Exact synthesis of unitaries generated by Clifford and T gates

Exact Synthesis of Unitaries Generated by Cliﬀord+T
Gates
Yi-Hsin Ma
Apr 2019
Yi-Hsin Ma Exact Synthesis of Unitaries Generated by Cliﬀord+T Gates Apr 2019 1 / 57

Reference
Vadym Kliuchnikov, Dmitri Maslov and Michele Mosca. Fast and
efficient exact synthesis of single qubit unitaries generated by Clifford
and T gates
Brett Giles, Peter Selinger. Exact synthesis of multiqubit Clifford+T
circuits

Outline
1 Introduction
2 Exact Synthesis of Single Qubit Gates
3 Appendix A
4 Exact Synthesis of Multiple Qubits Gates
5 Appendix B

Introduction
Introduction

Introduction
Introduction
Unitaries Synthesis Problems.
Classiﬁed by number of qubits.
Single Qubit Unitary Decomposition
Multiple Qubit Unitary Decomposition
Classiﬁed by Error
Exact Decomposition
Approximate Decomposition
Solovay-Kitaev Algorithm

Introduction The Solovay-Kitaev Theorem
The Solovay-Kitaev Theorem
Deﬁnition 1.1
An instruction set G for an n-qubits is a ﬁnite set of quantum gates
satisfying:
1 All gates g ∈ G are in SU(2n)
2 For each g ∈ G, the inverse g† is also in G
3 G is a universal set for SU(2n), i.e. the group generated by G is dense
in SU(2n).

Definition 1.1
An instruction set G for an n-qubits is a finite set of quantum gates
satisfying:
1 All gates g ∈ G are in SU(2n)
2 For each g ∈ G, the inverse g† is also in G
3 G is a universal set for SU(2n), i.e. the group generated by G is dense
in SU(2n).
Definition 1.2
Suppose S and W are subsets of SU(2). Then S is said to form an -net
for W , where > 0, if every point in W is within a distance of some
point in S.

Let G := {g ...g1|gi ∈ G} be the set of SU(2) which are implementable
with G with length .
Theorem 1.3 (Solovay-Kitaev Theorem)
Let G be an instruction set for a single qubit. Let > 0 be given. Then G
is an -net in SU(2) for = O(logc(1/ )), where c ≈ 4

What Solovay-Kitaev Algorithm Do?
To compile an quantum algorithm into an eﬃcient fault-tolerance
form using Hadamard, T, and C-NOT gates.
To produce sequence of gate with length O(logc(1
)), which requires
time O(logd (1
)), c, d ∈ Z.

What Solovay-Kitaev Algorithm Do?
To compile an quantum algorithm into an eﬃcient fault-tolerance
form using Hadamard, T, and C-NOT gates.
To produce sequence of gate with length O(logc(1
)), which requires
time O(logd (1
)), c, d ∈ Z.
It does not guarantee
Finding an exact decomposition if there is one.
Whether the exact decomposition exists.

Exact Synthesis of Single Qubit Gates
Exact Synthesis of Single Qubit Gates

Exact Synthesis of Single Qubit Gates Preliminaries
Some algebra
Let
ω = e
2πi
8 =
1
√
2
+
i
√
2
H =
1
√
2
1 1
1 −1
, T =
1 0
0 ω

Some algebra
Let
ω = e
2πi
8 =
1
√
2
+
i
√
2
H =
1
√
2
1 1
1 −1
, T =
1 0
0 ω
Z[ω] := a + bω + cω2
+ dω3
a, b, c, d ∈ Z
Z
1
√
2
, i :=
∞
n=0
an
1
√
2
n
+ bn
1
√
2
n
i an, bn ∈ Z

The Main Theorem (Single Qubit)
Theorem 2.1
The set of 2 × 2 unitaries over the ring Z[ 1√
2
, i] is equivalent to the set of
those unitaries implementable exactly as single-qubit circuits constructed
using H and T gates only.
Note that ”⊇” is straightforward, since both H and T is over the ring
Z[ 1√
2
, i]. The other side is more diﬃcult to proof.

General Form of the Unitaries
Any unitary U can be represented as
U =
z −w∗eiφ
w z∗eiφ ,
where det(U) = eiφ should belong to Z[ 1√
2
, i], which gives that eiφ = ωk.

