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Canonical Quantum Observables Approximated by Molecular Dynamics for Matrix Valued Potentials Anders Szepessy, KTH Stockholm Can molecular dynamics determine canonical quantum observables for any temperature? Which stress and heat flux in the

SLIDE 1

Canonical Quantum Observables Approximated by Molecular Dynamics for Matrix Valued Potentials

Anders Szepessy, KTH Stockholm

Can molecular dynamics determine canonical quantum observables for any temperature? Which stress and heat flux in the conservation laws?

SLIDE 2

Conservation of mass, momentum and energy1

∂tρ(y, t) + ∂k(ρuk) = 0 ∂t

ρuj) + ∂k(ρujuk − σjk) = 0

∂tE + ∂k(Euk − σkjuj + qk) = 0

Jokkfall in Kalix¨ alven

Stress tensor σ =? Heat flux q =?

1Euler (1752), Laplace (1816)

SLIDE 3

Stress tensor and heat flux from molecular dynamics:

˙ Xj

t = P j t

˙ P j

t = −

∂jVjk(Xt)

pair potential interaction

Figure 1: solid-liquid phase transformation, von Schwerin & Szepessy (2010)

SLIDE 4

and Irving & Kirkwood (1950), Hardy (1981) definition ρ(y, t) :=

η(y − Xj

t )f(X0, p0)dX0dP0

ρu(y, t) := p(y, t) :=

η(y − Xj

t )P j t f(X0, P0)

initial density

dX0dP0 E(y, t) :=

η(y − Xj

t )(|P j t |2

2 + 1 2

Vjk)fdX0dP0

R3 η(y)dy = 1

x ¡ Xj ¡ X1 ¡ X2 ¡ X3 ¡

Figure 2: support of η with red particle positions

SLIDE 5

yields stress tensor σ(y, t) = 1 2

(Xj

t − Xk t ) ⊗ ∂jVjk bjk f dX0dP0

−

η(y − Xj

t )

t − u(y, t)

f dX0dP0 ,

where bjk(Xt) := 1 η

y − (1 − s)Xj

t − sXk t

SLIDE 6

Potential

k Vjk =? Schr¨

dinger equation with Hamiltonian:

ˆ H = − 1 M ∆ + V (x) x ∈ RN nuclei coordinates, V : RN → Cd2

potential

d = 1: Schr¨

dinger observables for mass, momentum and energy

satisfy the conservation laws2.

2Irving and Zwanzig (1951) for scalar smooth potentials

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Constant temperature and several electron states

ˆ H = − 1 M ∆ + V (x) Schr¨

dinger: ˆ

HΦn = EnΦn Goal determine

nΦn, ˆ

AΦne−En/T Observable A and Temperature T

1 2 3 4 −2 2 4 6 8 t M = 12800, δ = M−0.25 ψt, V (Xt)ψt E λ+(Xt) λ−(Xt)

Electron eigenvalueproblem V (x)Ψj(x) = λj(x)Ψj(x)

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If λ2 − λ1 ≫ T

limτ→∞ τ

0 A(Xt, Pt)dt τ

approximates

nΦn, ˆ

AΦne−En/T

nΦn,Φne−En/T

using Langevin: ˙ Xt = Pt ˙ Pt = −∇λ1(Xt) − κPt + √ 2κT ˙ Wt . All T possible?

SLIDE 9

All T Theorem3 possible:

There holds

nΦn, ˆ

A e−H/TΦn

nΦn,

e−H/TΦn = lim

τ→∞ d

qk τ ˜ Akk(Zk

t )dt

τ , where Zk

t = (Xt, Pt) with λk ,

qk = ¯ qk d

i=1 ¯

qi , ¯ qk = lim

τ→∞

τ e−λk(X1

t )−λ1(X1 t ) T

dt τ , Ψ∗(x)H(x, p)Ψ(x) = ˜ H(x, p) diagonal, Ψ∗(x)A(x, p)Ψ(x) = ˜ A(x, p) diagonal, . . . + O(

1 M1/2T ) for e− ˆ H/T.

3C. Lasser, M. Sandberg, A. Szepessy, A. Kammonen in preparation

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Proof uses Weyl quantization:

ˆ Aφ(x) =

RN (M 1/2

2π )N

RN eiM1/2(x−y)·pA(x + y

2 , p)dp

L2-kernel

φ(y)dy,

V (x) = V (x),
|p|2

2 = − 1 2M ∆, so H(x, p) = |p|2 2 I + V (x), Weyl’s law:

Φn, ˆ AΦn = trace ˆ A = trace(L2-kernel) = (M 1/2 2π )N

R2N traceA(x, p)dxdp

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In fact also

Φn, ˆ A ˆ BΦn = (M 1/2 2π )N

R2N trace
A(z)B(z)
dz

Choosing B = e−H/T: trace(Ae−H/T) = trace( ˜ Ae− ˜

H/T)

=

d

˜ Akke− ˜

Hkk/T,

and ˜ Hkk(z) = |p|2 2 + λk(x),

e−H/T = e− ˆ

H/T + O(M −1/2T −1),

qk from normalization

SLIDE 12

The quantum density, momentum and energy observ- ables satisfy the conservation laws (Irving & Zwanzig, 1951)

ρ(y, t) := trace

ρtf( ˆ H)

Φn, ˆ ρtf( ˆ H)Φn ˆ ρ0 =

η(y − xj)

density operator

ˆ ρt = eit

√ M ˆ H ˆ

ρ0e−it

√ M ˆ H

time evolution

ˆ HΦn = EnΦn

Schr¨

dinger eigensolutions

ˆ p0 :=

j

η(y − xj)pj

momentum operator

ˆ E0 :=

j

η(y − xj)(|pj|2 2 + 1 2

Vjk)

(scalar) energy operator

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Why is quantum same as classical?

∂t ˆ At = i √ M[ ˆ H, ˆ At]

Heisenberg time evolution

If m ≤ 2: i √ M[ ˆ H, A(x)pm] =

{H, A(x)pm}

=

dtA(xt)pm

t

xt=x,pt=p

=     

A′(x)p

m = 0

A′(x)p2 − V ′A(x)

m = 1

A′(x)p3 − 2V ′A′(x)p m = 2

SLIDE 14

Matrix valued potential?

i √ M[ ˆ H, ˆ A] = O(M 1/2) = {H, A} = O(1) Seek ˆ At = ˆ Ψ(x) ˆ ˜ At ˆ Ψ(x)∗ then ∂t ˆ ˜ At = iM 1/2[ˆ Ψ∗(x) ˆ H ˆ Ψ(x)

diagonal?

, ˆ ˜ At] ˆ Ψ∗(x) ˆ H ˆ Ψ(x) = (Ψ∗HΨ + 1 4M ∇Ψ∗ · ∇Ψ)

Choose Ψ so that: diagonal + Ok(M −k), any k.