Transcript No Slide Title

Trial wave function
construction and the nodes
of trial and exact wave
functions in
Quantum Monte Carlo
Dario Bressanini
Universita’ dell’Insubria, Como, Italy
http://www.unico.it/~dario
ACS National Meeting 2003 – New York
Nodes and the Sign Problem
• Fixed-node QMC is efficient.
If only we
could have the exact nodes …
• … 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?
2
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.
3
Tiling Theorem (Ceperley)
Impossible for
ground state
Nodal regions must have the same shape
The Tiling Theorem does not say how
many nodal regions we should expect
4
Nodes and Configurations
A better Y does not mean better nodes
Why? What can we do about it?
It is necessary to get a better understanding how CSF
influence the nodes. Flad, Caffarel and Savin
6
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
Find a way to parametrize the nodes
using simple functions, and optimize
the nodes directly minimizing the FixedNode energy
7
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
8
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  Y  0
r1
r2
r1  r2  Y  0
9
Other He states: 1s2s 2 1S and 2 3S
• AlthoughY  Y(r1 , r2 , 12 ),
the node does not
depend on 12 (or does very weakly)
•
12
•
r2
r1
A very good approximation
of the node is
r14  r24  const
The second triplet has
similar properties
r15  r25  const
Surface contour plot of the node
13
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
• 1s3s 3S : node independent from r12
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Helium Nodes
YExact  N (R)e
f (R)
• Independent from r12
• More “symmetric” than the wave function
• Some are described by polynomials in
distances and/or coordinates
• The HF Y, sometimes, has the correct node,
or a node with the correct (higher) symmetry
• Are these general properties of nodal
surfaces ?
15
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
• Node has higher symmetry than Y
• How good is the RHF node?
YRHF is not very good, however its node is
surprisingly good
DMC(YRHF ) = -7.47803(5) a.u. Lüchow & Anderson JCP 1996
Exact
= -7.47806032 a.u. Drake, Hylleraas expansion
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Li atom: Study of Exact
Node
•
•
•
•
We take an “almost exact” Hylleraas expansion
The node seems to be
r1 = r3, taking different
cuts, independent from r2
or rij
250 term
r1
r3
r2
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 ?
17
Li atom: Study of Exact Node
• Li exact node is more symmetric than Y
n m l i j k  r1   r2  r3
ˆ
YHy  A r1 r2 r3 r12r13r23e
• At convergence, there is a delicate
cancellation in order to build the node
• Crude Y has a good node (r1-r3)Exp(...)
• Increasing the expansion spoils the node,
by including rij terms
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Nodal Symmetry Conjecture
• This observation is general:
If the symmetry of the nodes is higher
than the symmetry of Y, adding terms in
Y might decrease the quality of the nodes
(which is what we often see).
WARNING: Conjecture Ahead...
Symmetry of nodes of Y is
higher than symmetry of Y
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Beryllium Atom
YRHF  1s(r1 )2s(r2 )  1s(r3 )2s(r4 ) 
 HF predicts 4 nodal regions
Bressanini et al. JCP 97, 9200 (1992)
 Node: (r1-r2)(r3-r4) = 0
 Y factors into two determinants
each one “describing” a triplet
Be+2. The node is the union of the
two independent nodes.
The HF node is wrong
•DMC energy -14.6576(4)
•Exact energy -14.6673
Plot cuts of (r1-r2) vs (r3-r4)20
Be Nodal Topology
r1+r2
r1+r2
r3-r4
r3-r4
r1-r2
YHF  0
r1-r2
Y  1s 2 2s 2  c 1s 2 2 p 2
YCI  0
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Be nodal topology
• Now there are only two
nodal regions
• It can be proved that
the exact Be wave
function has exactly
two regions
Node is (r1-r2)(r3-r4) + ...
See Bressanini, Ceperley and Reynolds
http://www.unico.it/~dario/
http://archive.ncsa.uiuc.edu/Apps/CMP/
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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 ?
25
Nodal Regions
Nodal Regions
YHF YCI
Li
Be
B
C
Ne
Li2
2
4
4
4
4
4
2
2
2
2
2
2
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Nodal Topology Conjecture
WARNING: Conjecture Ahead...
The HF ground state of Atomic
and Molecular systems has 4
Nodal Regions, while the Exact
ground state has only 2
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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
• What's next?
r3-r4
r1-r2
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Be numbers
•
•
•
•
•
•
•
HF node
•
Including 1s2 ns ms or 1s2 np mp configurations
does not improve the Fixed Node energy...
GVB node
-14.6565(2) 1s2 2s2
same
1s1s' 2s2s'
Luechow & Anderson -14.6672(2)
+1s2 2p2
Umrigar et al.
-14.66718(3)
+1s2 2p2
Huang et al.
-14.66726(1)
+1s2 2p2 opt
Casula & Sorella
-14.66728(2)
+1s2 2p2 opt
Exact
-14.6673555
...Why?
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Be Node: considerations
•
... (I believe) they give the same contribution to the
node expansion
•
•
ex: 1s22s2 and 1s23s2 have the same node
ex: 2px2, 2px3px and 3px2 have the same structure
x1 x2 f1 (r1 ) f 2 (r2 )
•
The nodes of "useful" CSFs belong to higher and
different symmetry groups than the exact Y
(r1  r2 )(r3  r4 )
 
