Answer to Question #89544 in Quantum Mechanics for Sukritha p

Question

The ground state wave function of a particle in a 3D potential is given by ¥=Ae^{ar^3}, where A is the normalisation constant. The value of A is given by

Answer 1

Let a<0. Normalization condition:

$$\int \psi^2 r^2 dr=1$$

(integral limits: from 0 to infinity ).

$\int A^2 e^{2ar^3} r^2 dr=(A^2/6a)\int e^{2ar^3} d(2ar^3)=(A^2/6a)e^{2ar^3}$

Substituting the limits of the integral and equating the expression to 1:

$$-A^2/(6a)=1$$

$$A^2=-6a$$

$$A1=\sqrt{-6a}$$