Unordered Tuples
in Quantum Computation
Robert Furber
[email protected]
Bas Westerbaan
[email protected]
Radboud Universiteit Nijmegen
July 15, 2015
What we did
What we did
Computed algebras for several unordered
quantum types. (eg. unordered pair, cycles)
What we did
Computed algebras for several unordered
quantum types. (eg. unordered pair, cycles)
using representation theory of finite groups.
What we did
Computed algebras for several unordered
quantum types. (eg. unordered pair, cycles)
using representation theory of finite groups.
(After discussing paper of Pagani, Selinger, Valiron with Sam Staton.)
The heavy lifting
The heavy lifting
Schur
Weyl
Quantum types as algebras
type
algebra
Quantum types as algebras
type
qubit
algebra
Quantum types as algebras
type
qubit
algebra
M2 ∼
= B(C2)
Quantum types as algebras
type
qubit
bit
algebra
M2 ∼
= B(C2)
Quantum types as algebras
type
qubit
bit
algebra
M2 ∼
= B(C2)
C2
Quantum types as algebras
type
qubit
bit
(ordered) pair of bits
algebra
M2 ∼
= B(C2)
C2
Quantum types as algebras
type
qubit
bit
(ordered) pair of bits
algebra
M2 ∼
= B(C2)
C2
C 2 ⊗ C2 ∼
= C4
Quantum types as algebras
type
qubit
bit
(ordered) pair of bits
(ordered) pair of qubits
algebra
M2 ∼
= B(C2)
C2
C 2 ⊗ C2 ∼
= C4
Quantum types as algebras
type
qubit
bit
(ordered) pair of bits
(ordered) pair of qubits
algebra
M2 ∼
= B(C2)
C2
C 2 ⊗ C2 ∼
= C4
M2 ⊗ M2 ∼
= B(C4)
Quantum types as algebras
type
qubit
bit
(ordered) pair of bits
(ordered) pair of qubits
unordered pair of bits
algebra
M2 ∼
= B(C2)
C2
C 2 ⊗ C2 ∼
= C4
M2 ⊗ M2 ∼
= B(C4)
Quantum types as algebras
type
qubit
bit
(ordered) pair of bits
(ordered) pair of qubits
unordered pair of bits
algebra
M2 ∼
= B(C2)
C2
C 2 ⊗ C2 ∼
= C4
M2 ⊗ M2 ∼
= B(C4)
C3
Quantum types as algebras
type
qubit
bit
(ordered) pair of bits
(ordered) pair of qubits
unordered pair of bits
algebra
M2 ∼
= B(C2)
C2
C 2 ⊗ C2 ∼
= C4
M2 ⊗ M2 ∼
= B(C4)
C3 {00, 01 = 10, 11}
Quantum types as algebras
type
qubit
bit
(ordered) pair of bits
(ordered) pair of qubits
unordered pair of bits
unordered pair of qubits
algebra
M2 ∼
= B(C2)
C2
C 2 ⊗ C2 ∼
= C4
M2 ⊗ M2 ∼
= B(C4)
C3 {00, 01 = 10, 11}
?
Unordered pair of qubits
Unordered pair of qubits
I asked a physicist.
He replied:
Unordered pair of qubits
I asked a physicist.
He replied: “Fermions or Bosons?”
Unordered pair of qubits
I asked a physicist.
He replied: “Fermions or Bosons?”
1. Bosons: M3
Unordered pair of qubits
I asked a physicist.
He replied: “Fermions or Bosons?”
1. Bosons: M3
|00i, |11i, |01i + |10i
Unordered pair of qubits
I asked a physicist.
He replied: “Fermions or Bosons?”
1. Bosons: M3
|00i, |11i, |01i + |10i
2. Fermions: C
Unordered pair of qubits
I asked a physicist.
He replied: “Fermions or Bosons?”
1. Bosons: M3
|00i, |11i, |01i + |10i
2. Fermions: C
|01i − |10i
Unordered pair of qubits
I asked a physicist.
He replied: “Fermions or Bosons?”
1. Bosons: M3
|00i, |11i, |01i + |10i
2. Fermions: C
|01i − |10i
(Pauli exclusion principle)
CS Reflex
CS Reflex
1. Type should not depend on
implementation
CS Reflex
1. Type should not depend on
implementation
2. Type should come with a rule
CS Reflex
1. Type should not depend on
implementation
2. Type should come with a rule
So what about CoEq(id, swap)?
CS Reflex
1. Type should not depend on
implementation
2. Type should come with a rule
So what about CoEq(id, swap)?
f
t ⊗t →
− s
(f ◦ swap = f )
CoEq(id, swap) −
→
s
0
f
CoEq(id, swap)
CoEq(id, swap)
M3 ⊕ C
CoEq(id, swap)
M3 ⊕ C
(In fd-CStarop
cPsU )
CoEq(id, swap)
M3 ⊕ C
(In Selinger’s Q)
CoEq(id, swap)
M3 ⊕ C
(In CPMs )
CoEq(id, swap)
M3 ⊕ C
(In fd-CStarop
cPsU )
M3 comes from |00i, |11i and |10i + |01i.