General Form of the Unitaries
Any unitary U can be represented as
U =
z −w∗eiφ
w z∗eiφ ,
where det(U) = eiφ should belong to Z[ 1√
2
, i], which gives that eiφ = ωk.
Hence we have the general form of the matrix we interested as
U =
z −w∗ωk
w z∗ωk

Reducing Unitary Implementation to State Preparation
Goal: Find the implementation of unitary matrix U using only H and T
gates.

gates.
First, we reduce the problem into state preparation, which means to find
Û = U ...U1 such that Û|0 =
z
w
, where U1, ..., U ∈ {H, T}.

gates.
First, we reduce the problem into state preparation, which means to find
Û = U ...U1 such that Û|0 =
z
w
, where U1, ..., U ∈ {H, T}.
Since Û ∈ U(2) ∩ M2×2(Z[ 1√
2
, i]), let Û =
z −w∗ωk
w z∗ωk .
Hence, we can multiply Û by a power of T to get U
ÛTk−k
=
z −w∗ωk
w z∗ωk Tk−k
=
z −w∗ωk
w z∗ωk = U

Reducing unitary implementation to state preparation
Lemma 2.2
Any single-qubit state with entries in the ring Z[ 1√
2
, i] can be prepared
using only H and T gates given the initial state |0 .

Reducing unitary implementation to state preparation
Lemma 2.2
Any single-qubit state with entries in the ring Z[ 1√
2
, i] can be prepared
using only H and T gates given the initial state |0 .
Observation
Power of
√
2 in the
denominator of the entries is
the same.
Power of
√
2 in the
denominator of |zn|2 increases
by 1 after multiplication by
HT.

Smallest Denominator Exponent
Deﬁnition 2.3 (smallest denominator exponent)
The smallest denominator exponent, sde(z) of z ∈ Z[ 1√
2
, i] is the smallest
integer value k such that z
√
2
k
∈ Z[ω]. If there is no such k, the smallest
denominator exponent is inﬁnite.

Smallest Denominator Exponent
Deﬁnition 2.3 (smallest denominator exponent)
The smallest denominator exponent, sde(z) of z ∈ Z[ 1√
2
, i] is the smallest
integer value k such that z
√
2
k
∈ Z[ω]. If there is no such k, the smallest
denominator exponent is inﬁnite.
Example
For z1 = 1
√
2
5 (ω3 + 1), sde(z1) = 5.
For z2 = 1
√
2
5 (ω3 − ω) = − 1
√
2
4 , sde(z2) = 4.

Greatest Denominator Exponent
Deﬁnition 2.4 (greatest dividing exponent)
The greatest dividing exponent, gde(z, x), of base x ∈ Z[ω] with respect
to z ∈ Z[ω] is the integer value k such that xk divides z and x does not
divide quotient z
xk . If no such k exists, the greatest dividing exponent is
said to be inﬁnite.

Greatest Denominator Exponent
Deﬁnition 2.4 (greatest dividing exponent)
The greatest dividing exponent, gde(z, x), of base x ∈ Z[ω] with respect
to z ∈ Z[ω] is the integer value k such that xk divides z and x does not
divide quotient z
xk . If no such k exists, the greatest dividing exponent is
said to be inﬁnite.
Example
For z1 =
√
2
5
ω, gde(z1,
√
2) = 5.
For z2 =
√
2
4
(ω + ω3) =
√
2
4
(
√
2i), gde(z2,
√
2) = 5.

Exact Synthesis of Single Qubit Gates Some Important Lemmas
Some Important Lemmas
Lemma 2.5
Let u = z
w be a state with entries in Z[ 1√
2
, i] and sde(|z|2) ≥ 4. Then
there exits integer k ∈ {0, 1, 2, 3} such that:
sde(|(HTk
u)1|2
) = sde(|z|2
) − 1

Exact Synthesis of Single Qubit Gates Some Important Lemmas
Some Important Lemmas
Lemma 2.5
Let u = z
w be a state with entries in Z[ 1√
2
, i] and sde(|z|2) ≥ 4. Then
there exits integer k ∈ {0, 1, 2, 3} such that:
sde(|(HTk
u)1|2
) = sde(|z|2
) − 1
Lemma 2.6
Let u = z
w be a state over Z[ 1√
2
, i] such that sde(|z|2) ≥ 1 or
sde(|w|2) ≥ 1, then
sde(|z|2
) = sde(|w|2
), sde(z) = sde(w)

Exact Synthesis of Single Qubit Gates The Algorithm for Decomposition
Idea of Algorithm
Lemma 2.6 implies that sde(|(HTku)2|2) = sde(|w|2) − 1 too.

Idea of Algorithm
Both entries of the state have the same sde(| · |2), and their sde(| · |2)
decrease by 1 each time we apply HTk on condition that sde(|z|2) ≥ 4.

Idea of Algorithm
Both entries of the state have the same sde(| · |2), and their sde(| · |2)
decrease by 1 each time we apply HTk on condition that sde(|z|2) ≥ 4.
Iterating such procedure, at the end we can ﬁnd k1, ..., ks−3 ∈ {0, 1, 2, 3},
where s = sde(|z|2) such that
HTks−3
...HTk1
z
w
=
zs−3
ws−3
, sde(|zs−3|2
) = 3

Idea of Algorithm
Define sde|·|2
( Û) = sde(|u|2) where u is an entry of Û.

