Quantum Chemistry

Chapter06 - 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 - 42 - 43 - 44 - 45 - 46 - 47 - 48 - 49

Exercise. This is the solution to exercise 6.1 in the book.

Solution.

(a) If the three force constants are equal kx = ky = kz = k then the vibration frequency is the same along all directions:

νx = νy = νz = 1 2π k m

Hence, according to exercise 4.18  the potential energy operator is:

V ̂ = 2π2ν2m(x2 + y2 + z2) = 2π2ν2mr2

Thus in this case the potential energy is spherically symmetric. It was shown in the text (following the derivation of equations up to 6.16) that in this case the wave function can be written as a product of a radial function and another function dependent on θ and ϕ.

(b) Again, following the arguments leading to equation 6.16 in the text we get that:

G(θ,ϕ) = Y lm(θ,ϕ)

(c) f(r) must satisfy equation 6.17 in the text.

(d) In exercise 4.18  we found that:

ψ(x,y,z) = ψx(x)ψy(y)ψz(z)

For the isotropic harmonic oscillator in the ground state νx = νy = νz = ν and ψx = ψy = ψz = ψ0(x), where ψ0(x) is given by (according to 4.49 in the text):

ψ0(x) = c0eα2x2 α = 2πνm

The energy is:

E = 3 2hν

c0 is given by the normalization condition:

ψ 02(x)dx = 1

or

c02eαx2 dx = 1

c02π α = 1

or

c0 = α π14

Hence,

ψ(x,y,z) = α π34eα2(x2+y2+z2)

or

ψ = α π34eα2r2

Hence, in this case:

G(θ,ϕ) = 1

and

f(r) = α π34eα2r2

We can show that f(r) verifies equation 6.17 in the text for:

E = 3 2hνV ̂ = 2π2ν2mr2

when l = 0 (i.e., the ground state).

It is easy to show that:

f(r) = αrf(r)

and

f(r) = (α2r2 α)f(r)

Hence:

f(r) + 2 rf(r) = (α2r2 3α)f(r)

and then the left hand side of the equation is:

LHS = 3α2 2m f(r)

With energy:

E = 3 2hν = 3 2 α2 m

then indeed equation 6.17 is satisfied.