Minimally Entangled Typical Thermal States (METTS)

Minimally Entangled Typical Thermal States (METTS) Vijay B. Shenoy Centre for Condensed Matter Theory, IISc Bangalore shenoy@physics.iisc.ernet.in Quantum Condensed Matter Journal Club April 17, 2012 1 / 33

Acknowledgements 2 / 33

Acknowledgements Funding: DST, DAE 2 / 33

Acknowledgements Funding: DST, DAE 2 / 33

Acknowledgements Funding: DST, DAE Amal Medhi 2 / 33

Papers in Focus 3 / 33

Papers in Focus 3 / 33

Papers in Focus 3 / 33

Papers in Focus 3 / 33

Papers in Focus Main source: Talk by M. Stoudenmire at the KSK Discussion Meeting, March 19-21, 2012, MatScience, Chennai 3 / 33

Overview Part I: Difficulties with quantum many body simulations 4 / 33

Overview Part I: Difficulties with quantum many body simulations Part II: Ideas of entanglement and entanglement entropy 4 / 33

Overview Part I: Difficulties with quantum many body simulations Part II: Ideas of entanglement and entanglement entropy Part III: METTS Algorithm 4 / 33

Overview Part I: Difficulties with quantum many body simulations Part II: Ideas of entanglement and entanglement entropy Part III: METTS Algorithm Part IV: (Brief) Technical remarks and our plan 4 / 33

Overview Part I: Difficulties with quantum many body simulations Part II: Ideas of entanglement and entanglement entropy Part III: METTS Algorithm Part IV: (Brief) Technical remarks and our plan 4 / 33

I. Numerical Simulations of Quantum Many Body Systems Difficulties 5 / 33

A Simple Model - TFIM 6 / 33

A Simple Model - TFIM The transverse field Ising model (TFIM) 6 / 33

A Simple Model - TFIM The transverse field Ising model (TFIM) A lattice of N sites with a spin- 1 2 at each site 6 / 33

A Simple Model - TFIM The transverse field Ising model (TFIM) A lattice of N sites with a spin- 1 2 at each site Ferro-magnetic Ising interaction between neighbouring spins 6 / 33

A Simple Model - TFIM The transverse field Ising model (TFIM) A lattice of N sites with a spin- 1 2 at each site Ferro-magnetic Ising interaction between neighbouring spins A homogeneous transverse magnetic field h 6 / 33

A Simple Model - TFIM The transverse field Ising model (TFIM) A lattice of N sites with a spin- 1 2 at each site Ferro-magnetic Ising interaction between neighbouring spins A homogeneous transverse magnetic field h H = S z i Sz j + h ij i S x i 6 / 33

A Simple Model - TFIM The transverse field Ising model (TFIM) A lattice of N sites with a spin- 1 2 at each site Ferro-magnetic Ising interaction between neighbouring spins A homogeneous transverse magnetic field h H = S z i Sz j + h ij i S x i Z 2 symmetry (spin rotation by π about the x-axis) 6 / 33

A Simple Model - TFIM The transverse field Ising model (TFIM) A lattice of N sites with a spin- 1 2 at each site Ferro-magnetic Ising interaction between neighbouring spins A homogeneous transverse magnetic field h H = S z i Sz j + h ij i S x i Z 2 symmetry (spin rotation by π about the x-axis) Goal: Obtain the h T phase diagram of TFIM numerically (T is temperature) 6 / 33

A Simple Model - TFIM The transverse field Ising model (TFIM) A lattice of N sites with a spin- 1 2 at each site Ferro-magnetic Ising interaction between neighbouring spins A homogeneous transverse magnetic field h H = S z i Sz j + h ij i S x i Z 2 symmetry (spin rotation by π about the x-axis) Goal: Obtain the h T phase diagram of TFIM numerically (T is temperature) Size of Hilbert space M = 2 N = e N ln 2... 6 / 33

A Simple Model - TFIM The transverse field Ising model (TFIM) A lattice of N sites with a spin- 1 2 at each site Ferro-magnetic Ising interaction between neighbouring spins A homogeneous transverse magnetic field h H = S z i Sz j + h ij i S x i Z 2 symmetry (spin rotation by π about the x-axis) Goal: Obtain the h T phase diagram of TFIM numerically (T is temperature) Size of Hilbert space M = 2 N = e N ln 2...Hilbert space grows exponentially! 6 / 33

