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Pure State Quantum Statistical Mechanics

Akira ShimizuDepartment of Basic Science, The University of Tokyo, Komaba@Tokyo

(東京大学総合文化研究科)

Collaborators:

Sho Sugiura (D3)

Masahiko Hyuga (D1)

Kyota Fujikura (M2)

Kazumitsu Sakai (Assistant Prof. of CM Theory G@Komaba)

References:

S. Sugiura and AS, Phys. Rev. Lett. 108, 240401 (2012).


S. Sugiura and AS, Kinki Univ. Series on Quantum Computing 9, 245 (2014).

M. Hyuga, S. Sugiura, K. Sakai and AS, Phys. Rev. B 90, 121110(R) (2014).

K. Fujikura and AS, in preparation.

Contents

0. Definition of equilibrium states

1. Pure state quantum statistical mechanics

• Typicality• Thermal Pure Quantum (TPQ) formulation

2. Applications of the TPQ formulation to strongly-correlated systems

• 2d kagome Heisenberg antiferromagnet

• Hubbard model (1d and 2d)

3. Fundamental questions about quantum statistical mechanics

4. Practical formulas using microcanonical TPQ states

5. Squeezed equilibrium states

6. Summary

Definition of equilibrium states

清水「熱力学の基礎」(東大出版会, 2007)

(i) Isolated system

Consider an isolated macroscopic system. After a sufficiently long time, it

evoloves to a state s.t. all macroscopic variable is macroscopically constant;variation

value→ 0 in the thermodynamic limit (t.d.l).

Such a state is called a (thermal) equilibrium state.

(ii)Non-isolated system (subsystem)

Consider a macroscopic system that is not isolated from other systems. Sup-

pose that its state is macroscopically identical to an equilibrium state in the

above sense, i.e., for all macroscopic variable

its value

its value in an equilibrium state (of an isolated system)→ 1 in the t.d.l..

Such a state is also called a (thermal) equilibrium state.

♣ Other definitions ⇒ violation of many theorems of thermodynamics

Equlibrium Statistical Mechanics: Conventional

Ensemble formulation� �I. An equilibirum state is given by a mixed state called the Gibbs state;

ρens

microcanonical=

1

W

∑n

′|n⟩⟨n| or ρens

canonical=

1

Z

∑n

e−βEn|n⟩⟨n| or · · ·

II. Thermodynamic functions are given by von Neumann entropy or partition

functions;

S = −Tr [ρens ln ρens] or F = −T lnZ or · · ·� �

Questions about the Conventional Formulation

Ensemble formulation� �I. An equilibirum state is given by a mixed state called the Gibbs state;

ρens

microcanonical=

1

W

∑n

′|n⟩⟨n| or ρens

canonical=

1

Z

∑n

e−βEn|n⟩⟨n| or · · ·

II. Thermodynamic functions are given by von Neumann entropy or partition

functions;

S = −Tr [ρens ln ρens] or F = −T lnZ or · · ·� �QI. Are the Gibbs states the only representations of an equilibirum state?

AI. No. Macroscopic quantum systems have typicality:� �Almost all quantum state |ψ⟩ in the energy shell (E − ∆E,E] is the

equilibirum state specified by E.

⇒ A single pure state |ψ⟩ can represent the equilibirum state.� �

Almost all state in the energy shell is an equilibrium state

P. Bocchieri and A. Loinger, Phys. Rev. 114, 948 (1959).

Recently, Sugita (2006), Popescu et al. (2006), Goldstein et al. (2006), Reimann (2007)

• Hilbert space: HN (N : number of spins or particles)

• Energy shell: EuN : subspace of HN in (E −∆E,E]. (u ≡ E/N)

dim EuN = W = eNs (s ≡ S/N = O(1))

• Probability measure: a random vector in EuN ,

|ψrnd⟩ =∑i

′ci|i⟩

ci : random complex numbers drawn uniformly from the sphere∑i′|ci|2 = 1.

|i⟩ : aribitrary basis of EuN .

This measure is invariant under choice of the basis. → natural measure!

• Physical quantities to define an equilibrium state: mechanical variables

— Two types of macroscopic variables in equilibirum statistical mechanics —

Mechanical Variables and Genuine Thermodynamic Variables

Mechanical variables (latest definiton by Hyuga, Sugiura, Sakai and AS, 2014)

• Low-degree polynomials (i.e., their degree = O(1)) of local operators:

H, Mz =∑r

sz(r), (H)2, sx(r)sy(r′), · · ·

•We impose a physical condition:

|⟨A†A⟩| ≤ KN2m (⟨ · ⟩ : equilibrium value) ,

where K,m are constants independent of A and V .

Genuine thermodynamic variables

• Thermodynamic variables that cannot be represented as such operators:

T, µ, · · · , S, F, · · ·/∈ quantum-mechanical observables in the standard sense.

• All genuine thermodynamic variables can be derived from one of thermody-

namic functions, S(E,N, V ), F (T,N, V ), · · · . (⇔ second law!)

