Quantum Chemistry

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Exercise. This is the solution to exercise 4.1 in the book.

Solution.

(a)

ex2 dx = π

This is the Gaussian integral. Hence, the normalized function is 1 πex2 . If we treat this as a normlizable function then we must have:

ψ(x) = Cex2

The normalization condition requires:

ψ(x)ψ(x)dx = 1

Upon variable change and integration we find C = 2 π 14, thus the function is:

ψ(x) = 2 π14ex2

(b) ex is unbounded in (0,), hance it cannot be normalized.

(c) A normalizable function based on eiθ over the domain (0,2π) is

ψ(θ) = Ceiθψ(θ) = Ceiθ

Integrating

02πψ(θ)ψ(θ) = 2π|C|2 = 1

Hence, the normalized function is:

ψ(θ) = 1 2πeiθ

(d) A normalized function based on sinhx over the domain (0,) cannot be found because sinhx is unbounded at + .

(e) A normalized function based on xex is

ψ(x) = Cxex

Normalization requires that

0C2x2e2x = 1

or

0x2e2x = 1 C2

Evaluating the integral we have

0x2e2x = 1 4 = 1 C2

Hence

ψ(x) = 2xex

is normalized.