Transcript No Slide Title

The nodes of trial and exact wave
functions in Quantum Monte Carlo
Dario Bressanini
Universita’ dell’Insubria, Como, Italy
http://www.unico.it/~dario
Peter J. Reynolds
Office of Naval Research
CECAM 2002 - Lyon
Nodes and the Sign Problem
• So far, solutions to the sign problem have
not proven to be efficient
• Fixed-node approach is efficient.
could have the exact nodes …
If only we
• … or at least a systematic way to improve
the nodes ...
• … we could bypass the sign problem
• How do we build a Y with good nodes?

CI? MCSCF? Natural Orbitals (Lüchow) ?
Nodes
•
What do we know about wave function
nodes?

Very little ....
•
NOT fixed by (anti)symmetry alone.
Only a 3N-3 subset
•
•
Very very few analytic examples
Nodal theorem is NOT VALID

Higher energy states does not mean more
nodes (Courant and Hilbert )
•
They have (almost) nothing to do with
Orbital Nodes. It is possible to use
nodeless orbitals.
•
Tiling theorem
Tiling Theorem (Ceperley)
Impossible for
ground state
The Tiling Theorem does not say how
many nodal regions we should expect
Nodes and Configurations
It is necessary to get a better understanding how CSF influence
the nodes. Flad, Caffarel and Savin
The (long term) Plan of Attack
• Study the nodes of exact and good
approximate trial wave functions
• Understand their properties
• Find a way to sistematically improve the
nodes of trial functions, or...
• …find a way to parametrize the nodes
using simple functions, and optimize the
nodes directly minimizing the Fixed-Node
energy
The Helium Triplet
• First
3S
state of He is one of very few
systems where we know exact node
• For S states we can write
Y  Y(r1 , r2 , r12 )
•For the Pauli Principle
Y(r1 , r2 , r12 )  Y(r2 , r1 , r12 )
• Which means that the node is
r1  r2
or
r1  r2  0
The Helium Triplet
•
•
•
•
Independent of r12
The node is more symmetric
than the wave function itself
r1
r2
It is a polynomial in r1 and r2
Present in all 3S states of
two-electron atoms
• The wave function is not
factorizable
but
Y(r1 , r2 , r12 )   (r1 , r2 ) (r12 )
Y(r1 , r2 , r12 )  N (r1 , r2 )e f ( r1 ,r2 ,r12 )
r12
r1  r2  Y  0
r1
r2
r1  r2  Y  0
The Helium Triplet
Y(r1 , r2 , r12 )  N (r1 , r2 )e
•
•
•
•
•
•
•
f ( r1 , r2 , r12 )
This is NOT trivial
N = r1-r2 , Antisymmetric Nodal Function
f = unknown, totally symmetric
The HF function has the exact node
Which of these properties are present in other systems?
Are these be general properties of the nodal surfaces ?
For a generic system, what can we say about N ?
YExact  N (R)e
f (R)
Helium Singlet 1s2s 2 1S
• AlthoughY  Y(r1 , r2 , q12 ),
the node does not
depend on q12 (or does very weakly)
•
A very good approximation
of the node is
r  r  const
q12
•
r2
r1
Surface contour plot of the node
4
1
4
2
The second triplet has
similar properties
r15  r25  const
He: Other states
• 1s2s 3S : (r1-r2) f(r1,r2,r12)
• 1s2p 1P o : node independent from r12 (J.B.Anderson)
• 2p2 3P e : Y = (x1 y2 – y1 x2) f(r1,r2,r12)
• 2p3p 1P e : Y = (x1 y2 – y1 x2) (r1-r2) f(r1,r2,r12)
• 1s2s 1S : node independent from r12
Similar to

