Quantum 3-SAT is QMA1-complete
David Gosset (Institute for Quantum Computing, University of Waterloo)
Daniel Nagaj (University of Vienna)
Long version: arXiv: 1302.0290
Short version : Proceedings of FOCS 2013
Quantum k-SAT (Bravyi 2006)Each clause is a k-local projector and is satisfied by a state if .
The amount that violates a clause is
Quantum k-SAT (Bravyi 2006)Each clause is a k-local projector and is satisfied by a state if .
The amount that violates a clause is
Quantum k-SATGiven k-local projectors {. We are promised that either
(YES) There is a state which satisfies for each
(NO) for all states
and asked to decide which is the case.
Quantum k-SAT (Bravyi 2006)Each clause is a k-local projector and is satisfied by a state if .
The amount that violates a clause is
Quantum k-SATGiven k-local projectors {. We are promised that either
(YES) There is a state which satisfies for each
(NO) for all states
and asked to decide which is the case.
Exactly satisfies eachclause
Quantum k-SAT (Bravyi 2006)Each clause is a k-local projector and is satisfied by a state if .
The amount that violates a clause is
Quantum k-SATGiven k-local projectors {. We are promised that either
(YES) There is a state which satisfies for each
(NO) for all states
and asked to decide which is the case.
Exactly satisfies eachclause
Total violation is at least 1. Can be obtained from by repeating each term
Quantum k-SAT (Bravyi 2006)Each clause is a k-local projector and is satisfied by a state if .
The amount that violates a clause is
Quantum k-SATGiven k-local projectors {. We are promised that either
(YES) There is a state which satisfies for each
(NO) for all states
and asked to decide which is the case.
Exactly satisfies eachclause
Total violation is at least 1. Can be obtained from by repeating each term
Classical k-SAT is the special case where all projectors are diagonal
Quantum k-SAT is a special case of k-local Hamiltonian where the Hamiltonian is frustration-free for yes instances
k
k-local Hamiltonian problem
Quantum k-SAT
Classical k-SAT
Yes instances are frustration-free
All constraints are diagonal
k
k-local Hamiltonian problem
Quantum k-SAT
Classical k-SAT
Yes instances are frustration-free
All constraints are diagonal
Complexity of quantum k-SAT
k=2
πβ₯4
Contained in P
QMA1-complete
π=ππ=π
4
[Bravyi 2006]
also follows from [Kitaev 99])
k
k-local Hamiltonian problem
Quantum k-SAT
Classical k-SAT
Yes instances are frustration-free
All constraints are diagonal
Complexity of quantum k-SAT
k=2
πβ₯4
Contained in P
QMA1-complete
Contained in QMA1
NP-hard
π=ππ=π
4
[Bravyi 2006]
also follows from [Kitaev 99])
k
k-local Hamiltonian problem
Quantum k-SAT
Classical k-SAT
Yes instances are frustration-free
All constraints are diagonal
Complexity of quantum k-SAT
We prove quantum 3-SAT is QMA1-hard (and therefore QMA1-complete).
k=2
πβ₯4
Contained in P
QMA1-complete
Contained in QMA1
NP-hard
π=ππ=π
4
[Bravyi 2006]
also follows from [Kitaev 99])
k
k-local Hamiltonian problem
Quantum k-SAT
Classical k-SAT
Yes instances are frustration-free
All constraints are diagonal
Complexity of quantum k-SAT
k=2
πβ₯4
Contained in P
QMA1-complete
π=ππβ₯π
[Ambainis Kempe Sattath 2010][Arad Sattath 2013][Schwarz Cubitt Verstraete 2013]
Many authors have studied quantum SAT since Bravyiβs work
Quantum LovΓ‘sz Local Lemma
[Laumann LΓ€uchli Moessner Scardicchio Sondhi 2010][Laumann Moessner Scardicchio Sondhi 2010][Bravyi Moore Russell 2010][Hsu Laumann LΓ€uchli Moessner Sondhi 2013][Bardoscia Nagaj Scardicchio 2013]
Ensembles of randominstances of quantum k-SAT
[Eldar Regev 2008] Complexity of quantum 2-SAT with higher dimensional particles (qudits)
[Ji Wei Zeng 2011] Characterization of the groundspace of yes instances of quantum 2-SAT
[Sattath 2013] βAn almost sudden jump in quantum complexityβ
QMA1
If is a yes instance there exists (a witness) which is accepted with probability exactly 1.If is a no instance every state is accepted with probability at most
Wm-1Wm-2β¦W0
ΒΏπ β©ΒΏ0 β©βππ
QMA1 verification circuit
Because of the perfect completeness, the definition of QMA1 is gate-set dependent.It is not known whether or not QMA=QMA1; see
[Aaronson 2009] [Jordan, Kobayashi, Nagaj, Nishimura 2012][Kobayashi, Le Gall, Nishimura 2013] [Pereszlenyi 2013]
QMA1 is a one-sided error version of QMA. This is the relevant class becausequantum k-SAT is defined with one-sided error.
