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 energy1 tρ(y, t) + k (ρuk ) = 0 t ρu ) + k (ρu uk σk ) = 0 te + k (Euk σk u + qk ) = 0 Jokkfall in Kalixa lven Stress tensor σ =? Heat flux q =? 1 Euler (1752), Laplace (1816)
Stress tensor and heat flux from molecular dynamics: Ẋ t = P t P t = k V k (X t ) pair potential interaction Figure 1: solid-liquid phase transformation, von Schwerin & Szepessy (2010)
and Irving & Kirkwood (1950), Hardy (1981) definition ρ(y, t) := η(y X t )f(x 0, p 0 )dx 0 dp 0 ρu(y, t) := p(y, t) := E(y, t) := R 3 η(y)dy = 1 R 2N R 2N R 2N η(y X t )P t f(x 0, P 0 ) dx }{{} 0 dp 0 initial density η(y X t )( P t 2 + 1 2 2 k V k )fdx 0 dp 0 X2 x X1 X X3 Figure 2: support of η with red particle positions
yields stress tensor σ(y, t) = 1 2 where R 2N b k (X t ) := R 2N,k (X t X k t ) V k b k f dx 0 dp 0 η(y X t ) ( P t u(y, t) ) ( P t u(y, t) ) f dx 0 dp 0, 1 0 η ( y (1 s)x ) t sxt k ds
Potential k V k =? Schrödinger equation with Hamiltonian: Ĥ = 1 M + V (x) x R N nuclei coordinates, V : R N C d2 potential d = 1: Schrödinger 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 Ĥ = 1 M + V (x) Schrödinger: ĤΦ n = E n Φ n Goal determine n Φ n, ÂΦ n e E n/t Observable A and Temperature T 8 6 4 M = 12800, δ = M 0.25 ψ t, V (X t )ψ t E λ + (X t ) λ (X t ) 2 0 2 0 1 2 3 4 t Electron eigenvalueproblem V (x)ψ (x) = λ (x)ψ (x)
If λ 2 λ 1 T lim τ τ 0 A(X t, P t ) dt τ approximates n Φ n,âφ n e E n/t n Φ n,φ n e E n/t using Langevin: Ẋ t = P t P t = λ 1 (X t ) κp t + 2κT Ẇt. All T possible?
All T Theorem 3 possible: There holds where n Φ n, Â ê H/T Φ n n Φ n, ê H/T Φ n = lim τ Zt k = (X t, P t ) with λ k, q k q k = d i=1 q, i d τ q k k=1 0 Ã kk (Z k t ) dt τ, τ q k = lim τ 0 e λ k (X1 t ) λ 1 (X1 t ) T dt τ, Ψ (x)h(x, p)ψ(x) = H(x, p) Ψ (x)a(x, p)ψ(x) = Ã(x, p) diagonal,... + O( 1 M 1/2 T ) for e Ĥ/T. diagonal, 3 C. Lasser, M. Sandberg, A. Szepessy, A. Kammonen in preparation
Proof uses Weyl quantization: Âφ(x) = RN ( M 1/2 2π )N e im 1/2 (x y) p A( x + y, p)dp φ(y)dy, } R N {{ 2 } L 2 -kernel V (x) = V (x), p 2 2 = 1 2M, p 2 so H(x, p) = 2 I + V (x), Weyl s law: Φ n, ÂΦ n = trace  n = trace(l 2 -kernel) = ( M 1/2 2π )N tracea(x, p)dxdp R 2N
In fact also Φ n, Â ˆBΦ n = ( M 1/2 n 2π )N R 2N trace ( A(z)B(z) ) dz Choosing B = e H/T : and trace(ae H/T ) = trace(ãe H/T ) d = Ã kk e H kk /T, k=1 H kk (z) = p 2 2 + λ k(x), ê H/T = e Ĥ/T + O(M 1/2 T 1 ), q k from normalization
The quantum density, momentum and energy observables satisfy the conservation laws (Irving & Zwanzig, 1951) ρ(y, t) := trace (ˆρ t f(ĥ)) = n Φ n, ˆρ t f(ĥ)φ n ˆρ 0 = ( N η(y x ) ) =1 ˆρ t = e it MĤ ˆρ 0 e it MĤ density operator time evolution ĤΦ n = E n Φ n Schrödinger eigensolutions ˆp 0 := ( η(y x )p ) momentum operator Ê 0 := ( η(y x )( p 2 + 1 V k ) ) 2 2 k (scalar) energy operator
Why is quantum same as classical? If m 2: t  t = i M[Ĥ, Ât] Heisenberg i M[Ĥ, A(x)pm ] = {H, A(x)pm } d = dt A(x t)p m t = time evolution xt =x,p t =p  (x)p m = 0 A (x)p 2 V A(x) m = 1 A (x)p 3 2V A (x)p m = 2
Matrix valued potential? i M[Ĥ, Â] = O(M 1/2 ) {H, A} = O(1) Seek Ât = ˆΨ(x) ˆÃ t ˆΨ(x) then t ˆÃ t = im 1/2 [ ˆΨ (x)ĥ }{{ ˆΨ(x), } ˆÃ t ] diagonal? ˆΨ (x)ĥ ˆΨ(x) = (Ψ HΨ + 1 4M Ψ Ψ) Choose Ψ so that: diagonal + O k (M k ), any k.
Then ˆρ 0 := ˆΨ ( N η(y x )I ) ˆΨ =1 ˆp 0 := ˆΨ ( η(y x )p I ) ˆΨ Ê 0 := ˆΨ ( η(y x ) H ) ˆΨ with diagonal energy per particle partition N H = H =1 density operator Theorem 4 : The Schödinger 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