT |0, yi =|0, yi; CNOT |1,yi=|1, 1-yi
|0,0i = 1/21/2 (|00i +|11i)=|+i
Quantum Teleportation
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We are given a quantum state |i
Can we reproduce that state in another location?
We cannot just send them over…. And we cannot
copy them
Classical communication is possible
This problem seems hard since a single qubit
0 |0i+1 |1i might contain a lot of information
due to possibly irrational amplitudes
Quantum Teleportation
|i
Local operations
Classical Communication
Local operations
1/21/2 (|00i+|11i)
Quantum Teleportation
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[Bennett et al. 93]
Alice and Bob share an EPR pair
(Alice has the first qubit, Bob the other)
Alice has |i (1 qubit quantum state)
Alice applies CNOT to her qubit q0 with |i (control) and her
EPR qubit q1
Alice applies H to q0 and measures q0 and q1
Alice sends the result of the measurement (2 bits)
Bob applies a unitary transformation depending on the
message to his EPR qubit q2
Quantum Teleportation
q1, q2: EPR Pair; q0: |i;
q0,q1 with Alice, q2 with Bob
Measurement
Pauli Transformations
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X:
(NOT, Bit Flip)
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Z:
(Phase Flip)
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Y:
Quantum Teleportation
Create EPR Pair
Quantum Teleportation
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Then: CNOT on q0 and q1
x=0,1 x |xi y=0,1 1/21/2|y,yi
CNOT: x,y x/21/2 |x, x © y, yi
Measure q1: Result is a
Prob. for 0/1 is 0.5 each: |0|2/2+|1|2/2=1/2
Send a to Bob
a=0 ) Bob does nothing
a=1 ) Bob applies X-Gate (Bit Flip)
Result in both cases:
Quantum Teleportation
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State y y |y,yi
Problem: Bob’s qubit q2 is still entangled with q0
y y |y,yi
Quantum Teleportation
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Alice applies H to q0
Alice measures q0, assume the result is b, she sends
b to Bob
Applying H:
y y |y,yi  1/21/2 z,y y (-1)z¢ y |z,yi
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Measuring q0 (Result is b; 0/1 with Prob. 1/2):
y y (-1)b¢ y |yi on q2
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Bob corrects by applying a Z-Gate, if necessary
Quantum Teleportation
y y|yi
Remarks
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The EPR-pair is consumed
The original state |i is destroyed
[no cloning!]
States with many qubits can be teleported one by
one (we need to make sure entanglement between
the teleported qubits is preserved)
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