Bravyi proved quantum k-SAT is contained in QMA1 (verification circuit: choose one projector at random and measure it).
Bravyi proved quantum k-SAT is contained in QMA1 (verification circuit: choose one projector at random and measure it).
To prove QMA1-hardness of quantum 3-SAT we use a circuit-to-Hamiltonian mapping, i.e., we reduce from quantum circuit satisfiability.
If x is a yes instance there is an input state (witness) which makes the circuit output 1 with certainty. Ground energy of is zero.
If x is a no instance no input state makes the circuit output 1 with probability greater than Ground energy of is at least .
QMA1-hardness via circuit-to-Hamiltonian mapping
Wm-1Wm-2β¦W0ΒΏπ β©
ΒΏ0 β©βππ π» π₯=βπΞ π
QMA1 Verification circuit for Quantum 3-SAT Hamiltonian
Wm-1Wm-2β¦W0ΒΏπ β©ΒΏ0 β©βππ
Hilbert space
QMA1 verification circuit (n qubits, m gates)
|π§ β©|π‘ β© π§β {0,1 }π , π‘β{0,1,2 ,β¦,π }
Example part 1 [Kitaev 99]
Wm-1Wm-2β¦W0ΒΏπ β©ΒΏ0 β©βππ
Hilbert space
QMA1 verification circuit (n qubits, m gates)
|π§ β©|π‘ β© π§β {0,1 }π , π‘β{0,1,2 ,β¦,π }
π»π‘ ,π‘+1 (π π‘ )=12 ΒΏTransitionoperators
Example part 1 [Kitaev 99]
Wm-1Wm-2β¦W0ΒΏπ β©ΒΏ0 β©βππ
Hilbert space
Transitionoperators
QMA1 verification circuit (n qubits, m gates)
|π§ β©|π‘ β© π§β {0,1 }π , π‘β{0,1,2 ,β¦,π }
π»π‘ ,π‘+1 (π π‘ )=12 ΒΏ
π» πΉππ¦ππππ=βπ=1
ππ|1 β© β¨1|πββ¨0β©β¨ 0β¨ΒΏ+β
π‘=0
πβ1
π»π‘ ,π‘+1(π π‘)+|0 β© β¨0|ππ’π‘ββ¨πβ©β¨πβ¨ΒΏΒΏHamiltonian
Example part 1 [Kitaev 99]
Wm-1Wm-2β¦W0ΒΏπ β©ΒΏ0 β©βππ
Hilbert space
Transitionoperators
QMA1 verification circuit (n qubits, m gates)
|π§ β©|π‘ β© π§β {0,1 }π , π‘β{0,1,2 ,β¦,π }
π»π‘ ,π‘+1 (π π‘ )=12 ΒΏ
Hamiltonian
1βπ+1
(|π β©|0 β©+π 0|π β©|1 β©+π 1π 0|π β©|2 β©+β¦+ππβ1ππβ2β¦π 0β¨π β©β¨πβ©)Nullspace consists of βhistory statesβ
π» πΉππ¦ππππ=βπ=1
ππ|1 β© β¨1|πββ¨0β©β¨ 0β¨ΒΏ+β
π‘=0
πβ1
π»π‘ ,π‘+1(π π‘)+|0 β© β¨0|ππ’π‘ββ¨πβ©β¨πβ¨ΒΏΒΏ
Example part 1 [Kitaev 99]
Wm-1Wm-2β¦W0ΒΏπ β©ΒΏ0 β©βππ
Hilbert space
Transitionoperators
QMA1 verification circuit (n qubits, m gates)
|π§ β©|π‘ β© π§β {0,1 }π , π‘β{0,1,2 ,β¦,π }
π»π‘ ,π‘+1 (π π‘ )=12 ΒΏ
Hamiltonian
1βπ+1
(|π β©|0 β©+π 0|π β©|1 β©+π 1π 0|π β©|2 β©+β¦+ππβ1ππβ2β¦π 0β¨π β©β¨πβ©)Nullspace consists of βhistory statesβ
To have zero energy for the other two terms, we must have|π β©=|0 β©ππβ¨π β©A witness accepted with probability 1
π» πΉππ¦ππππ=βπ=1
ππ|1 β© β¨1|πββ¨0β©β¨ 0β¨ΒΏ+β
π‘=0
πβ1
π»π‘ ,π‘+1(π π‘)+|0 β© β¨0|ππ’π‘ββ¨πβ©β¨πβ¨ΒΏΒΏ
Example part 1 [Kitaev 99]
Wm-1Wm-2β¦W0ΒΏπ β©ΒΏ0 β©βππ
Hilbert space
QMA1 verification circuit (n qubits, m gates)
|π§ β©|π‘ β© π§β {0,1 }π , π‘β{0,1,2 ,β¦,π }
π»π‘ ,π‘+1 (π π‘ )=12 ΒΏ
has a zero energy ground state if and only if the QMA1 verification circuit accepts a witness with probability 1. However, itβs not local.