A Simple Model - TFIM The transverse field Ising model (TFIM) A lattice of N sites with a spin- 1 2 at each site Ferro-magnetic Ising interaction between neighbouring spins A homogeneous transverse magnetic field h H = S z i Sz j + h ij i S x i Z 2 symmetry (spin rotation by π about the x-axis) Goal: Obtain the h T phase diagram of TFIM numerically (T is temperature) Size of Hilbert space M = 2 N = e N ln 2...Hilbert space grows exponentially! Matrix diagonalization difficulty goes as M 3 e3 ln 2N 6 / 33

A Simple Model - TFIM The transverse field Ising model (TFIM) A lattice of N sites with a spin- 1 2 at each site Ferro-magnetic Ising interaction between neighbouring spins A homogeneous transverse magnetic field h H = S z i Sz j + h ij i S x i Z 2 symmetry (spin rotation by π about the x-axis) Goal: Obtain the h T phase diagram of TFIM numerically (T is temperature) Size of Hilbert space M = 2 N = e N ln 2...Hilbert space grows exponentially! Matrix diagonalization difficulty goes as M 3 e3 ln 2N Difficulty generic to all quantum problems! 6 / 33

TFIM in 1D 7 / 33

TFIM in 1D TFIM on a 1D chain is exactly soluble 7 / 33

TFIM in 1D TFIM on a 1D chain is exactly soluble At T = 0, there are two phases 7 / 33

TFIM in 1D TFIM on a 1D chain is exactly soluble At T = 0, there are two phases For h 1: Z 2 -broken ferromagnetic phase S z 0 7 / 33

TFIM in 1D TFIM on a 1D chain is exactly soluble At T = 0, there are two phases For h 1: Z 2 -broken ferromagnetic phase S z 0 For h 1: Z2 symmetric (unique) state S x 0 7 / 33

TFIM in 1D TFIM on a 1D chain is exactly soluble At T = 0, there are two phases For h 1: Z 2 -broken ferromagnetic phase S z 0 For h 1: Z2 symmetric (unique) state S x 0 Quantum phase transition at h = 1 2 7 / 33

TFIM in 1D TFIM on a 1D chain is exactly soluble At T = 0, there are two phases For h 1: Z 2 -broken ferromagnetic phase S z 0 For h 1: Z2 symmetric (unique) state S x 0 Quantum phase transition at h = 1 2 Critical point described by a gapless free fermion theory (CFT with central charge 1/2) 7 / 33

TFIM in 1D TFIM on a 1D chain is exactly soluble At T = 0, there are two phases For h 1: Z 2 -broken ferromagnetic phase S z 0 For h 1: Z2 symmetric (unique) state S x 0 Quantum phase transition at h = 1 2 Critical point described by a gapless free fermion theory (CFT with central charge 1/2)... 7 / 33

TFIM in 1D TFIM on a 1D chain is exactly soluble At T = 0, there are two phases For h 1: Z 2 -broken ferromagnetic phase S z 0 For h 1: Z2 symmetric (unique) state S x 0 Quantum phase transition at h = 1 2 Critical point described by a gapless free fermion theory (CFT with central charge 1/2)... Strategy: Use TFIM as a tool to understand METTS 7 / 33

TFIM in 1D TFIM on a 1D chain is exactly soluble At T = 0, there are two phases For h 1: Z 2 -broken ferromagnetic phase S z 0 For h 1: Z2 symmetric (unique) state S x 0 Quantum phase transition at h = 1 2 Critical point described by a gapless free fermion theory (CFT with central charge 1/2)... Strategy: Use TFIM as a tool to understand METTS All calculations shown here with N = 10 (M = 1024) done on this laptop!... Essentially impossible if N = 14 7 / 33

II. All is not lost -Entanglement and Entanglement Entropy 8 / 33

Difficulty: A Second Look 9 / 33

Difficulty: A Second Look Use the states s 1, s 2,..., s n the product states as the basis (s i = ± 1 2 ) 9 / 33

Difficulty: A Second Look Use the states s 1, s 2,..., s n the product states as the basis (s i = ± 1 2 ) A generic state ψ = ψ(s 1, s 2,..., s n ) s 1, s 2,..., s n 9 / 33

Difficulty: A Second Look Use the states s 1, s 2,..., s n the product states as the basis (s i = ± 1 2 ) A generic state ψ = ψ(s 1, s 2,..., s n ) s 1, s 2,..., s n...we need 2 N pieces of information to describe a typical state...looks complicated! 9 / 33

Difficulty: A Second Look Use the states s 1, s 2,..., s n the product states as the basis (s i = ± 1 2 ) A generic state ψ = ψ(s 1, s 2,..., s n ) s 1, s 2,..., s n...we need 2 N pieces of information to describe a typical state...looks complicated! Question: How complicated is a state ψ? Is there a way to quantify this? 9 / 33

