the bose-hubbard model is qma-complete

The Bose-Hubbard model is QMA-complete

Andrew M. ChildsDavid Gosset

Zak Webb

arXiv: 1311.3297

Institute for Quantum Computing University of Waterloo

Solving for the ground energy of a quantum system can be viewed as a quantum constraint satisfaction problem. How difficult is it?

Image source: http://www.condmat.physics.manchester.ac.uk/imagelibrary/

QMA1-complete for

Local Ground energy problem

Frustration-free

Stoquastic(no “sign problem”)

Quantum k-SAT (testing frustration-freeness)

k-local Hamiltonian problem

Stoquastic k-local Hamiltonian problem

Fermions or Bosons

Class of Hamiltonians ComplexityQMA-complete for

Contained in AMMA-hard

QMA-complete

[Kempe, Kitaev, Regev 2006]

[Bravyi et. al. 2006]

[Liu, Christandl, Verstraete 2007] [Wei, Mosca, Nayak 2010]

Contained in P for

[Bravyi 2006] [G. , Nagaj 2013 ]

The computational difficulty of computing the ground energy has been studied for many broad classes of Hamiltonians

[Kitaev 1999] [Kempe, Regev 2003]

h 𝑖𝑗

𝐻 𝑖 , 𝑖+1

�⃗�𝑖

2-local Hamiltonian on a 2D grid [Oliveira Terhal 2008]

2-local Hamiltonian on a line with qudits[Aharonov et. al 2009] [Gottesman Irani 2009]

Hubbard model on a 2D grid with site-dependent magnetic field [Schuch Verstraete 2009].

Versions of the XY, Heisenberg, and other models with adjustable coefficients[Cubitt Montanaro 2013] )

E.g.,

Systems with QMA-complete ground energy problems can be surprisingly simple

…However, the complexity of many natural models from condensed matter physics is still not understood, e.g., the XY model on a graph

)

…However, the complexity of many natural models from condensed matter physics is still not understood, e.g., the XY model on a graph

Unfortunately, proof techniques using perturbation theory * require coefficients which grow with system size, e.g., [Cubitt Montanaro 2013]

*[Kempe Kitaev Regev 2006], [Oliveira Terhal 2008]

)

Allowed to scale polynomially with n

)

…However, the complexity of many natural models from condensed matter physics is still not understood, e.g., the XY model on a graph

Unfortunately, proof techniques using perturbation theory * require coefficients which grow with system size, e.g., [Cubitt Montanaro 2013]

In this work we consider a model of interacting Bosons on a graph with no adjustable coefficients…

*[Kempe Kitaev Regev 2006], [Oliveira Terhal 2008]

)

Allowed to scale polynomially with n

)

Bosons move between adjacent vertices and experience an energy penalty if two or more particles occupy the same site.

Adjacency matrix (a symmetric 0-1 matrix)

The Bose-Hubbard model on a graph

Bosons move between adjacent vertices and experience an energy penalty if two or more particles occupy the same site.

𝐻𝐺(𝑡 ,𝑈 )=𝑡 ∑𝑖 , 𝑗 ∈𝑉

𝐴(𝐺)𝑖𝑗 𝑎𝑖† 𝑎 𝑗+𝑈 ∑𝑘∈𝑉

𝑛𝑘 (𝑛𝑘−1 )

Movement Repulsive on-site interaction

Adjacency matrix (a symmetric 0-1 matrix)

Conserves totalnumber of particles

The Bose-Hubbard model on a graph

annihilates a particle at site i counts the number of particles at site i

Bosons move between adjacent vertices and experience an energy penalty if two or more particles occupy the same site.

The Bose-Hubbard model on a graph

𝐻𝐺(𝑡 ,𝑈 )=𝑡 ∑𝑖 , 𝑗 ∈𝑉

𝐴(𝐺)𝑖𝑗 𝑎𝑖† 𝑎 𝑗+𝑈 ∑𝑘∈𝑉

𝑛𝑘 (𝑛𝑘−1 )

Movement Repulsive on-site interaction

Adjacency matrix (a symmetric 0-1 matrix)

(t,U)-Bose-Hubbard Hamiltonian problemInput: A graph number of particles , energy threshold , and precision parameter Problem: Is the ground energy of in the -particle sector at most or at least (promised that one of these conditions holds)

Conserves totalnumber of particles

𝑈

QMA-complete[our results]

𝑡

Theorem: (t,U)-Bose Hubbard Hamiltonian is QMA-complete for all .

Complexity of (t,U)-Bose Hubbard Hamiltonian

𝑈“Stoquastic”

QMA-complete[our results]

Contained in AMQMA[Bravyi et. al 2006]

𝑡

Theorem: (t,U)-Bose Hubbard Hamiltonian is QMA-complete for all .

