1 Statistical Mechanics and Multi- Scale Simulation Methods ChBE 591-009 Prof. C. Heath Turner...
26
1 Statistical Mechanics and Multi-Scale Simulation Methods ChBE 591-009 Prof. C. Heath Turner Lecture 24 • Some materials adapted from Prof. Keith E. Gubbins: http://gubbins.ncsu.edu • Some materials adapted from Prof. David Kofke: http:// www.cbe.buffalo.edu/kofke.htm
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Transcript of 1 Statistical Mechanics and Multi- Scale Simulation Methods ChBE 591-009 Prof. C. Heath Turner...
1
Statistical Mechanics and Multi-Scale Simulation Methods
ChBE 591-009
Prof. C. Heath Turner
Lecture 24
• Some materials adapted from Prof. Keith E. Gubbins: http://gubbins.ncsu.edu
• Some materials adapted from Prof. David Kofke: http://www.cbe.buffalo.edu/kofke.htm
2
Histogram Reweighting Method to combine results taken at different state conditions Microcanonical ensemble
Canonical ensemble
The big idea:• Combine simulation data at different temperatures to improve quality of all data via
their mutual relation to (E)
( , , ) 1microstates
E V N
1( , )N N Ee
Q r p Probability of a microstate
( ; ) ( )( )
EeE E
Q
Probability of an energy
Number of microstates having this energy
Probability of each microstate
3
In-class Problem 1. Consider three energy levels
What are Q, distribution of states and <E> at = 1?
0 0
1 1
2 2
1 0 ln1
100 2.3 ln10
1000 6.9 ln1000
E
E
E
4
In-class Problem 1A. Consider three energy levels
What are Q, distribution of states and <E> at = 1?
0 1 20 1 2
ln1 ln10 ln10001 100 1000
1 1 100 0.1 1000 0.001
12
E E EQ e e e
e e e
0
1
2
00
11
22
10.083
12
100.833
12
10.083
12
E
E
E
e
Q
e
Q
e
Q
0 0
1 1
2 2
1 0 ln1
100 2.3 ln10
1000 6.9 ln1000
E
E
E
3 07.3
9.633.0
3.233.0
033.0
ii EE
3
1
5
In-class Problem 1A. Consider three energy levels
What are Q, distribution of states and <E> at = 1?
And at = 3?
3
10 1 2
0 1 2
ln1 ln10 ln10001 100 1000
1 1 100 0.1 1000 0.001
12
E E EQ e e e
e e e
0
1
2
00
11
22
10.083
12
100.833
12
10.083
12
E
E
E
e
Q
e
Q
e
Q
0 0
1 1
2 2
1 0 ln1
100 2.3 ln10
1000 6.9 ln1000
E
E
E
3ln1 3ln10 3ln1000
3
1 100 1000
1 1 100 0.001 1000 1000
1.1
Q e e e
0
1
2
10.91
1.11
0.091.10.0
0.001.1
3
10 1 2
0 1 2
ln1 ln10 ln10001 100 1000
1 1 100 0.1 1000 0.001
12
E E EQ e e e
e e e
0
1
2
00
11
22
10.083
12
100.833
12
10.083
12
E
E
E
e
Q
e
Q
e
Q
3
07.3
9.633.0
3.233.0
033.0
ii EE
1.01
1.01
1.01
1.01
0.99
0.01
023.0
9.60.0
3.201.0
099.0
ii EE
3
1
6
Histogram Reweighting Approach Knowledge of (E) can be used to obtain averages at any temperature
Simulations at different temperatures probe different parts of (E) But simulations at each temperature provides information over a range of
values of (E) Combine simulation data taken at different temperatures to obtain better
information for each temperature
E
(E)1 2 3
7
In-class Problem 2. Consider simulation data from a system having three energy
levels• M = 100 samples taken at = 0.5
• mi times observed in level i
What is (E)?
0 0
1 1
2 2
4 0 ln1
46 2.3 ln100
50 9.2 ln10000
m E
m E
m E
8
In-class Problem 2. Consider simulation data from a system having three energy
levels• M = 100 samples taken at = 0.5
• mi times observed in level i
What is (E)?
0 0
1 1
2 2
4 0 ln1
46 2.3 ln100
50 9.2 ln10000
m E
m E
m E
( ) ( )( )
EeE E
Q
Reminder
9
In-class Problem 2. Consider simulation data from a system having three energy
levels• M = 100 samples taken at = 0.5
• mi times observed in level i
What is (E)?
0 0
1 1
2 2
4 0 ln1
46 2.3 ln100
50 9.2 ln10000
m E
m E
m E
( )( )
( )EE
E eQ
HintCan get only relative values!