Example part 1 [Kitaev 99]
Kitaev used a clock construction to convert it to a local Hamiltonianβ¦
Transitionoperators
Hamiltonian π» πΉππ¦ππππ=βπ=1
ππ|1 β© β¨1|πββ¨0β©β¨ 0β¨ΒΏ+β
π‘=0
πβ1
π»π‘ ,π‘+1(π π‘)+|0 β© β¨0|ππ’π‘ββ¨πβ©β¨πβ¨ΒΏΒΏ
Hilbert space HcompβHclock
m qubitsn qubits
Example part 2: Clock construction [Kitaev 99]
π» πΎππ‘πππ£=1ββπ=1
πβ1
|01 β© β¨ 01ΒΏπ ,π+1+π»π ππ
Hilbert space HcompβHclock
m qubitsn qubits
Hamiltonian
Example part 2: Clock construction [Kitaev 99]
A sum of 5-local projectors
π» πΎππ‘πππ£=1ββπ=1
πβ1
|01 β© β¨ 01ΒΏπ ,π+1+π»π ππ
Nullspace spanned by
Hilbert space HcompβHclock
m qubitsn qubits
|t β©π’=|111β¦1000β¦0 β© ,t=0 ,β¦,m
π‘ πβπ‘
Hamiltonian
Example part 2: Clock construction [Kitaev 99]
A sum of 5-local projectors
π» πΎππ‘πππ£=1ββπ=1
πβ1
|01 β© β¨ 01ΒΏπ ,π+1+π»π ππ
π» π ππ|HcompβSclock=π» πΉππ¦ππππ
Nullspace spanned by
is designed so that
This implies has the same nullspace as
Hilbert space HcompβHclock
m qubitsn qubits
|t β©π’=|111β¦1000β¦0 β© ,t=0 ,β¦,m
π‘ πβπ‘
Hamiltonian
Example part 2: Clock construction [Kitaev 99]
A sum of 5-local projectors
hπ‘ ,π‘+1 (π π‘ ) |HcompβS clock=π»π‘ ,π‘+1(π π‘)
π0|Sclock=|0 β© β¨ 0β¨ΒΏ
ππ|Sclock=|π β© β¨πβ¨ΒΏ
This is achieved βterm by termβ, by exhibiting projectors (acting on ) and projectors acting on such that
Example part 2: Clock construction [Kitaev 99]
hπ‘ ,π‘+1 (π π‘ ) |HcompβS clock=π»π‘ ,π‘+1(π π‘)
π0|Sclock=|0 β© β¨ 0β¨ΒΏ
ππ|Sclock=|π β© β¨πβ¨ΒΏ
This is achieved βterm by termβ, by exhibiting projectors (acting on ) and projectors acting on such that
Example part 2: Clock construction [Kitaev 99]
π» πΎππ‘πππ£=1ββπ=1
πβ1
|01 β© β¨ 01ΒΏπ ,π+1+βπ=1
ππ|1 β© β¨1|πβπ0+β
π‘=0
πβ1
hπ‘ , π‘+1(π π‘)+|0 β© β¨0|ππ’π‘βππ
hπ‘ ,π‘+1 (π π‘ ) |HcompβS clock=π»π‘ ,π‘+1(π π‘)
π0|Sclock=|0 β© β¨ 0β¨ΒΏ
ππ|Sclock=|π β© β¨πβ¨ΒΏ
This is achieved βterm by termβ, by exhibiting projectors (acting on ) and projectors acting on such that
Example part 2: Clock construction [Kitaev 99]
π» πΎππ‘πππ£=1ββπ=1
πβ1
|01 β© β¨ 01ΒΏπ ,π+1+βπ=1
ππ|1 β© β¨1|πβπ0+β
π‘=0
πβ1
hπ‘ , π‘+1(π π‘)+|0 β© β¨0|ππ’π‘βππ
A -local projector if is j-local
1-local projectors
hπ‘ ,π‘+1 (π π‘ ) |HcompβS clock=π»π‘ ,π‘+1(π π‘)
π0|Sclock=|0 β© β¨ 0β¨ΒΏ
ππ|Sclock=|π β© β¨πβ¨ΒΏ
This is achieved βterm by termβ, by exhibiting projectors (acting on ) and projectors acting on such that
Kitaevβs Hamiltonian is a sum of k-local projectors with for circuits made from 1- and 2-qubit gates.