CoEq(id, swap)
M3 ⊕ C
(In fd-CStarop
cPsU )
M3 comes from |00i, |11i and |10i + |01i.
C corresponds to |01i − |10i, which is
symmetric up to global phase.
The coequalizer is easy to describe:
E = {a; a ∈ M2 ⊗ M2; swap(a) = a}
The coequalizer is easy to describe:
E = {a; a ∈ M2 ⊗ M2; swap(a) = a}
Crux: E ∼
= M3 ⊕ C.
The coequalizer is easy to describe:
E = {a; a ∈ M2 ⊗ M2; swap(a) = a}
Crux: E ∼
= M3 ⊕ C.
Has simple 1/2-page proof, which led to . . .
Remainder of this talk
1. Unordered tuples
I
Sketch of proof
2. Cycles
3. Unordered words
Remainder of this talk
1. Unordered tuples
I
Sketch of proof
2. Cycles
3. Unordered words
Result 1: unordered tuples
Result 1: unordered tuples
Unordered n-tuples of d -level systems
M
Mmλ
λ∈Yn,d
Result 1: unordered tuples
Unordered n-tuples of d -level systems
M
Mmλ
λ∈Yn,d
where Yn,d denotes the n-block Young diagrams
Result 1: unordered tuples
Unordered n-tuples of d -level systems
M
Mmλ
λ∈Yn,d
where Yn,d denotes the n-block Young diagrams
of height at most d
Result 1: unordered tuples
Unordered n-tuples of d -level systems
M
Mmλ
λ∈Yn,d
where Yn,d denotes the n-block Young diagrams
of height at most d and mλ the dimension
Result 1: unordered tuples
Unordered n-tuples of d -level systems
M
Mmλ
λ∈Yn,d
where Yn,d denotes the n-block Young diagrams
of height at most d and mλ the dimension of the corresponding
representation of GL(d).
Result 1: unordered tuples
Unordered n-tuples of d -level systems
M
Mmλ
λ∈Yn,d
where Yn,d denotes the n-block Young diagrams
of height at most d and mλ the dimension of thecorresponding
representation of GL(d). Or explicitly: Yn,d =
n
λ; λ ∈ Nd ;
λ1 ≥ . . . ≥ λd ≥ 0 o
λ1 + . . . + λd = n
Result 1: unordered tuples
Unordered n-tuples of d -level systems
M
Mmλ
λ∈Yn,d
where Yn,d denotes the n-block Young diagrams
of height at most d and mλ the dimension of thecorresponding
representation of GL(d). Or explicitly: Yn,d =
and mλ =
Q
1≤i<j≤d
λi −λj +j−i
.
j−i
n
λ; λ ∈ Nd ;
λ1 ≥ . . . ≥ λd ≥ 0 o
λ1 + . . . + λd = n
Examples
Examples
Unordered triple of qutrits
Examples
Unordered triple of qutrits M10 ⊕ M8 ⊕ C
Examples
Unordered triple of qutrits M10 ⊕ M8 ⊕ C
Unordered pair of ququads
Examples
Unordered triple of qutrits M10 ⊕ M8 ⊕ C
Unordered pair of ququads M10 ⊕ M6
Examples
Unordered triple of qutrits M10 ⊕ M8 ⊕ C
Unordered pair of ququads M10 ⊕ M6
Unordered quad of qubits
Examples
Unordered triple of qutrits M10 ⊕ M8 ⊕ C
Unordered pair of ququads M10 ⊕ M6
Unordered quad of qubits M5 ⊕ M3 ⊕ C
Examples
Unordered triple of qutrits M10 ⊕ M8 ⊕ C
Unordered pair of ququads M10 ⊕ M6
Unordered quad of qubits M5 ⊕ M3 ⊕ C
d
2
3
4
5
6
7
8
M5
M15
M35
M70
M126
M210
M330
M3
M15
M45
M105
M210
M378
M630
C
M6
M20
M50
M105
M196
M336
M3
M15
M45
M105
M210
M378
C
M5
M15
M35
M70
Proof, setting up
Proof, setting up
Sn acts on H = (Cd )⊗n in the obvious way.
Proof, setting up
Sn acts on H = (Cd )⊗n in the obvious way.
Also on B(H) by π(a) = π −1aπ.
Proof, setting up
Sn acts on H = (Cd )⊗n in the obvious way.
Also on B(H) by π(a) = π −1aπ.
We wish to compute their equalizer
Proof, setting up
Sn acts on H = (Cd )⊗n in the obvious way.
Also on B(H) by π(a) = π −1aπ.
We wish to compute their equalizer
E = {a; a ∈ B(H); π −1aπ = a ∀π ∈ Sn }
Proof, crucial observation
Proof, crucial observation
E = {a; a ∈ B(H); π −1aπ = a ∀π ∈ Sn }
= RepSn (H, H)
Proof, crucial observation
E = {a; a ∈ B(H); π −1aπ = a ∀π ∈ Sn }
= RepSn (H, H)
The equalizer coincides with the
representation endomorphisms of H!
