Geometric Insights on Quantum Circuit Size Lower Bounds

1. A geometric approach to lower bounds on quantum circuit size Michael A. Nielsen School of Physical Sciences The University of Queensland To appear next week on quant-ph MAQIS, UQ, Jan 28, 2005

2. A supposedly hard problem The traveling salesman problem (TSP) 22 km 23 km 12 km 15 km 19 km 14 km 16 km Goal: Find the shortest tour through all the cities. It is widely believed that this problem requires exponential time to solve. Unfortunately, nobody can prove this. In fact, we have embarassing little idea how to prove this.

3. The P versus NP problem P is (roughly) the class of problems soluble in polynomial time on a classical computer. NP is (roughly) the class of problems whose solutions can be checked in polynomial time on a classical computer. TSP and factoring are examples of problems in NP, but believed not to be in P. Conjecture: P NP Smale (Fields medal, 1960): The most important mathematical conjecture made in the second half of the twentieth century. Clay Millenium Prize problem

4. The P versus NP problem It seems to be very difficult to prove that a problem like TSP is hard to solve. We expect TSP to take time 2n. In fact, the best lower bounds on running time essentially say “You must read the input”, i.e., they scale as the size of the problem.

5. Quantum computing 101: quantum circuits |0i Built up out of qubits: C2 |0i A single line – a quantum wire– represents a single qubit. |0i |0i Progress through time is from left to right. |0i |0i |0i Fix a standard computational basis,|0i, |1i, for the qubit. |0i Quantum computer is made up of n qubits: ­C2 Initially, the qubits are all in some standard state, say |0i­n. Dynamics: One- and two-qubit unitary gates. Conclude by measuring some of the qubits in their computational bases. Knill: Most unitaries require exponentially many gates to approximate. Church-Turing-Deutsch Principle (rough): If it can’t be done efficiently on a quantum computer, it can’t be done efficiently in nature.

6. Proving things are hard to do quantumly Let U be a unitary operation on n qubits. Find the minimal number of quantum gates needed to implement U exactly. Main point of the talk: To sketch a geometric program aimed at getting a handle on this minimal number of gates.

7. Idea 1: Replace discrete structures by smooth structures f(x) x To minimize a function f(x) defined on the integers we’ve got to solve the inequality f(x) · f(x-1),f(x+1) Suppose f(x) is defined on the reals. Powerful principle: Provided f is smooth, then f(x) is stationary at any minimum. Local minima can be found by solving a single equality, f’(x) = 0. Problem of finding minimal quantum circuits: it is tempting to think in terms of optimizing over a discrete space of quantum circuits. Better approach: Replace circuits by time-dependent Hamiltonians, and use the principle of stationarity (via calculus of variations) to identify local minima.

8. Idea 2: Find a metric space so the optimal Hamiltonian control is a minimal geodesic of that metric space Motivated by general relativity (GR). In GR a test particle follows geodesics of spacetime. Geodesics are paths which are (local) minima of length. Thus, the motion of a test particle minimizes a global functional – the length. This is equivalent to following a “local force law” known as the geodesic equation: acceleration determined by local shape of spacetime velocities

9. Idea 2: Find a metric space so the optimal Hamiltonian control is a minimal geodesic of that metric space Motivated by general relativity (GR). In GR a test particle follows geodesics of spacetime. Geodesics are paths which are (local) minima of length. Thus, the motion of a test particle minimizes a global functional – the length. This is equivalent to following a “local force law” known as the geodesic equation: Not special to GR: In any Finsler metric space, finding geodesics minimizing length is equivalent to following a local force law. The geodesic equation is a second order ODE, so given an initial position and velocity, subsequent motion is completely determined. (Compare, e.g., with being given part of an optimal circuit.)

10. Basic Picture SU(2n) We can think of U as being generated by an n-qubit time-dependent Hamiltonian: H(t) = (t)  V(0) = I; V(1) = U U I V(t) The Hamiltonian H(t) generates “small displacements” near U(t). We define a local metric: F(V,H) Length of the curve V: lF(V) ´s01 dt F(V(t),H(t)). Distance between I and U: dF(I,U) ´ inf lF(V) For any fixed V, F(V,¢) should be a norm: F(V,H) = 0 with equality iff H = 0 F(V,H) = || F(V,H) F(V,H1+H2) · F(V,H1)+F(V,H2) Notation:  = generalized Pauli = tensor product of Is, Xs, Ys, and Zs

11. Basic Picture SU(2n) We can think of U as being generated by an n-qubit time-dependent Hamiltonian: H(t) = (t)  V(0) = I; V(1) = U U I V(t) The Hamiltonian H(t) generates “small displacements” near U(t). We have to impose two additional conditions in order to obtain a suitable geodesic equation. 1. F(V,H) smooth away from from H = 0. F is a Finsler metric on SU(2n) 2. Let y be set of co-ordinates for H.

12. Basic Picture SU(2n) We can think of U as being generated by an n-qubit time-dependent Hamiltonian: H(t) = (t)  V(0) = I; V(1) = U U I V(t) The Hamiltonian H(t) generates “small displacements” near U(t). Key point: Operationally, a Finsler metric (on SU(2n)) corresponds to the type of metric for which we can write down a geodesic equation to study the minimal length paths.

