Fundamentals of Quantum information week 3

1. AP3421: Fundamentals of Quantum Information
Week 3
Version: 2019/10/04
Multiple qubits:
states, gates
& measurements.
Photo:A.Bruno

2. 2
Class announcements
● Homework #1 will be due at start of class on Friday.
TUD students:
- please staple all the papers you hand in.
Twente students:
- please submit in pdf format only.
- please send to TA list: TA_AP3421-TNW@tudelft.nl
Can you LaTeX your response? Yes, but that is not required!
If you handwrite your response, please write as legibly as you can!
● Quiz #1 graded back to you today.
- Please pick up yours during break.
- We gave 2 points to everyone to compensate for the confusion with
two of the questions.
- In the future, please ask any of teaching staff for clarification during
the quiz. That way, we can rectify on the spot.

3. 3
Quiz #1 results

4. 4
Outline of today’s lecture
2&4
Multi-qubit states ( )
1
2
10 10+
Multi-qubit gates
Multi-qubit measurement
mˆM
m
ˆM
1

5. 5
00
01
00 01 10 11
10
11
00 0 1 10 1 1
α
α
α α α α
α
α
 
 
 Ψ= + +
 
 
 
+ 
00 01 10 11, , , Cα α α α ∈
2 2 2 2
00 01 10 11 1α α α α+ + + =
Is there a simple, convenient geometric way to visualize
two-qubit states in analogy to the Bloch vector?
Not really.
Global phase is not relevant.
The state of two qubits
6 real parameters to fully specify 2-qubit state.

6. 6
Computational basis of 2 qubits
{ , , }0 10 0 1 , 10 1
A basis for the space of 2 qubits can be obtained as a composite
of individual qubit bases:
Examples:
the 2-qubit computational basis
(the most common)
{ , , , }++ −−+ −+ −
{ , , }0 1 ,0 1+ + − − the individual bases need not match.

7. 7
Quantum registers with n qubits
A collection of n qubits is called a quantum register of size n.
Conventions:
In a n=2 qubit register, the state “3” is
There are N=2n states of this kind, representing all binary strings of length n
or numbers from 0 to N -1.
3 11 11= ⊗ =
2 01 10= ⊗ =
and the state “2” as
number of states: N=2n
number of qubits: n
LSQMSQ

8. 8
n-qubit states
● How many real-valued coefficients does it take to specify the
quantum state of n qubits?
● Imagine you discretize these real coefficients using 4 bytes for
each. How much classical memory (in bytes) does it take to specify
the state of a register with n=100 bits?
Answer: 2(2 1)n
−
3
Answer: 2(2 1) 4 bytes 2 bytesn n+
− × ≈

9. 9
2-qubit superpositions
( ) ( )
1 1
2 3
2 2
1110+ = +
( )01 1
1
2
=⊗ +
( ) ( )
1 1
2 1
2 2
0110+ = +
( ) ( )
( ) ( )
1 1
0 1 2 3
2 2
1 1
0 1 0 1
2
0 0
0
2
1
1
1
0 1
+ + + = + + +
= + ⊗ +

10. 10
ϕ ψΨ = ⊗
Two qubits are entangled when their joint states cannot possibly
be separated into a product of individual qubit states
Some common terms:
Entangled = non-separable = non-product state
Unentangled = separable = product state
ϕψϕ ψΨ = ⊗ + ⊗ ′′vs
When are two qubits entangled?

11. 11
How to tell if two qubits are entangled or not
,0 1 ψα ψ βΨ ⊗ ⊗ ′= +
2 2
1α β+ =
You can always write the state of two qubits in the form
where , and are normalized,ψ ψ ′
Is ?ψ ψ= ′
no
entangled
yes
separable
( ) ( )
1 1 2 1
20 2
5 5 5 5
0 1 01 1= ⊗ + + ⊗ +
so not entangled!
( ) ( )0 1 0
1
1
1
2 2
5 5
= + ⊗ +
Example:
0 1
1 2 2 4
5 5 5
0 1
5
10 10+ + +