Idea of Algorithm
Define sde|·|2
Since the set S3 := { Û ∈ U(2)|sde|·|2
( Û) ≤ 3} is finite and small.
It takes little efforts to find Û ∈ S3 where its first column equals
zn−3
wn−3
.

Idea of Algorithm
Define sde|·|2
Since the set S3 := { Û ∈ U(2)|sde|·|2
( Û) ≤ 3} is finite and small.
It takes little efforts to find Û ∈ S3 where its first column equals
zn−3
wn−3
.
Hence,
zn−3
wn−3
= HTkn−3
...HTk1
z
w
= Û|0 ⇒ T−k1
H...T−kn−3
H Û|0 =
z
w
U = T−k1
H...T−kn−3
H ÛTk−k

Algorithm 1 Decomposition of a unitary matrix over Z[ 1√
2
, i]
Input: Unitary U =
z00 z01
z10 z11
over Z[ 1√
2
, i].
Output: Sequence Sout of H and T gates that implements U.
1: //S3 : set of all unitaries over Z[ 1√
2
, i] such that for all ˆU ∈ S3, sde|·|( ˆU) ≤ 3
2: queue Sout ← ∅
3: s ← sde(|z00|2)
4: while s > 3 do
5: state ← unfound
6: for all k ∈ {0, 1, 2, 3} do
7: while state = unfound do
8: z00 = [HT−k U]00
9: if sde(|z00|2) = s − 1 then
10: state ← found
11: add Tk H to Sout
12: s ← sde(|z00|2)
13: U ← HT−k U
14: end if
15: end while
16: end for
17: end while
18: lookup sequence Srem for U in S3
19: add Srem into Sout
20: return Sout

Appendix A
Appendix A

Appendix A
Appendix
The goal of the appendix is to show that if eiφ ∈ Z[ 1√
2
, i] and |eiφ|2 = 1,
then eiφ = ωk where k ∈ {0, 1, 2, · · · , 7}.

Appendix A
Appendix
2
, i] and |eiφ|2 = 1,
Let z = eiφ and z
√
2
sde(z)
= x ∈ Z[ω],
x can be expressed as x = x0 + x1ω + x2ω2 + x3ω3, where x0, x1, x2, x3 ∈ Z.

Appendix A
Appendix
2
, i] and |eiφ|2 = 1,
Let z = eiφ and z
√
2
sde(z)
= x ∈ Z[ω],
x can be expressed as x = x0 + x1ω + x2ω2 + x3ω3, where x0, x1, x2, x3 ∈ Z.
We can further expressed |x|2 = P(x) +
√
2Q(x), and P(x), Q(x) are:
P(x) := x2
0 + x2
1 + x2
2 + x2
3 ,
Q(x) := x0(x1 − x3) + x2(x1 + x3).

Appendix A
Appendix
Proposition 3.1
Let x ∈ Z[ω]. If gde(x,
√
2) = 0, then P(x), Q(x) have diﬀerent parity.
i.e. if one of P(x) or Q(x) is even, then another must be odd.
Moreover, gde(|x|2, 2) = 0.

Appendix A
Appendix
Proof.
gde(x,
√
2) = 0 implies gde(
√
2x, 2) = 0. Let x = x0 + x1ω + x2ω2 + x3ω3
⇒
√
2x = (x1 − x3) + (x0 + x2)ω + (x1 + x3)ω3 + (x2 − x0)ω3
Hence, at least x1 − x3 or x2 − x0 is odd.

Appendix A
Appendix
Proof.
gde(x,
√
2) = 0 implies gde(
√
2x, 2) = 0. Let x = x0 + x1ω + x2ω2 + x3ω3
⇒
√
2x = (x1 − x3) + (x0 + x2)ω + (x1 + x3)ω3 + (x2 − x0)ω3
WLOG. Let x1 − x3 is odd. For P(x) and Q(x) modulo two, we have
P(x) = (x1 + x3) + (x0 + x2) mod 2
Q(x) = (x1 + x3)(x0 + x2) mod 2
We conclude that P(x) and Q(x) have diﬀerent parity.

Appendix A
Appendix
Proof.
gde(x,
√
2) = 0 implies gde(
√
2x, 2) = 0. Let x = x0 + x1ω + x2ω2 + x3ω3
⇒
√
2x = (x1 − x3) + (x0 + x2)ω + (x1 + x3)ω3 + (x2 − x0)ω3
WLOG. Let x1 − x3 is odd. For P(x) and Q(x) modulo two, we have
P(x) = (x1 + x3) + (x0 + x2) mod 2
Q(x) = (x1 + x3)(x0 + x2) mod 2
We conclude that P(x) and Q(x) have diﬀerent parity.
The above immediately implies gde(|x|2, 2) = 0.

