Question

The quantum mechanical wave function for a particle is given by

Ax^3/2 e^-ax, x>0
ψ (x)= {
0 , x<0

Determine (i) the normalization constant A and (ii) the most probable positio

Answer 1

Answer on Question #61254-Physics-Quantum Mechanics

The quantum mechanical wave function for a particle is given by

$$\psi \left(x\right) = \left\{ \begin{array}{c} A x ^ {\frac {3}{2}} e ^ {- a x}, x > 0 \\ 0, x < 0 \end{array} \right.$$

Determine (i) the normalization constant A and (ii) the most probable position.

Solution

(i)

$$\begin{array}{l} 1 = \int_{-\infty}^{\infty} |\psi(x)|^2 dx = \int_{0}^{\infty} A^2 x^3 \, e^{-2ax} dx = A^2 \int_{0}^{\infty} x^3 \, e^{-2ax} dx = |y = 2ax| = \frac{A^2}{(2a)^4} \int_{0}^{\infty} y^3 \, e^{-y} dy \\ = \frac{A^2}{(2a)^4} \Gamma(4) = \frac{A^2}{(2a)^4} 3! = \frac{3A^2}{8(a)^4} \\ A = 2 \sqrt{\frac{2}{3}} a^2 \\ \end{array}$$

(ii)

$$\begin{array}{l} \langle x \rangle = \int_{-\infty}^{\infty} \psi^{*}(x) x \psi(x) dx = \int_{0}^{\infty} A^2 x^4 \, e^{-2ax} dx = \int_{0}^{\infty} \frac{8a^4}{3} x^4 \, e^{-2ax} dx = |y = 2ax| = \frac{1}{12a} \int_{0}^{\infty} y^4 \, e^{-y} dy \\ = \frac{1}{12a} \Gamma(5) = \frac{1}{12a} 4! = \frac{2}{a}. \\ \end{array}$$

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