Complexity of (t,U)-Bose Hubbard Hamiltonian

𝑈“Stoquastic”

Contained in QMA QMA-complete

[our results]

Contained in AMQMA[Bravyi et. al 2006]

𝑡

Theorem: (t,U)-Bose Hubbard Hamiltonian is QMA-complete for all .

Complexity of (t,U)-Bose Hubbard Hamiltonian

Complexity of (t,U)-Bose Hubbard Hamiltonian

In the limit of infinite repulsion (i.e., hard core bosons) there is only ever 0 or 1 particle at each vertex, and the Hamiltonian is equivalent to a spin model…

𝑈“Stoquastic”

Contained in QMA QMA-complete

[our results]

Contained in AMQMA[Bravyi et. al 2006]

𝑡

Theorem: (t,U)-Bose Hubbard Hamiltonian is QMA-complete for all .

A related class of 2-local Hamiltonians

Conserves total magnetization (Hamming weight)

XY Hamiltonian problemInput: A graph , magnetization , energy threshold , precision parameter Problem: Is the ground energy of in the sector with magnetization at most or at least (promised one of these conditions holds)

Theorem: XY Hamiltonian is QMA-complete.

Quantum Merlin Arthur

If is a yes instance there exists (a witness) which is accepted with high probability.If is a no instance every state has low acceptance probability.

Wm-1Wm-2…W0

¿𝜓 ⟩¿0 ⟩⊗𝑛𝑎

Every instance of a problem in QMA has a verification circuit

QMA: class of decision problems where yes instances can be efficiently verified on a quantum computer.

Ground energy problems are usually contained in QMA (witness ground state ; verification circuit energy measurement)

Proving QMA-hardness is more involved…

Proving QMA-hardness for ground energy problems

Wm-1Wm-2…W0¿𝜓 ⟩

¿0 ⟩⊗𝑛𝑎 𝐻 𝑥

Desired properties:Ground energy of is small Verification circuit accepts a state with high probability is a yes instance

QMA Verification circuit for Hamiltonian

Proving QMA-hardness for ground energy problems

Wm-1Wm-2…W0¿𝜓 ⟩

¿0 ⟩⊗𝑛𝑎 𝐻 𝑥

Desired properties:Ground energy of is small Verification circuit accepts a state with high probability is a yes instance

Key intermediate step: Design a Hamiltonian where each ground state encodes a quantum computation associated with the circuit, e.g., Feynman-Kitaev Hamiltonian has ground states

QMA Verification circuit for Hamiltonian

Proving QMA-hardness for ground energy problems

Wm-1Wm-2…W0¿𝜓 ⟩

¿0 ⟩⊗𝑛𝑎 𝐻 𝑥

Desired properties:Ground energy of is small Verification circuit accepts a state with high probability is a yes instance

Key intermediate step: Design a Hamiltonian where each ground state encodes a quantum computation associated with the circuit, e.g., Feynman-Kitaev Hamiltonian has ground states

QMA Verification circuit for Hamiltonian

In our case we show how to encode the history of an -qubit, -gate computation in the groundspace of the -particle Bose-Hubbard model on a graph with vertices

When the Hamiltonian is just the adjacency matrix of the graph. We use a variant of the Feynman-Kitaev circuit-to-Hamiltonian mapping which outputs a symmetric 0-1 matrix.

Encoding one qubit with one particle

H H HT (HT)† HT (HT)† H H

When the Hamiltonian is just the adjacency matrix of the graph. We use a variant of the Feynman-Kitaev circuit-to-Hamiltonian mapping which outputs a symmetric 0-1 matrix.

Example (building block for later on)

Encoding one qubit with one particle

Vertices are labeled(𝑧 , 𝑡 , 𝑗 ) 𝑧∈ {0,1 } 𝑡 , 𝑗∈ [8]

H H HT (HT)† HT (HT)† H H

Groundstates of the adjacency matrix:

When the Hamiltonian is just the adjacency matrix of the graph. We use a variant of the Feynman-Kitaev circuit-to-Hamiltonian mapping which outputs a symmetric 0-1 matrix.

Example (building block for later on)

Encoding one qubit with one particle

a specific state encodes the computation where the circuit is complex-conjugated

for

Vertices are labeled(𝑧 , 𝑡 , 𝑗 ) 𝑧∈ {0,1 } 𝑡 , 𝑗∈ [8]

More particles ()

𝐻𝐺(𝑡 ,𝑈 )=𝑡 ∑𝑖 , 𝑗 ∈𝑉

𝐴(𝐺)𝑖𝑗 𝑎𝑖† 𝑎 𝑗+𝑈 ∑𝑘∈𝑉

𝑛𝑘 (𝑛𝑘−1 )

0

= smallesteigenvalue of

More particles ()

𝐻𝐺(𝑡 ,𝑈 )=𝑡 ∑𝑖 , 𝑗 ∈𝑉

𝐴(𝐺)𝑖𝑗 𝑎𝑖† 𝑎 𝑗+𝑈 ∑𝑘∈𝑉

𝑛𝑘 (𝑛𝑘−1 )

0

If the ground energy is we say the groundspace is frustration-free. In this case the groundspace is the same for all>0!

