Matrix-Element Calculus in GUGA - univie.ac.at · 2009-12-15 · Principal References 1. I....

Matrix-Element Calculus in GUGA

Isaiah Shavitt

Department of ChemistryUniversity of Illinois at Urbana–Champaign

Principal References1. I. Shavitt, in Mathematical Frontiers in

Computational Chemical Physics, D. G. Truhlar, ed. (Springer, New York, 1988), pp. 300–349.

2. I. Shavitt, in The Unitary Group (Lecture Notes in Chemistry No. 22), J. Hinze, ed. (Springer, Berlin, 1981), pp. 51–99.

The Hamiltonian can be expressed in terms of the unitary-group generators as:

H = ∑ij hijEij + ½∑ijkl [ij;kl]eij,kl

where

hij = ⟨i|h| j⟩, [ij;kl] = ⟨i(1)k(2)|1/r12| j(1)l(2)⟩

and

eij,kl = Eij Ekl − δjk Eil = ekl,ij .

A matrix element of H between two CSF’sis obtained as

⟨m′|H|m⟩ = ∑ij hij⟨m′|Eij|m⟩ + ½∑ijkl [ij;kl] ⟨m′|eij,kl|m⟩

Thus the matrix elements of the generators Eij and eij,kl are the “coupling coefficients”needed for calculating the matrix elements of H.

The weight generator is just the number operator,

Eii|m⟩ = ni(m)|m⟩,where ni(m) is the occupation number of orbital i in |m⟩. Since this operator leaves |m⟩ unchanged, and since the CSFs are orthonormal, the matrix representation of the weight generator is diagonal,⟨m′|Eii|m⟩ = δm′mni(m) = δm′m(di − δd 2 − δd 3 ),

where di is the step number at level i in |m⟩.i i

The range of a raising generator Eij is the set of levels from i to j, inclusive. This range includes the nodes at levels ithrough j and the arcs leading up to them (but not the nodes at level i−1).

i i

i

j

When a raising generator Eij (i < j) operates on |m⟩,Eij|m⟩ = ∑m′|m′⟩⟨m′|Eij|m⟩

it changes the spin couplings in its range, but changes nothing outside that range. Therefore the walks for |m⟩ and |m′⟩ must coincide outside the range. Furthermore, because the operator removes an electron from level j, changing step number djfrom 1 or 2 to 0, or from 3 to 1 or 2, the j-level arc for |m′⟩ must be to the left of the arc for |m⟩, and thus

m′ < m.It therefore follows that the matrix representation of a raising generator is strictly upper triangular.

Nonzero contributions (i.e., nonzero raising-generator matrix elements ⟨m′|Eij|m⟩) can be obtained only for those values of m′ in the sum

Eij|m⟩ = ∑m′|m′⟩⟨m′|Eij|m⟩for which the arc step numbers dk in the range ofEij satisfy the occupancy relationshipsnj(m′) = nj(m) − 1, nk(m′) = nk(m) (i < k < j), ni(m′) = ni(m) + 1,allowing the following combinations:

{1, 0}, {2, 0}, {3, 1}, {3, 2}k = i{0, 0}, {1, 1}, {1, 2}, {2, 1}, {2, 2}, {3, 3}i < k < j

{0, 1}, {0, 2}, {1, 3}, {2, 3}k = j

Valid values for {dk(m′), dk(m)}Level

The matrix element ⟨m′|Eij|m⟩ (i < j) is represented on the graph by a loop in the range of Eij, and coincident upper and lower walks outside it.Its value depends on the loop and is independent of the upper and lower walks.

i

j

Some terminology:

i

jUpper walk

Lower walk

Loop

Loop head

Loop tail

For the lowering generators, which are the adjoints of the raising generators, the situation is analogous, with the roles of |m⟩and |m′⟩ reversed.

Thus the matrix representation of a generator is• Diagonal for a weight generator,• Strictly upper triangular for a raising generator,• Strictly lower triangular for a lowering generator.

So far we have obtained the following results:• A nonzero generator matrix element is

described by a pair of walks which coincide outside the range of the generator, and form a loop within that range.

• The value of the matrix element is independent of the coincident parts of the walks, and depends only on the loop.

To these we add another result, based on unitary group analysis:

• The value of the matrix element depends on the shape of the loop, i.e., on the step numbers in it, and on its position in the bdimension, but not on its position in the adimension or on its vertical position (i.e., its level).

We call this value the loop value.