Kitaevβs construction can be used to prove that quantum 5-SAT is QMA1-hard.
Example part 2: Clock construction [Kitaev 99]
π» πΎππ‘πππ£=1ββπ=1
πβ1
|01 β© β¨ 01ΒΏπ ,π+1+βπ=1
ππ|1 β© β¨1|πβπ1+β
π‘=0
π β1
hπ‘ ,π‘+1(π π‘)+|0 β© β¨0|ππ’π‘βππ
A -local projector if is j-local
1-local projectors
The first ingredient in our QMA1-hardness proof is a new clock construction (with different locality from Kitaevβs)β¦
Properties of the new clock construction
π»ππππππ
.
Hc lockSum of 3-local projectors Hamiltonian acting on
7N-3 qubits
Nullspace
ClockHamiltonian
Properties of the new clock construction
π»ππππππ
.
Hc lockSum of 3-local projectors Hamiltonian acting on
HcompβHclock
7N-3 qubits
Nullspace
ClockHamiltonian
Transitionoperators
act on
A -local projector if U is j-local
Properties of the new clock construction
π»ππππππ
.
Hc lockSum of 3-local projectors Hamiltonian acting on
HcompβHclock
7N-3 qubits
Nullspace
ClockHamiltonian
Transitionoperators
act on
Greater than/Less than operators πΆβ€ π
πΆβ€ π|Sclock = β1β€ π<π
|πΆ π β©β¨ πΆ πβ¨+ΒΏ12|πΆπ β©β¨ πΆπβ¨ΒΏΒΏ πΆβ₯ π|Sclock =
12|πΆπ β©β¨ πΆπβ¨+ β
π< π β€π|πΆ π β©β¨ πΆ πβ¨ΒΏΒΏ
act on Hclock
A -local projector if U is j-local
1-local projectors
3-local 2-local 4-local 2-local
Like Kitaevβs clock construction, ours could be used to emulate Feynmanβs Hamiltonian
This isnβt good enough for our purposesβit only shows that quantum 4-SAT is QMA1-hard (already known).