= smallesteigenvalue of

More particles ()

We design a class of graphs where the frustration-free -particle ground states encode -qubit computations. These graphs are built out of multiple copies of

𝐻𝐺(𝑡 ,𝑈 )=𝑡 ∑𝑖 , 𝑗 ∈𝑉

𝐴(𝐺)𝑖𝑗 𝑎𝑖† 𝑎 𝑗+𝑈 ∑𝑘∈𝑉

𝑛𝑘 (𝑛𝑘−1 )

0

If the ground energy is we say the groundspace is frustration-free. In this case the groundspace is the same for all>0!

= smallesteigenvalue of

4096 vertex graphmade from32 copies of

Graphs for two-qubit gates

Two qubit gate A graph shaped like this

𝑈

Single-particle ground states encode a qubit and one out of four possible locations

Two qubit gate A graph shaped like this

𝑈 Graphs for two-qubit gates

4096 vertex graphmade from32 copies of

Single-particle ground states encode a qubit and one out of four possible locations

Two qubit gate A graph shaped like this

𝑈 Graphs for two-qubit gates

4096 vertex graphmade from32 copies of

Single-particle ground states encode a qubit and one out of four possible locations

Two qubit gate A graph shaped like this

𝑈 Graphs for two-qubit gates

4096 vertex graphmade from32 copies of

Single-particle ground states encode a qubit and one out of four possible locations

Two qubit gate A graph shaped like this

𝑈 Graphs for two-qubit gates

4096 vertex graphmade from32 copies of

Single-particle ground states encode a qubit and one out of four possible locations

Two qubit gate A graph shaped like this

𝑈 Graphs for two-qubit gates

4096 vertex graphmade from32 copies of

Single-particle ground states encode a qubit and one out of four possible locations

Two qubit gate A graph shaped like this

𝑈 Graphs for two-qubit gates

4096 vertex graphmade from32 copies of

Single-particle ground states encode a qubit and one out of four possible locations

Two qubit gate A graph shaped like this

𝑈 Graphs for two-qubit gates

4096 vertex graphmade from32 copies of

Single-particle ground states encode a qubit and one out of four possible locations

Two qubit gate A graph shaped like this

𝑈 Graphs for two-qubit gates

4096 vertex graphmade from32 copies of

Single-particle ground states encode a qubit and one out of four possible locations

Two qubit gate A graph shaped like this

𝑈 Graphs for two-qubit gates

4096 vertex graphmade from32 copies of

Two-particle frustration-free ground states have the form

1√2

|both particles on the ¿𝜙 ⟩+ 1√2 ¿𝑈 𝜙 ¿¿

Single-particle ground states encode a qubit and one out of four possible locations

Two qubit gate A graph shaped like this

𝑈 Graphs for two-qubit gates

4096 vertex graphmade from32 copies of

1√2

|both particles on the ¿𝜙 ⟩+ 1√2 ¿𝑈 𝜙 ¿¿

Single-particle ground states encode a qubit and one out of four possible locations

Two-particle frustration-free ground states have the form

Two qubit gate A graph shaped like this

𝑈 Graphs for two-qubit gates

4096 vertex graphmade from32 copies of

Single-particle ground states encode a qubit and one out of four possible locations

1√2

|both particles on the ¿𝜙 ⟩+ 1√2 ¿𝑈 𝜙 ¿¿

Two qubit gate A graph shaped like this

Two-particle frustration-free ground states have the form

𝑈 Graphs for two-qubit gates

4096 vertex graphmade from32 copies of

Good news: there are two-particle ground states which encode computations

,𝑈 1𝜙,𝜙+¿ ,𝑈 1𝜙+¿,𝑈 1𝜙+¿ ,𝑈 1𝜙+¿ ,𝑈 2𝑈 1𝜙+¿

Constructing a graph from a verification circuit

𝑈 1 𝑈 2

Bad news: there are other two-particle ground states where the particles are in regions of the graph where they shouldn’t be.

To get rid of the bad states, we develop a method for enforcing “occupancy constraints”, i.e. penalizing certain configurations of the particles. To prove our results we establish spectral bounds without using perturbation theory.

Connect up the graphs for two-qubit gates to mimic the circuit, e.g.,

• Improvements to our construction? E.g., remove restriction to fixed particle number and/or consider simple graphs (no self loops).

• Complexity of other models of indistinguishable particles on graphs?– bosons or fermions with nearest-neighbor interactions– Attractive interactions– Negative hopping strength

• Complexity of other spin models on graphs?– Antiferromagnetic Heisenberg model– Models with only one type of 2-local term: for which matrices is the model

QMA-complete?

Open questions