10
In-class Problem 2A. Consider simulation data from a system having three energy
levels• M = 100 samples taken at = 0.5
• mi times observed in level i
What is (E)?
0 0
1 1
2 2
4 0 ln1
46 2.3 ln100
50 9.2 ln10000
m E
m E
m E
0 0
1 1
2 2
0.50
0.51
0.52
0.04 1 .04
0.46 100 4.6
0.50 10000 50
m UA AM
m UA AM
m UA AM
e Q Q Q
e Q Q Q
e Q Q Q
( 0.5)AQ Q
11
In-class Problem 3. Consider simulation data from a system having three energy
levels• M = 100 samples taken at = 0.5
• mi times observed in level i
What is (E)?
Here’s some more data, taken at = 1• what is (E)?
0 0
1 1
2 2
4 0 ln1
46 2.3 ln100
50 9.2 ln10000
m E
m E
m E
( 0.5)AQ Q
0 1 250 48 2m m m
0 0
1 1
2 2
0.50
0.51
0.52
0.04 1 .04
0.46 100 4.6
0.50 10000 50
m UA A AM
m UA A AM
m UA A AM
e Q Q Q
e Q Q Q
e Q Q Q
12
In-class Problem 3. Consider simulation data from a system having three energy
levels• M = 100 samples taken at = 0.5
• mi times observed in level i
What is (E)?
Here’s some more data, taken at = 1• what is (E)?
0 0
1 1
2 2
4 0 ln1
46 2.3 ln100
50 9.2 ln10000
m E
m E
m E
0 0
1 1
2 2
0.50
0.51
0.52
0.04 1 .04
0.46 100 4.6
0.50 10000 50
m UA A AM
m UA A AM
m UA A AM
e Q Q Q
e Q Q Q
e Q Q Q
( 0.5)AQ Q
0 1 250 48 2m m m
0
1
2
0.50 1 .50
0.48 100 48
0.02 10000 200
B B
B B
B B
Q Q
Q Q
Q Q
( 1.0)BQ Q
13
Reconciling the Data
We have two data sets
Questions of interest• what is the ratio QA/QB? (which then gives us A)
• what is the best value of 1/0, 2/0?
• what is the average energy at = 2?
In-class Problem 4• make an attempt to answer these questions
0
1
2
.04
4.6
50
A
A
A
Q
Q
Q
0
1
2
.50
48
200
B
B
B
Q
Q
Q
14
In-class Problem 4A. We have two data sets
What is the ratio QA/QB? (which then gives us A)
• Consider values from each energy level
What is the best value of 1/0, 2/0?
• Consider values from each temperature
What to do?
0
1
2
.04
4.6
50
A
A
A
Q
Q
Q
0
1
2
.50
48
200
B
B
B
Q
Q
Q
0 1 2
0 1 2
0.04 4.6 500.50 48 200
0.50 48 2000.04 4.6 50
12.5 10.4 4
A A A
B B B
A A A
B B B
Q Q QQ Q Q
Q Q QQ Q Q
1 2
0 0
1 2
0 0
4.6 500.04 0.04
48 2000.05 0.5
0.5 115 1250
1 96 400
A A
A A
B B
B B
Q QQ Q
Q QQ Q
15
Accounting for Data Quality
Remember the number of samples that went into each value
• We expect the A-state data to be good for levels 1 and 2
• …while the B-state data are good for levels 0 and 1
Write each as an average of all values, weighted by quality of result
0
1
2
0
1
2
.04
4.6
5
6
00
4
4
5
A
A
A
m
m
m
Q
Q
Q
0
2
0
1
2
1
.50
48
20
50
48
20
B
B
B
m
m
Q
Q
Q m
0, 0,0 0
0 0, 0,
A BA B
A B
est est estA A B B
m mU UA A B BM M
w w
w e Q w e Q
( ) ( )( )
EeE E
Q
16
Histogram Variance
Estimate confidence in each simulation result
Assume each histogram follows a Poisson distribution• probability P to observe any given instance of distribution
• the variance for each bin is
0
1
2
0
1
2
.04
4.6
5
6
00
4
4
5
A
A
A
m
m
m
Q
Q
Q
0
2
0
1
2
1
.50
48
20
50
48
20
B
B
B
m
m
Q
Q
Q m
31 21 2 3
1 2 31 2 3
[{ }] !!
