Canonical Quantum Observables Approximated by Molecular Dynamics for - PowerPoint PPT Presentation

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?

Conservation of mass, momentum and energy 1 ∂ t ρ ( y, t ) + ∂ k ( ρu k ) = 0 � ∂ t ρu j ) + ∂ k ( ρu j u k − σ jk ) = 0 ∂ t E + ∂ k ( Eu k − σ kj u j + q k ) = 0 Jokkfall in Kalix¨ alven Stress tensor σ =? Heat flux q =? 1 Euler (1752), Laplace (1816)

Stress tensor and heat flux from molecular dynamics: X j ˙ t = P j t � P j ˙ t = − ∂ j V jk ( X t ) pair potential interaction k Figure 1: solid-liquid phase transformation, von Schwerin & Szepessy (2010)

and Irving & Kirkwood (1950), Hardy (1981) definition � � η ( y − X j ρ ( y, t ) := t ) f ( X 0 , p 0 )d X 0 d P 0 R 2 N j � � η ( y − X j t ) P j ρu ( y, t ) := p ( y, t ) := t f ( X 0 , P 0 ) d X 0 d P 0 � �� � R 2 N j initial density � � � t )( | P j t | 2 + 1 η ( y − X j E ( y, t ) := V jk ) f d X 0 d P 0 2 2 R 2 N j k � R 3 η ( y )d y = 1 X 2 ¡ x ¡ X 1 ¡ X j ¡ X 3 ¡ Figure 2: support of η with red particle positions

yields stress tensor � � σ ( y, t ) = 1 ( X j t − X k t ) ⊗ ∂ j V jk b jk f d X 0 d P 0 2 R 2 N j,k � � � � � � η ( y − X j P j P j − t ) t − u ( y, t ) ⊗ t − u ( y, t ) f d X 0 d P 0 , R 2 N j where � 1 � � y − (1 − s ) X j t − sX k b jk ( X t ) := η d s t 0

Potential � k V jk =? Schr¨ odinger equation with Hamiltonian: H = − 1 ˆ M ∆ + V ( x ) V : R N → C d 2 x ∈ R N nuclei coordinates , potential d = 1 : Schr¨ odinger observables for mass, momentum and energy satisfy the conservation laws 2 . 2 Irving and Zwanzig (1951) for scalar smooth potentials

Constant temperature and several electron states H = − 1 ˆ M ∆ + V ( x ) odinger: ˆ Schr¨ H Φ n = E n Φ n Goal determine � n � Φ n , ˆ A Φ n � e − E n /T Observable A and Temperature T M = 12800, δ = M −0.25 8 � ψ t , V ( X t ) ψ t � E 6 λ + ( X t ) λ − ( X t ) 4 2 0 −2 0 1 2 3 4 Electron eigenvalueproblem V ( x )Ψ j ( x ) = λ j ( x )Ψ j ( x ) t

If λ 2 − λ 1 ≫ T � τ � n � Φ n , ˆ A Φ n � e − En/T 0 A ( X t , P t ) d t lim τ →∞ approximates � n � Φ n , Φ n � e − En/T τ using Langevin: ˙ X t = P t √ ˙ 2 κT ˙ P t = −∇ λ 1 ( X t ) − κP t + W t . All T possible?

All T Theorem 3 possible: There holds � � τ A � d � n � Φ n , ˆ e − H/T Φ n � t )d t ˜ A kk ( Z k = lim q k τ , � n � Φ n , � τ →∞ e − H/T Φ n � 0 k =1 where Z k t = ( X t , P t ) with λ k , q k ¯ q k = , � d i =1 ¯ q i � τ e − λk ( X 1 t ) − λ 1( X 1 d t t ) q k = lim ¯ τ , T τ →∞ 0 Ψ ∗ ( x ) H ( x, p )Ψ( x ) = ˜ H ( x, p ) diagonal , Ψ ∗ ( x ) A ( x, p )Ψ( x ) = ˜ A ( x, p ) diagonal , M 1 / 2 T ) for e − ˆ 1 H/T . . . . + O ( 3 C. Lasser, M. Sandberg, A. Szepessy, A. Kammonen in preparation