Appendix A
Appendix
For z = eiφ ∈ Z[ 1√
2
, i], let z = x
(
√
2)k
where k = sde(z).
|z|2 = 1 implies |x|2 = P(x) +
√
2Q(x) = 2k.
⇒ |x|2 =
1, if k = 0
0, otherwise
mod 2.

Appendix A
Appendix
2
, i], let z = x
(
√
2)k
where k = sde(z).
|z|2 = 1 implies |x|2 = P(x) +
√
2Q(x) = 2k.
⇒ |x|2 =
1, if k = 0
0, otherwise
mod 2.
Assume k > 0, then by Prop 3.1 ⇒ gde(|x|2, 2) = 0.
Which means |x|2 = 0 mod 2. Contradiction.

Appendix A
Appendix
2
, i], let z = x
(
√
2)k
where k = sde(z).
|z|2 = 1 implies |x|2 = P(x) +
√
2Q(x) = 2k.
⇒ |x|2 =
1, if k = 0
0, otherwise
mod 2.
Assume k > 0, then by Prop 3.1 ⇒ gde(|x|2, 2) = 0.
Which means |x|2 = 0 mod 2. Contradiction.
Hence k = 0 ⇒ z = eiφ ∈ Z[ω]. Let z = z0 + z1ω + z2ω2 + z3ω3.
⇒ |z|2 = (z2
0 + z2
1 + z2
2 + z2
3 ) +
√
2(z1z0 + z2z1 + z3z2 − z0z3) = 1
Which implies z = ωk, k ∈ {0, 1, · · · , 7}.

Exact Synthesis of Multiple Qubits Gates

The Main Theorem (Multiple Qubits)
Theorem 4.1
For n > 1, the set of 2n × 2n unitaries over Z[ 1√
2
, i] is equivalent to the
set of unitaries implementable exactly as circuits with Cliﬀord and T gates
built using (n+1) qubits, where the last qubit is an ancillary qubit which
is set to |0 at the beginning and is required to be |0 at the end.

The Ancillary Qubit May be Necessary
Consider, for example the controlled-T gates.




1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 ω




which has determinant ω.

The Ancillary Qubit May be Necessary
Consider, for example the controlled-T gates.




1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 ω




which has determinant ω.
However, any Cliﬀord gate and T gate viewed as matrices over a set of
two qubits has a determinant that is a power of the imaginary number i.
Hence it is impossible to implement the controlled-T without adding an
ancillay qubit.

Exact Synthesis of Multiple Qubits Gates Preliminaries
Cliﬀord Group
Pauli matrices
I =
1 0
0 1
, X =
0 1
1 0
, Y =
0 −i
i 0
, Z =
1 0
0 −1
.
The Pauli group on n qubits is
Pn = {±1, ±i} × {I, X, Y , Z}⊗n

Cliﬀord Group
Pauli matrices
I =
1 0
0 1
, X =
0 1
1 0
, Y =
0 −i
i 0
, Z =
1 0
0 −1
.
The Pauli group on n qubits is
Pn = {±1, ±i} × {I, X, Y , Z}⊗n
The Cliﬀord group on n qubits is the set of unitaries that normalize the
Pauli group.
Cn = {U : UPnU†
= Pn}

Cliﬀord Group
Theorem 4.2
Cn = Hi , Pi , CNOTi,j /U(1), where
H =
1
√
2
1 1
1 −1
, P =
1 0
0 i
, CNOT =




1 0 0 0
0 1 0 0
0 0 0 1
0 0 1 0





One- and Two- Level Matrix
Let U is a 2 × 2-matrix
U =
a b
c d
A two-level matrix U[j,l] of type U is a n × n-matrix, where j = l is
deﬁned by
U[j,l] =






I
a b
I
c d
I







Let U is a 2 × 2-matrix
U =
a b
c d
A two-level matrix U[j,l] of type U is a n × n-matrix, where j = l is
deﬁned by
U[j,l] =






I
a b
I
c d
I






If a is a scalar, a[j] for the one-level matrix
a[j] =


I
a
I



Each of one- and two-level matrices of types X, H, T, and ω can be
further decomposed into controlled-not gates and multiply-controlled X,
H, T, and ω-gates.

Each of one- and two-level matrices of types X, H, T, and ω can be
further decomposed into controlled-not gates and multiply-controlled X,
H, T, and ω-gates.
We ﬁrst decompose the unitary using one- and two-level matrix of type H,
T, X, and ω.

An Example for Unitary Decomposition
Consider the matrix
U =
1
√
2
3




−ω3 + ω − 1 ω2 + ω + 1 ω2 −ω
ω2 + ω −ω3 + ω2 −ω2 − 1 ω3 + ω
ω3 + ω2 −ω3 − 1 2ω2 0
−1 ω 1 −ω3




which is unitary and over Z[ 1√
2
, i].

