1. 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
Trento 2 Sep 2003

2. 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

3. Helium Clusters Small mass of helium atom Very weak He-He interaction
0.02 Kcal/mol 0.9 * 10-3 cm * 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

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

5. 3Hen Clusters Stability
What is the smallest 3Hem stable cluster ?
Liquid: stable
3Hem
3He2 dimer unbound
m = ? < m < 35
critically bound
Even less is known for mixed clusters. Is 3Hem4Hen stable ?

6. 3He4Hen Clusters Stability
Bonding interaction
Non-bonding interaction
3He4Hen All clusters up bound
3He4He2 Trimer bound
3He4He dimer unbound
3He4He2 E = (5) cm-1
4He3 E = (7) cm-1

7. 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 = (1) cm-1
3He4He3 E = (1) cm-1
3He24He2 E = (1) cm-1

8. Evidence of 3He24He2 Kalinin, Kornilov and Toennies

9. 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

10. Evidence of 4He43He3 Kalinin, Kornilov and Toennies

11. Mixed He clusters: 3Hem4Hen
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

12. 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

13. 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

14. 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