To proceed further, we use some basic results of Gel’fand and Tsetlin, as translated to the language of Paldus tableaux by Paldus. These results refer to the values of matrix elements of elementary generators.

An elementary raising generator has the form Ei,i+1. There are only eight possible shapes for loops representing such generators, and these are shown, with their values, on the following slide.

i – 1

i + 1

i

Step numbers are shown next to the arcs.Relative b values are also shown.Loop values are shown below the loops.

0

b

b b+1

2

2

0

b+1

Level

0

b

b b+1

3

2

1

b

1

b

b−1 b

2

3

0

b

1 3

b−1

3 1

b−1

b

b

b+2b+1

bb+11 1

i – 1

i + 1

i

0

b

b b−1

1

1

0

b−1

1

0

b

b b−1

3

1

2

b+1

1b+2b+1

b

b

b b

b

b+1 b

b+1

2 2

2

3

3

3

1 0

bb+1

Next we make the assumption (subject to verification) that we can express these values as products of segment values.The set of segment values shown in greensatisfy the assumption.

0

b

b b+1

2

2

0

b+1

0

b

b b =b+1

3

2

1

b

1

b

b−1 b

2

3

0

b

1 3

b−1

31

b−1

b

b

b+2b+1

bb+11 1

0

b

b b−1

1

1

0

b−1

1

0

b

b b =b−1

3

1

2

b+1

1b+2b+1

b

b

b b

b

b+1 b

b+1

2 2

2

3

3

3

1 0

bb+1

1

1 1

1

1

1

1

1

bb+1

′

b+1b

b +1b′′

"

b+2b+1

b+1b+2

b +1b +2""

Segment values are always expressed in terms of the b value at the top of their ket arc.

When expressed in terms of the bvalue at the top of its |m⟩ arc, the value of each segment shape is the same in all loops in which it appears. (The |m⟩ and ⟨m′|arcs are shown in red and blue, respectively.)

Top segments

Bottom segments

0

b

2

b b+1

1

1

b−1 b

3

b

bb+1

0

b

b b−1

1

1

b

2 3

b+1 b

b+2b+1

b

2 0

b

b−1

1

b

3 1

b−1

b−1

b+1b

b+1 b

b

1 0

1

b+1

b+1

23

b

b+1b+2

The choice of the segment values is not unique (e.g., multiplying all top-segment values by any constant and dividing all bottom-segment values by the same constant does not change the loop values, but more general variations are also possible).

The choice given here has turned out to be quite convenient, because it gives unit values to all segments involving dk = 0 arcs (empty orbitals).

The factorization of the elementary generator loops can be extended to all one-body loops by deriving segment values for all internalsegments, using the generator commutation relation in the form

Eij= EikEkj – EkjEikand the resolution of the identity, to give

⟨m′|Eij|m⟩ = ∑m″⟨m′|Eik|m″⟩⟨m″|Ekj|m⟩– ∑m″⟨m′|Ekj|m″⟩⟨m″|Eik|m⟩.

Example⟨m′|Eij|m⟩ = ⟨m′|Eik|m″⟩⟨m″|Ekj|m⟩ – ⟨m′|Ekj|m′″⟩⟨m″′|Eik|m⟩

(i < k < j)

= –

k

k – 1

1

1

Only one intermediate walk (m″ or m′″) can fit between m′ and m in each of the two terms on the r.h.s., and the result is

× – 1×1 =

b bb

bb−1b b b−1 b−1

b−1b−1 b−1

Level

1 1 12 2 23

0

b+1b

b+1b

b+1b

b+1b

1b

This result is independent of the shapes of the loops above and below the k-level segments, as long as these shapes are consistent in the three terms of the equation. Thus we can assign the value as the segment value for the segment

A similar procedure can be carried out for all other middle segment shapes, giving the results on the next slide.

b

b−1b

b−1

121b

1b

1

bb−1 bb−1 bb−1 bb−1

bb−1 bb−1

bb−1bb+1 bb+1 bb+1 bb+1 bb+1

bb+1 bb+1b−1b b−1b b+1bb+1b+2

1

–1 –1b+1bb−1b−2

–1 –1 –1 b+2

√(b−1)(b+1)b

Middle segments

0 0

0 0

1 1

1 1 1 1

2 2

2 2 2 2

3 3

33

1b

√(b+1)(b+3)b+2

Conditions for middle segmentsThe two walks for the raising-generator matrix element ⟨m′|Eij|m⟩ satisfy the condition