12. 12
How to tell if two qubits are entangled or not
2 2
1α β+ =
You can always write the state of two qubits in the form
where , and are normalized,ψ ψ ′
Is ?ψ ψ= ′
no
entangled
yes
separable
Note:
You can interchange the roles of red and green qubits if you like.
,0 1 ψα ψ βΨ ⊗ ⊗ ′= +

13. 13
( )
1
2
10 10−
( )0 1
1
2
10 0 1 10− + −
Quantifying 2-qubit entanglement
( ) 00 11 01 102C α α α α≡ −Ψ
00 01 10 110 10 0 10 11α α α αΨ= + + +
Two qubits in the state
are entangled if and only if (iff) they have nonzero concurrence
0C =
1C =
1C =( )0 1
1
2
10 0 1 10+ − +
( )0 0
3
0 1
1
1 1+ + 2 / 3C =

14. 14
The Bell states
( )
( )
( )
( )
0 1
0
0 1
0 1
1
2
1
2
1
2
1
2
1
1 00 1
11 00
+
−
+
−
Φ= +
Φ= −
Ψ= +
Ψ= −
These states form a basis for the 4-dimensional Hilbert space of 2 qubits.
This basis is called The Bell basis.
The singlet.
The even-parity Bell states.
The odd-parity Bell states.
C=1 for all these states.

15. 15
Definition: a transformation is unitary ifU † †
U U UU I= =
Ψ
Φ
′Ψ
′Φ
U
For unitary :U
†
U U′ ′ ′Φ Ψ = Φ Ψ = Φ Ψ
preservation of inner product
Also works in other direction, thus
U unitary iff inner product preserved.
U
Review: unitary transformations

16. 16
Catalog of 1-qubit gates
I
1 0ˆ
0 1
I
 
 
 
Identity
Z
1 0ˆ
0 1
Z
 
 − 
Pauli Z
0 1ˆ
1 0
X
 
 
 
XPauli X
0ˆ
0
i
Y
i
− 
 
 
YPauli Y
1 11ˆ
1 12
H
 
 − 
HHadamard
ˆ ˆ( ) cos( / 2) sin( / 2)nR I i nθ θ θ σ= − ⋅
 
( )nR θRotations
{ }ˆ ˆ ˆ, ,X Y Zσ =

/2
1 0ˆ
0 i
S
eπ
 
 
 
SS
/4
1 0ˆ
0 i
T
eπ
 
 
 
TT

17. 17
Products of single-qubit unitaries
V
U
Ψ V U⊗ Ψ
a b
V
c d
 
 
 

e f
U
g h
 
 
 

e f e f e f e f
g h g h g h g h
U
e f e fe f e f
g h g
a a b b
a b
a a b b
V
c c d d
c d
c c d hh g h dg
      
      
      
      
      
     
⊗  
in the composite basis.

18. 18
rg
1 0 0 0
0 1 0 0
0 0 0
C
1
0 0 1 0
-NOT
 
 
 
 
 
 

Controlled-NOT gate (C-NOT)
00
01
10
11
a
b b
a b⊕
2-qubit gates (Part 1)
00 11
gr
1 0 0 0
0 0 0 1
0 0 1
C
0
0 1 0 0
-NOT
 
 
 
 
 
 

a
b a b⊕
a
=
X
01 10
control
target

19. 19
How to make a Bell state
0 0 00
1 1 1 1
2 2 2 2
1 0 1
 
+ = + 
 
0
0 H
00 1
2
00
1 1
2
1+
How to make the other 3 Bell states?
?
? H
Starting from different
computational states produces
different Bell states!

20. 20
1 0 0 0
0 1 0 0
0 0 1
C PHAS
0
E
0
0 0 i
e
ϕ
ϕ
 
 
 
 
 
 
− 
b
a a
b
Controlled-Phase
A 2-qubit gates (Part 2)
=
Z
Z
=π
ϕ ( )i ab
eϕ
×
Note about convention: when the phase is not specified, it is π.