Difficulty: A Second Look Use the states s 1, s 2,..., s n the product states as the basis (s i = ± 1 2 ) A generic state ψ = ψ(s 1, s 2,..., s n ) s 1, s 2,..., s n...we need 2 N pieces of information to describe a typical state...looks complicated! Question: How complicated is a state ψ? Is there a way to quantify this? Idea: If the state ψ is made up by gluing states of different parts of the full system then the wave function is not complicated 9 / 33

Difficulty: A Second Look Use the states s 1, s 2,..., s n the product states as the basis (s i = ± 1 2 ) A generic state ψ = ψ(s 1, s 2,..., s n ) s 1, s 2,..., s n...we need 2 N pieces of information to describe a typical state...looks complicated! Question: How complicated is a state ψ? Is there a way to quantify this? Idea: If the state ψ is made up by gluing states of different parts of the full system then the wave function is not complicated Need to find a way to see if the state can be split up into pieces that reside on different (spatial) parts of the system 9 / 33

Splitting up a state 10 / 33

Splitting up a state 10 / 33

Splitting up a state Imagine a system broken up in to two parts A and B 10 / 33

Splitting up a state Imagine a system broken up in to two parts A and B The product states in A(B) are i A ( j B ) 10 / 33

Splitting up a state Imagine a system broken up in to two parts A and B The product states in A(B) are i A ( j B ) The state ψ of the system can written as ψ = ψ(i A, j B ) i A, j B, i A, j B = i A j B i A,j B since i A, j B span the Hilbert space of the full system 10 / 33

Splitting up a state Imagine a system broken up in to two parts A and B The product states in A(B) are i A ( j B ) The state ψ of the system can written as ψ = ψ(i A, j B ) i A, j B, i A, j B = i A j B i A,j B since i A, j B span the Hilbert space of the full system Question: Suppose I want to keep information about ψ using i A, what should I keep so that it most faithfully represents ψ? 10 / 33

The Reduced Density Matrix 11 / 33

The Reduced Density Matrix Most faithfully means: Let O A be an observable that is defined only on sub-system A..ask that ψ O A ψ to be reproduced exactly for every O A with the information that we keep! 11 / 33

The Reduced Density Matrix Most faithfully means: Let O A be an observable that is defined only on sub-system A..ask that ψ O A ψ to be reproduced exactly for every O A with the information that we keep! It is easy to show that the information about ψ that we need to keep is the reduced density matrix 11 / 33

The Reduced Density Matrix Most faithfully means: Let O A be an observable that is defined only on sub-system A..ask that ψ O A ψ to be reproduced exactly for every O A with the information that we keep! It is easy to show that the information about ψ that we need to keep is the reduced density matrix ρ A = i A,i A ρ A (i A, i A) i A i A, ρ A (i A, i A) = j B ψ (i A, j B)ψ(i A, j B ) which is an operator defined on A by tracing out the states of B in ψ 11 / 33

The Reduced Density Matrix Most faithfully means: Let O A be an observable that is defined only on sub-system A..ask that ψ O A ψ to be reproduced exactly for every O A with the information that we keep! It is easy to show that the information about ψ that we need to keep is the reduced density matrix ρ A = i A,i A ρ A (i A, i A) i A i A, ρ A (i A, i A) = j B ψ (i A, j B)ψ(i A, j B ) which is an operator defined on A by tracing out the states of B in ψ For every O A, ψ O A ψ = tr(ρ A O A ) 11 / 33

The Reduced Density Matrix Most faithfully means: Let O A be an observable that is defined only on sub-system A..ask that ψ O A ψ to be reproduced exactly for every O A with the information that we keep! It is easy to show that the information about ψ that we need to keep is the reduced density matrix ρ A = i A,i A ρ A (i A, i A) i A i A, ρ A (i A, i A) = j B ψ (i A, j B)ψ(i A, j B ) which is an operator defined on A by tracing out the states of B in ψ For every O A, ψ O A ψ = tr(ρ A O A ) ρ A is non-negative Hermitian 11 / 33

The Reduced Density Matrix Most faithfully means: Let O A be an observable that is defined only on sub-system A..ask that ψ O A ψ to be reproduced exactly for every O A with the information that we keep! It is easy to show that the information about ψ that we need to keep is the reduced density matrix ρ A = i A,i A ρ A (i A, i A) i A i A, ρ A (i A, i A) = j B ψ (i A, j B)ψ(i A, j B ) which is an operator defined on A by tracing out the states of B in ψ For every O A, ψ O A ψ = tr(ρ A O A ) ρ A is non-negative Hermitian trρ A = 1 11 / 33