Almost all state in the energy shell is an equilibrium state

Bocchieri & Loinger (1959), Sugita (2006), Popescu et al. (2006), Goldstein et al. (2006), Reimann (2007)

� �Theorem (typicality) : Take a random vector from the energy shell EuN ,

|ψrnd⟩ =∑i

′ci|i⟩ (|i⟩ : aribitrary basis of EuN ).

Then,

⟨ψrnd|A|ψrnd⟩P→ ⟨A⟩ens

(= Tr [ρensA]

)for every mechanical variable A uniformly, and exponentially fast, asN →∞.� �That is, for ∀ϵ > 0

P(∣∣∣⟨ψrnd|A|ψrnd⟩ − ⟨A⟩ens∣∣∣ ≥ ϵ

)≤ 1

ϵ2· O(⟨A

2⟩ens)dim EuN

≤ 1

ϵ2· O(N

2m)

eNs→ 0

for every mechanical variable A, as N →∞.

It’s not purification

An example of purification:

ρ = (2/3) |1⟩⟨1| + (1/3) |2⟩⟨2| : a mixed state on HN .By attaching an auxiliary system, enlarge HN to HN ⊗Haux, and consider

|Ψ⟩ =√

2/3 |1⟩ ⊗ |1′⟩ +√

1/3 |2⟩ ⊗ |2′⟩ : a pure state in HN ⊗Haux.

Then,⟨Ψ|(a⊗ 1)|Ψ⟩ = Tr [ρa] for all observables a on HN .

Purification� �It is always possible to represent a mixed state ρ on HN as a pure state |Ψ⟩in an enlarged space HN ⊗Haux. They are the same state on HN .� �ex. Thermo Field Dynamics (TFD) utilizes purification (Haux = HN ).

By contrast,

• |ψrnd⟩ is a pure state in HN .

• |ψrnd⟩⟨ψrnd| = ρens on HN (manifest by entanglement ← discuss later)

• But, |ψrnd⟩ and ρens are statistical-mechanically identical!

Answer to Question I: Typicality

QI. Are the Gibbs states ρens the only representations of an equilibirum state?

AI. No. Almost all |ψrnd⟩ represents the same equilibirum state. (typicality)

Contents










6. Summary

Answer to Question I, and further questions

QI. Are the Gibbs states ρens the only representations of an equilibirum state?

AI. No. Almost all |ψrnd⟩ represents the same equilibirum state. (typicality)

Problems and further questions:

1. If the conventional principle II is applied to ρ = |ψrnd⟩⟨ψrnd|,S = Tr [ρ ln ρ] = 0, Z is not obtained from |ψrnd⟩.→ Impossible to obtain genuine thermodynamic variables from |ψrnd⟩ .→ QII. How to obtain S, F, · · · from (another) |ψ⟩?

2. Generally, the canonical Gibbs state (specified by T ,N, V ) is much more

convenient than the microcanonical Gibbs state (specified by E,N, V ).

|ψrnd⟩ (specified by E,N, V ) corresponds to the microcanonical one.

→ QIII. Another |ψ⟩ specified by T ,N, V ?

3. Practically, |ψrnd⟩ is harder to obtain than ρens.

→ QIV. Another |ψ⟩ easier to obtain?

Our Solutions — TPQ formulation of statistical mechanics



M. Hyuga, S. Sugiura, K. Sakai and AS, Phys. Rev. B 90, 121110(R) (2014).

1. Generally define Thermal Pure Quantum (TPQ) state as a pure state that

represents an equilibirum state. ex. |ψrnd⟩ is one of TPQ states.

2. New types of TPQ states |β,N, V ⟩, |β, µ, V ⟩, ....→ Yes to QIII. Another |ψ⟩ specified by T ,N, V ?

3. Formulas for getting thermodynamic functions from them.

→ Solutions to QII. How to obtain S, F, · · · from (another) |ψ⟩?→ Replaces all the principles of the ensemble formulation!

4. |β,N, V ⟩, |β, µ, V ⟩, .... are much easier to obtain than ρens and |ψrnd⟩.→ Yes to QIV. Another |ψ⟩ easier to obtain?


Setup

Quantum system

• composed of N sites or particles, confined in a box of volume V .

• (irreducible) Hilbert space is HN . dimHN can be ∞.

• equilibrium state is specified by E,N, V,M , ... → abbreviated as E,N, V .

Assumptions : Ensemble formulation gives correct results, which are consis-

tent with thermodynamics in the t.d.l (E ∝ V ∝ N →∞).

• S(E,N, V )/N → s(u, v): entropy density, u ≡ E/N, v ≡ V/N .

• s(u, v) is a concave function, continuously differentiable even at phase tran-

sitions. see, e.g., 清水「熱力学の基礎」(東大出版会, 2007)

General definition : Thermal Pure Quantum (TPQ) state

� �A state |ψ⟩ (∈ HN ), which has a random variable, is called a TPQ state if

⟨A⟩ψN ≡⟨ψ|A|ψ⟩⟨ψ|ψ⟩

P→ ⟨A⟩ensN ≡ Tr [ρensA]

for every mechanical variable A uniformly, as N →∞.� �That is, for ∀ϵ > 0 there exists a function ηϵ(N) that vanishes as N →∞ and

P(∣∣∣⟨A⟩ψN − ⟨A⟩ensN ∣∣∣ ≥ ϵ

)≤ ηϵ(N)

for every mechanical variable A.