21. 21
H H
=
H H
=
Simplifying/realizing a quantum circuit
The process of simplifying a quantum circuit and/or finding a quantum circuit
that implements a particular unitary using a specific set of gates is called
quantum compiling.
Can compile C-NOT from CPHASE and
viceversa by ‘dressing’ with Hadamard
gates.
H Z H X=
H H I=
H X H Z=
To see this, refer to these identities:

22. 22
Universal set of gates
4.5.2
A set of gates that can be used to implement an arbitrary unitary operation on any number of
qubits is said to be universal.
An arbitrary operation on n qubits can be implemented using a circuit
containing O(n24n) single-qubit and C-NOT gates.
Examples:
,n θ

( )nR θ
(1) Arbitrary single-qubit rotations + C-NOT
any
(2) Arbitrary rotations around x and around y axes + C-NOT
θ
( )xR θ
any θ
( )yR θ
any

23. 23
Euler angle decomposition of single-qubit gates
( ) ( ) ( )ˆ ˆ ˆn m nU R R Rβ γ δ=
An arbitrary single-qubit gate may be written asU
up to an irrelevant global phase,
provided and are non-parallel unit vectors, and
for appropriate choices of .
ˆn ˆm
, ,β γ δ

24. 24
A universal discrete set of gates
T
C-NOT
1 0 0 0
0 1 0 0
C-NOT
0 0 0 1
0 0 1 0
 
 
 
 
 
 

A standard choice is:
Hadamard
1 11
1 12
H
 
 − 

/4
1 0
0 i
T
eπ
 
 
 

From just these three gates, you can do any unitary on n qubits
(Warning: it may require many of these three gates!)

25. 25
Rotations around non-parallel axes
( )nTHTH R θ= 
( )mHTHT R θ= 
cos ,sin ,cos
8 8 8
n
π π π       ∝       
      

cos , sin ,cos
8 8 8
m
π π π       ∝ −      
      

( )( )1 2
2cos cos / 8
0.174443 2
θ π
π
−
=
= ×
An irrational
multiple of 2π
x
y
z
Irrationality of theta is critical:
Can reach any desired rotation about
to any desired accuracy by repeating THTH
as many times as needed.
Can reach any desired rotation about
by repeating HTHT as many times as needed.
n

m


26. 26
Arbitrary rotations around either axis
For any , there exists an integer such thatε l
( ) ( )( ),
3
l
n nE R R
ε
δ θ < 
( )( , ) maxE U V U V
ψ
ψ≡ −
1
ε
1
( )ε −
Θ

27. 27
Solovay-Kitaev theorem
An arbitrary single-qubit gate may be approximated to an accuracy
using gates from a discrete set, where .ε
App.
3
1
(log ( ))c
ε −
Θ 2c ≈
Can do much better than that!
(Don’t worry about the proof!)

28. Coffee break.

29. 29
1 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0
0 0 1 0 0 0 0 0
0 0 0 1 0 0 0 0
0 0 0 0 1 0 0 0
0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 1
0 0 0 0 0 0 1 0
TOFFOLI
 
 
 
 
 
 
 
 
 
 
 
 

000 111
Controlled-Controlled-X (C-C-X)
X
Controlled-Controlled-Not (C-C-NOT)
…
Other names Other symbols
3-qubit gates: Toffoli gate
c
b
c
a bc⊕a
b
111
000
…

30. 30
H
=
H
S
T
†
T
T
†
T T †
T
†
T
Toffoli decomposition into 1- and 2-qubit gates
4
Claim:

31. 31
C-C-Phase decomposition into 1- and 2-qubit gates
4
00 01 10 1100 10 1 11 0α α α αψ ψ ψ ψΨ= ⊗ ′ ′′+ ⊗ + ⊗ + ⊗ ′′′
Without loss of generally, we can write
the input state of the three qubits as:
Let’s consider the action of this gate on each of the four terms:
= S
T
†
T
T
†
T T †
T
†
T
π
Claim:

32. 0
0
ψ
T
T†
T T †
T
S†
T
†
T
0
0
ψ
0
1
ψ ′
T
†
T T T
†
T
S†
T
†
T
X X
0
1
ψ ′
1
0
ψ ′′
T
T †
T
S†
T
†
T
X X
X X4
i
e T
π
−
/4
1i
e π+
/4
0i
e π−
ψ ′′T†
T
† 4
i
XT X e T
π
−
=