Entanglement Spectrum and Entanglement Entropy 12 / 33

Entanglement Spectrum and Entanglement Entropy We can diagonalize ρ A = α wα A α A α A (note 0 w α A 1) 12 / 33

Entanglement Spectrum and Entanglement Entropy We can diagonalize ρ A = α wα A α A α A (note 0 w α A 1) The spectrum w α A is called the entanglement spectrum 12 / 33

Entanglement Spectrum and Entanglement Entropy We can diagonalize ρ A = α wα A α A α A (note 0 w α A 1) The spectrum w α A is called the entanglement spectrum Back to the question: How complicated is ψ?... 12 / 33

Entanglement Spectrum and Entanglement Entropy We can diagonalize ρ A = α wα A α A α A (note 0 w α A 1) The spectrum w α A is called the entanglement spectrum Back to the question: How complicated is ψ?...if w α A is such that it is non-zero only for, say, α = 0, then ψ can be broken up...in fact, ψ = 0 A φ B...need much less 2 N numbers to describe ψ! 12 / 33

Entanglement Spectrum and Entanglement Entropy We can diagonalize ρ A = α wα A α A α A (note 0 w α A 1) The spectrum w α A is called the entanglement spectrum Back to the question: How complicated is ψ?...if w α A is such that it is non-zero only for, say, α = 0, then ψ can be broken up...in fact, ψ = 0 A φ B...need much less 2 N numbers to describe ψ! If there are many w α A 0, there is non trivial entanglement 12 / 33

Entanglement Spectrum and Entanglement Entropy We can diagonalize ρ A = α wα A α A α A (note 0 w α A 1) The spectrum w α A is called the entanglement spectrum Back to the question: How complicated is ψ?...if w α A is such that it is non-zero only for, say, α = 0, then ψ can be broken up...in fact, ψ = 0 A φ B...need much less 2 N numbers to describe ψ! If there are many w α A 0, there is non trivial entanglement We can quantify the amount of entanglement by the entanglement entropy S A = tr(ρ A log ρ A ) = α w α A log wα A 12 / 33

Entanglement Spectrum and Entanglement Entropy We can diagonalize ρ A = α wα A α A α A (note 0 w α A 1) The spectrum w α A is called the entanglement spectrum Back to the question: How complicated is ψ?...if w α A is such that it is non-zero only for, say, α = 0, then ψ can be broken up...in fact, ψ = 0 A φ B...need much less 2 N numbers to describe ψ! If there are many w α A 0, there is non trivial entanglement We can quantify the amount of entanglement by the entanglement entropy S A = tr(ρ A log ρ A ) = α w α A log wα A Example for the 1D chain shown: 12 / 33

Entanglement Spectrum and Entanglement Entropy We can diagonalize ρ A = α wα A α A α A (note 0 w α A 1) The spectrum w α A is called the entanglement spectrum Back to the question: How complicated is ψ?...if w α A is such that it is non-zero only for, say, α = 0, then ψ can be broken up...in fact, ψ = 0 A φ B...need much less 2 N numbers to describe ψ! If there are many w α A 0, there is non trivial entanglement We can quantify the amount of entanglement by the entanglement entropy S A = tr(ρ A log ρ A ) = α w α A log wα A Example for the 1D chain shown: Only one w α A 0 mean S A = 0...unentangled 12 / 33

Entanglement Spectrum and Entanglement Entropy We can diagonalize ρ A = α wα A α A α A (note 0 w α A 1) The spectrum w α A is called the entanglement spectrum Back to the question: How complicated is ψ?...if w α A is such that it is non-zero only for, say, α = 0, then ψ can be broken up...in fact, ψ = 0 A φ B...need much less 2 N numbers to describe ψ! If there are many w α A 0, there is non trivial entanglement We can quantify the amount of entanglement by the entanglement entropy S A = tr(ρ A log ρ A ) = α w α A log wα A Example for the 1D chain shown: Only one w α A 0 mean S A = 0...unentangled If all w α A are nearly equal...s A L...badly entangled 12 / 33

Entanglement Spectrum and Entanglement Entropy We can diagonalize ρ A = α wα A α A α A (note 0 w α A 1) The spectrum w α A is called the entanglement spectrum Back to the question: How complicated is ψ?...if w α A is such that it is non-zero only for, say, α = 0, then ψ can be broken up...in fact, ψ = 0 A φ B...need much less 2 N numbers to describe ψ! If there are many w α A 0, there is non trivial entanglement We can quantify the amount of entanglement by the entanglement entropy S A = tr(ρ A log ρ A ) = α w α A log wα A Example for the 1D chain shown: Only one w α A 0 mean S A = 0...unentangled If all w α A are nearly equal...s A L...badly entangled 12 / 33

