8/3/2019 ex_2_13

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Szabo-Ostlund. Exercise 2.13

Evaluate the matrix element r

a|O1|s

b, where O1 is a sum of one-electronoperators h(i).Background

Slater determinant short hand notation for normalized determinant, includ-ing the normalization constant, only showing diagonal elements, (Szabo-Ostlundeq 2.38):

(x1, x2, ...,xN) = |i(x1)j(x2) k(xN)

Electron labels always chosen to be in order x1, x2, ...,xN, further shorteningof notation (Szabo-Ostlund eq 2.39):

(x1, x2, ...,xN) = |ij k

Interchange of two columns changes sign of determinant (Szabo Ostlund eq2.40):

| m n = | n m

Definition of Hartree-Fock ground state (Szabo-Ostlund eq 2.64):

|0 = |12 ab N

Definition of singly excited state (Szabo-Ostlund eq 2.65):

|ra = |12 rb N

Example (p.69): To use the rules from tables 2.3/2.4 and 2.5/2.6, the deter-minants |K and |L need to be in maximum conincidence:

1 = |abcd

2 = |crds

2 can be rearranged by switching columns and introducing minus sign:

2 = |crds = |crsd = |srcd

then 1|O1|2 = 0 (case 3, table 2.3).

Exercise : Now ra|O1|sb.

Ifa = b and r = s:The matrix element is 0 according to case 3 of table 2.3.

Ifa = b and r = s:The matrix element is written as rb |O1| as . Now, on the

left side, I switch the two spin-orbitals, to bring the spin-orbitals a and b intomaximum conicidende, since from the definition they are equal and evaluate thematrix element using case 2 from table 2.3. This introduces a minus sign:

rb |O1| as = br |O1| as

= r|h|s (Case 2)

However, this should be r|h|s, without a minus. Where am I going wronghere???

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