Monte Carlo and cold gases Lode Pollet - Max Planck Societybec10/Pollet_QuoVadis_1.pdf · 2010. 8....

Monte Carlo and cold gases

Lode [email protected]

1Monday, August 2, 2010

mailto:[email protected]:[email protected]

Outline

•The Monte Carlo trick•Markov chains•Metropolis algorithm•Ising model•critical slowing down

Classical Monte Carlo

Quantum Monte Carlo

Applications

•diffusion Monte Carlo•worm algorithm•sign problem


What I will not do

•high performing computing•parallelization•programming language (C++)•other numerical methods

Required knowledge

•statistical mechanics•quantum mechanics•some knowledge about many-body physics


Integration

f(x)

I =

� b

af(x)dx = h

N�

i=1

f(a+ ih) +O(1/N)

h =b− aN

a b

discrete sum:

higher order integrators

I = h

�1

2f(a) +

N−1�

i=1

f(a+ ih) +1

2f(b)

�

I =h

3

�f(a) +

N−1�

i=1

(3− (−1)i)f(a+ ih) + f(b)�

trapezoidal

Simpson(N even)

+O(1/N4)

+O(1/N2)


Integration

Simpson rule with M points per dimension

The error in one dimension is

In d dimensions we need N = Md pointsthe error is

O(M−4)

O(M−4) = O(N−4/d)

Integration becomes inefficient in higher dimensions

In statistical mechanics the integrals are 6N-dimensional(3N positions, 3N momenta)


I =

∫D

g(x)dx

identical and independent random samples uniformly drawn from D

Îm =1

m

(

g(x(1)) + · · · + g(x(m)))

P

[

limm→∞

|Îm − I| = 0]

= 1

law of the large numbers

The Monte Carlo trick

D

g(x)


The Monte Carlo trick

limm→∞

Cdf√

m(

Îm − I)

= CdfN(0, σ2)

central limit theorem

convergence is

O(m−1/2)

regardless of the dimensionality of D


random numbers

Pseudo random numbers•not random at all, but generated by an algorithm•probably good enough•reproducible

Linear congruential generatorsxn+1 = f(xn)

xn+1 = (ax+ c) mod mGGLa = 16807, c = 0,m = 231 − 1problem is periodicitywith 500 million random numbers per second : 4sshould not be used any more


Lagged Fibonacci generatorxn = xn−p ⊗ xn−q mod m

0 < p < q , ⊗ = +,−,×,XORInitialization using linear congruential generatorvery long period (large block of seeds)very fast complex mathematics

see: http://beta.boost.org/doc/libs/1_42_0/libs/random/random-generators.html


http://beta.boost.org/doc/libs/1_42_0/libs/random/random-generators.htmlhttp://beta.boost.org/doc/libs/1_42_0/libs/random/random-generators.html

Independent sampling

Known geometryconverges to π/4


demonstration

1e-05

0.0001

0.001

0.01

0.1

1

10

1 10 100 1000 10000 100000 1e+06 1e+07 1e+08

! - <

MC

>

Nsteps

1./sqrt(x)

Segmentation fault


Importance sampling

draw random numbers that are exponentially distributed, then

I =

∫∞

0

g(x)e−xdx

Îm =1

m

(

g(x(1)) + · · · + g(x(m)))

how?

u ∈ [0, 1[

p = −lnu p ∈ [0,∞[

inverse transformation is needed

standard random number generator


Importance sampling

�f� = 1V

�f(x)dx =

1

V

�f(x)

p(x)p(x)dx ≈ 1

N

N�

i=1

f(xi)

p(xi)

∆ =

�Varf/p

N

•imagine function f is sharply peaked•then the variance can be reduced by finding p(x) such that p(x) is close to f(x) and that it is easy to generate random numbers according to p(x)


change of variables

Box-Mueller (gaussian)

exponential

general

p = − ln(u)

p1 = R cos(θ) =�

−2 ln(u1) cos(2πu2)

p2 = R sin(θ) =�

−2 ln(u1) sin(2πu2)

y = F (y) =

� y

0f(u)du f(x) = x

x = F−1(y)y = F (x) =

� x

0x�dx� =

x2

2

x = 2√y


〈Q〉 =1

ZTr

[

Qe−βH]

=Tr[Qe−βH ]

Tr[e−βH ]

W (x) = e−βE(x)unnormalized weights

Statistical mechanics :

how do we get random variables that are distributed according to W(x) ?


?

Small stepsrandom walk

through configuration space

at each time : measure

Rejection : stay at same configuration, update clock and measure

transition function

Markov chains and rejections


• A Markov chain is a sequence of random variables X1, X2, X3, ... with the property that the future state depends only on the past via the present state.

P [Xn+1 = x|Xn = xn, . . . , X1 = x1, X0 = x0] = P [Xn+1 = x|Xn = xn]

transition functionT (x, y)

∑

y

T (x, y) = 1conservation of probability

Markov chains

irreducible : it must be possible to reach any configuration x from any other configuration y in a finite number of steps.