Instead, we use our clock construction in a different wayβ¦
1βπ»ππππππ+ 1 +β
π=1
ππ|1 β©β¨ 1β¨πβπΆβ€ 1+β
π‘hπ‘ ,π‘+1 (π π‘ )+|0 β© β¨ 0β¨ππ’π‘βπΆβ₯π+1
Two clock registers
We map the verification circuit to a Hamiltonian acting on a Hilbert space with one n-qubit computational register and two clock registers:
2D grid of zero energy clock states
1βπ»ππππππ β1+1β1βπ»πππππ
π
βInitialβ βFinalβ
Two clock registers
We map the verification circuit to a Hamiltonian acting on a Hilbert space with one n-qubit computational register and two clock registers:
1βπ»ππππππ β1+1β1βπ»πππππ
π +π» ππππ
Every zero energy groundstate encodes the history ofa computation
Two clock registers
We map the verification circuit to a Hamiltonian acting on a Hilbert space with one n-qubit computational register and two clock registers:
for 1-local U
1βπ»ππππππ β1+1β1βπ»πππππ
π +π» ππππ
is built out of 3-local projectors such as
h π ,π+1 (π )β1
1βπΆβ₯πβπΆβ€ π
|0 β© β¨ 0β¨πβh π ,π+1β11βhπ , π+1βπΆβ€ π
Every zero energy groundstate encodes the history ofa computation
(writing )
Two clock registers
We map the verification circuit to a Hamiltonian acting on a Hilbert space with one n-qubit computational register and two clock registers:
1βπ»ππππππ β1+1β1βπ»πππππ
π +π» ππππ+βπ=1
ππ|1 β© β¨ 1β¨πβπΆβ€ 1βπΆβ€ 1+|0 β© β¨ 0β¨ππ’π‘βπΆβ₯πβπΆβ₯π
Enforce initialization of ancillasand correct output of circuit
for 1-local U
is built out of 3-local projectors such as
h π ,π+1 (π )
1βπΆβ₯πβπΆβ€ π
|0 β© β¨ 0β¨πβh π ,π+1β11βhπ , π+1βπΆβ€ π (writing )
Two clock registers
We map the verification circuit to a Hamiltonian acting on a Hilbert space with one n-qubit computational register and two clock registers:
1βπ»ππππππ β1+1β1βπ»πππππ
π +π» ππππ+βπ=1
ππ|1 β© β¨ 1β¨πβπΆβ€ 1βπΆβ€ 1+|0 β© β¨ 0β¨ππ’π‘βπΆβ₯πβπΆβ₯π
Enforce initialization of ancillasand correct output of circuit
I will now show you how to construct for the case where the verification circuit is a specific two-qubit gate (warning: gadgetry ahead)β¦
for 1-local U
is built out of 3-local projectors such as
h π ,π+1 (π )
1βπΆβ₯πβπΆβ€ π
|0 β© β¨ 0β¨πβh π ,π+1β11βhπ , π+1βπΆβ€ π (writing )
Zero energy ground states
1 2 3 4 5 6 7 8 91
2
3
4
5
6
7
8
9
1β1βπ» πππππ9 +1βπ»πππππ
9 β1
Two clock registers: Example
Zero energy ground states is a vertex in the above graph
1 2 3 4 5 6 7 8 91
2
3
4
5
6
7
8
9
Two clock registers: Example
1β1βπ» πππππ9 +1βπ»πππππ
9 β1+1βπΆβ₯ 3βπΆβ€1
Zero energy ground states
1 2 3 4 5 6 7 8 91
2
3
4
5
6
7
8
9
is a vertex in the above graph
Two clock registers: Example
+1βπΆβ€ 1βπΆβ₯ 3+1βπΆβ₯ 3βπΆβ€11β1βπ» πππππ9 +1βπ»πππππ
9 β1
1 2 3 4 5 6 7 8 91
2
3
4
5
6
7
8
9
|π β©ππ|Ξ β©=|π β©ππ βπ , πβ Ξ
|πΆπ β©β¨πΆ πβ© where is a connected component of the graph
Zero energy ground states
Two clock registers: Example
+1βh12βπΆβ€2+1βπΆβ€ 1βπΆβ₯ 3+1βπΆβ₯ 3βπΆβ€11β1βπ» πππππ9 +1βπ»πππππ
9 β1
1 2 3 4 5 6 7 8 91
2
3
4
5
6
7
8
9
|π β©ππ|Ξ β©=|π β©ππ βπ , πβ Ξ
|πΆπ β©β¨πΆ πβ© where is a connected component of the graph
Zero energy ground states
Two clock registers: Example
+1βh12βπΆβ€2+1βπΆβ€ 1βπΆβ₯ 3+1βπΆβ₯ 3βπΆβ€11β1βπ» πππππ9 +1βπ»πππππ