How complicated is a state?...and Area Law 13 / 33

How complicated is a state?...and Area Law To study this vary the subsystem size L and obtain S A (L) 13 / 33

How complicated is a state?...and Area Law To study this vary the subsystem size L and obtain S A (L) If S A (L) grows as L the state is highly entangled! 13 / 33

How complicated is a state?...and Area Law To study this vary the subsystem size L and obtain S A (L) If S A (L) grows as L the state is highly entangled! Bad news: 13 / 33

How complicated is a state?...and Area Law To study this vary the subsystem size L and obtain S A (L) If S A (L) grows as L the state is highly entangled! Bad news: A typical state in the Hilbert space is highly entangled! 13 / 33

How complicated is a state?...and Area Law To study this vary the subsystem size L and obtain S A (L) If S A (L) grows as L the state is highly entangled! Bad news: A typical state in the Hilbert space is highly entangled! Give up? 13 / 33

How complicated is a state?...and Area Law To study this vary the subsystem size L and obtain S A (L) If S A (L) grows as L the state is highly entangled! Bad news: A typical state in the Hilbert space is highly entangled! Give up? No: 13 / 33

How complicated is a state?...and Area Law To study this vary the subsystem size L and obtain S A (L) If S A (L) grows as L the state is highly entangled! Bad news: A typical state in the Hilbert space is highly entangled! Give up? No: Ground states are special! 13 / 33

How complicated is a state?...and Area Law To study this vary the subsystem size L and obtain S A (L) If S A (L) grows as L the state is highly entangled! Bad news: A typical state in the Hilbert space is highly entangled! Give up? No: Ground states are special! Area Law The entanglement entropy of the ground states of nice Hamiltonians in d-dimensions satisfies the area law S A (L) L d 1 d where L is the volume of the subsystem A! Ground states are such that information shared between subsystems is proportional to the interfacial area! 13 / 33

How complicated is a state?...and Area Law To study this vary the subsystem size L and obtain S A (L) If S A (L) grows as L the state is highly entangled! Bad news: A typical state in the Hilbert space is highly entangled! Give up? No: Ground states are special! Area Law The entanglement entropy of the ground states of nice Hamiltonians in d-dimensions satisfies the area law S A (L) L d 1 d where L is the volume of the subsystem A! Ground states are such that information shared between subsystems is proportional to the interfacial area! In 1D this means that S A (L) = Const! This is because the interface is a set of two points! 13 / 33

How complicated is a state?...and Area Law To study this vary the subsystem size L and obtain S A (L) If S A (L) grows as L the state is highly entangled! Bad news: A typical state in the Hilbert space is highly entangled! Give up? No: Ground states are special! Area Law The entanglement entropy of the ground states of nice Hamiltonians in d-dimensions satisfies the area law S A (L) L d 1 d where L is the volume of the subsystem A! Ground states are such that information shared between subsystems is proportional to the interfacial area! In 1D this means that S A (L) = Const! This is because the interface is a set of two points! Crux: It is much easier to describe ground states! 13 / 33

Area Law in Action 14 / 33

Area Law in Action Check the area law in TFIM 14 / 33

Area Law in Action Check the area law in TFIM Exact diagonalization with N = 10 14 / 33

Area Law in Action Check the area law in TFIM Exact diagonalization with N = 10 0.8 S (Entanglement Entropy) 0.6 0.4 L 2 3 4 0.2 5 6 0 0 0.25 0.5 0.75 1 h(transverse magnetic field) 14 / 33

Area Law in Action Check the area law in TFIM Exact diagonalization with N = 10 0.8 S (Entanglement Entropy) 0.6 0.4 L 2 3 4 0.2 5 6 0 0 0.25 0.5 0.75 1 h(transverse magnetic field) Away from the critical h, area law holds exceedingly well! 14 / 33

Area Law in Action Check the area law in TFIM Exact diagonalization with N = 10 0.8 S (Entanglement Entropy) 0.6 0.4 L 2 3 4 0.2 5 6 0 0 0.25 0.5 0.75 1 h(transverse magnetic field) Away from the critical h, area law holds exceedingly well! Critical ground states seem to be more complicated! 14 / 33

Critical Ground States 15 / 33

Critical Ground States Critical ground states are more complicated 15 / 33

Critical Ground States Critical ground states are more complicated Good news: they are not highly entangled 15 / 33