Nk(m′) = Nk(m) + 1 (i ≤ k < j, Nk = 2ak + bk).Furthermore, for the intermediate walks (m″ or m′″) in the resolution-of-the-identity method for determining middle-segment values to exist, we must have

Sk(m′) = Sk(m) ± ½ (i ≤ k < j, Sk = ½bk). Therefore the nodes in that range must satisfy either of the two conditions:• ak(m′) = ak(m) and bk(m′) = bk(m) + 1 (∆k = –1 case)• ak(m′) = ak(m) + 1 and bk(m′) = bk(m) – 1 (∆k = 1 case)where ∆k = bk(m) – bk(m′). “wide top”

“narrow top”

Notation for segment classesWe introduce the following notation for the different classes of segments:

• R = raising-generator top segment

• L = lowering-generator top segment

• R = raising-generator middle segment

• L = lowering-generator middle segment

• R = raising-generator bottom segment

• L = lowering-generator bottom segment

• W = weight-generator segment

Using this notation we can tabulate the segment values in terms of the step numbers of the two arcs of the segment, d′ and d for ⟨m′| and |m⟩, respectively, and the difference ∆ = b – b′between the b values at the top of the two arc, as functions of the b value at the top of the |m⟩ arc.To simplify the tables we define the following functions:

C(p) = √(b+p–1)(b+p+1)b+p

B(p) = 1b+p

A(p,q) = b+pb+q

233122111000Wd′d

–1–1–1–133C(1)–B(1)

––11

∆ = 1

–1–

B(1)C(1)

1∆ = –1

L

–1B(0)

–C(0)

1∆ = 1

C(2)22–21

–B(2)12–111100

∆ = –1

Rd′d

A(0,1)A(2,1)

11L

A(2,1)23A(0,1)13

102101Rd′d

A(2,1)A(0,1)

11L

A(1,2)32A(1,0)31

120110Rd′d

One-body segment-value tables– –

– –

Two-body generator matrix elementsMatrix elements of products of two generators can be computed by the same resolution of the identity method that was used to derive the single generator segment values,

⟨m′|EijEkl|m⟩ = ∑m″⟨m′|Eij|m″⟩⟨m″|Ekl|m⟩.

However, unlike the one-body case, multiple intermediate walks m″ can contribute to the sum. This aspect complicates the factorization in terms of segment values.

i

k

j

lR

RR

RR

RR

RR

R

–

–

–

–

⟨m′| |m⟩

0

0

0

3 1

1

1

1

1

22

2

2

2

2

2

2

1

11

1

1

Example for the factorization of the generator product EijEklfor i < k < j < l(i.e., two overlapping raising generators).

There are 5 intermediate walks in this case.

1

With each intermediate arc we associate the product of the two one-body segments with which it is associated. For the colored segment we have such a product for each of the three intermediate arcs.

R

RR

RR

RR

RR

R

–

–

–

–

⟨m′| |m⟩

0

0

0

3 1

1

1

1

2

2

2

2

2

2

2

1

1 1

1

11

2

l

j

k

i

i

k

j

lR

RR

RR

RR

RR

R

–

–

–

–

⟨m′| |m⟩

0

0

0

3 1

1

1

1

1

22

2

2

2

2

2

2

1

11

1

1

Proceeding from the top of the loop down, We accumulate partial products for the different intermediate states, forking from a singlepartial product to two, or merging two to one (by adding the two partial products), etc.

A1 A2

C1 C2

B12

B21

B11 B22

C1 = A1B11 + A2B21

C2 = A1B12 + A2B22

⇓

C = AB

(A and C are row-vectors)

At each two-body level of the loop we have a partial product associate with each of the (one or two) intermediate nodes. In principle, we could have four intermediate segments (associated with the Bjk’s) connecting two adjacent levels, as shown schematically in this diagram. In practice there are no more than three for any segment. The cumulative partial products can be obtained by vector × matrix multiplication.

RR RL LL

RR RL RR RL––

––– ––

b−1b b+1 b

b+1 b b+1b+2

22

2

12

b−1b b+1 b−1b b+1 b

b−1b b+1 bb−1 b−2

11

1 1

1

2

22

1 12

3 0 21

2 0 1 3 10

b+1 b b b−1 b b−1 b

b−1b b+1 b b−1b b+1 b−1b b−1b

√(b+1)(b+3)b+2

√ b(b+2)b+1

−0−

(b+1)(b+2)−1 0

1

b(b+2)b+1

1b+2

1b+1

b+2

1b+2−

bb+1 √(b+1)(b+2)

−1 1

1

1 b+1b

b⎧⎪⎩

⎫⎪⎭

⎧⎪⎩

⎫⎪⎭

⎧⎪⎩

⎫⎪⎭

⎧⎩

⎫⎭

⎧⎩

⎫⎭

⎧⎩⎫⎭

⎧⎩

⎫⎭

Here are some examples of segment-value matrices of two-generator products.Segment types are indicated. Step numbers are shown next to the arcs and b values are shown next to the nodes.

