Basic Monte Carlo(chapter 3)

Algorithm

Detailed Balance

Other points

2

Molecular Simulations

Molecular dynamics: solve equations of motion

Monte Carlo: importance sampling

r1

MD

MC

r2rn

r1

r2

rn

3

Does the basis assumption lead to something that is consistent with classical

thermodynamics?

1ESystems 1 and 2 are weakly coupled such that they can exchange energy.

What will be E1?

( ) ( ) ( )1 1 1 1 2 1,E E E E E EΩ − = Ω ×Ω −

BA: each configuration is equally probable; but the number of states that give an energy E1 is not know.

2 1E E E= −

4

( ) ( ) ( )1 1 1 1 2 1,E E E E E EΩ − = Ω ×Ω −

( ) ( ) ( )1 1 1 1 2 1ln , ln lnE E E E E EΩ − = Ω + Ω −

( )1 1

1 1

1 ,

ln ,0

N V

E E E

E

⎛ ⎞∂ Ω −=⎜ ⎟

∂⎝ ⎠

∂lnΩ1 E1( )∂E1

⎛

⎝⎜

⎞

⎠⎟

N1 ,V1

+∂lnΩ2 E −E1( )

∂E1

⎛

⎝⎜

⎞

⎠⎟

N2 ,V2

=0

( ) ( )1 1 2 2

1 1 2 1

1 2, ,

ln ln

N V N V

E E E

E E

⎛ ⎞ ⎛ ⎞∂ Ω ∂ Ω −=⎜ ⎟ ⎜ ⎟

∂ ∂⎝ ⎠ ⎝ ⎠

Energy is conserved!dE1=-dE2

This can be seen as an equilibrium

condition

( ),

ln

N V

E

Eβ

⎛ ⎞∂ Ω≡ ⎜ ⎟

∂⎝ ⎠

1 2β β=

5

Canonical ensemble

Consider a small system that can exchange heat with a big reservoir

iE iE E− ( ) ( ) lnln lni iE E E E

E

∂ ΩΩ − = Ω − +

∂L

( )( )

ln i i

B

E E E

E k T

Ω −= −

Ω

1/kBT

Hence, the probability to find Ei:

( ) ( )( )

( )( )

exp

expi i B

i

j j Bj j

E E E k TP E

E E E k T

Ω − −= =

Ω − −∑ ∑( ) ( )expi i BP E E k T∝ −

Boltzmann distribution

7

Statistical Thermodynamics

( )NVTQF ln−=β

( ) ( )[ ]NNNN

NVTNVT

UANQ

A rexprdr!

113

β−Λ

= ∫€

QNVT =1

Λ3N N!drN∫ exp −βU rN

( )[ ]

Partition function

Ensemble average

Free energy

( ) ( ) ( ) ( )3

1 1N r dr' r' r exp r' exp r

!N N N N N N

NNVT

U UQ N

δ β β⎡ ⎤ ⎡ ⎤= − − ∝ −⎣ ⎦ ⎣ ⎦Λ ∫

Probability to find a particular configuration

8

Monte Carlo simulation

9

Ensemble averageA

NVT=

1QNVT

1Λ3NN!

drN∫ A rN( )exp −βU rN( )⎡⎣ ⎤⎦

= drN A rN( )P rN( )∫ =drN A rN( )P rN( )∫

drN P rN( )∫

=drN A rN( )C∫ exp −βU rN( )⎡⎣ ⎤⎦

drNC exp −βU rN( )⎡⎣ ⎤⎦∫=

drN A rN( )∫ exp −βU rN

( )⎡⎣ ⎤⎦drN exp −βU rN

( )⎡⎣ ⎤⎦∫Generate configuration using MC:

PMC rN( ) =CMC exp −βU rN( )⎡⎣ ⎤⎦

r1

N , r2N , r3

N , r4N L ,rM

N{ } A =1M

Ai=1

M

∑ riN( )

with

=drN A rN( )PMC rN( )∫

drN PMC rN( )∫

=drN A rN

( )C MC exp −βU rN( )⎡⎣ ⎤⎦∫

drNC MC exp −βU rN( )⎡⎣ ⎤⎦∫

=drN A rN

( )exp −βU rN( )⎡⎣ ⎤⎦∫

drN exp −βU rN( )⎡⎣ ⎤⎦∫

( ) ( )[ ]!

rexpr

3 NQ

UP

NNVT

NN

Λ

−=

β

10

Monte Carlo simulation

11

12

13

Questions

• How can we prove that this scheme generates the desired distribution of configurations?

• Why make a random selection of the particle to be displaced?

• Why do we need to take the old configuration again?

• How large should we take: delx?

14

Detailed balance

acc(o → n)acc(n→ o)

=N(n) ×α(n→ o)N(o) ×α(o→ n)

=N(n)N(o)

K(o → n) =K (n→ o)

K(o → n) =N(o) ×α(o→ n) ×acc(o→ n)

o n

K(n→ o) =N(n) ×α(n→ o) ×acc(n→ o)

15

NVT-ensemble

acc(o → n)acc(n→ o)

=N(n)N(o)

N (n) ∝exp −βU n( )⎡⎣ ⎤⎦

acc(o→ n)acc(n→ o)

=exp −β U n( )−U o( )⎡⎣ ⎤⎦⎡⎣

⎤⎦

16

17

Questions

• How can we prove that this scheme generates the desired distribution of configurations?

• Why make a random selection of the particle to be displaced?

• Why do we need to take the old configuration again?

• How large should we take: delx?

