Zero-Point Energy Leakage in Quantum Thermal Bath Molecular

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Zero-point Energy Leakage in Quantum Thermal Bath Molecular Dynamics Simulations Fabien Brieuc, Yael Bronstein, Hichem Dammak, Philippe Depondt, Fabio Finocchi, and Marc Hayoun J. Chem. Theory Comput., Just Accepted Manuscript • DOI: 10.1021/acs.jctc.6b00684 • Publication Date (Web): 21 Oct 2016 Downloaded from http://pubs.acs.org on October 24, 2016

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Journal of Chemical Theory and Computation is published by the American Chemical Society. 1155 Sixteenth Street N.W., Washington, DC 20036 Published by American Chemical Society. Copyright © American Chemical Society. However, no copyright claim is made to original U.S. Government works, or works produced by employees of any Commonwealth realm Crown government in the course of their duties.

Page 1 of 40

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Journal of Chemical Theory and Computation

† § ‡ § ∗ † ¶ ‡

∗ ‡ ¶

! "# $ % & '

! $ ( ) *+ ,,- !

! " # $%&' ! #! ( !

!! " # $ ‡ %& ' ( ) *"!"! $ ¶ ' + , -. //!0 $ § , 1 ∗

†

ACS Paragon Plus Environment


1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

Page 2 of 40

! ! ) $%&' ! $%&' ! * + # , " $%&' , - #

$%&' , * ) ) ! ! *

!

" # $ % # # &' ' # # #

(

# # $ ) ) *


Page 3 of 40

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60


# #

+ # , - # ) # ./ 0 ! ./

,

$ # # ./) # 1 ) 2 3 # # $ ./ # 2 1

4 ./ 5) 2

6&7- 2 # "

6&7- 2 # ) ) " 6&7- # ! - ) 8

9 ./ 6&7- #

# / ) 5 2 $ 6&7- ./ 2 # 4 6&7- # ./) # / ) 5 2 # # $ )

2 2 : 2

6&7- 2 #

$ 6&7- ./) ;



1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

Page 4 of 40

# ./ < 6&7- 3 # ./ 1 = ) - )8 / %3 '

./ # , - ! / - ./ x m # f (x) =

m¨ x = f (x) − mγ x˙ + R(t)

) # # R(t) # : 5 =

< R(t) > = 0 < R(t)R(t + τ ) > =

* +∞ −∞

IR (ω, T ) −iωτ

ω . 2π

;

7 ; ()>
&4 IR (ω, T ) T # ?


Page 5 of 40

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60


&4 &4 Ix # < ) =

χ˜ (ω) =

ω Ix (ω, T ) 2kB T

?

˜ (ω) # χ(ω) ˜ 1 χ ˜ x ˜(ω) 1 R(ω) =

˜ x˜(ω) = χ(ω) ˜ R(ω)

@

1 # # &4 Ix &4 IR = 2 Ix (ω, T ) = |χ(ω)| ˜ IR (ω, T )

A

< ) # =

2kB T χ˜ (ω) IR (ω, T ) = . 2 ω |χ(ω)| ˜ ' ω0

B

@

#

χ(ω) ˜ =

m [ω02

1 . − ω 2 + iγω]

C

/ B &4 # =

IR (ω, T ) = 2mγ kB T

∀ω.

/ ()> ;

D

- # ' @



1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

Page 6 of 40

# kB T ω ' # #

⎤

⎡

1 1 ⎦ θ(ω, T ) = ω ⎣ + 2 exp ω − 1 kB T

E

./ &4 D # # E # < ) # F ( # > # =

χ ˜ (ω) =

ω Ix (ω, T ) 2θ(ω, T )

()> &4 R(t) ./ =

IR (ω, T ) = 2mγ θ(ω, T )

*

' - IR ω R(t) # # )! ω # ) ./ = 0 γ )! ω # ( - 0 γ # , # # / +

ω # # ) A


Page 7 of 40

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60


ω γ ' 2 # ω 2ω ω # ' ./) 0 γ D *

' # ./) ) 2 # ./) ) 4G # # $

H= 1 1 1 1 H = mx˙ 21 + mω12 x21 + mx˙ 22 + mω22 x22 + C3 (x1 − x2 )3 + C4 (x1 − x2 )4 2 2 2 2

;

x1 x2 ω1 ω2 ) m C3 C4 $ H #

˜ = H/ω1 = H 2 2 2 2 ˜ = q˙1 + q1 + q˙2 + Ω2 q2 + c3 (q1 − q2 )3 + c4 (q1 − q2 )4 H 2 2 2 2

?