Critical Ground States Critical ground states are more complicated Good news: they are not highly entangled In a 1D critical ground state (roughly) S A (L) c log(l) 15 / 33

Critical Ground States Critical ground states are more complicated Good news: they are not highly entangled In a 1D critical ground state (roughly) S A (L) c log(l) c is the central charge of the CFT describing the critical point! 15 / 33

Excited States? 16 / 33

Excited States? Does area law hold for excited states? 16 / 33

Excited States? Does area law hold for excited states? For h = 0.25, vary L = 2, 3, 4, 5 (N = 10) 16 / 33

Excited States? Does area law hold for excited states? For h = 0.25, vary L = 2, 3, 4, 5 (N = 10) 3 2.5 S(Ent. Entropy) 2 1.5 1 0.5 0 0 250 500 750 1000 EigenState Number 16 / 33

Excited States? Does area law hold for excited states? For h = 0.25, vary L = 2, 3, 4, 5 (N = 10) 3 2.5 S(Ent. Entropy) 2 1.5 1 0.5 0 0 250 500 750 1000 EigenState Number 16 / 33

Excited States? Does area law hold for excited states? For h = 0.25, vary L = 2, 3, 4, 5 (N = 10) 3 2.5 S(Ent. Entropy) 2 1.5 1 0.5 0 0 250 500 750 1000 EigenState Number 16 / 33

Excited States? Does area law hold for excited states? For h = 0.25, vary L = 2, 3, 4, 5 (N = 10) 3 2.5 S(Ent. Entropy) 2 1.5 1 0.5 0 0 250 500 750 1000 EigenState Number S ground and highest state (and their neighbours) are unaltered...consistent with area law (if H is nice, so is H) 16 / 33

Excited States? Does area law hold for excited states? For h = 0.25, vary L = 2, 3, 4, 5 (N = 10) 3 2.5 S(Ent. Entropy) 2 1.5 1 0.5 0 0 250 500 750 1000 EigenState Number S ground and highest state (and their neighbours) are unaltered...consistent with area law (if H is nice, so is H) But S of a typical excited state grows with L...very bad news for finite temperature simulations! 16 / 33

Excited States? Does area law hold for excited states? For h = 0.25, vary L = 2, 3, 4, 5 (N = 10) 3 2.5 S(Ent. Entropy) 2 1.5 1 0.5 0 0 250 500 750 1000 EigenState Number S ground and highest state (and their neighbours) are unaltered...consistent with area law (if H is nice, so is H) But S of a typical excited state grows with L...very bad news for finite temperature simulations! 16 / 33

A Question 17 / 33

A Question Ground state obeys area law Excited states are highly entangled... Question: Is there a critical excited state? Is there an Anderson Transition somewhere in the spectrum? My prejudice: YES! 17 / 33

18 / 33

Summary of Part II 18 / 33

Summary of Part II Ideas of entanglement, entanglement entropy 18 / 33

Summary of Part II Ideas of entanglement, entanglement entropy Area Law: ground states are easy(easier) to describe...the amount of information required to describe a ground state scales as N (linear scaling) 18 / 33

Summary of Part II Ideas of entanglement, entanglement entropy Area Law: ground states are easy(easier) to describe...the amount of information required to describe a ground state scales as N (linear scaling) Critical ground states are harder 18 / 33

Summary of Part II Ideas of entanglement, entanglement entropy Area Law: ground states are easy(easier) to describe...the amount of information required to describe a ground state scales as N (linear scaling) Critical ground states are harder Excited states...tough! 18 / 33

Summary of Part II Ideas of entanglement, entanglement entropy Area Law: ground states are easy(easier) to describe...the amount of information required to describe a ground state scales as N (linear scaling) Critical ground states are harder Excited states...tough! Not a good omen for finite temperature calculations! We may be stuck in the cold confines of the ground state! 18 / 33

III. The way to warmth: METTS 19 / 33

Finite T Party Line 20 / 33

Finite T Party Line Hamiltonian H = n E n Ψ n Ψ n 20 / 33

Finite T Party Line Hamiltonian H = n E n Ψ n Ψ n Any observable O (β = 1/T) O = tr(e βh O) tre βh = n e βen Z Ψ n O ψ n Not possible in practise since it is very hard to describe (and find) Ψ n 20 / 33

METTS Idea 21 / 33

METTS Idea 21 / 33

METTS Idea Idea of METTS is to find a (complete) set of states with minimal entanglement that describe the Boltzmann density matrix 21 / 33