r12  r34
iˆ34 (1s 2 2 s 2 )  1s 2 2 s 2
iˆ34 (1s 2 2 p 2 )  1s 2 2 p 2
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The effect of d orbitals
1s2 2s2 = -14.6565(2)
-14.6668
+ 1s2 2p2
-14.6669
Energy
-14.6670
+ 1s2 3d2
-14.6671
-14.6672
-14.6673
Exact
-14.6674
0
0.002
0.004
Time step
0.006
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Be numbers
•
•
•
•
•
•
•
•
HF
GVB node
-14.6565(2) 1s2 2s2
same
1s1s' 2s2s'
Luechow & Anderson -14.6672(2)
+1s2 2p2
Umrigar et al.
-14.66718(3)
+1s2 2p2
Huang et al.
-14.66726(1)
+1s2 2p2 opt
Casula & Sorella
-14.66728(2)
+1s2 2p2 opt
Bressanini et al.
-14.66733(7)
Exact
-14.6673555
+1s2 3d2
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CSF nodal conjecture
WARNING: Conjecture Ahead...
If the basis is sufficiently
large, only configurations
built with orbitals of
different angular momentum
and symmetry contribute to
the shape of the nodes
This explains why single excitations are not useful
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Carbon Atom: Topology
HF
GVB
1s 2 2s 2 2 p 2
4 Nodal Regions
Aˆ 1s1s2s 2s2 p 2 p        
4 Determinants
4 Nodal Regions
Adding determinants might not be
sufficient to change the topology
CI
1s 2 2s 2 2 p 2  c 1s 2 2 p 4
2 Nodal Regions
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Carbon Atom: Energy
• CSFs
Det.
Energy
• 1 1s22s2 2p2
1
-37.8303(4)
• 2 + 1s2 2p4
2
-37.8342(4)
• 5 + 1s2 2s 2p23d
18 -37.8399(1)
• 83 1s2 + 4 electrons in 2s 2p 3s 3p 3d shell
422 -37.8387(4)
adding f orbitals
•7
(4f2 + 2p34f)
Exact
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-37.8407(1)
-37.8450
Where is the missing energy? (g, core, optim..)
36
Li2 molecule, large basis
Adding CFS with a larger basis ... (1sg2 1su2 omitted)
2
2
s
• HF
g
+8 nσ g2
%CE
-14.9919(1)
97.2(1)
-14.9914(1)
96.7(1)
• GVB 8 dets
-14.9907(6)
96.2(6)
•
-14.9933(1)
98.3(1)
-14.9933(1)
98.3(1)
-14.9952(1)
99.8(1)
2
2
1


1

+
ux
uy
2
2
n


n

+4
ux
uy
•+
2 p z s u2  3 p z s 2g
Estimated n.r. limit -14.9954
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O2
• Small basis
1 Det.
-150.268(1)
Filippi & Umrigar
7 Det.
-150.277(1)
.....................
• Large basis
1 Det.
-150.2850(6)
Tarasco, work in progress
2 Det.
-150.2873(7)
..................................
Exact
-150.3268
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Conclusions
• Exact or good nodes (at least for simple systems)
seem to



depend on few variables
have higher symmetry than Y itself
resemble simple functions
• Possible explanation on why HF nodes are
quite good: they “naturally” have these
properties
• Use large basis, until HF nodes are converged
• Include "different" CSFs
• Has the ground state only 2 nodal volumes?
40
Acknowledgments.. and a suggestion
Silvia Tarasco Peter Reynolds
Gabriele Morosi Carlos Bunge
Take a look at your nodes
41