Question

Refer to a particle of massdescribed by the wave function

$\mathbf{\psi }\left(x\right)\mathbf{=}\{\begin{array}{l}2\sqrt{a^3{\mathrm{xe}}^{-\mathrm{ax}}}X>0\\ 0X<0\end{array}$

Verify that the normalization constant$\mathbf{2}\sqrt{{\mathbf{a}}^{\mathbf{3}}}$ is correct.

Short answer

The integral of the square of the wave function over all the space is properly normalized. The normalization constant is indeed equal to $2\sqrt{{\mathrm{a}}^3}$.

Step-by-step solution

Concept of wave function

$\mathrm{\Psi }(\mathrm{x},\mathrm{t})=\mathrm{Acos}(\mathrm{kx}-\mathrm{wt})$is the general equation for a moving wave.The amplitude is equal to A. The wavelength is determined by multiplying k by x, and the location of the peak is determined by t.

Equals the wave function to 1.

The area under the square of the wave function is equal to 1 if the wave function is properly normalized.

$\mathbf{1}\mathbf{=}{\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}\mathbf{\Psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{·}\mathbf{x\Psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{dx}$

$\mathbf{=}{\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{0}}\mathbf{0}\mathbf{dx}\mathbf{+}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\mathbf{4}{\mathbf{a}}^{\mathbf{3}}{\mathbf{x}}^{\mathbf{2}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{ax}}\mathbf{dx}$

$\mathbf{=}\mathbf{4}{\mathbf{a}}^{\mathbf{3}}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}{\mathbf{x}}^{\mathbf{2}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{ax}}\mathbf{dx}$

$\mathbf{=}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{a}\mathbf{x}}(2a^2{x}^2+2ax+{\left.1\right|}_0^{\infty }$

Verify the normalization constant

We need to take the limit as $x$approaches $\infty$and the limit as $x$approaches 0.

$$\begin{gathered} \underset{\mathbf{x}\mathbf{\to }\mathbf{\infty }}{\mathbf{lim}}\left(-e^{-2\mathrm{ax}}\left(2a^2{x}^2+2\mathrm{ax}+1\right)\right)\mathbf{=}\left(0\underset{x\to \infty }{\lim }\left(2a^2{x}^2+2\mathrm{ax}+1\right)\right) \\ \mathbf{=}\mathbf{0} \end{gathered}$$

$$\begin{gathered} \underset{\mathbf{x}\mathbf{\to }\mathbf{\infty }}{\mathbf{lim}}\mathbf{(}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{ax}}\mathbf{(}\mathbf{2}{\mathbf{a}}^{\mathbf{2}}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{ax}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{)}\mathbf{=}\mathbf{\left(}\left(-1\right)\mathbf{(}\mathbf{2}{\mathbf{a}}^{\mathbf{2}}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{ax}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{\right)} \\ \mathbf{=}\mathbf{1} \end{gathered}$$

By evaluating the integral using the limits, we get:

${\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}\mathbf{\Psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{·}\mathbf{x\Psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{dx}\mathbf{=}\mathbf{0}\mathbf{-}\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}$

=1

Hence, the integral of the square of the wave function over all the space is properly normalized. The normalization constant is indeed equal to role="math" localid="1656109477459" $2\sqrt{{\mathrm{a}}^3}$.