Remark: It’s not purification of ρens (|ψ⟩ ∈ HN ).

Independent variables

• β,N, V : canonical TPQ (cTPQ) state |β,N, V ⟩ (abbreviated as |β,N⟩).

• β, µ, V : grand-canonical TPQ (gTPQ) state |β, µ, V ⟩.

Canonical TPQ (cTPQ) state : |β,N⟩Latest version applicable to dimHN =∞. (our PRL (2013) assumed dimHN < +∞).

Let

• {|ν⟩}ν : an aribitrary basis of HN (c.f. {|i⟩}i ∈ EuN of previous works)

• xν, yν : real random variables, obeying the standard normal distribution

• zν ≡ (xν + iyν)/√2. (c.f. our PRL (2013) used cν s.t.

∑ν |cν|2 = 1)

Then,

|β,N⟩ ≡∑ν

zν exp[−βH/2]|ν⟩

is the canonical TPQ (cTPQ) state, specified by β,N . That is,� �

⟨A⟩TPQβ,N ≡⟨β,N |A|β,N⟩⟨β,N |β,N⟩

P→ ⟨A⟩ensβ,N ≡1

ZTr [e−βHA]

for every mechanical variable A uniformly, in the t.d.l.� �

Probability of Error

Let f (T ;N) ≡ F/N (free energy density) → f (T ) as N →∞.� �For ∀ϵ > 0,

P(∣∣∣⟨A⟩TPQβ,N − ⟨A⟩

ensβ,N

∣∣∣ ≥ ϵ)

≤ 1

ϵ2·⟨(∆A)2⟩ens2β,N + (⟨A⟩ens2β,N − ⟨A⟩

ensβ,N )2

exp[2Nβ{f (1/2β;N)− f (1/β;N)}]≤ 1

ϵ2· N

2m

eO(N)→ 0.

� �This shows that

⟨A⟩TPQβ,NP→ ⟨A⟩ensβ,N for every mechanical variable A unifromly.

Therefore,� �• |β,N⟩ is the canonical TPQ (cTPQ) state.

• Its single realization gives the equilibrium values of mechanical variables,

with exponentially small probability of error, as ⟨A⟩TPQβ,N .� �

Formula for Thermodynamic Function

Let Z(β,N) ≡ Tr e−βH (partition function).� �

P

(∣∣∣∣⟨β,N |β,N⟩Z(β,N)− 1

∣∣∣∣ ≥ ϵ

)≤ 1

ϵ2· 1

exp[2Nβ{f (1/2β;N)− f (1/β;N)}]≤ 1

ϵ2· 1

eO(N).

� �This shows ⟨β,N |β,N⟩ P→ Z(β,N).� �∴ A single realization of |β,N⟩ gives F with exponentially small probability

of error, by

−βF = ln⟨β,N |β,N⟩.� �All genuine thermodynamic variables can be calculated from F .� �∴ Only a single realization of the TPQ state gives all variables of statistical-

mechanical interest.� �

Self-Validation

Using F (= Nf ) obtained from this formula, one can estimate the upper bounds

of errors of F itself and ⟨A⟩TPQβ,N .� �• Our formulas are almost self-validating.

• This is particularly useful in practical applications!� �

|β,N⟩ is well-defined even when dimHN = ∞

When dimHN =∞,

|β,N⟩ =∑ν

zν exp[−βH/2] |ν⟩

= exp[−βH/2]∑ν

zν |ν⟩.∑ν

zν|ν⟩ is ill-defined because its norm∑ν

|zν|2→∞.

By contrast, (almost every) |β,N⟩ has finite norm,

⟨β,N |β,N⟩ P→ Z(β,N) < +∞ for finite V .

• |β,N⟩ is well-defined even when dimHN =∞.

•∑ν

zν and exp[−βH/2] cannot be inverted.

TPQ formulation of Statistical Mechanics

TPQ formulation� �I. An equilibirum state is given by a pure state called the TPQ state;

|β,N⟩ =∑ν

zν e−βH/2|ν⟩ or |β, µ, V ⟩ =

∑ν

zν e−β(H−µN)/2|ν⟩ or · · ·

canonical TPQ state grand-canonical TPQ state

II. Thermodynamic function (S, F, J, · · · ) is given by its norm;

−βF = ln⟨β,N |β,N⟩ or −βJ = ln⟨β, µ, V |β, µ, V ⟩ or · · ·� �And, many practical formulas

⇒ Can obtain TPQ states by simply multiplying H repeatedly O(N) times.

Solid theoretical basis

⇒ Unnecessary for gray wisdom.� �Possible to analyze systems which are hardly analyzable with other methods.� �

Summary of Part 1


• Typicality (previous works)

A random vector in the energy shell is an equilibrium state.

It gives equilibrium values of all mechanical variables.