# =

ω2 Ω= , ω1

ξ=

, mω1

xi qi = , ξ

C3 ξ 3 c3 = , ω1

C4 ξ 4 c4 = , ω1

t∗ = ω1 t,

q˙i =

dqi . @ dt∗

q1 q2 Ω ω1 > ω2 ) # B


Page 8 of 40

Ω=0.5

0.5 0.4 0.3 0.2 0

2

4

6

8 104 x c3

0.6 average energy

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

average energy


10

12

14

Ω=0.25

0.4 0.2 0 0

5

10

QTB

15

20 4 10 x c4

QTB

ε1

25

30

QTB

ε2

εc

35

Exact

1 = 1 2 c # ./) c3 c4 = # c3 = 0 c4 = 0 Ω = 0.5 - = c4 = 0 c3 = 0 Ω = 0.25 / c = 0 # 6&7- ./ ' < # 1 2 =

1 =

q˙12 2

+

q12 2

,

2 =

q˙22 2

2

+Ω

q22 2

.

A

./) 0 γ = 4 × 10−4 ω1

)! ω = 2ω1 δt = 0.05ω1−1 ;E " 107 Ω EE@HEC c3 c4 EH25 × 10−4 EH40 × 10−4 , kB T = 0.03 ω1 T ∼ 60 > ω1 = 2π × 40 $5 # # 5) C


Page 9 of 40

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60


1 0.8

ζ

(a)

4

10 x c3 = 2.4 9.7 14.6 19.4

0.6 0.4 0.2 0 0.48

0.5

0.52

0.54

0.56

0.58

0.6

Ω 1

0.6

ζ

(b)

104 x c4 = 3.9 7.7 15.4 30.8

0.8

0.4 0.2 0 0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

Ω

1 *= 6) 2 , ζ B

Ω @ = # c3 = 0 c4 = 0 - = c4 = 0 c3 = 0 5) 9 1 = 0.5 2 = Ω/2 1

# ./ =

Ω = 0.5 # Ω = 0.25 c3 = 0 c4 = 0 ./ 5) ' c3 c4 ./) =

* 6&7-

' 6&7- Ω c3 c4 , $ 1 ζ 6&7-=

Δ − Δ ( − ) − ( − ) 1 2 1 2 ζ= = Δ 1 − 2

B

( , 2 ζ = 1 D



-4

(a)

arbitrary units 0

ω1-ω2

2ω2

0.5

1

γ=4x10 ω1

ω1+ω2

1.5

2

0

-4

(b)

-3

(c)

γ=4x10 ω1

0.5

1

2 -3

(d)

γ=4x10 ω1

1.5

γ=4x10 ω1

arbitrary units

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

Page 10 of 40

0

0.5

1

1.5

2

0

0.5

ω/ω1 oscillator 1

1

ω/ω1

1.5

2

oscillator 2

1 ;= I# * # # ./) # Ω = 0.5 c3 = 2.4 × 10−4

Ω = 0.2 c4 = 15.4 × 10−4 0= γ = 4 × 10−4 ω1 γ = 4 × 10−3 ω1

= = ' 2 ζ = 0 1 2 1,2 = 1,2 ' , * # ζ Ω !

c3 c4 % 6&7- Ω ' # Ω = 0.5 , * ' # 2 # # # 2ω , # # ./) # , ; = 2ω2 ω1 − ω2 ω1 + ω2 # Ω = 0.5 # (ω1 ; 2ω2) (ω2 ; ω1 − ω2 ) 4 # 3ω 9 6&7 # Ω = 1/3 , *# ( , 6&7- Ω 1 ;# Ω = 0.2 c4 = 15.4 × 10−4 ω1 ω2 E


