Universita degli Studi dellInsubria Quantum Monte Carlo Simulations
Universita’ degli Studi dell’Insubria Quantum Monte Carlo Simulations of Mixed 3 He/4 He Clusters Dario Bressanini dario. bressanini@uninsubria. it http: //www. unico. it/~dario Göttingen 24/05/2002
Overview n Introduction to quantum monte carlo methods ´ n VMC, QMC, advantages and drawbacks Stability and structure of small 3 He/4 He mixed clusters ´ Trimers Dario Bressanini – Göttingen 24/05/2002 2
Monte Carlo Methods n How to solve a deterministic problem using a Monte Carlo method? n Rephrase the problem using a probability distribution n “Measure” A by sampling the probability distribution Dario Bressanini – Göttingen 24/05/2002 3
Monte Carlo Methods n The points Ri are generated using random numbers This is why the methods are called Monte Carlo methods n Metropolis, Ulam, Fermi, Von Neumann (-1945) n We introduce noise into the problem!! ´ Our results have error bars. . . ´ . . . Nevertheless it might be a good way to proceed Dario Bressanini – Göttingen 24/05/2002 4
VMC: Variational Monte Carlo n To solve H Y = E Y start from the Variational Principle n Translate it into Monte Carlo language Dario Bressanini – Göttingen 24/05/2002 7
VMC: Variational Monte Carlo n E is a statistical average of the local energy EL over P(R) n Recipe: ´ take an appropriate trial wave function ´ distribute N points according to P(R) ´ compute the average of the local energy Dario Bressanini – Göttingen 24/05/2002 8
The Metropolis Algorithm ? n How do we sample n Use the Metropolis algorithm (M(RT)2 1953). . . and a powerful computer Anyone who consider n The algorithm is a random arithmetical methods of producing random digits walk (markov chain) in is, of course, in a state of sin. configuration space John Von Neumann Dario Bressanini – Göttingen 24/05/2002 9
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The Metropolis Algorithm move Ri Call the Oracle Rtry reject accept Ri+1=Ri Ri+1=Rtry Compute averages Dario Bressanini – Göttingen 24/05/2002 11
The Metropolis Algorithm The Oracle if p 1 /* accept always */ accept move If 0 p 1 /* accept with probability p */ if p > rnd() accept move else reject move Dario Bressanini – Göttingen 24/05/2002 12
VMC: Variational Monte Carlo n No need to analytically compute integrals: complete freedom in the choice of the trial wave function. n Can use explicitly correlated wave functions n Can satisfy the cusp conditions r 12 r 2 He atom Dario Bressanini – Göttingen 24/05/2002 13
VMC advantages n Can compute lower bounds n Can go beyond the Born-Oppenheimer approximation, with ANY potential, in ANY number of dimensions. Ps 2 molecule (e+e+e-e-) in 2 D and 3 D M+m+M-m- as a function of M/m Dario Bressanini – Göttingen 24/05/2002 14
First Major VMC Calculations n n Mc. Millan VMC calculation of ground state of liquid 4 He (1964) Generalized for fermions by Ceperley, Chester and Kalos PRB 16, 3081 (1977). Dario Bressanini – Göttingen 24/05/2002 18
VMC drawbacks n Error bar goes down as N-1/2 n It is computationally demanding n The optimization of Y becomes difficult as the number of nonlinear parameters increases n It depends critically on our skill to invent a good Y n There exist exact, automatic ways to get better wave functions. Let the computer do the work. . . Dario Bressanini – Göttingen 24/05/2002 19
Diffusion Monte Carlo n VMC is a “classical” simulation method Nature is not classical, dammit, and if you want to make a simulation of nature, you'd better make it quantum mechanical, and by golly it's a wonderful problem, because it doesn't look so easy. Richard P. Feynman n Suggested by Fermi in 1945, but implemented only in the 70’s Dario Bressanini – Göttingen 24/05/2002 20