• Thermal Pure Quantum (TPQ) formulation (our works)

Another random vector in the Hilbert space is also an equilibrium state.

It gives all variables of statistical-mechanical interest, including S, F, · · · .Various TPQ states, micro-canonical, canonical, grand-canonical, ....

Contents










6. Summary

Application : spin-1/2 kagome Heisenberg antiferromagnet

One of the hardest systems to analyze

(ex. Sign problem is fatal to quantum Monte Carlo method.)

Numerical diagonalization : N ≲ 18 → double peaks in the specific heat?

cTPQ : N = 27-30 → the lower peak disappears!

N=30N=27N=18bN=18a

0.0 0.2 0.4 0.6 0.8 1.0

0.05

0.10

0.15

0.20

T

c

Thermodynamic functions f = F/N and s = S/N of KHA

sHΒ;NLf HΒ;NL

0.0 0.5 1.0 1.5 2.0-1.5

-1

-0.5

0

0

0.5

1

1.5

T

f HΒ;NL sHΒ;NL

45% of the total entropy (= N ln 2) remains at low T (= 0.2J).

→ typical to frustrated systems.

→ hard with most other methods.

→ But, favorable to TPQ formulation because error ∼ 1/eNs.

From our formulas for errors, errors ≤ 1% for T ≥ 0.1J .

Application : Hubbard model (1d and 2d)

Hubbard model (t = 1)

H = −∑⟨r,r′⟩

(c†rσcr′σ + h.c.) + U

∑r

(nr↑ − 1/2)(nr↓ − 1/2)

Grand-canonical TPQ (gTPQ) state

|β, µ, V ⟩ =∑ν

zν e−β(H−µN)/2|ν⟩

⟨β, µ,N |β, µ,N⟩ P→ Ξ(β, µ,N) : grand partition function

⋆ V ≤ 16 in our workstation

→ error ∝ 1/eV s is not small enough at low T

→ averge over realizations of {zν}→ error ∝ 1/

√# of realizations

c.f. average was not necessary for kagome Heisenberg antiferromagnet, because

V = 27-30 and s is large due to frustration

Application : Hubbard model on a 1d chain

Number density vs. T

# of realizations = 18-22 ∝ (error)−1

1

1.1

1.2

1.3

1.4

1.5

1.6

1.7

0 0.5 1 1.5 2 2.5 3

Nu

mb

er D

ensi

ty n

Temperature T

U = 1, µ = 0.5U = 8, µ = 2

U = 8, µ = 3

U = 1, µ = 2

exact (L = ∞)TPQ (L = 15)

Specific heat vs. T


One of the most challenging quantities

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0 0.5 1 1.5 2 2.5 3

Spec

ific

Hea

t c

Temperature T

U = 1, µ = 0 (half-filled)


U = 8, µ = 3

U =

1, µ

= 2

exact (L = ∞)

TPQ (L = 15)

Charge and staggered spin correlations, ϕ+ and ϕ−# of realizations = 21 ∝ (error)−1

Exact results are unknown

0.96

0.98

1

1.02

1.04

1.06

1.08

0 1 2 3 4 5 6 7

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Distance i

L = 14, U = 8, µ = 0 (half-filled)

φ+

φ-

T = 0.1T = 0.2T = 1.0

Applicatin: Hubbard model on 2d triangular lattice


U = 3, µ = 1 ⇒ slightly doped ⇒ sign problem in QMC method

1

1.02

1.04

1.06

1.08

1.1

1.12

1.14

0 2 4 6 8 10

Num

ber

Den

sity

n

Temperature T

ND (V = 8)TPQ (V = 8)

TPQ (V = 15)

Specific heat of Hubbard model on 2d triangular lattice


U = 3, µ = 1 ⇒ slightly doped ⇒ sign problem in QMC method

0

0.2

0.4

0.6

0.8

1

1.2

0 0.5 1 1.5 2 2.5 3

Spec

ific

Hea

t c

Temperature T

ND (V = 8)TPQ (V = 8)

TPQ (V = 15)

V=

8V = 15

Contents










6. Summary

Fundamental Questions about Quantum Statistical Mechanics

•Which gives better results, cTPQ or gTPQ state?

• TPQ state = energy eigenstate → time evolution?

• TPQ state is a pure state → thermal fluctuation?

• Entanglement?

•Which micostate is realized in experiments, Gibbs state or TPQ state?

• How to study time evolution in statistical mechanics

Which gives better results, cTPQ or gTPQ state?

Q. They give identical results for V →∞. When V < +∞?

Specific heat of Hubbard chain [# of realizations = 22-128 ∝ (error)−1 ]

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0 0.5 1 1.5 2 2.5 3

Spec

ific

Hea

t c

Temperature T


gTPQ & exact

cTPQ gTPQ (L = 8)

gTPQ (L = 14)

exact (L = ∞)

cTPQ (L = 14)

cTPQ (L = 8)

• Even for L = 14,

(results of the gTPQ state of finite L) ≃ (exact results for L→∞)

• gTPQ state gives much better results than cTPQ state:∣∣∣(results of the gTPQ state of finite L)− (exact results for L→∞)∣∣∣

≪∣∣∣(results of the cTPQ state of finite L)− (exact results for L→∞)

∣∣∣� �To study properties of infinite systems using numerical computation:

• If you employ the TPQ formulation, use the gTPQ state.