Ω=0.5

0.5

2x10-2 1x10-2

0.4

4x10

-3

4x10

-4

4x10-3

0.3

1x10-2 -2 2x10

0

average energy

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60


average energy

Page 11 of 40

2

4

6

8 4 10 x c3

10

12

14

Ω=0.25

0.5 0.4 0.3 0.2 0.1 0

5

10

15

20 4 10 x c4

25

30

35

1 ?= 7! 0 γ ω1 A = # c3 = 0 c4 = 0 Ω = 0.5 - = c4 = 0 c3 = 0 Ω = 0.25 # # ./) $ 2 , 6&7- Ω < 1/3 γ

0

4 × 10−4 ω1 2 × 10−2 ω1 ( 6&7- = Ω = 0.5 # Ω = 0.25 , * 1 ? 6&7- γ ' γ 2 # ' c3 = 2.4 × 10−4

γ 4 × 10−3 ω1 0 6&7- ζ = 0.08 1 ; # # γ = 6&7- # 2 2ω2 ω1 − ω2 ω1 + ω2 # 6&7- γ # 2

γ/2π -



γ 6&7- , ? 1 ; γ

Ω = 0.2 c4 = 15.5 × 10−4 0 # ! ' ttr # JI7

ttr

# # * 1 @ ttr Ω = 0.5 # c3 6&7- γ ω = 1/ttr 1

c3 = 4 × 10−4 , ttr ∼ 400ω1−1 ω ∼ 2.5 × 10−3ω1 , ? γ = 10−2 ω1 2 6&7- γ ω Ω=0.5 1500

-1

ttr (ω1 )

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

Page 12 of 40

1000

500 1

1.5

2

2.5

3

3.5

4

-4

10 x c3

1 @= 7 # ttr ω1−1 # c3 Ω = 0.5 JI7 $ c4 = 0 ' =

# , ! 6&7- # γ # # ( *


Page 13 of 40

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60


( ) 3 3 # #

# # % %H$ ) # 8 )# =

V (r) =

−b(r−r0 )

a(r−r0 ) u0 a e − 1 + b e − 1 − u0 a + bea(r−r0 )

C

r %H$ u0 # r0 # %H$ a b = r0 = 0.96 K # %$− a 7.11 K−1

b 2.00 K−1 u0 = 2.73 I %H$ ν

V 100 $5 % %H% # # =

2 V (R) = C0 1 − e−α0 (R−R0 ) − C0

D

C0 α0 R0 %H% # = C0 = 3.81 I R0 = 2.88 K

α0 %H% ν # 10 60 $5 ./) 0.1 # # * " 3 # V (r) + V (R − r)

#) ( , L L %H$ # ∼ 1 K

L # L ∼ 1.9 K 5 # %)$ ;



1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

Page 14 of 40

! 6&7- )# ) ν2 %H% ) ν1 %H$ ' %H$

) %H% ' < Ω = ν2 /ν1 0 γ 6&7- T = 600 >

ν2 α0 D ν , 100 $5 ν1 ./) &' # P = 20 1 ./) 2 2

# &' ' 2 2 , ) ) ! T T $ % , 2 = N 1 kB T (i) = E 2 N i=1 k

N 1 (i) kB T = E , 2 N i=1 k

*E

(i)

N = 3 N = 3 # $ % Ek

2 i ' 2 # = ! = ) 2 ! ./ &' 1 , A 2 !

?


Page 15 of 40

2500 effective temperature (K)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60


H - PIMD ZPEL

2000

H - QTB-MD O - QTB-MD

ZPEL

1500

O - PIMD

1000 0

1000

2000 time (ps)

3000

4000

1 A= 7! T T ./) &' T = 600K $ Ω = 0.5 γ = 0.2 $5 6&7- ' 6&7- # , =

ζ=

(T − T )() − (T − T )() (T − T )()

.