[{ , , }] !! ! !
imi
ii
mm m
P m Mm
P m m m Mm m m
u1 u2 u3
piInstance
2im i im M
17
Variance in Estimate of Formula for estimate of
Variance
0, 0,0 00
A BA B
A B
m mE EestA A B BM M
w e Q w e Q
0 02 20, 0,0
0 02 2
0 00 00 0
0 0
2 22 2 2 2 2 2 21 1
2 22 2 2 21 10, 0,
2 22 2 2 21 1
2 21 10 0
A Best A B
A B
A B
A B
E EA BA B
A A B B
A B
A B
E EA A m B B mM M
E EA A A A B B B BM M
e eE EA A B BM Q M Q
E EA A B BM M
w e Q w e Q
w e Q M w e Q M
w e Q w e Q
w e Q w e Q
( ) ( )( )
EeE E
Q
18
Optimizing Weights Variance
Minimize with respect to weight, subject to normalization• In-class Problem 5
Do it!
0 0
0
2 2 21 10 0
A Best
A B
E EA A B BM M
w e Q w e Q
19
Optimizing Weights Variance
Minimize with respect to weight, subject to normalization• Lagrange multiplier
Equation for each weight is Rearrange
Normalize
0
2 1est aMin w
002 aa
a
QEa M
w e
( ) ( )( )
Eie m
E EQ M
0
0
12
Eaa
a
e Ma Q
w
0
0 0
/Eaa a
E EA Be M e MA BQ QA B
e M Qaw
0 0
0
2 2 21 10 0
A Best
A B
E EA A B BM M
w e Q w e Q
20
Optimal Estimate
Collect results
Combine
0
0 0
/Eaa a
E EA Be M e MA BQ QA B
e M Qaw
0, 0,0 00
A BA B
A B
m mE EestA A B BM M
w e Q w e Q
0 0 0 00, 0,0 0
0 0
1
0
1
0, 0,
E E E EA B A BA BA B A BA B
A B A A B B
E EA BA B
A B
m me M e M e M e ME EestA BQ Q Q M Q M
e M e MA BQ Q
e Q e Q
m m
21
Calculating Formula for
In-class Problem 6• explain why this formula cannot yet be used
0 01
0 0, 0,
E EA BA B
A B
e M e MestA BQ Q
m m
22
Calculating Formula for
We do not know the Q partition functions
One equation for each Each equation depends on all Requires iterative solution
0 01
0 0, 0,
E EA BA B
A B
e M e MestA BQ Q
m m
0 1 20 1 2
a a aE E EaQ e e e
0 0
0 01 2 1 20 1 2 0 1 2
1
0 0, 0,
E EA BA B
E EE E E EA BA A B B
e M e MestA B
e e e e e em m
23
In-class Problem 7 Write the equations for each using the example values
0 0
0 01 2 1 20 1 2 0 1 2
1
0 0, 0,
E EA BA B
E EE E E EA BA A B B
e M e MestA B
e e e e e em m
0
1
2
0.5
4
46
50
m
m
m
0
1
2
0 ln1
2.3 ln100
9.2 ln10000
E
E
E
0
1
2
1
50
48
2
m
m
m
100A BM M
24
In-class Problem 7 Write the equations for each using the example values
Solution
0
1
2
0.5
4
46
50
m
m
m
0
1
2
0 ln1
2.3 ln100
9.2 ln10000
E
E
E
0
1
2
1
50
48
2
m
m
m
100A BM M
0 1 2 0 1 2
10.1 100 0.01 100
1 1 0.1 0.01 1 0.01 0.000146 48est
0 1 2 0 1 2
11 100 1 100
0 1 0.1 0.01 1 0.01 0.00014 50est
0 1 2 0 1 2
10.01 100 0.0001 100
2 1 0.1 0.01 1 0.01 0.000150 2est
1 2
0 094.7 943.2
0 0
0 01 2 1 20 1 2 0 1 2
1
0 0, 0,
E EA BA B
E EE E E EA BA A B B
e M e MestA B
e e e e e em m
25
In-class Problem 7A. Solution
Compare
“Exact” solution
Free energy difference
1 2
0 094.7 943.2
1 2
0 0
1 2
0 0
4.6 500.04 0.04
48 2000.05 0.5
0.5 115 1250
1 96 400
A A
A A
B B
B B
Q QQ Q
Q QQ Q
0
1
2
0.5
4
46
50
m
m
m
0
1
2
1
50
48
2
m
m
m
1 2
0 0100 1000
0 1 2
0 1 2
1 0.1 0.019.74
1 0.01 0.0001A
B
Q
Q
“Design value” = 10
(?)
26
Extensions of Technique
Method is usually used in multidimensional form Useful to apply to grand-canonical ensemble
Can then be used to relate simulation data at different temperature and chemical potential
Many other variations are possible
31
!
( , , )
NN N N E
h NN
N E
N U
e d d e
U V N e e
r p