Decomposition into Two-Level Matrices
Lemma 4.3 (Matrix Decomposition)
Let U be a unitary n × n-matrix with entries in D[ω]. Then there exists a
sequence U1, ..., Uh of one- and two-level unitary matrices of types
X, H, T, and ω such that U = U1...Uh.

Proof.
It suﬃces to show that there exist one- and two-level unitary matrices
V1, . . . , Vh of types X, H, T, and ω such that Vh · · · V1U = I.

Proof.
It suffices to show that there exist one- and two-level unitary matrices
V1, . . . , Vh of types X, H, T, and ω such that Vh · · · V1U = I.
First use the column lemma (Lemma4.10) to find V1, . . . , Vh1 such that
the leftmost column of Vh1 · · · V1U is e1. Because Vh1 · · · V1U is unitary, it
is of the form
1 0
0 U
Now recursively find row operations to reduce U to the identity matrix.

Proof of Theorem4.1
Theorem 4.1
2

Proof of Theorem4.1
Theorem 4.1
2
Proof.
(⊇) This way is trivial.

Proof of Theorem4.1
Theorem 4.1
2
Proof.
(⊆) Let U is a unitary 2 × 2 matrix over D[ω]. By Lemma4.3(Matrix
Decomposition) U can be decomposed into one- and two-level matrices of
types X, H, T and ω.

Proof of Theorem4.1
Theorem 4.1
2
Proof.
(⊆) Let U is a unitary 2 × 2 matrix over D[ω]. By Lemma4.3(Matrix
Decomposition) U can be decomposed into one- and two-level matrices of
types X, H, T and ω.
Each of such matrix can be further decomposed into controlled-not gates
and multiply-controlled X, H, T and ω-gates. These gates have
well-known exact representations in Cliﬀord+T with ancillas.
This ﬁnishes the proof.

Preliminaries
Let
D := Z
1
2
=
a
2n
a ∈ Z, n ∈ N
D[ω] := a + bω + cω2
+ dω3
a, b, c, d ∈ D
It is easy to show that D[ω] = Z 1√
2
, i .

Preliminaries
Deﬁnition 4.4 (Residue Map)
The residue map ρ is a surjective ring homomorphism: Z[ω] → Z2[ω],
deﬁned by
ρ(a + bω + cω2
+ dω3
) = ¯a + ¯bω + ¯cω2
+ ¯dω3
where ¯(·) is the parity function, ¯a = a mod 2.

Preliminaries
Definition 4.4 (Residue Map)
The residue map ρ is a surjective ring homomorphism: Z[ω] → Z2[ω],
defined by
ρ(a + bω + cω2
+ dω3
) = ¯a + ¯bω + ¯cω2
+ ¯dω3
where ¯(·) is the parity function, ¯a = a mod 2.
Definition 4.5 (k-residue)
Let t ∈ D[ω], k ∈ N such that
√
2
k
t ∈ Z[ω].
(We call k is a denominator exponent for t)
The k-residue of t, denoted by ρk(t) is defined to be
ρk(t) = ρ(
√
2
k
t)

Preliminaries
Deﬁnition 4.6 (Reducibility)
x ∈ Z2[ω] is reducible if it is of the form
√
2y, for some y ∈ Z2[ω].
Moreover, we say that x is twice reducible iﬀ x = 0000.

Preliminaries
Deﬁnition 4.6 (Reducibility)
x ∈ Z2[ω] is reducible if it is of the form
√
2y, for some y ∈ Z2[ω].
Moreover, we say that x is twice reducible iﬀ x = 0000.
Lemma 4.7
For x ∈ Z2[ω], these are equivalent:
(I) x is reducible;
(II) x ∈ {0000, 0101, 1010, 1111};
(III)
√
2x = 0000;
(IV) x†x = 0000;

Table 2. Some operation on residues

Exact Synthesis of Multiple Qubits Gates Decomposition into Two-Level Matrices
Row Operation
Lemma 4.8 (Row Operation)
Let u1, u2 ∈ D[ω] be entries of a column. Let u = (u1, u2)T is of
denominator exponent k > 0 and k-residue ρk(u) = (x1, x2), such that
x†
1x1 = x†
2x2. Then there exist a sequence of matrices U1, ..., Uh, each of
which is H or T, such that v = U1...Uhu has denominator exponent k − 1,
or equivalently, ρk(v) is deﬁned and reducible.

Row Operation
Lemma 4.8 (Row Operation)
Let u1, u2 ∈ D[ω] be entries of a column. Let u = (u1, u2)T is of
denominator exponent k > 0 and k-residue ρk(u) = (x1, x2), such that
x†
1x1 = x†
2x2. Then there exist a sequence of matrices U1, ..., Uh, each of
which is H or T, such that v = U1...Uhu has denominator exponent k − 1,
or equivalently, ρk(v) is deﬁned and reducible.
This lemma is used to ﬁnd a sequence of two-level matrices of type H and
T such that after applying the sequence to a column, the power of
√
2 at
the denominator can vanish.