*

ζ = 0 2 0 < ζ < 1 2 Ω , B 4 3 ; 6&7- Ω ∼ 1/2 1 B 6&7- # # # γ % 6&7- # Ω < 0.2= ! Ω

1 C #

d d Ω = 0.5 ./) &' ) ' , C d # ! # 6&7- % d # 6&7-= ./) # # 2 6&7- $ 6&7- # @



1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

1

Page 16 of 40

γ = 0.2 THz γ = 2 THz γ = 5 THz γ = 10 THz γ = 20 THz

0.8 0.6 ζ

0.4 0.2 0 0.2

0.3

0.4

0.5

0.6

0.7

Ω

1 B= 6&7- , # ζ , *

Ω γ ./) d # &'

(

γ # 9 2 # 2 γ , ; 1 D # ) Ω = 0.5 ! γ 0.2 10 $5 1 γ = 0.2 $5 6&7- γ = 10 $5 6&7 , B ( 0 # 2 $ 2 , # γ 7 # #

' ! 2 # # 6&7- 6&7- # # F 0 6&7- ' 9 A


Page 17 of 40

(a) : = 0.5

-1

probability density (Å )

2 Standard-MD PIMD

1.5

QTB-MD (0.2 THz) QTB-MD (20 THz)

1 0.5 0 0.8

1 d

1.2

OH

-1

10

1.4

(Å)

12 probability density (Å )

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60


: = 0.5

(b)

8 6 4 2 0 2.7

2.8 d

2.9 (Å)

3

3.1

OO

1 C= &# # # %H$ %H% # ) &' ./) 0 γ 0.2 20 $5 Ω = 0.5 # # #

( ! 0 γ 6&7-

M ./) / ) N5 . 6&7- T = 10 > # - )8 /kB = 1450.6 > σ = 2.54 K ! = 1.37σ = 3.49 K ' ) ) # ./ # B



0.7

γ=2 THz γ=10 THz

0.6 0.5 arbitrary units

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

Page 18 of 40

0.4 0.3 0.2 0.1 0 20

40

60

80

100

120

frequency (THz)

1 D= I# # # ./) Ω = 0.5 0= γ = 0.2 $5 # γ = 10 $5 # 2 ( ./) 1

! γ ( , γ = 0.9 $5 ./) # # , E ' # # # $ γ E $5 # ./) # θ(ν, T )

E #

, E 6&7- 5 # ./ , E # 5 # # γ 0 6&7- γ = 9 $5 1 γ 2 # # # ./) ./ - # # # 2 # ) %4 1 γ γ Δω 2 %4 γ , # F

γ # # Δν = Δω/2π = γ/2π C


Page 19 of 40

J = 0.9 THz J = 10 THz

200

150 1

slope / QD slope

K

B

2 E /k (K)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60


100

50

0.8 0.6 0.4 0.2 0

0

2

0

4

2

4

6

6

J (THz)

8

10

8

frequency (THz)

1 E= 4 2 # T = 10 > # ./) γ ED E $5 )! 20 $5 # . θ(ν, T ) E ν = ω/2π '= γ # 5 # ,

' - %4

# # 5 1 #

kB T ' ./) kB T # # θ(ω, T ) E 1

6&7- %4 # ./) " ) ' ,

# 6&7-

%4 # # ./) " # ) )

/ %3 /% 5 # ! # < ' ) ) = # + # % # F 7 ,

D



(a)

Standard-MD (10 THz) QTB-MD (10 THz)

(b)

Standard-MD (2 THz) QTB-MD (2 THz)

DOS (ps)

500 400 300 200 100 0

500

DOS (ps)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

Page 20 of 40

400 300 200 100 0

0

5

10

15

20

frequency (THz)

1 = I# %4 ! γ # 1 ) 5 # kB T - # θ(ν, T ) E ./) 0 γ = 10 $5

# 2 $5

*E


Page 21 of 40

5 4 # ./ #

./)

> *BE >

- # / /% # ! $ ,) ) $ 0 γ E@ A $5 )! ν @ $5 $ γ ) +)% %) )F 1 *

0.03

reduced polarization

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60


PIMD QTB (16 THz)

0.02 R

C

C

200

T O

T (K)

0.01

T

O

300

100

R 0

0 100

1

J(THz)

