Improved Simulation of Stabilizer Circuits Scott Aaronson (UC Berkeley) Joint work with Daniel Gottesman (Perimeter) ZI +IX +XI +IZ

Quantum Computing: New Challenges for Computer Architecture Cant speculate on a measurement and then roll back Cache coherence protocols violate no- cloning theorem How do you design and debug circuits that you cant even simulate efficiently with existing tools?

Our Approach: Start With A Subset of Quantum Computations Stabilizers (Gottesman 1996): Beautiful formalism that captures much (but not all) of quantum weirdness – Linear error-correcting codes – Teleportation – Dense quantum coding – GHZ (Greenberger-Horne-Zeilinger) paradox

Gates Allowed In Stabilizer Circuits 1. Controlled-NOT |00 |00, |01 |01, |10 |11, |11 |10 2. Hadamard |0 (|0 +|1 )/ 2 |1 (|0 -|1 )/ 2 H 3. Phase = |0 |0, |1 i|1 P 100i100i 4. Measurement of a single qubit AMAZING FACT These gates are NOT universalGottesman & Knill showed how to simulate them quickly on a classical computer To see why we need some group theory…

X 2 =Y 2 =Z 2 =IXY=iZYZ=iXZX=iY XZ=-iYZY=-iXYX=-iZ Unitary matrix U stabilizes a quantum state | if U| = |. Stabilizers of | form a group X stabilizes |0 +|1 -X stabilizes |0 +|1 Y stabilizes |0 +i|1 -Y stabilizes |0 -i|1 Z stabilizes |0 -Z stabilizes | I =Z = X = 0-i i0 Y = Pauli Matrices: Collect Em All

If | can be produced from the all-0 state by just CNOT, Hadamard, and phase gates, then | is stabilized by 2 n tensor products of Pauli matrices or their opposites (where n = number of qubits) So the stabilizer group is generated by log(2 n )=n such tensor products Indeed, | is then uniquely determined by these generators, so we call | a stabilizer state Gottesman-Knill Theorem

Goal: Using a classical computer, simulate an n- qubit CNOT/Hadamard/Phase computer. Gottesman & Knills solution: Keep track of n generators of the stabilizer group Each generator uses 2n+1 bits: 2 for each Pauli matrix and 1 for the sign. So n(2n+1) bits total Example: But measurement takes O(n 3 ) steps by Gaussian elimination +X X -ZZ |01 +|11 |01 +|10 CNOT(1 2) +XI -IZ Updating stabilizers takes only O(n) steps OUCH!

Our Faster, Easier-To-Implement Tableau Algorithm Idea: Instead of n(2n+1) bits, store 2n(2n+1) bits n stabilizers S 1,…,S n, 2n+1 bits each n destabilizers D 1,…,D n Together generate full Pauli group Maintain the following invariants: D i s commute with each other S i anticommutes with D i S i commutes with D j for i j

Destabilizers Stabilizers State: |00 +XI +IX +ZI +IZ x ij bitsz ij bitsr i bits I: x ij =0, z ij =0+ phase: r i =0 X: x ij =1, z ij =0- phase: r i =1 Y: x ij =1, z ij =1 Z: x ij =0, z ij =1 S1S2S1S2 D1D2D1D2

Destabilizers Stabilizers State: |00 +XI +IX +ZI +IZ Hadamard on qubit a: For all i {1,…,2n}, swap x ia with z ia, and set r i := r i x ia z ia

Destabilizers Stabilizers State: |00 +|10 +ZI +IX +XI +IZ

Destabilizers Stabilizers State: |00 +|10 +ZI +IX +XI +IZ CNOT from qubit a to qubit b: For all i {1,…,2n}, set x ib := x ib x ia and z ia := z ia z ib

Destabilizers Stabilizers State: |00 +|11 +ZI +IX +X X +Z Z

Destabilizers Stabilizers State: |00 +|11 +ZI +IX +X X +Z Z Phase on qubit a: For all i {1,…,2n}, set r i := r i x ia z ia, then set z ia := z ia x ia

Destabilizers Stabilizers State: |00 +i|11 +ZI +IY +X Y +Z Z

Destabilizers Stabilizers State: |00 +i|11 +ZI +IY +X Y +Z Z Measurement of qubit a: If x ia =0 for all i {n+1,…,2n}, then outcome will be deterministic. Otherwise 0 with ½ probability and 1 with ½ probability.

Destabilizers Stabilizers State: |11 +X Y +IY -ZI +Z Z Random outcome: Pick a stabilizer S i such that x ia =1 and set D i :=S i. Then set S i :=Z a and output 0 with ½ probability, and set S i :=-Z a and output with ½ probability, where Z a is Z on a th qubit and I elsewhere. Finally, left-multiply whatever rows dont commute with S i by D i

Novel part: How to obtain deterministic measurement outcomes in only O(n 2 ) steps, without using Gaussian elimination? Z a must commute with stabilizer, so for a unique choice of c 1,…,c n {0,1}. If we can determine c i s, then by summing corresponding S h s we learn sign of Z a. Now So just have to check if D i commutes with Z a, or equivalently if x ia =1

CHP: An interpreter for quantum assembly language programs that implements our scoreboard algorithm Example: Quantum Teleportation H H HH | |0 | Alices Qubits Bobs Qubits Prepare EPR pair Alices partBobs part h 1 c 1 2 c 0 1 h 0 m 0 m 1 c 0 3 c 1 4 c 4 2 h 2 c 3 2 h 2 CHP Code

Performance of CHP Average time needed to simulate a measurement after applying βnlogn random unitary gates to n qubits, on a 650MHz Pentium III with 256MB RAM

Simulating Stabilizer Circuits is L-Complete L = class of problems reducible in logarithmic space to evaluating a circuit of CNOT gates Natural, well-studied subclass of P Conjectured that L P. So our result means stabilizer circuits probably arent even universal for classical computation! Simulating L with stabilizer circuits: Obvious Simulating stabilizer circuits with L : Harder

How many n-qubit stabilizer states are there? Fun With Stabilizers Can we represent mixed states in the stabilizer formalism? YES Can we efficiently compute the inner product between two stabilizer states? YES

Future Directions Measurements (at least some) in O(n) steps? Apply CHP to quantum error-correction, studying conjectures about entanglement in many-qubit systems… Efficient minimization of stabilizer circuits? Superlinear lower bounds on stabilizer circuit size? Other quantum computations with efficient classical simulations: bounded entanglement (Vidal 2003), matchgates (Valiant 2001)…