Row Operation
Proof.
It can be seen from the table that x†
1x1 is either 0000, 1010, or 0001.
Case 1: x†
1x1 = x†
2x2 = 0000. In this case, ρk(u) is already reducible.

Row Operation
Proof.
1x1 is either 0000, 1010, or 0001.
Case 1: x†
1x1 = x†
Case 2: x†
1x1 = x†
2x2 = 1010. From Table 2, we know
x1, x2 ∈ {0011, 0110, 1100, 1001}. Then there exists m ∈ {0, 1, 2, 3}
such that x1 = ωmx2.

Row Operation
Proof.
1x1 is either 0000, 1010, or 0001.
Case 1: x†
1x1 = x†
Case 2: x†
1x1 = x†
2x2 = 1010. From Table 2, we know
x1, x2 ∈ {0011, 0110, 1100, 1001}. Then there exists m ∈ {0, 1, 2, 3}
such that x1 = ωmx2.
Let v = HTmu. Then
ρk(
√
2v) = ρk
1 1
1 −1
1 0
0 ωm
u1
u2
= ρk
u1 + ωmu2
u1 − ωmu2
=
x1 + ωmx2
x1 − ωmx2
=
0000
0000
⇒ ρk(
√
2v) is trice reducible⇒ ρk(v) is reducible.

Row Operation
Proof (Cont.)
Case 3: x†
1x1 = x†
2x2 = 1010. From Table 2, x1, x2 ∈ A B, where
A = {0001, 0010, 0100, 1000}, B = {1110, 1101, 1011, 0111}.

Row Operation
Proof (Cont.)
Case 3: x†
1x1 = x†
A = {0001, 0010, 0100, 1000}, B = {1110, 1101, 1011, 0111}.
There exists an m such that x1 + ωmx2 = 1111. Let u = HTmu.
ρk(
√
2u ) = ρk
u1 + ωmu2
u1 − ωmu2
=
x1 + ωmx2
x1 − ωmx2
=
1111
1111

Row Operation
Proof (Cont.)
Case 3: x†
1x1 = x†
A = {0001, 0010, 0100, 1000}, B = {1110, 1101, 1011, 0111}.
ρk(
√
2u ) = ρk
u1 + ωmu2
u1 − ωmu2
=
x1 + ωmx2
x1 − ωmx2
=
1111
1111
By Lemma4.7 this is reducible ⇒ u has denominator exponent k.
Let ρk(u ) = (y1, y2). Because
√
2 y1 =
√
2 y2 = 1111,
we see from Table 2 that y1, y2 ∈ {0011, 0110, 1100, 1001} and
y†
1 y1 = y†
2 y2 = 1010. Therefore, u satisﬁes Case 2.

Row Operation
Proof (Cont.)
Case 3: x†
1x1 = x†
A = {0001, 0010, 0100, 1000}, B = {1110, 1101, 1011, 0111}.
ρk(
√
2u ) = ρk
u1 + ωmu2
u1 − ωmu2
=
x1 + ωmx2
x1 − ωmx2
=
1111
1111
By Lemma4.7 this is reducible ⇒ u has denominator exponent k.
Let ρk(u ) = (y1, y2). Because
√
2 y1 =
√
2 y2 = 1111,
we see from Table 2 that y1, y2 ∈ {0011, 0110, 1100, 1001} and
y†
1 y1 = y†
2 y2 = 1010. Therefore, u satisﬁes Case 2.
Proceeding as in Case 2, we ﬁnd m such that
v = HTm u = HTm HTmu has denominator exponent k − 1.

Column Lemma
Lemma 4.9
Consider a vector u ∈ Z[ω]n. If u 2 = 1, then
ui =
ωk, if i = j
0, otherwise
Proof.
Let ui = ai ω3 + bi ω2 + ci ω + di ⇒ u 2 = i ui
2 = i u†
i ui
= i a2
i + b2
i + c2
i + d2
i + (ci di + bi ci + ai bi − di ai )
√
2
Since 1 is an integer ⇒ i (ci di + bi ci + ai bi − di ai ) = 0
⇒ i a2
i + b2
i + c2
i + d2
i = 1

Column Lemma
Lemma 4.10 (Column Lemma)
Consider an unit vector u ∈ D[ω]n. Then there exist a sequence U1...Uh of
one- and two-level unitary matrices of types X, H, T, and ω such that
U1...Uhu = e1, the ﬁrst standard basis vector.

Column Lemma
Lemma 4.10 (Column Lemma)
Consider an unit vector u ∈ D[ω]n. Then there exist a sequence U1...Uh of
one- and two-level unitary matrices of types X, H, T, and ω such that
U1...Uhu = e1, the ﬁrst standard basis vector.
The Column Lemma is used for reducing a column into e1.