METTS Idea Idea of METTS is to find a (complete) set of states with minimal entanglement that describe the Boltzmann density matrix Minimum entanglement means that the entanglement entropy of any such state should follow the area law as closely as possible 21 / 33

METTS Idea Idea of METTS is to find a (complete) set of states with minimal entanglement that describe the Boltzmann density matrix Minimum entanglement means that the entanglement entropy of any such state should follow the area law as closely as possible White provides a prescription to do this...he calls the state METTS 21 / 33

Schrödinger vs. Feynmann...battle in Steven s mind...schrödinger s seems to win! 22 / 33

METTS formulation 23 / 33

METTS formulation Say we have a complete set of states c 23 / 33

METTS formulation Say we have a complete set of states c O = 1 c e βh O c = 1 c e βh 2 Oe βh 2 c = φ(c) O φ(c) Z Z c c c where φ(c) = e βh 2 Z c is an unnormalized state 23 / 33

METTS formulation Say we have a complete set of states c O = 1 c e βh O c = 1 c e βh 2 Oe βh 2 c = φ(c) O φ(c) Z Z βh where φ(c) = e 2 Z Normalize φ(c) c c c is an unnormalized state φ(c) φ(c) = P(c) = c e βh c, φ(c) = 1 φ(c) Z P(c) c 23 / 33

METTS formulation Say we have a complete set of states c O = 1 c e βh O c = 1 c e βh 2 Oe βh 2 c = φ(c) O φ(c) Z Z βh where φ(c) = e 2 Z Normalize φ(c) c c c is an unnormalized state φ(c) φ(c) = P(c) = c e βh c, φ(c) = 1 φ(c) Z P(c) We can rewrite this as c O = c P(c) φ(c) O φ(c) 23 / 33

METTS formulation Say we have a complete set of states c O = 1 c e βh O c = 1 c e βh 2 Oe βh 2 c = φ(c) O φ(c) Z Z βh where φ(c) = e 2 Z Normalize φ(c) c c c is an unnormalized state φ(c) φ(c) = P(c) = c e βh c, φ(c) = 1 φ(c) Z P(c) We can rewrite this as c O = c P(c) φ(c) O φ(c) Key requirement φ(c) s should have as little entanglement as possible! 23 / 33

METTS Formulation 24 / 33

METTS Formulation METTS idea I: Choose c to be the product states s 1, s 2,..., s N 24 / 33

METTS Formulation METTS idea I: Choose c to be the product states s 1, s 2,..., s N These c have zero entanglement...and represent the typical high energy states of the system 24 / 33

METTS Formulation METTS idea I: Choose c to be the product states s 1, s 2,..., s N These c have zero entanglement...and represent the typical high energy states of the system The state φ(c) will also be expected to have minimal amount of entanglement...(this can certainly be checked a posteriori) 24 / 33

METTS Formulation METTS idea I: Choose c to be the product states s 1, s 2,..., s N These c have zero entanglement...and represent the typical high energy states of the system The state φ(c) will also be expected to have minimal amount of entanglement...(this can certainly be checked a posteriori) Since O = c P(c) φ(c) O φ(c) we now need to device a strategy to sample the state c with the probability distribution P(c) 24 / 33

METTS Formulation METTS idea I: Choose c to be the product states s 1, s 2,..., s N These c have zero entanglement...and represent the typical high energy states of the system The state φ(c) will also be expected to have minimal amount of entanglement...(this can certainly be checked a posteriori) Since O = c P(c) φ(c) O φ(c) we now need to device a strategy to sample the state c with the probability distribution P(c) Develop a transition matrix W(c c) such that (Markov stationary condition) W(c c)p(c) = P(c ) c 24 / 33

METTS Formulation 25 / 33

METTS Formulation METTS idea II: Transition matrix W(c c) = c φ(c) 2 25 / 33

METTS Formulation METTS idea II: Transition matrix W(c c) = c φ(c) 2 Stated in words: The next classical state is selected by collapsing the state φ(c) via an S z measurement at every site! 25 / 33

METTS Formulation METTS idea II: Transition matrix W(c c) = c φ(c) 2 Stated in words: The next classical state is selected by collapsing the state φ(c) via an S z measurement at every site! This automatically satisfies the detailed balance condition W(c c)p(c) = W(c c )P(c ) and the Markov stationary condition is automatically satisfied! 25 / 33

METTS Formulation METTS idea II: Transition matrix W(c c) = c φ(c) 2 Stated in words: The next classical state is selected by collapsing the state φ(c) via an S z measurement at every site! This automatically satisfies the detailed balance condition W(c c)p(c) = W(c c )P(c ) and the Markov stationary condition is automatically satisfied! Note that there is no rejection! 25 / 33