Proof, basic representation theory
Proof, basic representation theory
d ⊗n
H = (C )
∼
=
M
m
Uλ λ
λ
where Uλ distinct irreducible representations.
Proof, basic representation theory
d ⊗n
H = (C )
∼
=
M
m
Uλ λ
λ
where Uλ distinct irreducible representations.
Schur’s lemma:
(
C µ=λ
Rep(Uλ, Uµ) =
0 µ 6= λ
Proof, putting it together
Proof, putting it together
E = RepSn (H, H)
Proof, putting it together
E = RepSn (H, H)
L
m
m
∼
= λ,µ RepSn (Uλ λ , Uµ µ )
Proof, putting it together
E = RepSn (H, H)
L
m
m
∼
= λ,µ RepSn (Uλ λ , Uµ µ )
L
∼
Mm
=
λ
λ
Proof, putting it together
E = RepSn (H, H)
L
m
m
∼
= λ,µ RepSn (Uλ λ , Uµ µ )
L
∼
Mm
=
λ
λ
What are the irreducible representations Uλ
and their multiplicities mλ?
Proof, putting it together
E = RepSn (H, H)
L
m
m
∼
= λ,µ RepSn (Uλ λ , Uµ µ )
L
∼
Mm
=
λ
λ
What are the irreducible representations Uλ
and their multiplicities mλ?
Answer is given by Schur-Weyl duality.
1. Unordered tuples
I
Sketch of proof
2. Cycles
3. Unordered words
3-cycle
3-cycle
A 3-cycle of bits is a 4dit:
3-cycle
A 3-cycle of bits is a 4dit:
{000,
3-cycle
A 3-cycle of bits is a 4dit:
{000, 001 = 010 = 100,
3-cycle
A 3-cycle of bits is a 4dit:
{000, 001 = 010 = 100, 011 = 101 = 110,
3-cycle
A 3-cycle of bits is a 4dit:
{000, 001 = 010 = 100, 011 = 101 = 110, 111}
3-cycle
A 3-cycle of bits is a 4dit:
{000, 001 = 010 = 100, 011 = 101 = 110, 111}
What about a 3-cycle of qubits?
3-cycle
A 3-cycle of bits is a 4dit:
{000, 001 = 010 = 100, 011 = 101 = 110, 111}
What about a 3-cycle of qubits?
(= coequalizer of obvious action of C3 on B(C2 ⊕ C2 ⊕ C2 ).)
Quantum 3-cycle
Quantum 3-cycle
M4 ⊕ M2 ⊕ M2
Quantum 3-cycle
M4 ⊕ M2 ⊕ M2
I
|001i + |010i + |100i
Quantum 3-cycle
M4 ⊕ M2 ⊕ M2
I
|001i + |010i + |100i
I
|001i + e
2πi
3
+ |010i e
4πi
3
|100i
Quantum 3-cycle
M4 ⊕ M2 ⊕ M2
I
|001i + |010i + |100i
I
|001i + e
2πi
3
+ |010i e
4πi
3
|100i
I
|001i + e
4πi
3
+ |010i e
2πi
3
|100i
Arbitrary cycles
Arbitrary cycles
Schur-Weyl does not apply.
Arbitrary cycles
Schur-Weyl does not apply.
How to compute multiplicities?
Arbitrary cycles
Schur-Weyl does not apply.
How to compute multiplicities?
By computing the character table.
Result 2: arbitrary cycles
Result 2: arbitrary cycles
mk =
X
0≤j<n
e
2πijk
n
d gcd(j,n)
Result 2: arbitrary cycles
mk =
X
e
2πijk
n
0≤j<n
With some number theory:
d gcd(j,n)
Result 2: arbitrary cycles
mk =
X
e
2πijk
n
d gcd(j,n)
0≤j<n
With some number theory:
φ(`)
1X n `
.
mk =
d `µ
`
n
gcd(`, k) φ gcd(`,k)
`|n
1. Unordered tuples
I
Sketch of proof
2. Cycles
3. Unordered words
Result 3: quantum unordered words
Result 3: quantum unordered words
Q
L d ⊗n
S
acts
on
B(
n n
n (C ) ).
Result 3: quantum unordered words
L d ⊗n
S
acts
on
B(
n n
n (C ) ).
With care we can compute the coequalizer:
Q
Result 3: quantum unordered words
L d ⊗n
S
acts
on
B(
n n
n (C ) ).
With care we can compute the coequalizer:
Q
2
B(` ) ⊕
Y
M mλ .
λ∈Y ∗
Y ∗: Young diagrams of height at least 2.
Recap
1. Algebras for unordered types are given
by coequalizers.
Recap
1. Algebras for unordered types are given
by coequalizers.
2. They are more interesting than
expected.
Recap
1. Algebras for unordered types are given
by coequalizers.
2. They are more interesting than
expected.
3. Representation theory of finite groups is
a perfect fit to study them.
Thanks!
Thanks!
Questions?