Modern Quantum Chemistry

Chapter02 - Home
Exercises: 1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10 - 11 - 12 - 13 - 14 - 15 - 16 - 17 - 18 - 19 - 20 - 21 - 22 - 23 - 24 - 25 - 26 - 27 - 28 - 29 - 30 - 31 - 32 - 33 - 34 - 35 - 36 - 37 - 38 - 39 - 40 - 41

Exercise. This is the solution to exercise 2.1 in the book.

Solution. Consider the following five cases.

Case 1. Both i and j are even and ij, i = 2k, and j = 2m. Then

χi(x)χj(x)dx =drψkβ(r)ψ mβ(r)dωβ2(ω) =drψkβ(r)ψ mβ(r) 1 = 0

because ψkβ(r) and ψmβ(r) are orthonormal.

Case 2. Both i and j are even and i = j, i = 2k, and j = 2k. Then

χi(x)χj(x)dx =drψkβ(r)ψ kβ(r)dωβ2(ω) =drψkβ(r)ψ kβ(r) 1 = 1

because ψkβ(r) are normalized.

Case 3. Both i and j are odd and ij, i = 2k 1, and j = 2m 1. Then

χi(x)χj(x)dx =drψkα(r)ψ mα(r)dωα2(ω) =drψkα(r)ψ mα(r) 1 = 0

because ψkα(r) and ψmα(r) are orthonormal.

Case 4. Both i and j are odd and i = j, i = 2k, and j = 2k. Then

χi(x)χj(x)dx =drψkα(r)ψ kα(r)dωα2(ω) =drψkα(r)ψ kα(r) 1 = 1

because ψkα(r) are normalized.

Case 5. If i is odd and j is even, for example i = 2k 1 and j = 2m, then

χi(x)χj(x)dx =drψkα(r)ψ mβ(r)dωα(ω)β(ω) =drψkα(r)ψ mβ(r)dωα(ω)β(ω) = Skm0 = 0

because α(ω) and β(ω) are orthonormal.

Therefore

χi(x)χj(x)dx = δij

so the set {χi} of 2K spin orbitals is orthonormal.