• If you employ the ensemble formulation, use the grand-canonical ensemble.� �Or, convert the results of cTPQ state to those of gTPQ state by∑

N

⟨βNV |βNV ⟩eβµN P→ Ξ(β, µ, V ).

TPQ state is invariant under time evolution

Q. Not an energy eigenstate → time evolution?

By taking the energy eigenstates {|n⟩}n as the arbitrary basis {|ν⟩}ν used in

the construction,

e−iHt/ℏ|β,N⟩ = e−iHt/ℏ∑n

zne−βH/2|n⟩

=∑n

(e−iEnt/ℏzn

)e−βH/2|n⟩

= another realization of |β,N⟩� �An TPQ state traverses various realizations of the same TPQ state.

⇒ It is macroscopically stationary.

⇒ Consistent with thermodynamics.� �� Macroscopic time-invariance does not necessarily imply the microscopic time-

invariance.� �

Where is Thermal fluctuation?

The Gibbs state ρens =∑nwn|n⟩⟨n| (wn = e−βEn/Z) has both quantum

and thermal fluctuations:

quantum fluctuation : ⟨(∆A)2⟩ensq ≡∑n

wn⟨n|(A− ⟨n|A|n⟩)2|n⟩,

thermal fluctuation : ⟨(∆A)2⟩ensth ≡∑n

wn⟨n|A|n⟩2 −(∑

n

wn⟨n|A|n⟩)2,

total fluctuation : ⟨(∆A)2⟩ens = ⟨(∆A)2⟩ensq + ⟨(∆A)2⟩ensth .

The TPQ state has only the quantum fluctuation:

quantum fluctuation : ⟨(∆A)2⟩TPQq ≡ ⟨(A− ⟨A⟩TPQ)2⟩TPQ,thermal fluctuation : ⟨(∆A)2⟩TPQth = 0,

total fluctuation : ⟨(∆A)2⟩TPQ = ⟨(∆A)2⟩TPQq .

Q. Contradiction?

NO contradiction because

• The quantum and thermal fluctuations are actually ill-defined because

ρens =∑n

wn|n⟩⟨n| =∑λ

wλ|λ⟩⟨λ| = · · · (infinitely many ways)

ex. ρ =2

3| ↑⟩⟨↑ | + 1

3| ↓⟩⟨↓ |

=1

2|+⟩⟨+| + 1

2|−⟩⟨−|

(|±⟩ ≡

√2

3| ↑⟩ ±

√1

3| ↓⟩; non-orthogonal

)• They cannot be distinguished experimentally!

• They are, separately, not observable quantities!

• The total fluctuation⟨(∆A)2⟩ens = ⟨(∆A)2⟩TPQ

is the only experimentally-observable quantity!� �In the TPQ formulation, ‘thermal fluctuation’ is completely included as quan-

tum fluctuation!� �

Entanglement

S. Sugiura and AS, Kinki Univ. Series on Quantum Computing 9, 245 (2014).

Q. For mechanical variables, TPQ state = Gibbs state. Entanglement?

Example: T ≫ J

ρens ≃ 1 ⇒ no entanglement.

TPQ states have almost maximum (exponentially large) entanglement.

ò

ò

ò

ò

ò

ò

ô

ô

ô

ô

ô

ô low

energy high

ô Average Value for Random Vector

ò Minimum Value

TPQ states

1 2 3 4 5 6

0.02

0.05

0.10

0.20

0.50

q

TrHΡ

q2 L

� �An equilibrium state can be represented either by a TPQ state with exponen-

tially large entanglement or by a mixed state with much less entanglement.� �Their difference can be detected only by high-order correlations of local opera-

tors, which are not of statistical-mechanical interest. (and almost metaphysical)

Various Representations of the Same Equilibrium State

macro state micro state (von Neumann entropy)

an equilibrium state ⇐⇒

• Gibbs state ρens (SvN = STD)

• TPQ state |β,N⟩ (SvN = 0)

• many other states (0 ≤ SvN ≤ STD)

microscopically different, macroscopically identical.

→ identical statistical-mechanical results.

Q. Which micostate is realized in experiments?

A. Depends on how you prepare the equilibrium state.

Example:

If initial state = ρens,

then final state = ρens

because SvN = STD. SvN = STD SvN < STD

How to Study Time Evolution in Statistical Mechanics

Simple application of the ensemble formulation gives wrong result!

If initial state = ρens

Then, final state = ρens

→ S = STD ???SvN = STD SvN < STD

Simple application of the TPQ formulation also gives wrong result!

∵ −βF = ln⟨β,N |β,N⟩ is conserved.

Correct way:

• Initial state = ρens or |β,N⟩ or ...

• Calculate time evolution.

• For the final state, calculate ⟨H⟩.

• Insert it into the Boltzmann formula; S = lnW (⟨H⟩, N, V ).