18

19

Questions

• How can we prove that this scheme generates the desired distribution of configurations?

• Why make a random selection of the particle to be displaced?

• Why do we need to take the old configuration again?

• How large should we take: delx?

20

Mathematical

π(o→ n) =α(o→ n) ×acc(o→ n)Transition probability:

π(o→ n)

n∑ =1

π(o→ o) =1− π(o→ n)

n≠o∑

Probability to accept the old configuration: ≠0

21

Keeping old configuration?

22

Questions

• How can we prove that this scheme generates the desired distribution of configurations?

• Why make a random selection of the particle to be displaced?

• Why do we need to take the old configuration again?

• How large should we take: delx?

23

Not too small, not too big!

24

25

Periodic boundary conditions

26

Lennard Jones potentials

uLJ r( ) =4εσr

⎛⎝⎜

⎞⎠⎟

12

−σr

⎛⎝⎜

⎞⎠⎟

6⎡

⎣⎢⎢

⎤

⎦⎥⎥

u r( ) =uLJ r( ) r ≤rc

0 r > rc

⎧⎨⎩

u r( ) =uLJ r( )−uLJ rc( ) r ≤rc

0 r > rc

⎧⎨⎩

•The truncated and shifted Lennard-Jones potential

•The truncated Lennard-Jones potential

•The Lennard-Jones potential

27

Phase diagrams of Lennard Jones fluids

28

Non-Boltzmann sampling A

NVT1=

1QNVT1

1Λ3NN!

drN∫ A rN( )exp −β1U rN( )⎡⎣ ⎤⎦

=drN A rN( )∫ exp −β1U rN( )⎡⎣ ⎤⎦

drN exp −β1U rN( )⎡⎣ ⎤⎦∫

=drN A rN

( )∫ exp −β1U rN( )⎡⎣ ⎤⎦exp β2U rN

( ) − β2U rN( )⎡⎣ ⎤⎦

drN exp −β1U rN( )⎡⎣ ⎤⎦∫ exp β2U rN

( ) − β2U rN( )⎡⎣ ⎤⎦

=drN A rN( )∫ exp β2U rN( ) − β1U rN( )⎡⎣ ⎤⎦exp −β2U rN( )⎡⎣ ⎤⎦

drN∫ exp β2U rN( ) − β1U rN( )⎡⎣ ⎤⎦exp −β2U rN( )⎡⎣ ⎤⎦

=Aexp β2 − β1( )U⎡⎣ ⎤⎦ NVT2

exp β2 − β1( )U⎡⎣ ⎤⎦ NVT2

We perform a simulation at T=T2 and

we determine A at T=T1

T1 is arbitrary!

We only need a single

simulation!

Why are we not using this?

29

T1

T2

T5

T3

T4

E

P(E

)

Overlap becomes very small

30

How to do parallel Monte Carlo

• Is it possible to do Monte Carlo in parallel– Monte Carlo is sequential!– We first have to know the fait of the current

move before we can continue!

31

Parallel Monte CarloAlgorithm (WRONG):1. Generate k trial configurations in parallel2. Select out of these the one with the lowest

energy

3. Accept and reject using normal Monte Carlo rule:

P n( ) =exp −β Un( )⎡⎣ ⎤⎦

exp −β U j( )⎡⎣

⎤⎦j=1

g∑

acc o→ n( ) =exp −β Un −Uo( )⎡⎣ ⎤⎦

32

Conventional acceptance rule

Conventional acceptance rules leads to a bias

33

Why this bias?Detailed balance!

acc( ) ( ) ( ) ( )

acc( ) ( ) ( ) ( )

o n N n n o N n

n o N o o n N o

αα

→ × →= =

→ × →

( ) ( )K o n K n o→ = →( ) ( ) ( ) acc( )K o n N o o n o nα→ = × → × →( ) ( ) ( ) acc( )K n o N n n o n oα→ = × → × →

34

Detailed balance

acc(o → n)acc(n→ o)

=N(n) ×α(n→ o)N(o) ×α(o→ n)

=N(n)N(o)

K(o → n) =K (n→ o)

K(o → n) =N(o) ×α(o→ n) ×acc(o→ n)

o n

K(n → o) =N(n) ×α(n→ o) ×acc(n→ o)

?

35

K(o→ n) =N(o) ×α(o→ n) ×acc(o→ n)

α(o→ n) =exp −β Un( )⎡⎣ ⎤⎦

exp −β U j( )⎡⎣

⎤⎦j=1

g∑

α(n→ o) =exp −β Uo( )⎡⎣ ⎤⎦

exp −β U j( )⎡⎣

⎤⎦j=1

g∑

α(o→ n) =exp −β Un( )⎡⎣ ⎤⎦

W n( )

α(n→ o) =exp −β Uo( )⎡⎣ ⎤⎦

W o( )

36

acc(o→ n)acc(n→ o)

=N(n) ×α(n→ o)N(o) ×α(o→ n)

=N(n)N(o)

( )( )

exp( )

n

n

Uo n

W

βα

⎡ ⎤−⎣ ⎦→ =( )

( )exp

( )o

o

Un o

W

βα

⎡ ⎤−⎣ ⎦→ =

acc(o → n)acc(n→ o)

=

N(n) ×exp −β Uo( )⎡⎣ ⎤⎦

W o( )

N(o) ×exp −β Un( )⎡⎣ ⎤⎦

W n( )

=W(n)W(o)

37

Modified acceptance rule

Modified acceptance rule remove the bias exactly!