ExperimentalExperimental demonstrationdemonstration ofof anyonicanyonic statisticsstatistics withwith photonsphotons

JKP Christian Schmid Witlef Wieczorek Reinhold Pohlner Nikolai Kiesel Harald Weinfurter

KITP, March 2007 Overview • In condensed matter anyons appear in ground or excited states of two dimensional systems: – Superconducting electrons in a strong magnetic field (Fractional Quantum Hall Effect) – Lattice systems (Kitaev’s toric code/hexagonal lattice, Wen’s models, Ioffe’s model, Freedman-Nayak- Shtengel model…) • Energy gap protects anyons: - if I get anyonic statistics I do not need gap. • Relatively large systems: - employ largest implementable system. • Close the gap between theory and experiment. Overview • Anyonic statistics is a property of a (highly entangled) wavefunction. – Engineer states rather than cool – same effect.

•Employ the toric code model. – One plaquette: one anyon and path of another. – No Hamiltonian: is like algorithmic encoding.

•How to generate, manipulate, measure anyons? •The control manipulations are exactly the same with Hamiltonian or larger system. • Future work: H ≠ 0, L >>1 Anyons

Anyons have non-trivial statistics.

Bosons Ψ → Ψ 3D Fermions Ψ → ei2π Ψ

i2ϕ 2D Ψ → e Ψ Ψ →U Ψ Anyons Consider as composite particles of fluxes and charges. Then phase is like the Aharonov-Bohm effect. Anyons: do they live among us?

Create two localized “things” with effective charge and magnetic field. Anyons: do they live among us?

Create two localized “things” with effective charge and magnetic field. Braid them -> PHASE FACTOR: Effective gauge theory! Toric Code (also ECC)

Consider the lattice Hamiltonian p

H = −∑ Z1Z2Z3Z4 − ∑ X1 X 2 X 3 X 4 s s p s p p p Spins live on the vertices. s s

There are two different types of p plaquettes, p and s, which support ZZZZ or XXXX X1 interactions respectively.

The four spin interactions involve X4 s X2 spins at the same plaquette.

X3 Toric Code

Consider the lattice Hamiltonian p

H = −∑ Z1Z2Z3Z4 − ∑ X1 X 2 X 3 X 4 s s p s It is easy to find the ground p p p state of this Hamiltonian. s s First observe that p [H, Z1Z2Z3Z4 ] = 0,[H, X1 X 2 X 3X 4 ] = 0

[X1 X 2 X 3X 4 , Z1Z2Z3Z4 ] = 0 X1 2 2 (X1 X 2 X 3X 4 ) = (Z1Z2Z3Z4 ) =1 X4 s X2 Eigenvalues of XXXX and ZZZZ terms are 1 and -1 X3 Toric Code

Consider the lattice Hamiltonian p

H = −∑ Z1Z2Z3Z4 − ∑ X1 X 2 X 3 X 4 s s p s Hence, the ground state is: p p p

s s ξ = ∏(Ι + X1 X 2 X 3 X 4 ) p 00...0 s p The |00…0> state is a ground state of the ZZZZ terms and the (I+XXXX) X1 term projects that state to the ground s state of the XXXX term. X4 X2

[F. Verstraete, et al. PRL, 96, 220601 (2006)] X3 Toric Code

• Excitations are produced by Z or X rotations of one spin. p Z s s • These rotations anticommute p p p with the X or Z part of the Hamiltonian, respectively. s X s

p • Z excitations on s plaquettes.

• X excitations on p plaquettes. X1

X4 s X2 • X and Z excitations behave as anyons with respect to each other. X3 One plaquette

It is possible to demonstrate the anyonic properties with one s plaquette only. Then the Hamiltonian takes the form

H = −X1 X 2 X 3 X 4

− Z1Z2 − Z2Z3 − Z3Z4 − Z4Z1 1 The following state is the 2 ground state s 4 1 ξ = (Ι + X1 X 2 X 3 X 4 ) 01020304 2 3 1 = ( 0 0 0 0 + 11 1 1 ) 2 1 2 3 4 1 2 3 4 GHZ state! One plaquette

One can demonstrate the anyonic statistics with only this plaquette. First create excitation with Z rotation at one spin: 1 Z1 Z = Z1 ξ = ( 01020304 − 11121314 ) 2 1 Energy of ground state 2 s 4 H ξ = −5 ξ Z Energy of excited state 3 H Z = −3 Z One plaquette

One can demonstrate the anyonic statistics with only this plaquette. First create excitation with Z rotation at one spin: 1 Z1 Z = Z1 ξ = ( 01020304 − 11121314 ) 2 1 Now we want to move an X anyon around the Z one. What we really 2 s 4 want is the path that it traces and Z this can be spanned on the spins 1,2,3,4. 3 Note that the second anyon from the Z rotation is outside the system. One plaquette

One can demonstrate the anyonic statistics with only this plaquette. First create excitation with Z rotation at one spin: 1 Z1 Z = Z1 ξ = ( 01020304 − 11121314 ) 2 1 Assume there is an X anyon outside the system. With 2 s 4 successive X rotations it can be Z transported around the plaquette. 3 One plaquette