Markov chain

• irreducible• aperiodic

convergence to the stationary distribution W

transition kernel has one eigenvalue 1, while all other eigenvalues satisfy

|λj | < 1, j = 2, . . . N

The second largest eigenvalue determines the correlations in the

Markov process

}


Detailed balance

A transition rule T(x,y) leaves the target distribution W(x) invariant if

∑

x

W (x)T (x, y) ∼ W (y)

This will certainly be the case if detailed balance is fulfilled,

W (x)T (x, y) = W (y)T (y, x)


Metropolis algorithm

we cannot construct a transition kernel T that fulfills

detailed balance.

proposal function P(x,y)

acceptance factor

q = min

[

1,W (y)P (y, x)

W (x)P (x, y)

]

go to the proposed configuration y with prob q, otherwise stay in x


thermalization

energy landscapeInitial configuration

Discard the first 20% of the Markov steps !

Markov Chain should be sufficiently long

|λ2|

λ2τint


recapitulation

W (X) = e−βE(X)

W (Y ) = e−βE(Y )

N particles interacting via Lennard-Jones interactions

P (X → Y ) = 1N

1δ2

P (Y → X) = 1N

1δ2

qX→Y = min�1, e−β(E(Y )−E(X))

�

δ


autocorrelations

Markov chains trivially correlate measurements

autocorrelations decay exponentially

integrated autocorrelation time

number of independent measurements is reduced, but central limit theorem still holds


Binning analysis

• Markov chain correlates measurements• if chain is long enough, then the configuration is

independent of the initial one

21 m

identically and independentτint(l) =

lσ2(l)

2σ2

How to get correct error bars?

http://arxiv.org/abs/cond-mat/9707221 appendix


http://arxiv.org/abs/cond-mat/9707221http://arxiv.org/abs/cond-mat/9707221

Jackknife analysis

R(0)

R(j), j = 1, k

Bias = (k-1)(Rav - R(0))

δR =√

k − 1

1

k

k∑

j

(R(j))2 − (Rav)2

1/2

R = R(0) - Bias


2d Ising model

H = −J∑

σiσj

W (x)T (x, y) = W (y)T (y, x)

q = min

[

1,W (y)P (y, x)

W (x)P (x, y)

]

Select random spin

P (x, y) = P (y, x) =1

L2

W (y)

W (x)= exp

−2βJσi∑

σj

q = min

[

1, e−2βJσi

∑

σj

]


Critical slowing down

Tc =ln(1 +

√

2)

2

Magnetization 1e-06

1e-05

0.0001

0.001

0.01

0.1

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

1/T

error(M)

divergence of correlation

length, critical fluctuations

∆m = ±2


improved estimator

domain size ~ susceptibility

�m2� = 1L2

�

i,j

�σiσj� =1

L�S(cluster)�

There exist more formal ways and also other ways to solve

the problem of critical slowing down


cluster algorithm

select like spins with probability p

p = 1 − exp[−2β]

then update always accepted

(exercise : show that this is true)

(why is this not the same as flipping many spins at the same time?)


e−β(c1−c2)(1− p)c2qX→Y = e−β(c2−c1)(1− p)c1

min�1,

e−β(c2−c1)(1− p)c1e−β(c1−c2)(1− p)c2

�


1e-06

1e-05

0.0001

0.001

0.01

0.1

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

1/T

local

Wolff

Error on magnetization

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

1/T

cluster size in Wolff

phase transitions and critical phenomena


phase transitions and critical phenomena

P.S. you need to know about finite size scaling

critical exponents have to be calculated numerically


Wang-Landau sampling

Cluster algorithms do not help near first order transitions


Wang-Landau sampling

Z =�

E

ρ(E)e−E/kT

Density of states is however unknown. probability minimum disappears however if we choose

p(E) ∼ 1/ρ(E)

- start with ρ(E) = 1 and f =1- repeat

- stop when

- reset histogram H(E) = 0

- when histogram is flat, reduce f →�

f

- perform simulations by picking random site and Metropolis updates p(E) ∼ 1/ρ(E)H(E) → H(E) + 1

ρ(E) → ρ(E) · f

f ≈ 1 + 10−8


References

• J. S. Liu, Monte Carlo strategies in scientific computing, Springer Verlag

• W. Krauth, “Statistical Mechanics : algorithms and computations”, Oxford University Press

• ETH Zurich professor scripts : http://www.ifb.ethz.ch/education/IntroductionComPhys ( and references therein)


http://www.ifb.ethz.ch/education/IntroductionComPhyshttp://www.ifb.ethz.ch/education/IntroductionComPhyshttp://www.ifb.ethz.ch/education/IntroductionComPhyshttp://www.ifb.ethz.ch/education/IntroductionComPhys

Homework

calculate (mean value, no error analysis) :

by : 1. direct integration (analytical/Monte Carlo by exponentially distributed random numbers)2. Markov chain Monte Carlo : choose a step size d (wisely), and update the current position according to the

Metropolis algorithm by choosing a random step of. Don’t forget that in every step you can move to larger or smaller x values. Show that you satisfy detailed balance.

3. modification (advanced). Suppose the following. Suppose you start with moving to the right. If you accept the move, then always choose moving to the right in the next step. If you reject the move, then start moving to the left. So, you keep moving in one direction until you have a rejection (bounce), and then you change the direction and keep moving in this direction until there is another bounce after which you change the direction again. Does this Monte Carlo Markov chain produce the right answer ?

I =� 6

0exp(−x/2)dx


Monte Carlo and cold gases Lode Pollet - Max Planck Societybec10/Pollet_QuoVadis_1.pdf · 2010. 8....

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Transcript of Monte Carlo and cold gases Lode Pollet - Max Planck Societybec10/Pollet_QuoVadis_1.pdf · 2010. 8....