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Basic Monte Carlo(chapter 3)
Algorithm
Detailed Balance
Other points
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Molecular Simulations
Molecular dynamics: solve equations of motion
Monte Carlo: importance sampling
r1
MD
MC
r2rn
r1
r2
rn
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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= −
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( ) ( ) ( )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=
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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
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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
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Monte Carlo simulation
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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
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Monte Carlo simulation
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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?
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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)
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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( )⎡⎣ ⎤⎦⎡⎣
⎤⎦
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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?
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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?
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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
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Keeping old configuration?
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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?
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Not too small, not too big!
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Periodic boundary conditions
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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
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Phase diagrams of Lennard Jones fluids
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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?
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T1
T2
T5
T3
T4
E
P(E
)
Overlap becomes very small
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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!
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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( )⎡⎣ ⎤⎦
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Conventional acceptance rule
Conventional acceptance rules leads to a bias
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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→ = × → × →
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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)
?
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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( )
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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)
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Modified acceptance rule
Modified acceptance rule remove the bias exactly!