Transcript No Slide Title

Universita’ dell’Insubria, Como, Italy
The Fascinating Helium
Dario Bressanini
http://scienze-como.uninsubria.it/bressanini
Crit05, Dresden 2005
The Beginning
• First discovered in the Sun by
Pierre Janssen and Norman
Lockyer in 1868
•
•
First liquefied by Kamerlingh
Onnes in 1908
First calculations by Egil
Hylleraas and John Slater
in 1928
2
Helium studies
• Thousands of theoretical and experimental
papers
Hˆ n (R)  En n (R)
have been published on Helium, in its various forms:
Atom
Small Clusters
Droplets
Bulk
3
Plan of the Talk
• Nodes of the Helium Atom:
(R)=0
• Stability of mixed 3Hem4Hen clusters
• Geometry of 4He3 (if time permits)
4
Nodes
Nodes are region of N-dimensional space where (R)=0
• Why study Nodes of wave functions?

They are very interesting mathematical

Very little is known about them

They have practical relevance
especially in
Quantum Monte Carlo Simulations
5
Nodes are relevant
• Levinson Theorem:

the number of nodes of the zero-energy
scattering wave function gives the number of
bound states
• Fractional quantum Hall effect
• Quantum Chaos
Integrable system
Chaotic system
6
Nodes and QMC

If we knew the exact nodes of , we could
exactly simulate the system by QMC methods
+

-
We restrict random walk to a positive
region bounded by (approximate) nodes.
7
Common misconception on nodes
• Nodes are not fixed by antisymmetry
alone, only a 3N-3 sub-dimensional subset
8
Common misconception on nodes
• They have (almost) nothing to do
with Orbital Nodes.

It is (sometimes) possible to use nodeless
orbitals
9
Common misconceptions on nodes
• A common misconception is that on a
node, two like-electrons are always close.
This is not true
0
0
1
2
2
0
1
10
Common misconceptions on nodes
•
Nodal theorem is NOT VALID in N-Dimensions


Higher energy states does not mean more nodes (Courant and
Hilbert )
It is only an upper bound
11
Common misconceptions on nodes
• Not even for the same symmetry species
3
2.5
2
1.5
1
0.5
Courant counterexample
0
0
0.5
1
1.5
2
2.5
3
12
The Helium triplet
• First
3S
state of He is one of very few
systems where we know the exact node
• For S states we can write
  (r1 , r2 , r12 )
•For the Pauli Principle
(r1 , r2 , r12 )  (r2 , r1 , r12 )
• Which means that the node is
r1  r2
or
r1  r2  0
14
The Helium triplet node
• Independent of r12
• The node is more
symmetric than the
wave function itself
• It is a polynomial in r1
and r2
• Present in all 3S
states of two-electron
atoms
r1
r12
r2
r1  r2    0
r1
r2
r1  r2    0
15
Helium 1s2p 3P o
( P )  z1 f (r1 , r2 , r12 )  z 2 f (r2 , r1 , r12 )
3
0
The Wave function (J.B.Anderson 1987) is
( P )  ( g ( z1 , r1 )  g ( z 2 , r2 )) (r1 , r2 , r12 )
3
0
•node independent from r12 (numerical proof)
16
Other He states: 1s2s 2 1S
• Although  (r1 , r2 , 12 ),
the node does not
depend on 12 (or does very weakly)
•
A very good approximation
of the node is
4
4
r1  r2  const
12
r2
r1
Surface contour plot of the node
17
Casual similarity ?
First unstable antisymmetric stretch orbit along with
the symmetric Wannier orbit r1 = r2 and various
equipotential lines
18
Other He states: 2 3S
• The second triplet has similar properties
"Almost"
r15  r25  const
19
He: Other states
• Other states have similar properties
• Breit (1930) showed that
(P e)= (x1 y2 – y1 x2) f(r1,r2,r12)