RR RL LL

RR RL RR RL––

––– ––

b−1b b+1 b

b+1 b b+1b+2

22

2

12

b−1b b+1 b−1b b+1 b

b−1b b+1 bb−1 b−2

11

1 1

1

2

22

1 12

3 0 21

2 0 1 3 10

b+1 b b b−1 b b−1 b

b−1b b+1 b b−1b b+1 b−1b b−1b

√(b+1)(b+3)b+2

√ b(b+2)b+1

−0−

(b+1)(b+2)−1 0

1

b(b+2)b+1

1b+2

1b+1

b+2

1b+2−

bb+1 √(b+1)(b+2)

−1 1

1

1 b+1b

b⎧⎪⎩

⎫⎪⎭

⎧⎪⎩

⎫⎪⎭

⎧⎪⎩

⎫⎪⎭

⎧⎩

⎫⎭

⎧⎩

⎫⎭

⎧⎩⎫⎭

⎧⎩

⎫⎭

For different segment shapes we can get 2 × 2, 2 × 1, 1 × 2matrices and some scalars, depending on the number of intermediate nodes at each level. The total matrix element is obtained by multiplying all segment values in sequence, including matrices and scalars.

eij,jl = EijEjl – Eil (i < j < l)–

–

20 0

b–1

b−1 b

RRb

0 0

b−1 b

Rb–1 b

–

1 × 1 – 1 = 0

j

231

b+1

b−1b

b

21

b−1b

b+1 b

–

× – = 1

j

b+1b+2

b+1b+2

–1b+2

When i < j = k < l, the j-level segments for matrix elements of eij,kl = EijEkl – δkjEjl

(RR-type) or of elk,ji = ElkEji – δkjEli

(LL-type) have to include the second (δ ) term. Since the rest of the loop is the same for both terms, it is enough to modify the j-level segment value. Examples are shown here and in the next slide.

––

––

elj,ji = EljEji – Eli (i < j < l)

––

10 0

b–1

b−1b

LLb

0 0

b−1b

Lb–1b

–

1 × 1 – 1 = 0

j

2 3 1

b+1

b−1b

b

2 1

b−1 b

b+1b

–

× – = 11b+1

j

b+2b+1

b+2b+1

Additional examples involving the δ term are shown here.

As seen, in some cases subtraction of this term results in a zero matrix element.

The scheme based on the multiplication of matrix-valued segment values works well, and is practical, but it was shown by Drake and Schlesinger and by Paldus and Boyle, using the diagrammatic methods of spin algebras, that a simpler solution exists. This solution represents the product of the segments in the two-body part of the loop as a sum of two scalar products, identified by an index x having two possible values, x = 0 for intermediate singlet coupling and x = 1 for intermediate triplet coupling.

Knowing this, it was possible to derive this result totally within GUGA using matrix transformations.

Just as in the case of one-body segments, the choice of segment values is not unique, as long as all values are modified in compatible ways.

For the two-body case we can use this flexibility to transform all 2 × 2 matrices to diagonal form.

The product of diagonal matrices is just a diagonal matrix, and each of its two diagonal elements is a product of scalars, leading to the desired result.

√ b √ b+2

√ b+2 –√ b

⎫⎪⎭

⎧⎪⎩√ 2(b+1)

1U(0,b) = V(0,b) =

U(±2,b) = (0 1); V(±2,b) =01⎧⎪⎩⎫⎪⎭

U(0,b)V(0,b) = ; U(±2,b)V(±2,b) = 1⎧⎪⎩

⎫⎭⎪1 0

0 1

To generate the desired transformations we define the matrices U(∆,b) and V(∆,b), where ∆ = b – b′ andb and b′ are the b values at the top of the ket (|m⟩) and bra (⟨m′|) segments, respectively (for two-body segments ∆ can only take the values 0, ±2):

These matrices and vectors have the property

Therefore these products can be inserted freely between the segment values matrices (choosing insertions with compatible dimensions).