Proof uses Weyl quantization: � � R N ( M 1 / 2 R N e iM 1 / 2 ( x − y ) · p A ( x + y ˆ 2 π ) N Aφ ( x ) = , p )d p φ ( y )d y, 2 � �� � L 2 -kernel � V ( x ) = V ( x ) , � so H ( x, p ) = | p | 2 | p | 2 = − 1 2 M ∆ , 2 I + V ( x ) , 2 Weyl’s law: � � Φ n , ˆ A Φ n � = trace ˆ A n = trace( L 2 -kernel ) � = ( M 1 / 2 2 π ) N R 2 N trace A ( x, p )d x d p

In fact also � � � � B Φ n � = ( M 1 / 2 � Φ n , ˆ A ˆ 2 π ) N R 2 N trace A ( z ) B ( z ) d z n Choosing B = e − H/T : Ae − ˜ trace( Ae − H/T ) = trace( ˜ H/T ) d � A kk e − ˜ ˜ H kk /T , = k =1 and H kk ( z ) = | p | 2 ˜ 2 + λ k ( x ) , � H/T + O ( M − 1 / 2 T − 1 ) , e − H/T = e − ˆ q k from normalization

The quantum density, momentum and energy observ- ables satisfy the conservation laws (Irving & Zwanzig, 1951) � � � ρ t f ( ˆ ρ t f ( ˆ ρ ( y, t ) := trace ˆ H ) = � Φ n , ˆ H )Φ n � n N � � � � η ( y − x j ) ρ 0 = ˆ density operator j =1 √ √ H ˆ M ˆ M ˆ ρ t = e it ρ 0 e − it H ˆ time evolution ˆ H Φ n = E n Φ n Schr¨ odinger eigensolutions � � η ( y − x j ) p j � � p 0 := ˆ momentum operator j � � � η ( y − x j )( | p j | 2 � + 1 � ˆ E 0 := V jk ) (scalar) energy operator 2 2 j k

Why is quantum same as classical? √ ∂ t ˆ M [ ˆ H, ˆ A t = i A t ] Heisenberg time evolution If m ≤ 2 : √ H, � � M [ ˆ A ( x ) p m ] = { H, A ( x ) p m } i � � d � dtA ( x t ) p m = � t x t = x,p t = p  �  A ′ ( x ) p m = 0  � A ′ ( x ) p 2 − V ′ A ( x ) = m = 1   � A ′ ( x ) p 3 − 2 V ′ A ′ ( x ) p m = 2

Matrix valued potential? √ A ] = O ( M 1 / 2 ) � = � M [ ˆ H, ˆ i { H, A } = O (1) Ψ( x ) ˆ Seek ˆ A t = ˆ A t ˆ ˜ Ψ( x ) ∗ then ∂ t ˆ , ˆ A t = iM 1 / 2 [ˆ ˜ Ψ ∗ ( x ) ˆ H ˆ ˜ Ψ( x ) A t ] � �� � diagonal? 1 4 M ∇ Ψ ∗ · ∇ Ψ) Ψ ∗ ( x ) ˆ ˆ H ˆ Ψ( x ) = (Ψ ∗ H Ψ + � Choose Ψ so that: diagonal + O k ( M − k ) , any k .

Then N � � � � ˆ ρ 0 := ˆ η ( y − x j )I Ψ ∗ ˆ Ψ density operator j =1 � � � � ˆ p 0 := ˆ η ( y − x j ) p j I Ψ ∗ ˆ Ψ j � � H j � � ˆ E 0 := ˆ ˆ η ( y − x j ) ˜ Ψ ∗ Ψ j with diagonal energy per particle partition N � H j = ˜ ˜ H j =1 Theorem 4 : The Sch¨ odinger observables for the density, momentum and energy solve the conservation laws O k ( M − k ) accurately, any k . 4 also in preparation: M. Sandberg and A. Szepessy