13. Basic Picture SU(2n) We can think of U as being generated by an n-qubit time-dependent Hamiltonian: H(t) = (t)  V(0) = I; V(1) = U U I V(t) The Hamiltonian H(t) generates “small displacements” near U(t). We are minimizing lF(V) = s01 dt F(V(t),H(t)) Introduce co-ordinates xj on SU(2n) Let yj = dxj/dt be the corresponding natural “velocity” co-ordinates for the tangent space. Euler-Lagrange equations from the calculus of variations tell us that any minimizing curve must satisfy:

14. Basic Picture SU(2n) We can think of U as being generated by an n-qubit time-dependent Hamiltonian: H(t) = (t)  V(0) = I; V(1) = U U I V(t) The Hamiltonian H(t) generates “small displacements” near U(t). With some algebra and using the properties of F, the Euler-Lagrange equations may be recast in the form: Main point: this is a second order ODE, and so once the initial position and velocity are set we have a unique solution. Functions of local metric (both position and velocity)

15. The relationship between geometry and circuits SU(2n) Universality: We can build up U out of gates of the form exp(-i), where || · 1. U Let m(U) be the minimal number of gates of this form necessary to build up U. I We can construct a control function (t) which replicates this path. Suppose F(V,) · 1 for all V and . Key condition on F. Then the distance along the induced path is at most: |1|+|2|+…+|m(U)| · m(U) It follows that: dF(I,U) · m(U)

16. Example SU(2n) H =  U I V(t) This is an example of a Riemannian metric on SU(2n) – a special type of Finsler metric, arising from a quadratic form. We have F2(V,) = 1, so dF2(I,U) · m(U). Unfortunately, dF2(I,U) ·2, and so the best lower bound on m(U) we can hope for is constant. Proof: Choose H so that exp(-iH) = U, and all eigenvalues of H are between - and . Then the curve U(t) = exp(-iHt) has length at most tr(H2)/2n·2.

17. Better examples Penalty function SU(2n) H =  U I V(t) Both satisfy F(V,) = 1, so dF(I,U) · m(U). F1 is not Finsler, and so needs to be appoximated by a Finsler metric. Idea of approximation: For small , this becomes a good approximation to F1. The idea is to choose F1 so that “unit vectors” H satisfy g(H) = 1. F1 defined in this way is a Finsler metric.

18. Summary SU(2n) We’ve constructed a metric F1(V,H) on SU(2n) which gives lower bounds on circuit size, dF1(I,U) · m(U) U I V(t) We write F1(H) = F1(V,H) F1 can be approximated by a family of Finsler metrics, F1 Minimal length curves for F1 must be solutions of the geodesic equation. The geodesic equation is not trivial to compute, nor to solve! Everything I say about F1 for the remainder of the talk is also true of Fq, provided q is chosen appropriately.

19. Finding geodesics on the sphere Let’s choose a starting point. And a starting velocity. The geodesic equation tells us there is a unique geodesic motion determined by these choices. Pick the pole going through the the starting point. Consider the plane spanned by the pole and the initial velocity. If we reflect the sphere through that plane, lengths are not changed. So the reflected geodesic is still a geodesic. The reflected geodesic has same starting point and velocity ) it’s the same geodesic! Must be a great circle.

20. Finding geodesics on SU(2n) SU(2n) Choose a starting point. And a starting Hamiltonian, H, diagonal in the computational basis. The geodesic equation tells us there is a unique geodesic motion determined by these choices. I Let  be a Pauli matrix. The map U ! U  is an isometry of F1. If V(t) is the geodesic, then  V(t)  is also a geodesic. If  is a product of Is and Zs, then  V(t)  has the same starting point and starting Hamiltonian. V(t) is diagonal in the computational basis. A simple computation based on the geodesic equation shows that V(t) = exp(-i Ht).

21. Pauli geodesics on SU(2n) SU(2n) H diagonal in the computational basis. U Geodesic is V(t) = exp(-i Ht). I Given any U diagonal in computational basis, we can find a Pauli geodesic through U: H = i ln(U) ) V(1) = U Given such a Pauli geodesic, the length between I and U is: lF1 = F1(H) For a given U, there are typically many Pauli geodesics from I through U. exp(-iH’t), where H’ = H-J, and J is diagonal in the computational basis, with entries which are integer multiples of 2. Q1: Which Pauli geodesic through a given U is the shortest? Q2: Is the shortest Pauli geodesic also the shortest geodesic?

22. Freedom in the Pauli geodesics SU(2n) U diagonal in computational basis U Fix H such that U = exp(-iH) • Find J such that: • J diagonal in comp basis, with entries2 £ integers. • J minimizes F1(H-J) I The set of allowed J forms a lattice – that is, it is closed under integer linear combinations: n1 J1 + n2 J2 + … The problem of minimizing F(H-J) is thus the problem of finding the closest lattice element J to H, under the norm F1. Main point: Finding the minimum length Pauli geodesic is an instance of solving a problem well known in computer science, the closest vector to a lattice problem.

23. Consequences A volume argument shows that all but a tiny fraction of U have exponentially long minimal Pauli geodesics. Q: Can we explicitly construct a family of unitaries U that haveexponentially long minimal length Pauli geodesics? When is the shortest Pauli geodesic from I to U the shortest among all geodesics? A: In general, I don’t know. But we do have… Proposition: Suppose U is diagonal in the computational basis. If the shortest geodesic from I to U is unique then it must be a Pauli geodesic. Proof: Similar argument to that used earlier – if it is unique, then it must commute with Paulis which are products of I and Z, and so must be diagonal in computational basis ) Pauli geodesic.