150

200

10

250

300

T (K)

1 *= 5 ) / %3 # # ./) γ = 16 $5 &' P = 16 I # +)% %) )F 0 γ ./ 5 # # &' 5 # # ./)

γ = 16 $5 1 ./) = +)%))F &'

# P = 16 = AE >

DE > *@@ > # # &' A; >

DC > *@C >

, *

γ ./) 1 γ # # *



1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

Page 22 of 40

' γ ./ = ! 6&7- #

( 5) 2 6&7- ./) # ./ ! ( 6&7- # # # $ γ , 2 2 # /% # # ./ ! # ./

# < ) ./ ) ) 6&7- ) ) = # # # # ./ ' # ./ # # γ # # ! #

$ !

6&7- ./) γ -

**


Page 23 of 40

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60


) F ! # # , # ./) # # 2 # ' # ./) # # γ γ # 2 %)$ # )# γ % # # ! # ./ 0 0 6&7- # ' ./ = #

# 2 # ./)

( 2 # 2 # :O : /% # ( 2 8 )- / -8 # P/ 2 , F +O 3Q) )1 ' % 2 $&F LOL F 4 O # F 4 O FJ+4

*;



1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

Page 24 of 40

< + # # # 2

. #

# . , $ : R(t) =

R(t) = 0 R(t)R(t + τ ) =

** +∞

−∞

IR (ω, T )−iωτ

ω 2π

*;

()> R(t) IR (ω, T ) ' ./) IR # * tn = nδt : # # : # Xj Wj +∞

Rn ≡ R(tn ) =

*?

Wj Xj+n

j=−∞

# Xj 5 1 *; *? Wj # =

+∞

Wj Wj−l =

j=−∞

+∞ −∞

IR (ω, T )−iωtl

ω 2π

*@

˜ (ω) , 1 W (t)= tl = lδt W Wj ≡ W (tj ) =

+∞

−∞

*?

˜ (ω)−iωtj ω W 2π


*A

Page 25 of 40

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60


M *A δt → 0= +∞

Wj Wj−l

j=−∞

∞

˜ (ω)W ˜ (−ω)−iωtl ω W 2π −∞

1 = δt

*B

˜ (ω) # F W (t) W *@ *B=

˜ (ω) = W

=

√ Wj = δt

δtIR (ω, T )

ω IR (ω, T )−iωtj 2π

+∞

−∞

*C

*D

' 5 1

1 Wj = √ N 1 Xj = √ N

N/2

˜ l −i2πjl/N W

;E

˜ l −i2πjl/N X

;

l=−N/2+1 N/2 l=−N/2+1

N # / *D ;E=

1 Wj = √ Nδω

+∞ −∞

˜ (ω)−iωtj ω W

;*

Nδtδω = 2π #

˜l = √ 1 W Nδt ωl = lδω 1 ;E ;

Rn =

IR (ωl )

;;

˜ −l X ˜ l −i2πnl/N W

;?

Rn N/2 l=−N/2+1

*@



1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

Page 26 of 40

;; Rn , = N/2 1 ˜ l −i2πnl/N Rn = √ IR (ωl )X Nδt l=−N/2+1

;@

˜ l - , 1 R 1 Rn = √ N

N/2

˜ l −i2πnl/N R

;A

l=−N/2+1

#

˜l = R

IR (ωl ) ˜ Xl δt

;B

˜ l # : # X ˜ + iN ˜l ˜ l = Ml √ X 2

;C

˜ l N ˜l : # 5 M ˜ l M ˜l = M ˜ −l # X ˜l = −N˜−l 1 = N ˜l = R

IR (ωl ) ˜ ˜l Ml + iN 2δt

;D

' Rn # =

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Page 32 of 40

Page 33 of 40

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Journal of Chemical Theory and ComputationPage 34 of 40

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20


Page 35 of 40

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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17


Page 37 of 40Journal of Chemical Theory and Computation

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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17


Page Journal 39 of of 40Chemical Theory and Computation

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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17


Zero-Point Energy Leakage in Quantum Thermal Bath Molecular

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