91影视

$\mathrm{唯}(\mathrm{x},\mathrm{t})=\mathrm{Acos}(\mathrm{kx}-\mathrm{wt})$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.

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

$\mathbf{1}\mathbf{=}{\mathbf{鈭�}}_{\mathbf{-}\mathbf{鈭�}}^{\mathbf{鈭�}}\mathbf{唯}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{路}\mathbf{虫唯}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{dx}$

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

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

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

We need to take the limit as $x$approaches $鈭�$and the limit as $x$approaches 0.

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

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

By evaluating the integral using the limits, we get:

${\mathbf{鈭�}}_{\mathbf{-}\mathbf{鈭�}}^{\mathbf{鈭�}}\mathbf{唯}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{路}\mathbf{虫唯}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{dx}\mathbf{=}\mathbf{0}\mathbf{-}\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}$

=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" $2\sqrt{{\mathrm{a}}^3}$.