1 00 1

b

b -1

bb

b

b -1

b -1

b -1

11

1

11

2

22

2

2

2

0

0

RR–

RR

RR–

√(b–1)(b+1)b

1b

⎧⎩

⎫⎭U(0,b–1) = 11

√2⎧⎩

b–1b+1

⎫⎭

V(0,b–1)–√(b–1)(b+1)

b

b(b+1)–1

⎧

⎪⎩

⎪ 0

–√b(b+2)b+1

⎫⎪⎭⎪U(0,b) =

–1

0

0

(b–1)(b+2)b(b+1)

⎧

⎪⎩

⎪–

⎫⎪⎭⎪

V(0,b) =⎧

⎪⎩

⎪ b+1–1

√(b(b+2)b+1

⎧

⎪⎩

⎪⎫⎪⎭⎪

b2(b+1)b+2

2(b+1)

⎫⎪⎭⎪

We inserted the products U(0,b–1)V(0,b–1) =

and U(0,b)V(0,b) = between the segment-

value matrices. As a result, the 2 × 2 matrix became diagonal, and the entire product became a sum of two contributions:

x = 0(singlet-coupled)

component

x = 1(triplet-coupled)

component

The same process diagonalizes all the 2 × 2segment-value matrices!

⎧⎪⎩

⎫⎭⎪1 0

0 11 00 1

⎧⎪⎩

⎫⎭⎪1 0

0 1

1√2⎧⎩

b–1b+1

⎫⎭1

–1

0

0

(b–1)(b+2)b(b+1)

⎧

⎪⎩

⎪–

⎫⎪⎭⎪

⎧

⎪⎩

⎪b

2(b+1)b+2

2(b+1)

⎫⎪⎭⎪

bb+1

12= – – (b–1)(b+2)

2b(b+1)b

b+1

Here are three examples of the transformation of two-bodysegment-value matrices. Three more examples are given in the next slide.

U(0,b) = √2 0⎧⎩

⎫⎭

b+2b+1

bb+1

⎫⎭

⎧⎩

⎧

⎪⎩

⎪b

2(b+1)b+2

2(b+1)

⎫⎪⎭⎪–

⎧

⎪⎩

⎪ b+1–1

√b(b+2)b+1

⎫⎪⎭⎪

V(0,b) =

⎧⎩ ⎭

1 00 1

⎫⎧⎩ ⎭

1 00 1

⎫V(0,b) U(0,b) =

RR

RR

RR

––

–

b

b–1 bb+1b

b–1

b–1 bb+1b

b–1 bb+1b

b–1 bb+1b

b

0

0

0 0 0 0

31

1

2

12

b+1b+3

1b+2⎧⎩ ⎭

⎫V(–2,b) U(0,b+1) =

RR

RR

RR

b–2 b–1bb–1

b–1

b–2 bb–1

b–3

bb+1b+2

b b+1b+2b+1

b–2

1 1

11

22

11

1

2

1 1

–√ b(b+2)b+1

⎫⎪⎭⎪

⎧

⎪⎩

⎪ b(b+1)–1–√(b–1)(b+1)

bV(0,b) U(0,b–1) =0 (b–1)(b+2)

b(b+1)

⎫⎪

⎭⎪

⎧

⎪⎩

⎪ 0

0 –

–1 ⎪

V(2,b) U(2,b–1) =(b–2)(b+1)(b–1)b (b–2)(b+1)

(b–1)b

⎫⎪

⎭⎪

⎧

⎪⎩

⎪ 0

0

0 ⎪

–1b+2 2

(b+2)(b+3)

⎫⎪

⎭⎪

⎧

⎪⎩

⎪ 0

0

0 ⎪

b–1 bb+1b

Tables of two-body segment valuesComplete tables of two-body segment values are given in the following slides. As before, ∆ = b – b′ and b and b′ are theb values at the top of the ket (|m⟩) and bra (⟨m′|) segments, respectively.Segments with ∆ = "2 can have only x = 1(triplet-coupled) components, because their ket-bra spin difference is ½∆ = "1.

Segments with ∆ = 0 can have both x = 0and x = 1 components.

The following abbreviations are used in the tables:

t = 1/√2

D(p) = (b+p–1)(b+p+2)(b+p)(b+p+1)

E(p) = 2(b+p)(b+p+2)

C(p) = √(b+p–1)(b+p+1)b+p

B(p) = 1b+p

A(p,q) = b+pb+q

When a " or Kappears in the tables, the lower sign refers to the generator pair marked by an asterisk.