Contents










6. Summary

Practical Formulas using Microcanonical TPQ (mTPQ) States

Practical computations: a cutoff to make dimHN < +∞. Then,

• Take an arbitrary basis {|ν⟩}, such as | ± ± ± · · ·⟩.

• Using random numbers {zν}, generate a random vector ∈ HN by

|0⟩ ≡∑ν

zν|ν⟩ ( = previous work’s |ψrnd⟩ ∈ EuN )

• h ≡ H/N has the maximum eigenvalue emax.

⇒ Take an aribitray number l such that l ≥ emax.

• Compute

|k⟩ ≡ (l − h)k|0⟩iteratively for k = 0, 1, 2, · · · .

We can show: u ≡ E/N has a sharp peak around u ≃ ⟨k|h|k⟩/⟨k|k⟩⇒ a mTPQ state ( = previous work’s |ψrnd⟩)

In terms of normalized mTPQ states,

|ψk⟩ ≡ |k⟩/√⟨k|k⟩,

|β,N⟩ can be expanded as

eNβl/2|β,N⟩ = eNβ(l−h)/2|0⟩ =∞∑k=0

Rk|ψk⟩

• uniformly convergent on any finite interval of β.

• Rk takes significant values only for k s.t. ⟨ψk|h|ψk⟩ = ⟨h⟩TPQβ,N +O(1/V ).

� �|β,N⟩ is superpositon of |ψk⟩’s which represent the same equilibirum state.� �• Terminate the sum at kterm, which depends on the largest β of interest, βmax.

• βmax = O(1) ⇒ kterm = O(N).� �One can obtain |β,N⟩ by multiplying (l−h) with a random vector |0⟩ ∈ HN ,

repeatedly O(N) times.� �♣ Larger l ⇒ better results but larger kterm.

Advantages as a Numerical (and Analytic?) Method

• No limitation on models, free from the sign problem

Any spatial dimensions, frustrated systems, fermion systems, ...

• All quantities of statistical-mechanical interest, e.g., correlation functions

• A wide range of finite temperature: from T ≪ J to T ≫ J .

T > 0 ⇒ exponentially large number of relevant states ⇒ accurate results

• Almost self-validating ⇒ Can estimate the upper bounds of errors.

• Can obtain TPQ states by simply multiplying h repeatedly Θ(N) times.

• Only two vectors in the memory, to obtain the results for all β ≤ βmax.

• Orthogonality, ⟨k|k′⟩ = 0 for k = k′, is not necessary at all.

• Solid theoretical basis.� �Possible to analyze systems which could not be analyzed with other methods.� �

Contents










6. Summary

Summary



• Typicality (previous works)

A random vector in the energy shell is an equilibrium state.

It gives equilibrium values of all mechanical variables.

• Thermal Pure Quantum (TPQ) formulation (our works)

Another random vector in the Hilbert space is also an equilibrium state.

It gives all variables of statistical-mechanical interest, including S, F, · · · .Various TPQ states, micro-canonical, canonical, grand-canonical, ....



Drastic changes of the ‘double peaks’ of specific heat with increasing N


Reliable results over wide ranges of T, µ, U .


•Which gives better results, cTPQ or gTPQ state?

The gTPQ state gives much better results.

• TPQ state = energy eigenstate → time evolution?

Its time evolution is exponentailly small.

• TPQ state is a pure state → thermal fluctuation?

Thermal fluctuation is included in quantum fluctuation.

• Entanglement?

entanglement of TPQ states ≫ entanglement of Gibbs states.

•Which micostate is realized in experiments, Gibbs state or TPQ state?

Depends on detailed experimental conditions.

• How to study time evolution in statistical mechanics

Neither S=−Tr [ρens ln ρens] nor −βF =ln⟨β,N |β,N⟩ evolves correctly.


Possible to analyze systems which could not be analyzed with other methods.


•What quantum state is the equilibrium state when you look at A?

A squeezed equilibrium state.

• How ⟨A(t)⟩ evolves in this state.

An explicit formula is obtained.

• Is the Onsager’s assumption on ⟨A(t)⟩ correct for quantum systems?

No, because quantum effects play crucial roles in general.

Remarks on Convergence

For general observable A,

∀ϵ > 0, P(∣∣∣⟨A⟩TPQβ,N − ⟨A⟩

ensβ,N

∣∣∣ ≥ ϵ)≤ 1

ϵ2· O(⟨A

†A⟩ens)eO(N)

•When A is a mechanichal variable, ⟨A†A⟩ens = ⟨A2⟩ens ≤ KN2m,

r.h.s ≤ 1

ϵ2· O(N

2m)

eO(N)→ 0 uniformly for all A.

•When A is a Weyl-type operator, A = eiξB (x ∈ R, B† = B),

r.h.s ≤ 1

ϵ2· O(1)eO(N)

→ 0 uniformly for all A.

• Similarly for time correations, A(t)B(t′) · · · .⇒ Non-equilibrium statistical mechanics based on TPQ states

• However, there exist other A’s s.t. r.h.s → 0, i.e., ⟨A⟩TPQβ,N,V

P→ ⟨A⟩ensβ,N,V .

ex. Entanglement is maximally different!