METTS Formulation METTS idea II: Transition matrix W(c c) = c φ(c) 2 Stated in words: The next classical state is selected by collapsing the state φ(c) via an S z measurement at every site! This automatically satisfies the detailed balance condition W(c c)p(c) = W(c c )P(c ) and the Markov stationary condition is automatically satisfied! Note that there is no rejection!?scheme fails for the classical Ising model! Product states should not diagonalize the Hamiltonian 25 / 33

METTS Algorithm 26 / 33

METTS Algorithm In a picture (Schollwock 2009) 26 / 33

METTS Algorithm In a picture In Fortran (Schollwock 2009) 26 / 33

TIFM with METTS 27 / 33

TIFM with METTS METTS simulations with h = 0.6 27 / 33

TIFM with METTS METTS simulations with h = 0.6 e βh 2 c is obtained exactly (cheating) 27 / 33

TIFM with METTS METTS simulations with h = 0.6 e βh 2 c is obtained exactly (cheating) 1.5 2 Exact METTS Energy 2.5 3 3.5 0.25 0.5 0.75 1 T (Temperature) 27 / 33

TIFM with METTS METTS simulations with h = 0.6 e βh 2 c is obtained exactly (cheating) 1.5 2 Exact METTS Energy 2.5 3 3.5 0.25 0.5 0.75 1 T (Temperature) Algorithm works remarkably well (even with a few hundred steps)! 27 / 33

M of METTS How entangled are the METTS? 0.24 S (Entanglement Entropy of METTS) 0.23 0.22 0.21 0.2 0.19 0.18 L 2 3 4 h=0.6 250 500 750 1000 MTTES Step Remarkably little entanglement (lower than the ground state even!) 28 / 33

Part III Summary: METTS 29 / 33

Part III Summary: METTS Need to use minimally entangled states to obtain partition function trace 29 / 33

Part III Summary: METTS Need to use minimally entangled states to obtain partition function trace Minimally entangled states are φ(c) e βh 2 c where c s are the product state 29 / 33

Part III Summary: METTS Need to use minimally entangled states to obtain partition function trace Minimally entangled states are φ(c) e βh 2 c where c s are the product state The induced probability distribution on c is sampled by a collapsing scheme 29 / 33

Part III Summary: METTS Need to use minimally entangled states to obtain partition function trace Minimally entangled states are φ(c) e βh 2 c where c s are the product state The induced probability distribution on c is sampled by a collapsing scheme The most difficult step: Calculation of e βh 2 c 29 / 33

IV. Brief Remarks for Experts and Plan for Future Work 30 / 33

MPS Implementation of METTS Matrix product states (MPS) play a crucial role in practical implementation of METTS Operators can also be represented a matrix products Calculation of e βh 2 c involves Breaking up of H = n H n...h n acts locally say at most on two sites Trotter decompose e τh = n e τhn (approximation) Each e τh n has a simple MPO representation Apply e τh sufficient number of times to generate e βh 2 Scales as Nm 3 β where m is the size of matrices in the matrix products 31 / 33

Plan for Future Work Strongly correlated fermions Work in grand canonical ensemble Product states are site basis with four states per site Develop efficient imaginary time evolution strategies Do high temperature calculations and check with virial expansion results 32 / 33

Summary Part II Summary: Entanglement ideas 33 / 33

Summary Part II Summary: Entanglement ideas Ideas of entanglement, entanglement entropy 33 / 33

Summary Part II Summary: Entanglement ideas Ideas of entanglement, entanglement entropy Area Law: ground states are easy(easier) to describe...the amount of information required to describe a ground state scales as N 33 / 33

Summary Part II Summary: Entanglement ideas Ideas of entanglement, entanglement entropy Area Law: ground states are easy(easier) to describe...the amount of information required to describe a ground state scales as N Critical ground states are harder 33 / 33

Summary Part II Summary: Entanglement ideas Ideas of entanglement, entanglement entropy Area Law: ground states are easy(easier) to describe...the amount of information required to describe a ground state scales as N Critical ground states are harder Excited states...tough! 33 / 33

Summary Part II Summary: Entanglement ideas Ideas of entanglement, entanglement entropy Area Law: ground states are easy(easier) to describe...the amount of information required to describe a ground state scales as N Critical ground states are harder Excited states...tough! Part III Summary: METTS Need to use minimally entangled states to obtain partition function trace Minimally entangled states are φ(c) e βh 2 c where c s are the product state The induced probability distribution on c is sampled by a collapsing scheme Lots of interesting things to do! 33 / 33