33. 1
1
ψ ′′′
T
T†
T T †
T
S†
T †
TX X
X XX X iZ−
/4
1i
e π+
/4
1i
e π+
iZ ψ− ′′′4
i
e T
π
−
4
i
e T
π
−
4
i
e T
π
−
1
1
Z ψ ′′′

34. 34
1 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0
0 0 1 0 0 0 0 0
0 0 0 1 0 0 0 0
0 0 0 0 1 0 0 0
0 0 0 0 0 1 0 0
0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 1
U
 
 
 
 
 
 
 
 
 
 
 
− 

It works!
C-C-Phase with 6 2-qubit C-NOTs

35. 35
Problem 4.4: (Minimal Toffoli construction) (Research)
(1) What is the smallest number of two qubit gates that can
be used to implement the Toffoli gate?
(2) What is the smallest number of one-qubit gates and C-NOT gates
that can be used to implement the Toffoli gate?
(3) What is the smallest number of one qubit gates and
controlled-Z gates that can be used to implement the Toffoli gate?
Lanyon et al., Nature Phys (2008)
Is this the most efficient decomposition?

36. 36Quant. Inf. Comp. 9:461 (2009)
It’s official!

37. 37
Summary of Part 1
You should be able to:
● Understand the concepts universal set of gates, and discrete set of universal gates.
● Easily determine whether a two-qubit state is entangled or not.
● Compile simple quantum circuits involving 1-qubit and CNOT/CPHASE gates.
● Analyze (not design) quantum circuits involving 1-qubit and CNOT/CPHASE gates.

38. 38
The generalized Born rule
Ψ
jλ
jm λ=
2
jα
jϕ
with probabilityˆM
ˆ jjjM λ λ λ=where
where form the basis for measurement,
are normalized but not necessarily orthogonal,
and .
j j j
j
α ϕ λΨ =∑
● Without loss of generality, can always write 2-qubit state
jϕ
2
1j
j
α =∑
jλ

39. 39
Example: measuring one qubit of two
( )1
2
0 1+ + −
jλ
jm λ=
jϕ
( )1
2
0 1+ + −
Z
( ) ( )1 1 1 1
2 2 2 2
0 1 00 110 1= + + −
( )1
2
0 1= + + −
( ) ( )1 1 1 1
2 2 2 2
0 0 11 110 0= + + −
1
2
p =
1
2
p =
1
1m = −
0
1m = +
+ −

40. 40
Example: measuring both qubits
jλ
1
ˆM 1 jm λ=
jλ ′
2
ˆM 2 jm λ ′′=
jϕ
2
ˆ jjjM λλ λ′ ′ ′=
j j j
j
α ϕ λΨ =∑

41. 41
Example: measuring both qubits
( )1
2
0 1+ + −
( )1
2
0 1+ + −
Z
( )1
2
0 1= + + −
X
1Bp =
0+
0Bp =
+
1Bm = + 1Bm = −
−
0−
1
2
Ap =
1
2
Ap =
+ −
0
1Am = + 1Am = −
1
0Bp =
1+
1Bp =
+
1Bm = + 1Bm = −
−
1−
Am
Bm

42. 42
Example: measuring both qubits
( )1
2
0 1+ + −
Z
( )1
2
0 1= + + −
X
1Ap =
0+
0Ap =
0
1Am = + 1Am = −
1
1+
1
2
Bp =
1
2
Bp =
0 1
+
1Bm = + 1Bm = −
−
0Ap =
0−
1Ap =
0
1Bm = + 1Bm = −
1
1−
Am
Bm
( )1
2
0 1+ + −

43. 43
Example: measuring both qubits
jλ
1
ˆM
Ψ 1 jm λ=
jλ ′
2
ˆM 2 jm λ ′′=
11
ˆ ˆIm M=Ψ Ψ
22
ˆ ˆMm I=Ψ Ψ
22 11
ˆ ˆm m M M=Ψ Ψ
(since order
doesn’t matter)
● As previous example shows, the statistics of measurement outcomes do not
depend on the order of measurements.
● Expectation values of measurement outcomes:

44. 44
Summary of Part 2
You should be able to:
● Apply the generalized Born rule to calculate probabilities of measurement
outcomes and the post-measurement states when some or all qubits in a register
are measured.

45. End of Tuesday’s lecture.