Stability

Any quantum state representing an equilibrium state should be stable against

weak external perturbations.

Using the results of- AS and T. Miyadera, Phys. Rev. Lett. 89, 270403 (2002)

- A. Sugita and AS, J. Phys. Soc. Jpn. 74, 1883 (2005)

we can show that the TPQ states do have such stability (against weak noises,

perturbations from environments, and local measurements).

Also, using- AS and T. Morimae, Phys. Rev. Lett. 95 (2005) 090401.

- Y. Matsuzaki, T. Morimae and AS, to be prepared.

we can show that e−Nβh also has such stability.� �An equilibrium state can be represented by various microstates, including

e−Nβh and |β,N⟩. All of them are consistent with macroscopic physics.� �

Interpretations of ρens =∑n

′ 1W|n⟩⟨n|

Interpretation 1 : The system is in |n⟩ with probability 1/W .

However, ρ (for a mixed state) can be decomposed in infinitely many ways.

This general fact is trivial for ρens:

ρens =∑n

′ 1W|n⟩⟨n| = 1

W1shell

=∑i

1

W|i⟩⟨i| for ∀orthogonal basis {|i⟩}shell

Interpretation 2 : The system is in |i⟩ with probability 1/W .

=∑j

ξjW|j⟩⟨j| for ∀non-orthogonal basis {|j⟩}shell,

∑j

ξj|j⟩⟨j| = 1shell.

Interpretation 3 : The system is in |j⟩ with probability ξj/W .

� �∴ Infinitely many interpretations are possible!� �

Non-Uniquness of Interpretation of a Mixed State

ρ =∑j

pj|ψj⟩⟨ψj|

Interpretation 1 : The system is in |ψj⟩ with probability pj.

However, ρ (for a mixed state) can be decomposed in infinitely many ways:

ρ =∑j

pj|ψj⟩⟨ψj| =∑j

p′j|ψ′j⟩⟨ψ

′j| =

∑j

p′′j |ψ′′j ⟩⟨ψ

′′j | = · · · · · ·

ex. ρ =2

3| ↑⟩⟨↑ | + 1

3| ↓⟩⟨↓ |

=1

2|+⟩⟨+| + 1

2|−⟩⟨−|

(|±⟩ ≡

√2

3| ↑⟩ ±

√1

3| ↓⟩; non-orthogonal

)Interpretation 2 : The system is in |ψ′j⟩ with probability p′j.� �

∴ Infinitely many interpretations are possible for the same mixed state.� �

ρens is a mixed quantum state

Any quantum state can be represented by a density operator ρ, where

⟨A⟩ = Tr [ρA] for every observable A.

There always exist {pj, |ψj⟩}j (not unique) s.t.

ρ =∑j

pj|ψj⟩⟨ψj|, 0 ≤ pj ≤ 1,∑j

pj = 1.

Interpretation : ρ is a classical miture of |ψj⟩’s with probability pj.

Definition (rough) :

ρ is pure if there exists |ψ⟩ s.t. ρ = |ψ⟩⟨ψ|. (p1 = 1, other pj = 0)

Otherwise, ρ is mixed. (all pj < 1)

ρens is a mixed state because

ρens = (1/Z)e−βH =∑n

(e−βEn/Z

)|n⟩⟨n|.

Interpretation : ρens is a classical miture of |n⟩’s with probability e−βEn/Z.

Treatment of phase transitions

ex. a first-order transition

Thermodynamics:

s(T ;∞) is discontinuous at T = Ttr.

→ u(T ;∞) is discontinuous at T = Ttr.

→ c = T∂s/∂T = ∂u/∂T diverges at T = Ttr.

In contrast, β(u;∞) is continuous even at T = Ttr.

→ c = −β2/(∂β/∂u)N =∞ corresponds to β = constant.

Phase transition viewed in the entropy representation

Ref. 清水明「熱力学の基礎」(東大出版会, 2007)

• G(T, P,N) : T, P,N ↛ equilibrium state

• S(U,N, V ) : U,N, V → equilibrium state

水

水蒸気

1気圧

1気圧

1気圧

(i) (ii) (iv)

1気圧

(iii)

水

水蒸気水蒸気

水を１気圧に保ったまま温める

u(T, P ) is discontinuous T (u, P ) has a plateau

Spin systems

V → Mz, P → hzu(T, P ) → u(T, hz) : canonical formalism

T (u, P ) → T (u, hz) : our formalism

Canonical

T is an independent variable

→ calculate a discontinuous function u(T ;∞).

→ calculation is hard around T = Ttr.

Microcanonical

u is taken as an independent variable

→ calculate β(u;∞), which is continuous even at T = Ttr.

→ calculation is easy (e.g., interpolation is effecive)

→ calculate c from c = −β2/(∂β/∂u)N .→ c =∞ corresponds to β = constant; easily identified!

� �Microcanonical formalism is advantageous to finding phase transitions.� �

Zero temperature vs. Finite temperature, which is important?