Column Lemma
Proof.
By induction. Let k = maxi (sde(ui )), u = (u1, ..., un)T .

Column Lemma
Proof.
Base case: k = 0. ⇒ u ∈ Z[ω]n. We know u = 1, hence by
Lemma4.9 only one entry uj of u is equal to ω−m, otherwise 0.

Column Lemma
Proof.
Let u = X[1,j]u if j = 1 otherwise u = u. ⇒ u = (ω−m, 0, ..., 0)T .
We have ωm
[1]u = e1.

Column Lemma
Proof.
We have ωm
[1]u = e1.
Induction Case: k > 0. ⇒ Let v =
√
2
k
u ∈ Z[ω]n, x = ρk(u) = ρ(v).
Since u 2 = 1 ⇒ v 2 = i v†
i vi = 2k ⇒ i x†
i xi = 0000.
⇒ There are an even number of j such that x†
j xj = 0001, 1010
respectively.

Column Lemma
Proof.
We have ωm
[1]u = e1.
Induction Case: k > 0. ⇒ Let v =
√
2
k
u ∈ Z[ω]n, x = ρk(u) = ρ(v).
Since u 2 = 1 ⇒ v 2 = i v†
i vi = 2k ⇒ i x†
i xi = 0000.
⇒ There are an even number of j such that x†
j xj = 0001, 1010
respectively.
Apply Lemma4.8(Row operation), then we can ﬁnd a sequence U over
one-, two-level unitary matrices of H and T such that Uu is reducible.

Now, We are ready to demonstrate the decomposition of a unitary over
Z[ 1√
2
, i] using the techniques shown in the proofs of previous lemmas.
Consider the matrix
U =
1
√
2
3




−ω3 + ω − 1 ω2 + ω + 1 ω2 −ω
ω2 + ω −ω3 + ω2 −ω2 − 1 ω3 + ω
ω3 + ω2 −ω3 − 1 2ω2 0
−1 ω 1 −ω3





It has denominator exponent 3. Its 3-, 4-, and 5-residues are:
ρ3(U) =




1011 0111 0100 0010
0110 1100 0101 1010
1100 1001 0000 0000
0001 0010 0001 1000





ρ3(U) =




1011 0111 0100 0010
0110 1100 0101 1010
1100 1001 0000 0000
0001 0010 0001 1000




ρ4(U) =




1010 0101 1010 0101
1111 1111 0000 0000
1111 1111 0000 0000
1010 0101 1010 0101





ρ3(U) =




1011 0111 0100 0010
0110 1100 0101 1010
1100 1001 0000 0000
0001 0010 0001 1000




ρ4(U) =




1010 0101 1010 0101
1111 1111 0000 0000
1111 1111 0000 0000
1010 0101 1010 0101




ρ5(U) = 0

We start with the ﬁrst column u of U:
u =
1
√
2
3




−ω3 + ω − 1
ω2 + ω
ω3 + ω2
−1




ρ3(u) =




1011
0110
1100
0001



 =




x1
x2
x3
x4









x†
1x1
x†
2x2
x†
3x3
x†
4x4





=




0001
1010
1010
0001




.

u =
1
√
2
3




−ω3 + ω − 1
ω2 + ω
ω3 + ω2
−1




ρ3(u) =




1011
0110
1100
0001



 =




x1
x2
x3
x4









x†
1x1
x†
2x2
x†
3x3
x†
4x4





=




0001
1010
1010
0001




.
x2 and x3 satisfy case 2 of Lemma 4.8.⇒ Apply H[2,3]T3
[2,3]

u =
1
√
2
3




−ω3 + ω − 1
ω2 + ω
ω3 + ω2
−1




ρ3(u) =




1011
0110
1100
0001



 =




x1
x2
x3
x4









x†
1x1
x†
2x2
x†
3x3
x†
4x4





=




0001
1010
1010
0001




.
x2 and x3 satisfy case 2 of Lemma 4.8.⇒ Apply H[2,3]T3
[2,3]
x1 and x4 satisfy case 3 of Lemma 4.8.⇒ Apply H[1,4]T[1,4]H[1,4]T2
[1,4]