•1s3s 3S
4
2
: node independent from r12
Similar to

r  r  const  0
4
1
(r15  r25  const)(r1  r2 )  0
Helium Nodes
• Independent from r12
• More “symmetric” than the wave function
• Some are described by polynomials in
distances and/or coordinates
• The same node is present in different states
(as if Helium were separable)
• The HF Y, sometimes, has the correct node,
or a node with the correct (higher) symmetry
Lithium Atom Ground State
YRHF  1s(r1 )1s(r2 )2s(r3 )  1s(r1 )2s(r3 )  1s(r3 )2s(r1 ) 1s(r2 )
• The RHF node is r1 = r3
if two like-spin electrons are at the same
distance from the nucleus then Y =0
• This is the same node we found in the He 3S
• Again, node has higher symmetry
• How good is the RHF node?
YRHF is not very good, however its node is
surprisingly good (might it be the exact one?)
DMC(YRHF ) = -7.47803(5) a.u. Lüchow & Anderson JCP 1996
Exact
= -7.47806032 a.u. Drake, Hylleraas expansion
Li atom: Study of Exact
Node
n m l i j k  r1   r2  r3
ˆ
YHy  A r1 r2 r3 r12r13r23e
•
•
We take an “almost exact” Hylleraas expansion
Energy YHy = -7.478059 a.u.
Exact
= -7.4780603 a.u.
250 terms
How different is its node from r1 = r3 ??
•
The node seems to be
r1 = r3, taking different
cuts, independent from r2
or rij
r1
r2
r3
Li atom: Study of Exact
Node
Numerically, we found only very small
deviations from the HF node, or artifacts
of the linear expansion
•
•
a DMC simulation with r1 = r3 node and
good Y to reduce the variance gives
DMC -7.478061(3) a.u.
Exact -7.4780603 a.u.
Is r1 = r3 the exact node of Lithium ?
Beryllium Atom Ground State
Y(R)  Y(1,2,3,4)  Y( x1, y1, z1, x2 , y2 , z2 , x3 , y3 , z3 , x4 , y4 , z4 )
12 D
Factor external angles
9D
Node Y(R) = 0
8-dimensional hypersurface
YRHF  1s(r1 )2s(r2 )1s(r2 )2s(r3 )
After spin assignment
YRHF  1s(r1 )2s(r2 ) 1s(r2 )2s(r1 )1s(r3 )2s(r4 ) 1s(r4 )2s(r3 )
Beryllium Atom
YRHF  1s (r1 )2s(r2 )  1s(r3 )2s(r4 ) 
 Y factors into two determinants each one “describing”
a triplet Be+2. The node is the union of the two
independent nodes.
 HF predicts 4 nodal regions
Bressanini et al. JCP 97, 9200 (1992)
 Node: (r1-r2)(r3-r4) = 0
The HF node is wrong
•DMC energy -14.6576(4)
•Exact energy -14.6673
Hartree-Fock Nodes
YHF  1,2,...N   N   1,..., N   N  J (rij )
•
YHF has always, at least, 4 nodal regions
for 4 or more electrons
• It might have N! N! Regions
• Ne atom: 5! 5! = 14400 possible regions
• Li2 molecule: 3! 3! = 36 regions
How Many ?
Clustering Algorithm
YHF
Nodal Regions
2
Li
1s 2 2s
Be
2
1s 2s
B
1s 2 2s 2 2 p
4
C
1s 2 2s 2 2 p 2
4
Ne
2
2
1s 2s 2 p
1
Li2
2


g
4
6
4
4
Be: beyond Restricted Hartree-Fock
• Hartree-Fock Y is not the most general
single particle approximation
• Try a GVB wave function (each electron in its
own orbital)
Aˆ 1s(r1 )2s(r2 )1s(r3 )2s(r4 )     
1s(r1 )2s(r2 )1s(r3 )2s(r4 )  term s....
YGVB(Be) = sum of 4 Determinants
Be: beyond Hartree-Fock
• YGVB(Be) = sum of 4 Determinants
• VMC energy improves
• s2(H) improves
• …but still the same node (r1-r2)(r3-r4) = 0
Aˆ  f1 (r1 ) f 2 (r2 ) f3 (r3 ) f 4 (r4 )(  )(  )
(r1-r2)(r3-r4) = 0 for any f1, f2, f3, f4
Be: CI expansion
• What happens to the HF
node in a CI expansion?
YCI   cn 1s 1s ns ms
•
n
Still the same topology
and the same node
Plot cuts of (r1-r2) vs (r3-r4)
(same in lithium)
In DMC, CI is not necessarily better than HF
•
Of course , one would first use Y  1s 2 2 s 2  c 1s 2 2 p 2
Be: CI expansion
•
What happens to the HF node in a good CI expansion?
Plot cuts of (r1-r2) vs (r3-r4)
•
In 9-D space, the direct product structure
“opens up”
Node is (r1-r2)(r3-r4) + ...
Be Nodal Topology
r1+r2
r1+r2
r3-r4
r3-r4
r1-r2
YHF  0
r1-r2
YCI  0
Be nodal topology
• The clustering algorithm
confirms that now there
are only two nodal regions
• It can be proved that the
exact Be wave function
has exactly two regions
See Bressanini, Ceperley and Reynolds
http://www.unico.it/~dario/
http://archive.ncsa.uiuc.edu/Apps/CMP/
Node is (r1-r2) (r3-r4) + ???
Fitting Nodes
• We would like a simple analytical
approximation of the node
• Ultimate goal: parametrize the node with few
parameters, and directly optimize them
• Useful also for diagnostic. To see if the node
changes, and how, by changing basis, or
expansion, or functional form.
• How can we model the implicit surface Y=0 ?