• Zero temperature ⇒ primary interest of many-body physics

– Ground state

– Low-energy excitations (whose density → 0 as V →∞)

Expectation: 0 K properties would survive at low-enough temperature.

But, is such low temperature important to applications?

• Finite temperature

chemical reactions, biochemical phenomena, ...

Important to applications.

But, many-body problems at finite temperature are much harder to solve.

– DMRG method : applicable only to 1d systems.

– Quantum Monte Carlo method : a fatal problem for fermion systems.� �TPQ formulation will make it possible!� �

Outline of Proof

We use a Markov-type inequality:

For ∀ϵ > 0, P (|x− y| ≥ ϵ) ≤ (x− y)2/ϵ2.

Taking x = ⟨A⟩TPQβ,N , y = ⟨A⟩ensβ,N , (x− y)2 = (⟨A⟩TPQβ,N − ⟨A⟩ensβ,N )2 ≡ DN (A)2,

P(∣∣∣⟨A⟩TPQβ,N − ⟨A⟩

ensβ,N

∣∣∣ ≥ ϵ)≤ DN (A)2/ϵ2.

Using c∗νcξ = δν,ξ, c∗νcξc

∗ηcζ = δν,ξδη,ζ + δν,ζδη,ξ, etc, and dropping smaller-

order terms, we find

DN (A)2 ≤⟨(∆A)2⟩ens2β,N + (⟨A⟩ens2β,N − ⟨A⟩

ensβ,N )2

exp[2Nβ{f (1/2β;N)− f (1/β;N)}]≤ N2m

eO(N)

where

⟨(∆A)2⟩ensβ,N≡⟨(A− ⟨A⟩ensβ,N )2⟩ensβ,N ,

f (T ;N) ≡ F/N (free energy density)→ f (T ) as N →∞,f (1/2β;N)− f (1/β;N) = O(1) > 0 because s = −∂f/∂T = O(1) > 0.

More Practical Formulas # 1

Let

{A}′β,N ≡∞∑k=0

(Nβ)2k

(2k)!⟨k|A|k⟩ +

∞∑k=0

(Nβ)2k+1

(2k + 1)!⟨k|A|k + 1⟩,

{A}TPQβ,N ≡ {A}′β,N/{1}

′β,N .

We can show that

{1}′β,NP→ Z(β,N),

{A}TPQβ,NP→ ⟨A⟩ensβ,N ,

exponentially fast and uniformly.

Useful because one needs only to calculate ⟨k|A|k⟩ and ⟨k|A|k + 1⟩ for all

k ≤ kterm to obtain the results for all β ≤ βmax.

More Practical Formulas # 2

When computer resources are not sufficient to treat large enough N

⇒ eNs is not large enough.

⇒ One can reduce errors by averaging over many realizations of the cTPQ

states because1

Vln ⟨β,N |β,N⟩ = 1

Vln {1}′β,N = −βf (1/β;N).

⟨β,N |A|β,N⟩⟨β,N |β,N⟩

={A}′β,N{1}′β,N

= ⟨A⟩ensβ,N ,

Averaging over M realiztions reduces the error, as measured by the standard

deviation, by the factor of 1/√M .

ρens is a mixed state

Any quantum state can be represented by a density operator ρ, where

⟨A⟩ = Tr [ρA] for every observable A.

(A rough) definition : pure/mixed� �

• Every vector state |ψ⟩ =∑n

cn|n⟩ (cn ∈ C) is a pure state, for which

ρ = |ψ⟩⟨ψ| classical−→ ρcl(q, p) = δ(q − q0, p− p0)

• Other quantum states are mixed states, whose ρ can be decomposed as

ρ =∑j

pj|ψj⟩⟨ψj|classical−→ ρcl(q, p) =

∑j

pjδ(q − qj, p− pj)

using some {pj, |ψj⟩}j (not unique) s.t. 0 ≤ pj<1,∑j pj = 1.

� �

∴ ρens =1

Z

∑n

e−βEn|n⟩⟨n| is a mixed state.

(|ψj⟩ = |n⟩, pj =

e−βEn

Z

)

Summary — TPQ formulation of statistical mechanics

1. Generally define Thermal Pure Quantum (TPQ) state as a pure state that

represents an equilibirum state.

2. Various types of TPQ states |β,N, V ⟩, |β, µ, V ⟩, ....

3. Formulas for getting thermodynamic functions from them.

TPQ formulation� �I. An equilibirum state is given by a pure state called the TPQ state;

|β,N⟩ =∑ν

zν e−βH/2|ν⟩ or |β, µ, V ⟩ = · · · or · · ·.

canonical TPQ state grand-canonical TPQ state

II. Thermodynamic function (S, F, · · · ) is given by its norm;

−βF = ln⟨β,N |β,N⟩. or −βJ = · · · or · · ·.� �Can obtain TPQ states by simply multiplying H repeatedly O(N) times.

⇒ Many advantages when applied to practical computations.

akira shimizu - osaka universityseminar/pdf_2014_kouki/... · 2019-03-29 · pure state quantum...

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