2p2
3P e
: f( ) symmetric
node = (x1 y2 – y1 x2) = 0
2p3p 1P e : f( ) antisymmetric
node = (x1 y2 – y1 x2) (r1-r2) = 0
20
He: Hyperspherical Approximation
• In the Hyperspherical approximation:
(R, )  F (R)(R, )
R  r r
2
1
2
2
• which means the first few S excited states
have circular nodes..
1s2s 3S
1s2s 1S
1s3s 1S
1s4s 3S
They have the correct topology, and a shape close to the
exact, which is more similar to r1k  r2k  Const
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Helium Nodes
Exact  N (R)e
f (R)
• Independent from r12
• Higher symmetry than the wave function
• Some are described by polynomials in
distances and/or coordinates
• Are these general properties of nodal
surfaces ?
• Is the Helium wave function separable in
some (unknown) coordinate system?
25
Nodal Symmetry Conjecture
WARNING: Conjecture Ahead...
Symmetry of (some) nodes of 
is higher than symmetry of 
• Other systems apparently show this feature:
Li atom, Be Atom, He2+ molecule
26
Be Nodal Topology
r1+r2
r1+r2
r3-r4
r3-r4
r1-r2
HF  0
r1-r2
  1s 2 2s 2  c 1s 2 2 p 2
CI  0
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A (Nodal) song...
He deals the cards to find the answers
the secret geometry of chance
the hidden law of a probable outcome
the numbers lead a dance
Sting: Shape of my heart
32
Helium

Helium as an elementary particle. A weakly interacting
hard sphere.

Interatomic potential is known very accurately
 3He (fermion: antisymmetric trial function, spin 1/2)
 4He (boson: symmetric trial function, spin zero)
Highly non-classical systems. No equilibrium structure.
ab-initio methods and normal mode analysis useless
High resolution spectroscopy
Superfluidity
Low temperature chemistry
33
4He
n
and
3He
n
Clusters Stability
4He
n
4He dimer exists
2
 4He3
All clusters
bound
Liquid: stable
bound. Efimov effect?
3He
3He
2
dimer unbound
m
m = ? 20 < m < 33
critically bound.
Probably m=32
(Guardiola & Navarro)
Liquid: stable
35
Questions
• When is 3Hem4Hen stable?
• What is the spectrum of the
3He
impurities?
• Can we describe it using simple
models (Harmonic Oscillator,
Rotator,...) ?
• What is the structure of these
clusters?
• What excited states do they have ?
36
3He 4He
m
n
3He
4He
n
0 1 2 3 4
m
Stability Chart
5 6
7 8
9 10 11
0
Bound L=0
1
Unbound
2
3
Unknown
4
L=1 S=1/2
5
L=1 S=1
Terra Incognita
Bound
32
3He 4He
2
2
L=0 S=0
3He 4He
2
4
L=1 S=1
3He 4He
3
8
3He 4He
3
4
L=0 S=1/2
L=1 S=1/2
37
3He4He :
n
energies
-1)
(cm
Total energies
-1
E (cm )
0
n=5
-2
n = 9 The p state appears at n=5
-4
The d state appears at n=9
L=0
L=1
L=2
-6
The f state (not shown) at n=19
-8
-10
2
4
6
8
10
12
n
n
39
3He4He :
n
Total energies (cm-1)
-44.5
3He4He
energies
30
-45
g
He30
-45.5
f
Spectrum similar to the
rigid rotator. Different
than harmonic oscillator
(sometimes used in the
literature)
-46
d
-46.5
l=2
p
s
1s2s
-47
0
L (angular momentum)
2
4
6
l=1
l=0
40
3He4He :
n
4He
0.01
3He
3He
s
p
d
f
0.008
0.006
Structure
stays on the surface.
Pushed outside as L
increases
3He4He
7
: L = 1 state
0.004
0.002
0
0
10
20
30
40
41
3He 4He
2
n

Clusters Stability
Now put two 3He
3He 4He
2
3He 4He
2
2
Trimer unbound
Tetramer bound
5 out of 6 unbound pairs
4He
4
3He4He
3
3He 4He
2
2
3He 4He
2
n
All clusters
up bound
E = -0.3886(1) cm-1
E = -0.2062(1) cm-1
E = -0.071(1) cm-1
42
Evidence of 3He24He2 Kalinin, Kornilov and Toennies
43
3He 4He
2
n
: energies relative to 4Hen
0
1P l = 1 ______
l = 0 ______
l = 1 ______
l = 0 ______
1S l = 0 ______
S
0
1
1
0
1
0
s2
sp
sp
-1
E (cm-1)
3P
-0.5
L
-1.5
-2
-2.5
The
1P
and
3P
0
4
states appear for n=4
8
12
16
20
n
The energy of 3He24Hen is roughly equal to the 4Hen energy plus the
3He orbital energies.
44
What is the shape of
4He
3
?
The Shape of the Trimers
Ne trimer
r(Ne-center of mass)
He trimer
r(4He-center of mass)
47
Ne3 Angular Distributions
a
b
Ne trimer
b
b
a
a
48
4He
3
a
Angular Distributions
b
b
b
a
a
49
Acknowledgments.. and a suggestion
Peter Reynolds
Silvia Tarasco Gabriele Morosi
Take a look at your nodes
50