9 β1
Continuing in this way,we can design a Hamiltonian with ground states described by a more complicated graphβ¦
Built out of terms likeh π ,π+1βπΆβ€π
πΆβ€πβ hπ ,π+1πΆβ₯ πβπΆβ€ π
|π β©ππ|Ξ β©=|π β©ππ βπ , πβ Ξ
|πΆπ β©β¨πΆ πβ© where is a connected component of the graph
1 2 3 4 5 6 7 8 91
2
3
4
5
6
7
8
9
Zero energy ground states
Two clock registers: Example
Commutes with
1 2 3 4 5 6 7 8 91
2
3
4
5
6
7
8
9
Zero energy ground states
+ΒΏ 0β©β¨ 0β¨πβ (h34+h67 )β1+ΒΏ1β©β¨ 1β¨πβ1β (h34+h67 )
Two clock registers: Example
sector1 2 3 4 5 6 7 8 9
1
2
3
4
5
6
7
8
9
1 2 3 4 5 6 7 8 91
2
3
4
5
6
7
8
9
sector
|0β©πβ¨π β©π|Ξ β© is a connected component
|1β©πβ¨π β©π|Ξ β© is a connected component
Zero energy ground states Zero energy ground states
+ΒΏ 0β©β¨ 0β¨πβ (h34+h67 )β1+ΒΏ1β©β¨ 1β¨πβ1β (h34+h67 )
Two clock registers: Example
sector1 2 3 4 5 6 7 8 9
1
2
3
4
5
6
7
8
9
1 2 3 4 5 6 7 8 91
2
3
4
5
6
7
8
9
sector
|0β©πβ¨π β©π|Ξ β© is a connected component
|1β©πβ¨π β©π|Ξ β© is a connected component
Zero energy ground states Zero energy ground states
+ΒΏ 0β©β¨ 0β¨πβ (h34+h67 )β1+ΒΏ1β©β¨ 1β¨πβ1β (h34+h67 )
Two clock registers: Example
sector1 2 3 4 5 6 7 8 9
1
2
3
4
5
6
7
8
9
|0 β©π|π β©π|β©
1 2 3 4 5 6 7 8 91
2
3
4
5
6
7
8
9
sector
|0 β©π|π β©π|β© |1 β©π|π β©π|β© |1 β©π|π β©π|β©+ others + others
Zero energy ground states Zero energy ground states
+ΒΏ 0β©β¨ 0β¨πβ (h34+h67 )β1+ΒΏ1β©β¨ 1β¨πβ1β (h34+h67 )
Two clock registers: Example
sector1 2 3 4 5 6 7 8 9
1
2
3
4
5
6
7
8
9
1 2 3 4 5 6 7 8 91
2
3
4
5
6
7
8
9
sector
π π
ππ ππ
π π
Zero energy ground states Zero energy ground states
+h45 (ππ)β1+h45ΒΏ
Acts on first clock register and qubit b
Acts on second clock register and qubit b
[π ,π ]β 0
+ΒΏ 0β©β¨ 0β¨πβ (h34+h67 )β1+ΒΏ1β©β¨ 1β¨πβ1β (h34+h67 )
Two clock registers: Example
sector1 2 3 4 5 6 7 8 9
1
2
3
4
5
6
7
8
9
1 2 3 4 5 6 7 8 91
2
3
4
5
6
7
8
9
sector
+h45 (ππ)β1+h45ΒΏ
π π
ππ ππ
π π
Acts on first clock register and qubit b
Acts on second clock register and qubit b
[π ,π ]β 0
Zero energy ground states Zero energy ground states
+ΒΏ 0β©β¨ 0β¨πβ (h34+h67 )β1+ΒΏ1β©β¨ 1β¨πβ1β (h34+h67 )
Two clock registers: Example
The point is that every zero energy ground state encodes the history of a two-qubit computation
|π β©ππ|πΆ1 β©|πΆ1 β©+β¦+π|π β©ππβ¨πΆ9 β©β¨πΆ9β©
where π=|0 β© β¨ 0β¨βππ+ΒΏ1β©β¨ 1β¨βππ
(An entangling two-qubit unitary for suitably chosen )
+h45 (ππ)β1+h45ΒΏ
Acts on first clock register and qubit b
Acts on second clock register and qubit b
[π ,π ]β 0
+ΒΏ 0β©β¨ 0β¨πβ (h34+h67 )β1+ΒΏ1β©β¨ 1β¨πβ1β (h34+h67 )
Two clock registers: Example
This was achieved without using the transition operator
Remarks and open questions
β’ Are there simpler βclause-by-clauseβ reductions for quantum k-SAT? In the classical case there is a clause-by-clause way to map a (k+1)-SAT instance to a k-SAT instance, for .
β’ Other applications for our new clock construction?
β’ βFrustration-freeβ gadgetry has the advantage over perturbation theory methods that one can avoid large (system size dependent) terms in the Hamiltonian.