133022011000

WWd′d

–2–2–2–233C(1)–B(1)

––10

∆ = 1

–1–

B(1)C(1)

0∆ = –1

WL, LW

–1B(0)

–C(0)

0∆ = 1

C(2)22–21

–B(2)12–111000

∆ = –1

WR, RWd′d

A(0,1)A(2,1)

00

WL, LW

A(2,1)23A(0,1)13

002001

WR, RWd′d

A(2,1)A(0,1)

00

WL, LW

A(1,2)32A(1,0)31

020010

WR, RWd′d– –– –

–– –

–

011022

1−1−21

−1−112

100111

111133

000000

∆ = 1∆ = −1∆ = 1∆ = −1

LL , LLRR , RR

d′d

––

––

–– –

–

A(0,1)A(2,1)03

∆ = 1∆ = −1

RL , LR

d′d

– –– –

A(1,0)A(1,2)30

∆ = 1∆ = −1

LR , RL

d′d

– –– –

0√203

x = 1x = 0

RR , LL

d′d–

––––

0√230

x = 1x = 0

RR , LL

d′d–

––––

∆ = 2

"A(–1,0)KtA(0,2)–t–"tA(2,0)–t"A(3,2)023

–"tA(2,0)–t"A(3,2)"A(–1,0)0KtA(0,2)–t13

–KtA(0,1)tA(2,1)"1"10"tA(3,2)tA(1,2)02

"1"tA(2,1)tA(0,1)–KtA(–1,0)tA(1,0)"1001

x = 1x = 1x = 0x = 1x = 1x = 0x = 1x = 0

∆ = 0∆ = –2∆ = 1∆ = –1

LL, LL*RR, RR*

d′d

– – – –

∆ = 2 ∆ = 1

"A(2,1)0"tA(3,1)–t–KtA(0,2)–t"A(1,2)32

KtA(–1,1)–t"A(0,1)0"A(1,0)"tA(2,0)–t–31

"tA(2,1)tA(0,1)"10"1KtA(0,1)tA(2,1)–20

"10KtA(0,1)tA(2,1)–"tA(2,1)tA(0,1)"110

x = 1x = 0x = 1x = 0x = 1x = 1x = 0x = 1

∆ = 0∆ = 0∆ = –2

LL, LL*RR, RR*

d′d

– –– –

1111111133

D(0)–D(1)–111–D(1)–1D(2)22

∆ = 0∆ = 2 ∆ = –2

B(0)B(1)0–B(–1)B(0)0–21

–B(0)0B(1)–B(1)0B(2)12

1–D(0)–1D(1)D(–1)–D(0)–1111

1111111100

x = 1x = 1x = 0x = 0x = 1x = 1x = 0x = 1∆ = 0∆ = 0∆ = –2

LL, LLRR, RR

d′d

∆ = 2

–tA(0,1)–tA(2,1)–A(1,2)–A(1,0)0–tA(3,2)tA(1,2)32

–A(1,0)–tA(2,1)–tA(0,1)–tA(–1,0)tA(1,0)–A(1,2)031

C(0)–tA(0,2)–t–tA(2,0)t1020

–tA(2,0)–tC(2)10–tA(0,2)t10

x = 1x = 1x = 0x = 1x = 1x = 0x = 1x = 0

∆ = 0∆ = –2∆ = 1∆ = –1

RL, LRRL, LR

d′d

– –– –

∆ = 2 ∆ = 1

–tA(2,1)tA(0,1)–A(2,1)0–A(–1,0)tA(0,1)–tA(2,1)–23

–A(0,1)0tA(0,1)tA(2,1)––tA(2,1)–tA(0,1)–A(3,2)13

10tA(3,1)t––tA(0,2)–tC(2)02

–tA(–1.1)t10C(0)tA(2,0)–t–01

x = 1x = 0x = 1x = 0x = 1x = 1x = 0x = 1

∆ = 0∆ = 0∆ = –2

RL, LRRL, LR

d′d

– –– –

x = 1x = 1x = 0x = 1

–C(0)D(1)1–C(2)22

–√2/b–E(0)0–21

––E(0)0–√2/(b+2)12

–C(0)D(0)1–C(2)11

111133

111100

∆ = 2∆ = 0∆ = –2RL , LRRL , LR

d′d

–tA(0,2)–ttA(3,1)t22

C(0)01021

C(2)01012

tA(2,0)–t–tA(–1,1)t11

0–√20√233

000000

x = 1x = 0x = 1x = 0

RL , LRRL , LRd′d

–– –––– ––

– –

–

–