The studied (simple) systems suggest polynomials
of distances (plus x, y and z for L0 states)
Fitting Nodes
Model Node:
f (R, c)   ci Pi (R)  0
i
Pi (R ) Polynomials of ri and rij (or other simple
functions) with the correct spin-space
symmetry.
•To find c:
min  (c)   f (R, c) dl
c
Node
2
2
f(Node,c)
line integral
Node
Fitting Nodes
Linear Fit:
2


c
P
(
R
)
min  i i
c
i
•
Collect points on the nodal
surface during a DMC or VMC
walk
•
Minimize least square deviation
2
•
Discard null solution ci = 0
Be model node
(r1  r2 )( r3  r4 )  c(r132  r142  r232  r242 )  0
 
(r1  r2 )( r3  r4 )  c r12  r34  0
• Second order approx.
• Gives the right
topology and the right
shape
r1+r2
r3-r4
r1-r2
Be Node: considerations
•
•
•
•
•
HF and GVB give the wrong topology
•
The nodes of 1s22s2 + 1s22p2 belong to different
symmetry groups
In CI, it seems useless to include 1s2 ns ms CSFs
1s22s2 + 1s22p2 give already the right topology
Exact Y has only 2 nodal volumes
The nodes of the individual CSFs belong to higher
symmetry groups than the exact Y
(r1  r2 )(r3  r4 )  0
 
r12  r34  0
iˆ34 (1s 2 2s 2 )  1s 2 2s 2
iˆ34 (1s 2 2 p 2 )  1s 2 2 p 2
CSFs Nodal Conjecture
• If we consider the nodes of the individual
CSFs as a basis for the node of the exact Y,
IF we can generalize from Be, it seems
necessary (maybe not sufficient) to
Include configurations built
with orbitals of different
angular momentum and
symmetry
?
Boron Atom
Is it possible to change the topology of the nodal
hypersurfaces by adding particular CSFs or do they
merely generate deformations? Flad, Caffarel and Savin
Both!
HF
GVB
1s 2 2 s 2 2 p
Aˆ 1s1s2s 2s2 p      
4 Determinants
CI
4 Nodal Regions
1s 2 2s 2 2 p  c 1s 2 2 p 3
4 Nodal Regions
2 Nodal Regions
Li2 molecule
HF
1s g21s u2 2s g2
4 Nodal Regions
GVB
8 determinants
2 Nodal Regions
1s g21s u2 2s g2  c1s g21s u2 2s u2
2 Nodal Regions
CI
But energy not different than HF
But energy not different than HF
Nodal Topology Conjecture
The HF ground state of Atomic
and Molecular systems has 4
Nodal Regions, while the Exact
ground state has only 2
?
Node Optimization ? LiH
LiH YExact  N (R)e
S (R)
N  (r1Li  r2 Li )  c(r1H  r2 H )(r3Li  r4 Li )  c(r3H  r4 H )
• LiH Exact
• Simple node
• Higher terms
• However
-8.0702
–8.0673(6)
optimized with derivatives
-8.0693(6)
optimized with derivatives

HF nodes gives practically the exact result

Big fluctuations, due to poor S(R)
Conclusions
•
Algorithms to study topology and shape of nodes
•
“Nodes are weird” M. Foulkes. Seattle meeting 1999
•
“...maybe not” Bressanini, CECAM workshop 2002
Exact or good nodes (at least for simple systems) seem to



depend on few variables
have higher symmetry than Y itself
resemble polynomial functions
•
Possible explanation on why HF nodes are quite good:
they “naturally” have these properties
•
•
•
Hints on how to build compact MultiDet. expansions
It seems possible to optimize nodes directly
Has the ground state only 2 nodal volumes?
Acknowledgments
Peter Reynolds
Gabriele Morosi
Mose’ Casalegno
Silvia Tarasco