Show that the quite general wave group given in equation (6-21) is a solution of the free-particle Schrödinger equation, provided that each plane wave's w does satisfy the matter-wave dispersion relation given in (6-23).

wave function of a quantum-mechanical system.

Schrödinger equation is given by,

_{$\mathbf{-}\frac{{\mathbf{h}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{\Psi }\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}}{\mathbf{\partial }{\mathbf{x}}^{\mathbf{2}}}\mathbf{=}\mathbf{ih}\frac{\mathbf{\partial }\mathbf{\Psi }\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}}{\mathbf{\partial }\mathbf{t}}$} ….. (1)

Where, h is Plank’s constant, m is the mass of the object, x is the position of the object, t is the time, and$\mathit{\Psi }$is the Wave function

Wave Group (6-21) is,

$\mathbf{\Psi }\left(x,t\right)\mathbf{=}{\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{+}\mathbf{\infty }}\mathbf{A}\left(k\right){\mathbf{e}}^{\mathbf{i}\left(\mathrm{kx}-\mathrm{\omega t}\right)}\mathbf{dk}$ ….. (2)

Where, k is the wave number, A is the amplitude of the wave, and $\mathbf{\omega }$is theangular frequency.

Matter Wave dispersion relation (6-23) is,

_{$\mathbf{\omega }\left(k\right)\mathbf{=}\frac{{\mathbf{hk}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}$} ….. (3)

$$\begin{gathered} -\frac{{\mathrm{h}}^2}{2\mathrm{m}}\frac{{\partial }^2}{\partial {\mathrm{x}}^2}{\int }_{-\infty }^{+\infty }\mathrm{A}\left(\mathrm{k}\right){\mathrm{e}}^{-\mathrm{i}\left(\mathrm{kx}-\mathrm{\omega t}\right)}\mathrm{dk}=\mathrm{ih}\frac{\partial }{\partial \mathrm{t}}{\int }_{-\infty }^{+\infty }\mathrm{A}\left(\mathrm{k}\right){\mathrm{e}}^{-\mathrm{i}\left(\mathrm{kx}-\mathrm{\omega t}\right)}\mathrm{dk} \\ -\frac{{\mathrm{h}}^2}{2\mathrm{m}}{\int }_{-\infty }^{+\infty }\left(-{\mathrm{k}}^2\right)\mathrm{A}\left(\mathrm{k}\right){\mathrm{e}}^{-\mathrm{i}\left(\mathrm{kx}-\mathrm{\omega t}\right)}\mathrm{dk}=\mathrm{ih}{\int }_{-\infty }^{+\infty }\left(-\mathrm{i\omega }\right)\mathrm{A}\left(\mathrm{k}\right){\mathrm{e}}^{-\mathrm{i}\left(\mathrm{kx}-\mathrm{\omega t}\right)}\mathrm{dk} \end{gathered}$$

Substituting $\frac{{\mathrm{hk}}^2}{2\mathrm{m}}$for $\mathrm{\omega }\left(\mathrm{k}\right)$ in above Equation, and you have

$$\begin{gathered} -\frac{{\mathrm{h}}^2}{2\mathrm{m}}{\int }_{-\infty }^{+\infty }\left(-{\mathrm{k}}^2\right)\mathrm{A}\left(\mathrm{k}\right){\mathrm{e}}^{-\mathrm{i}\left(\mathrm{kx}-\mathrm{\omega t}\right)}\mathrm{dk}=\mathrm{ih}{\int }_{-\infty }^{+\infty }\left(-\mathrm{i}\frac{{\mathrm{hk}}^2}{2\mathrm{m}}\right)\mathrm{A}\left(\mathrm{k}\right){\mathrm{e}}^{-\mathrm{i}\left(\mathrm{kx}-\mathrm{\omega t}\right)}\mathrm{dk} \\ -\frac{{\mathrm{h}}^2}{2\mathrm{m}}{\int }_{-\infty }^{+\infty }\left(-{\mathrm{k}}^2\right)\mathrm{A}\left(\mathrm{k}\right){\mathrm{e}}^{-\mathrm{i}\left(\mathrm{kx}-\mathrm{\omega t}\right)}\mathrm{dk}=-\frac{{\mathrm{h}}^2}{2\mathrm{m}}{\int }_{-\infty }^{+\infty }\left(-{\mathrm{k}}^2\right)\mathrm{A}\left(\mathrm{k}\right){\mathrm{e}}^{-\mathrm{i}\left(\mathrm{kx}-\mathrm{\omega t}\right)}\mathrm{dk} \end{gathered}$$

You can see in the above equation, LHS = RHS.

Hence, the given general wave equation (6-21) is a solution of the free-particle Schrödinger equation.