Diffusion equation analogy n The time dependent Schrödinger equation is similar to a diffusion equation n The diffusion equation can be “solved” by directly simulating the system Time evolution Diffusion Branch Can we simulate the Schrödinger equation? Dario Bressanini – Göttingen 24/05/2002 21
Imaginary Time Sch. Equation n The analogy is only formal ´ n Y is a complex quantity, while C is real and positive If we let the time t be imaginary, then Y can be real! Imaginary time Schrödinger equation Dario Bressanini – Göttingen 24/05/2002 22
Y as a concentration n Y is interpreted as a concentration of fictitious particles, called walkers n The schrödinger equation is simulated by a process of diffusion, growth and disappearance of walkers Ground State Dario Bressanini – Göttingen 24/05/2002 23
Diffusion Monte Carlo SIMULATION: discretize time • Diffusion process • Kinetic process (branching) Dario Bressanini – Göttingen 24/05/2002 24
The DMC algorithm Dario Bressanini – Göttingen 24/05/2002 25
The Fermion Problem n Wave functions for fermions have nodes. ´ Diffusion equation analogy is lost. Need to introduce positive and negative walkers. The (In)famous Sign Problem n Restrict random walk to a positive region bounded by nodes. Unfortunately, the exact nodes are unknown. n Use approximate nodes from a trial Y. Kill the walkers if they cross a node. Dario Bressanini – Göttingen 24/05/2002 + - 26
Helium Clusters 1. Small mass of helium atom 2. Very weak He-He interaction 0. 02 Kcal/mol 0. 9 * 10 -3 cm-1 0. 4 * 10 -8 hartree 10 -7 e. V Highly non-classical systems. No equilibrium structure. ab-initio methods and normal mode analysis useless Superfluidity High resolution spectroscopy Low temperature chemistry Dario Bressanini – Göttingen 24/05/2002 29
The Simulations n Both VMC and DMC simulations n Potential = sum of two-body TTY pair-potential n Three-body terms not important for small clusters n Standard Dario Bressanini – Göttingen 24/05/2002 30
4 He Clusters Stability n 4 He dimer exists 2 n All clusters bound Liquid: stable 4 He bound. Efimov effect? 3 Dario Bressanini – Göttingen 24/05/2002 31
Pure 4 Hen Clusters DMC gives exact results. The quality of the VMC simulations decreases as the cluster increases Dario Bressanini – Göttingen 24/05/2002 32
3 He Clusters Stability n n What is the smallest 3 Hem stable cluster ? 3 He m 3 He 2 dimer unbound n m = ? 20 < m < 35 critically bound Liquid: stable Even less is known for mixed clusters. Is 3 Hem 4 Hen stable ? Dario Bressanini – Göttingen 24/05/2002 34
3 He 4 He Clusters Stability n Bonding interaction Non-bonding interaction 3 He 4 He dimer unbound 3 He 4 He 2 Trimer bound 3 He 4 He n All clusters up bound 3 He 4 He E = -0. 00984(5) cm-1 2 4 He -1 3 E = -0. 08784(7) cm Dario Bressanini – Göttingen 24/05/2002 35
Mixed 3 He 4 Hen Clusters (m, n) = 3 Hem 4 Hen Bressanini et. al. J. Chem. Phys. 112, 717 (2000) 4 He is n destabilized by substituting a 4 He with a 3 He Dario Bressanini – Göttingen 24/05/2002 36
Helium Clusters: energy (cm-1) Dario Bressanini – Göttingen 24/05/2002 37
3 He/4 He Distribution Functions Pair distribution functions 3 He 4 He 5 Dario Bressanini – Göttingen 24/05/2002 38
3 He/4 He Distribution Functions Distributions with respect to the center of mass 3 He 4 He 5 c. o. m Dario Bressanini – Göttingen 24/05/2002 39
Distribution Functions c. o. m. = center of mass r(4 He-C. O. M. ) Similar to pure clusters Dario Bressanini – Göttingen 24/05/2002 r(3 He-C. O. M. ) 3 He is pushed away 41
