Quantum Monte Carlo Simulations of Mixed 3He/4He Clusters

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Quantum Monte Carlo Simulations of Mixed 3He/4He Clusters Universita’ degli Studi dell’Insubria Quantum Monte Carlo Simulations of Mixed 3He/4He Clusters Dario Bressanini dario.bressanini@uninsubria.it http://www.unico.it/~dario Trento 2 Sep 2003

Helium A helium atom is an elementary particle. A weakly interacting hard sphere. Interatomic potential is known more accurately than any other atom. Two isotopes: 3He (fermion: antisymmetric trial function, spin 1/2) 4He (boson: symmetric trial function, spin zero) The interaction potential is the same

Helium Clusters Small mass of helium atom Very weak He-He interaction 0.02 Kcal/mol 0.9 * 10-3 cm-1 0.4 * 10-8 hartree 10-7 eV Highly non-classical systems. No equilibrium structure. ab-initio methods and normal mode analysis useless High resolution spectroscopy Superfluidity Low temperature chemistry

4Hen Clusters Stability Liquid: stable 4Hen All clusters bound 4He2 dimer exists 4He3 bound. Efimov effect?

3Hen Clusters Stability What is the smallest 3Hem stable cluster ? Liquid: stable 3Hem 3He2 dimer unbound m = ? 20 < m < 35 critically bound Even less is known for mixed clusters. Is 3Hem4Hen stable ?

3He4Hen Clusters Stability Bonding interaction Non-bonding interaction 3He4Hen All clusters up bound 3He4He2 Trimer bound 3He4He dimer unbound 3He4He2 E = -0.00984(5) cm-1 4He3 E = -0.08784(7) cm-1

3He24Hen Clusters Stability Now put two 3He. Singlet state. Y is positive everywhere 3He24Hen All clusters up bound 3He24He Trimer unbound 3He24He2 Tetramer bound 5 out of 6 unbound pairs 4He4 E = -0.3886(1) cm-1 3He4He3 E = -0.2062(1) cm-1 3He24He2 E = -0.071(1) cm-1

Evidence of 3He24He2 Kalinin, Kornilov and Toennies

3He34Hen Clusters Stability Adding a third fermionic helium, introduces a nodal surface into the wave function that destabilizes the system What is the smallest 3He34Hen stable cluster ? For L=0 n = 9, for L=1 n = 4

Evidence of 4He43He3 Kalinin, Kornilov and Toennies

Mixed He clusters: 3Hem4Hen 0 1 2 3 4 5 6 7 8 9 10 11 Bound L=0 1 2 3 4 5 Unbound Unknown Maybe L=1 S=1/2 L=1 S=1 35 3He34He4 L=0 S=1/2 3He24He2 L=0 S=0 3He34He4 L=1 S=1/2 3He24He2 L=1 S=1

3Hem4He10 m = 0,1,2,3 L=0 3He distribution with respect to the center of mass 3He4He10 r(3He-C.O.M.) c.o.m 3He24He10 3He34He10

a outside b outside on opposite sides 3Hem4He10 m = 0,1,2,3 L=0 3He- 3He distributions r(3He- 3He) a outside b outside on opposite sides b outside a inside a outside a inside a b 4He10 The (tentative) picture: two 3He outside (a, b) and one a inside, pushed away from the other a 3He

3He34He10 L=0 a b 4He10 Why ? It is a Nodal Effect. The wave function is zero if the two a 3He are at the same distance from the b 3He. 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. a b