Apply V = H[1,4]T[1,4]H[1,4]T2
[1,4]H[2,3]T3
[2,3] to u

Apply V = H[1,4]T[1,4]H[1,4]T2
[1,4]H[2,3]T3
[2,3] to u
⇒ v(1) = Vu = 1
√
2
2




0
0
ω2 + ω
−ω + 1



 , ρ2(v(1)) =




0000
0000
0110
0011



 ⇒

Apply V = H[1,4]T[1,4]H[1,4]T2
[1,4]H[2,3]T3
[2,3] to u
⇒ v(1) = Vu = 1
√
2
2




0
0
ω2 + ω
−ω + 1



 , ρ2(v(1)) =




0000
0000
0110
0011



 ⇒
Apply H[3,4]T[3,4] to v(1) ⇒ v(2) = H[3,4]T[3,4]v(1) = 1√
2




0
0
ω
ω2





Apply V = H[1,4]T[1,4]H[1,4]T2
[1,4]H[2,3]T3
[2,3] to u
⇒ v(1) = Vu = 1
√
2
2




0
0
ω2 + ω
−ω + 1



 , ρ2(v(1)) =




0000
0000
0110
0011



 ⇒
Apply H[3,4]T[3,4] to v(1) ⇒ v(2) = H[3,4]T[3,4]v(1) = 1√
2




0
0
ω
ω2




Apply H[3,4]T3
[3,4] to v(2) ⇒ v(3) = H[3,4]T3
[3,4]v(2) =




0
0
0
ω





Apply ω7
[1]X[1,4] to v(3) ⇒ v(4) = ω7
[1]X[1,4]v(3) =




1
0
0
0





Apply ω7
[1]X[1,4] to v(3) ⇒ v(4) = ω7
[1]X[1,4]v(3) =




1
0
0
0




With W1 = ω7
[1]X[1,4]H[3,4]T[3,4]H[1,4]T[1,4]H[1,4]T2
[1,4]H[2,3]T3
[2,3]
and apply W1 to U gives:
W1U =
1
√
2
3




√
2
3
0 0 0
0 ω3 − ω2 + ω + 1 −ω2 − ω − 1 ω2
0 0 ω3 + ω2 − ω + 1 ω3 + ω2 − ω − 1
0 ω3 + ω2 + ω + 1 ω2 ω3 − ω2 + 1





Continue with the rest of the columns, we ﬁnd
W2 = ω6
[2]H[2,4]T3
[2,4]H[2,4]T[2,4]
W3 = ω4
[3]H[3,4]T3
[3,4]H[3,4], W4 = ω5
[4]
We then have U = W †
1 W †
2 W †
3 W †
4

Exact Synthesis of Multiple Qubits Gates The No-Ancilla Case
The No-Ancilla Case
Theorem 4.11
Under the hypotheses of Theorem 4.1, assume that det U = 1. Then U
can be exactly represented by a Cliﬀord+T circuit with no ancillas.
W =
ω 0
0 ω−1

The No-Ancilla Case
Proof.
In order to decompose the unitary without the use of ancillary qubits. We
use the two-level matrices of types iX, T−m(iH)Tm, W and one-level
matrix of type ωm to implement the unitary.
Note that these matrices can be further decomposed into Cliﬀord and T
gates without ancillary qubits.

The No-Ancilla Case
Proof.
Replacing HTm with T−m(iH)Tm in the procedure we do in Lemma4.8.
We have
√
2T−m
(iH)Tm u1
u2
= ω2 u1 + u2ωm
u1ω−m − u2
We further substitute iX[i,j] for X[i,j], W[k,k+1] for ω[k] where k < 2n.

The No-Ancilla Case
Proof.
Replacing HTm with T−m(iH)Tm in the procedure we do in Lemma4.8.
We have
√
2T−m
(iH)Tm u1
u2
= ω2 u1 + u2ωm
u1ω−m − u2
We further substitute iX[i,j] for X[i,j], W[k,k+1] for ω[k] where k < 2n.
For k = 2n we must use ω[2n] since we cannot use W[2n,2n+1]. Because the
unitary U as well as the matrices we use to implement are all of
determinant 1, we have det(ωm
[2n]) = 1. Which means it is no need to use
any ω[2n] gates.

The No-Ancilla Case
Corollary 4.12
Let U be a unitary 2n × 2n matrix. Then the following are equivalent:
(a) U can be exactly represented by a quantum circuit over the
Cliﬀord+T gate set on n qubits with no ancillas.
(b) The entries of U belong to the ring
det U = 1, if n ≥ 4;
det U ∈ {−1, 1}, if n = 3;
det U ∈ {i, −1, −i, 1}, if n = 2;
det U ∈ {ω, i, ω3
, −1, ω5
, −i, ω7
, 1}, if n = 1.

Appendix B
Appendix B

Appendix B
Appendix B
Objective To show that any arbitrary two-level matrix of type U can be
built from single qubit and CNOT gates.

Appendix B
Appendix B
Objective To show that any arbitrary two-level matrix of type U can be
built from single qubit and CNOT gates.
Example We wish to implement the two-level matrix U of type ˜U
U =












a 0 0 0 0 0 0 c
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
b 0 0 0 0 0 0 d












Where ˜U =
a c
b d
is a unitary matrix.

Appendix B
Appendix B
In this example, only states |000 and |111 is aﬀected by U.
We use Gray code to connect these two states

Exact synthesis of unitaries generated by Clifford and T gates

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