A magnonic logic gate in the:
open Heisenberg chain

Gabriel T. Landi

Universidade Federal do ABC

In collaboration with Dragi Karevski from the Université de Lorraine

Magnonic devices

load["Figures/mag1.png",0.6]

Row[{
load["Figures/mag2.png",0.4],
load["Figures/mag3.png",0.4]}]

Open quantum systems

The 1D Heisenberg chain is described by the Hamiltonian

H = (J/2)(∑_( i = 1)^( N-1)) (Subscript[σ, i])·(Subscript[σ, i+1])

Our goal is to describe this quantum system in contact with an external environment.

Describe the injection and absorption of excitations (magnons).

This can be done using Lindblad dissipators

((dρ)/dt)=-i[H,ρ]+(Subscript[D, L])(ρ)+(Subscript[D, R])(ρ)

We choose (Subscript[D, L])(ρ) to be a perfect injector of magnons, or a magnon pump.

(Subscript[D, L])(ρ) = γ (2(Subscript[σ, 1]^(+))ρ (Subscript[σ, 1]^(-))-{(Subscript[σ, 1]^(-))(Subscript[σ, 1]^(+)),ρ})

It injects γ magnons/second at site 1.

Similarly, (Subscript[D, R])(ρ) is a perfect magnon drain.

(Subscript[D, R])(ρ)=γ(2 (Subscript[σ, N]^(-))ρ (Subscript[σ, N]^(+))-{(Subscript[σ, N]^(+))(Subscript[σ, N]^(-)),ρ})

It absorbs magnons with the same rate at site N

load["Figures/drawing0.png",0.5]

Our goal is to compute spin current J.

This is a fully quantum-mechanical problem.

It may therefore present novel effects not observed in semi-classical calculations.

Exact solution for the steady-state

These types of many-body problems are usually very difficult to solve.

Analytically: maybe 3 or 4 spins

Numerically (without DMRG): maybe 10 spins.

Numerically with DMRG: maybe 100. Very difficult.

This case is a nice exception.

load["Figures/prl.png",0.82]

An exact solution was found for any chain size in terms of matrix product states.

The solution is not a simple formula, but a recipe to compute J numerically for any size

In this talk I want to focus on the physics.

load["Figures/prb.png",0.8]

Ballistic vs. sub-diffusive

Spin current J vs. the pumping rate γ for different chain sizes.

load["Figures/h0.pdf",0.5]

Ohm’s law: J ∝ 1/N -> diffusive current.

Low γ -> low magnon density -> ballistic spin flux

Magnons propagate freely (they do not collide).

J independent of N

High γ -> sub-diffusive spin flux

Magnon scattering events hinder the flux.

J ∝ 1/(N^2)

Transition occurs at (SuperStar[γ]) = 1/N

Boundary fields

It was also possible to obtain a solution when the spins at the boundaries are subject to magnetic fields at opposite directions.

H = (1/2)(∑_( i = 1)^( N-1)) (Subscript[σ, i])·(Subscript[σ, i+1])+h((Subscript[σ, 1]^z) -(Subscript[σ, N]^z))

Column[{
load["Figures/drawing1.pdf",0.4],
load["Figures/drawing2.pdf",0.4]}]

Boundary fields

In this case we obtain a quite interesting result:

Row[{
Labeled[load["Figures/h1.pdf",0.35],Text[Style["high γ","Times",23]],Top],
Labeled[load["Figures/h2.pdf",0.35],Text[Style["low γ","Times",23]],Top]
}]

At low γ, as we change the boundary fields, we observe an abrupt transition:

Ballistic inside the plateau.

Sub-diffusive outside.

Magnon density

This can also be seen in the density of magnons along the chain:

Inside the plateau -> flat density -> no accumulation of magnons.

Outside -> accumulation of magnons -> strong magnon-magnon interaction.

load["Figures/density.pdf",0.4]

Physical explanation

The boundary fields act as scattering barriers which confine the magnons inside the chain.

Low γ -> low magnon density.

If h is low, the magnons propagate freely -> ballistic flux.

If h is large, it confine the magnons -> more scattering -> sub-diffusive flux.

