2. A system at initial time t= 0 with wave function (x, 0) = Ae¬alxl propagates freely (no forces). Normalize p(x, 0), i.e., find A. Let k = and determine (x, 0) in momentum space, i.e., p(k,0). а. b.

The initial time of the wave function is given as, $t = 0$

The wave function is given as, $\Psi (x,0) = A{e}^{-a\left|x\right|}$

The value of $k$ is given as, $k = \frac{p}{h}$

TO DETERMINE:

(a) Normalize $\Psi \left(x,0\right)$ and find the value of $A$

(b) $\Psi \left(x,0\right)$ in momentum space, that is, ${\Psi }_p\left(k,0\right)$

(a) Normalize $\mathit{\Psi }(x,0)$ and find the value of $\mathit{A}$:

The normalization of wave function is given as,

${\int }_{-\infty }^{\infty }{\left|\Psi \left(x,0\right)\right|}^{2}dx = 1$ ---------------------- (1)

Substituting the values in equation (1), we have,

${\int }_{-\infty }^{\infty }{\left|A{e}^{-a\left|x\right|}\right|}^{2}dx = 1$

${\int }_{-\infty }^{\infty }{\left|A\right|}^2{e}^{-2a\left|x\right|}dx = 1$ (or)

${\left|A\right|}^2{\int }_{-\infty }^{\infty }{e}^{-2a\left|x\right|}dx = 1$ -------------------- (2)

The integral is an even function for the symmetric function. Therefore, equation (2) is given as,

$2{\left|A\right|}^2{\int }_0^{\infty }{e}^{-2a\left|x\right|}dx = 1$

Integrating the above equation, we get,

$2{\left|A\right|}^2{\left[\frac{e^{-2a\left|x\right|}}{-2a}\right]}_0^{\infty } = 1$

Applying the limits, we get,

$2{\left|A\right|}^2\left[\frac{-1}{-2a}\right] = 1$

$\frac{{\left|A\right|}^2}{a} = 1$ (or) ${\left|A\right|}^2 = a$

The value of $A$ is calculated as,

$\mathit{A}\mathbf{ }\mathbf{=}\mathbf{ }\sqrt{\mathbf{a}}$