4 What is the shape of He 3 ?
What is the shape of 4 He 3 ? n Some people say is an equilateral triangle. . . n . . . some say it is linear (almost). . . n . . . some say it is both. Pair distribution function We find NO sign of double peak Dario Bressanini – Göttingen 24/05/2002 43
What is the shape of 4 He 3 ? Dario Bressanini – Göttingen 24/05/2002 44
The Shape of the Trimers Ne trimer r(Ne-center of mass) He trimer r(4 He-center of mass) Dario Bressanini – Göttingen 24/05/2002 45
Ne 3 Angular Distributions a b Ne trimer b b a a Dario Bressanini – Göttingen 24/05/2002 46
4 He Angular Distributions 3 b a b b a a Dario Bressanini – Göttingen 24/05/2002 47
3 He 4 He Angular Distributions 2 a b b b a Dario Bressanini – Göttingen 24/05/2002 a 48
3 He 4 He Clusters Stability 2 n n Now put two 3 He. Singlet state. Y is positive everywhere 3 He 4 He 2 Trimer unbound 3 He 4 He 2 2 Tetramer bound 5 out of 6 unbound pairs 3 He 4 He 2 n All clusters up bound 4 He -1 4 E = -0. 3886(1) cm 3 He 4 He E = -0. 2062(1) cm-1 3 3 He 4 He E = -0. 071(1) cm-1 2 2 Dario Bressanini – Göttingen 24/05/2002 49
3 He 4 He Clusters Structure 2 n n The two 3 He atoms stay mainly on the surface of the 4 He cluster 3 He 4 He 2 10 Dario Bressanini – Göttingen 24/05/2002 50
3 He 4 He Clusters Stability 3 n n Adding a third fermionic helium, introduces a nodal surface into the wave function that destabilizes the system n What is the smallest 3 He 34 Hen stable cluster ? n 3 He is bound, so 3 He 4 He should be bound. n < 32 35 3 32 Dario Bressanini – Göttingen 24/05/2002 51
The Wave Function n The total wave function must be antisymmetric with respect to the fermionic helium n Consider the doublet spin eigenfunction (two a and one b) ´ The 4 He-4 He and 4 He-3 He functions are symmetric ´ The 3 He-3 He part is antisymmetric Dario Bressanini – Göttingen 24/05/2002 52
3 He 4 He m = 0, 1, 2, 3 Energies m n Dario Bressanini – Göttingen 24/05/2002 53
Work in progress: 3 Hem 4 Hen 3 Hem 0 1 2 3 4 5 6 7 8 9 10 11 0 1 2 Bound Unbound 3 Unknown 4 Maybe 5 Unlikely 35 First 3 He 24 Hen bound First 3 He 34 Hen bound
3 He 4 He m = 0, 1, 2, 3 m 10 4 He distribution with respect to the center of mass r(4 He-C. O. M. ) c. o. m The 4 He distribution is unchanged with 0, 1, 2 or 3 3 He Dario Bressanini – Göttingen 24/05/2002 55
3 He 4 He m = 0, 1, 2, 3 m 10 3 He distribution with respect to the center of mass 3 He 4 He 10 r(3 He-C. O. M. ) c. o. m 3 He 4 He 2 10 Dario Bressanini – Göttingen 24/05/2002 3 He 4 He 3 10 56
3 He 4 He m = 0, 1, 2, 3 m 10 3 He distribution with respect to the center of mass r(3 He-C. O. M. ) One a 3 He is pushed inside the cluster, the other two (a, b) outside Dario Bressanini – Göttingen 24/05/2002 57
3 He 4 He m = 0, 1, 2, 3 m 10 3 He- 3 He distributions r(3 He- 3 He) a outside b outside on opposite sides a outside a inside b outside a inside The (tentative) picture: two 3 He outside (a, b) and one a inside, pushed away from the other a 3 He Dario Bressanini – Göttingen 24/05/2002 a a b 4 He 10 58
3 He 4 He 3 10 a a 4 He 10 b Why ? It is a Nodal Effect. The wave function is zero if the two a 3 He are at the same distance from the b 3 He. For this reason the three atoms are not free to move on the surface of the cluster. One is pushed inside to avoid the wave function node. Dario Bressanini – Göttingen 24/05/2002 59
More Flexible Wave Function n The standard form is not very flexible n Difficult to optimize n Difficult to reproduce the shell structure Dario Bressanini – Göttingen 24/05/2002 61
Different wave function form 0. 60 f(r) 0. 40 0. 20 0. 00 Dario Bressanini – Göttingen 24/05/2002 2. 00 4. 00 6. 00 r (a. u. ) 8. 00 10. 00 62
Spline Wave Function SF 6 He 39 Knots of the He-SF 6 spline function Dario Bressanini – Göttingen 24/05/2002 63
Shell Structure Standard r(4 He-SF 6) Spline Dario Bressanini – Göttingen 24/05/2002 64
Work in Progress and Future n Various impurities embedded in a Helium cluster (suggestions welcome!) n Different functional forms for Y (splines) ´ anisotropy n Analysis of 3 He 34 Hen n What about 3 He 44 Hen and 3 He 54 Hen ? Dario Bressanini – Göttingen 24/05/2002 67
Acknowledgments Gabriele Morosi Mose’ Casalegno Giordano Fabbri Matteo Zavaglia Dario Bressanini – Göttingen 24/05/2002 69
A reflection. . . A new method for calculating properties in nuclei, atoms, molecules, or solids automatically provokes three sorts of negative reactions: Æ A new method is initially not as well formulated or understood as existing methods Æ It can seldom offer results of a comparable quality before a considerable amount of development has taken place Æ Only rarely do new methods differ in major ways from previous approaches Nonetheless, new methods need to be developed to handle problems that are vexing to or beyond the scope of the current approaches (Slightly modified from Steven R. White, John W. Wilkins and Kenneth G. Wilson) Dario Bressanini – Göttingen 24/05/2002 70
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