The situation where we found an exact solution is peculiar, but the physical principle is quite general:

Use non-uniform magnetic fields to confine the magnons.

Tuning the field amplitude, you can tune the spin current.

load["Figures/h2.pdf",0.35]

By tuning the field around the transition, you can get huge variations in the spin current.

This is a very efficient magnonic logic gate.

And this is a genuinely quantum mechanical effect.

Why large sizes are important

At sufficiently large sizes we reach the scaling region.

The bahvior of  J becomes predictable

load["Figures/h2.pdf",0.35]

At small sizes, we get “finite-size effects”.

load["Figures/small.pdf",0.35]

Conclusions

Open quantum systems can describe magnonic circuits.

Density of magnons determines the flux regime.

Magnetic fields confine magnons -> induces scattering effects.

The main results of this presentation are contained in

G. T. Landi and D. Karevski, Phys. Rev. B. 91 174422 (2015)

For more details see:

D. Karevski, V. Popkov, G. M. Schütz, Phys. Rev. Lett. 110 047201 (2013)
	V. Popkov, D. Karevski, G. M. Schütz, Phys. Rev. E. 88 062118 (2013)	
	G. T. Landi, et. al., Phys. Rev. E. 90 042142 (2014)

Thank you very much.

Matrix product solution

Quick answer

The spin flux reads

J = ((2γ)/((γ^2)+(h^2)))((Z(N-1))/(Z(N)))

Where Z(N) is the (0,0) element of a matrix B raised to the power N

Z(N) = (Subscript[((B^N)), 00])

(Subscript[B, k,l]) = 2|p-k((|)^2)(Subscript[δ, k,l])+(l^2)(Subscript[δ, k,l-1])+|2p-k((|)^2)(Subscript[δ, k,l+1])

p = (i/(2(γ-i h)))

Thus, to find J the procedure is:

Construct this N×N matrix B

Multiply it by itself N times (there are quick ways to do this)

Take the (0,0) entry.

Detailed solution

Our goal is to find the solution of

i[H,ρ] = (Subscript[D, L])(ρ)+(Subscript[D, R])(ρ)

First we decompose

ρ = ((S†S)/(tr(S†S)))

We then write

S = ⟨ϕ|(Ω^(⊗N))|ψ⟩

where Ω is an operator valued 2×2 matrix

Ω = (Subscript[S, z])(Subscript[σ, z])+(SubPlus[S])(SubPlus[σ])+(SubMinus[S])(SubMinus[σ])

The operators (Subscript[S, a]) act on an auxiliary space.

By taking the inner product with ⟨ϕ| and |ψ⟩ we then recover S in the Hilbert space of the N spins.

From the bulk structure of the Hamiltonian we find that the (Subscript[S, a]) must obey the SU(2) algebra

In the XXZ model this generalizes to the quantum (Subscript[U, q])[SU(2)] algebra

[(Subscript[S, z]),(Subscript[S, ±])]=± (Subscript[S, ±])
[(SubPlus[S]),(SubMinus[S])]=2(Subscript[S, z])

We then choose a irreducible representation of this algebra as

(Subscript[S, z]) = (∑_( n = 0)^( ∞)) (p-n)  |n⟩⟨n|

The boundary structure then fixes

p = (i/(2(γ-i h)))

|ϕ⟩=|ψ⟩ = |0⟩

Which completes the formal solution.

Auxiliary functions

SetDirectory[NotebookDirectory[]];
<<"LinLib`";
<<"CustomTicks`";
load[filename_,size_]:=Show[Import[filename],ImageSize->Scaled[size]];

SetOptions[Plot,Frame->True,Axes->False,BaseStyle->20,ImageSize->400,PlotStyle->{Black}];

SetOptions[InputNotebook[], 
DefaultNewCellStyle -> "Item",
ShowCellLabel->"False",
CellGrouping->Manual,
FontFamily->"Times",
DefaultNewCellStyle->{"Text",FontFamily->"Times"},
BaseStyle->{FontFamily->"Times"},
MultiLetterItalics->False,
SingleLetterItalics->Automatic
]