One can demonstrate the anyonic statistics with only this plaquette. First create excitation with Z rotation at one spin: Z1 1 X1 Z = Z1 ξ = ( 01020304 − 11121314 ) 2 1 Assume there is an X anyon outside the system. With 2 s 4 successive X rotations it can be Z transported around the plaquette. 3 One plaquette

One can demonstrate the anyonic statistics with only this plaquette. First create excitation with Z rotation at one spin: Z1 1 X1 Z = Z1 ξ = ( 01020304 − 11121314 ) 2 1

Assume there is an X anyon X2 outside the system. With 2 s 4 successive X rotations it can be Z transported around the plaquette. 3 One plaquette

One can demonstrate the anyonic statistics with only this plaquette. First create excitation with Z rotation at one spin: Z1 1 X1 Z = Z1 ξ = ( 01020304 − 11121314 ) 2 1

Assume there is an X anyon X2 outside the system. With 2 s 4 successive X rotations it can be Z transported around the plaquette. 3 X3 One plaquette

One can demonstrate the anyonic statistics with only this plaquette. First create excitation with Z rotation at one spin: Z1 1 X1 Z = Z1 ξ = ( 01020304 − 11121314 ) 2 1

Assume there is an X anyon X2 outside the system. With 2 s 4 X4 successive X rotations it can be Z transported around the plaquette. The final state is given by: 3 X3 Final = X1 X 2 X 3 X 4 Z = 1 − ( 0 0 0 0 − 11 1 1 ) = − Initial 2 1 2 3 4 1 2 3 4 One plaquette

Final = X1 X 2 X 3 X 4 Z = 1 − ( 0 0 0 0 − 11 1 1 ) = − Initial 2 1 2 3 4 1 2 3 4

After a complete rotation of an X anyon around a Z anyon (two successive exchanges) the resulting state gets a phase π (a minus sign): hence ANYONS!

A property we used is that X1 X 2 X 3 X 4 ξ = ξ which is true. A crucial point is that these properties can be demonstrated without the Hamiltonian!!! An interference experiment can reveal the presence of the phase factor. Interference Experiment

Create state 1 ξ = ( 0 0 0 0 + 11 1 1 ) 2 1 2 3 4 1 2 3 4 1/ 2 With half of an Z rotation on spin 1, Z 1 , one can create the superposition between an Z anyon and the vacuum: −iϕ 1/ 2 e Z1 ξ = ( ξ + i Z ) / 2 for ϕ = 3 π / 4 . Then the X anyon is rotated around it:

X1 X 2 X 3 X 4 ( ξ + i Z ) / 2 = ( ξ − i Z ) / 2 Then we make the inverse rotation

iϕ −1/ 2 e Z1 ( ξ − i Ζ ) / 2 = Ζ Interference Experiment

That we obtained the Z state is due to the minus sign produced from the anyonic statistics. If it wasn’t there then we would have returned to the vacuum state ξ . Distinguish between ξ and Z states:

H ⊗4 ξ has even number of 1’s. ⊗4 H Z has odd number of 1’s.

H ⊗4 ξ ∝ 0000 + 0011 + 0101 + 0110 + 1001 + 1010 + 1100 + 1111

H ⊗4 Z ∝ 0001 + 0010 + 0100 + 1000 + 0111 + 1011 + 1101 + 1110 Experimental process (preliminary) Qubit states 0 and 1 are encoded in the polarization, V and H, of four photonic Vacuum modes. 1

Counts ξ = ( 0000 + 1111 ) The states that come from 2 this setup are of the form: Ψ = a GHZ + b EPR ⊗ EPR = a( HHHH + VVVV ) + b( VHVH + HVVH + VHHV + HVHV ) Anyon 1

Counts Z = ( 0000 − 1111 ) Measurements and 2 manipulations are repeated over all modes. Experimental process (preliminary)

Consider correlations: Vacuum tr[(cosγσ x + sin γσ y )⊗4 ξ ξ ]= + cos(4γ )

x y ⊗4 tr[](cosγσ + sin γσ ) Z Z = −cos(4γ ) γ Visibility > 64% Correlations x displacement: -7° and +2° Fidelity: 2 2 * * F =| a1 | + | a16 | +a1 a16 + a1a16 > 70% Anyon Witness for genuine 4-qubit entanglement: 1 γ WGHZ = 1− GHZ4 GHZ4 4 2 Correlations ⇒ tr(W ρ) < 0 GHZ4 y displacement ~ EPRxEPR Conclusions • Invariance of vacuum w.r.t. to closed paths: XXXX ξ = ξ 4 qubit GHZ stabilizers Zi Z j ξ = ξ

Z = Zi ξ •Fusion rules: Zi Z j ξ = ξ e×e =1 Z Z Z = Z 1×e = e Useful for: i j • quantum anonymous broadcasting, • quantum error correction, • topological quantum memory (?) … Non-abelian statistics can be detected similarly. Implement Hamiltonian and larger systems. Annals of Physics, IJQI