Question

Show that$\frac{\mathbf{d}}{\mathbf{d}\mathbf{t}}{\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}{{\mathit{\Psi }}_{\mathbf{1}}}^{\mathbf{*}}{\mathit{\Psi }}_{\mathbf{2}}\mathit{d}\mathit{x}\mathbf{=}\mathbf{0}$

For any two solution to the Schrodinger equation${\mathbf{\Psi }}_{\mathbf{1}}$ and${\mathbf{\Psi }}_{\mathbf{2}}$ .

Short answer

The solutions for ${\mathrm{\Psi }}_1.$and ${\mathrm{\Psi }}_2$is $\frac{\mathrm{d}}{\mathrm{dt}}{\int }_{-\infty }^{\infty }{{\mathrm{\Psi }}_1}^{*}{\mathrm{\Psi }}_2\mathrm{dx}=0$

Step-by-step solution

Step 1: Define the Schrodinger equation

• A differential equation that describes matter in quantum mechanics in terms of the wave-like properties of particles in a field.
• Its answer is related to a particle's probability density in space and time.

Determine the solutions for Ψ1 and Ψ2.

$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{dt}}{\int }_{-\infty }^{\infty }{{\mathrm{\Psi }}_1}^{*}{\mathrm{\Psi }}_2\mathrm{dx}={\int }_{-\infty }^{\infty }\frac{\partial }{\partial t}\left({{\mathrm{\Psi }}_1}^{*}{\mathrm{\Psi }}_2\right)dx \\ ={\int }_{-\infty }^{\infty }\left({\mathrm{\Psi }}_2\frac{\partial {\mathrm{\Psi }}_1^{*}}{\partial \mathrm{t}}+{{\mathrm{\Psi }}^{*}}_1\frac{\partial {\mathrm{\Psi }}_2}{\partial \mathrm{t}}\right)dx \end{gathered}$$

But, from Schrodinger equation

$\frac{\partial {\mathrm{\Psi }}_2}{\partial \mathrm{t}}=\frac{i\sout{h}}{2m}\frac{{\partial }^2{\mathrm{\Psi }}_2}{\partial x^2}-\frac{i}{\sout{h}}V{\Psi }_2$

And

$\frac{\partial {\mathrm{\Psi }}_1^{*}}{\partial \mathrm{t}}=-\frac{i\sout{h}}{2m}\frac{{\partial }^2{\mathrm{\Psi }}_2}{\partial x^2}-\frac{i}{\sout{h}}V{\Psi }_1^{*}$

Thus,

localid="1658552483464" $$\begin{gathered} \frac{d}{dt}{\int }_{-\infty }^{\infty }{{\Psi }_{\mathit{1}}}^{\mathit{*}}{\Psi }_{\mathit{2}}dx={\mathit{\int }}_{\mathit{-}\mathit{\infty }}^{\mathit{\infty }}\mathit{(}{\mathit{\Psi }}_{\mathit{2}}\left[-\frac{i\sout{h}}{2m}\frac{{\partial }^2{\Psi }_1^{*}}{\partial x^2}+\frac{i}{\sout{h}}\right]\mathit{+}{\mathit{\Psi }}_{\mathit{1}}^{\mathit{*}}\mathit{[}\mathit{-}\frac{\mathit{i}\sout{\mathit{h}}}{\mathit{2}\mathit{m}}\frac{{\mathit{\partial }}^{\mathit{2}}{\mathit{\Psi }}_{\mathit{2}}}{\mathit{\partial }{\mathit{x}}^{\mathit{2}}}\mathit{+}\frac{\mathit{i}}{\sout{\mathit{h}}}\mathit{]}\mathit{V}{\mathit{\Psi }}_{\mathit{2}}\mathit{)}dx \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}=-\frac{i\sout{h}}{\mathit{2}m}{\int }_{-\infty }^{\infty }\left({\mathrm{\Psi }}_{\mathit{2}}\frac{{\mathit{\partial }}^{\mathit{2}}{\mathrm{\Psi }}_{\mathit{1}}^{\mathit{*}}}{\mathit{\partial }{\mathrm{x}}^{\mathit{2}}}-{\mathrm{\Psi }}_{\mathit{1}}^{\mathit{*}}\frac{{\mathit{\partial }}^{\mathit{2}}{\mathrm{\Psi }}_{\mathit{2}}}{\mathit{\partial }{\mathrm{x}}^{\mathit{2}}}\right)\mathrm{dx} \\ =-\frac{i\sout{h}}{\mathit{2}m}{\int }_{-\infty }^{\infty }\left({\mathrm{\Psi }}_{\mathit{2}}\frac{{\mathit{\partial }}^{\mathit{2}}{\mathrm{\Psi }}_{\mathit{1}}^{\mathit{*}}}{\mathit{\partial }{\mathrm{x}}^{\mathit{2}}}-{\mathrm{\Psi }}_{\mathit{1}}^{\mathit{*}}\frac{{\mathit{\partial }}^{\mathit{2}}{\mathrm{\Psi }}_{\mathit{2}}}{\mathit{\partial }{\mathrm{x}}^{\mathit{2}}}\right)\mathrm{dx} \\ =\mathit{-}\frac{i\sout{h}}{\mathit{2}m}{\int }_{-\infty }^{\infty }\frac{\partial }{\partial \mathrm{x}}\left({\mathrm{\Psi }}_{\mathit{2}}\frac{\mathit{\partial }{\mathrm{\Psi }}_{\mathit{1}}^{\mathit{*}}}{\mathit{\partial }\mathrm{x}}-{\mathrm{\Psi }}_{\mathit{1}}^{\mathit{*}}\frac{\mathit{\partial }{\mathrm{\Psi }}_{\mathit{2}}}{\mathit{\partial }\mathrm{x}}\right)\mathrm{dx} \\ \frac{\mathrm{d}}{\mathrm{dt}}{\int }_{-\infty }^{\infty }{{\mathrm{\Psi }}_{\mathit{1}}}^{\mathit{*}}{\mathrm{\Psi }}_{\mathit{2}}\mathrm{dx}=\mathit{-}\frac{i\sout{h}}{\mathit{2}m}{\overline{)\mathit{(}{\Psi }_{\mathit{2}}\frac{\mathit{\partial }{\Psi }_{\mathit{1}}^{\mathit{*}}}{\mathit{\partial }x}\mathit{-}{\Psi }_{\mathit{1}}^{\mathit{*}}\frac{\mathit{\partial }{\Psi }_{\mathit{2}}}{\mathit{\partial }x}\mathit{)}}}_{\mathit{\infty }}^{\mathit{\infty }} \end{gathered}$$

Where this must equal zero because ${\mathrm{\Psi }}_1.$and ${\mathrm{\Psi }}_2$are normalized.

Hence, the solutions for ${\mathrm{\Psi }}_1.$and ${\mathrm{\Psi }}_2$is$\frac{\mathrm{d}}{\mathrm{dt}}{\int }_{-\infty }^{\infty }{\mathrm{\Psi }}_1^{*}{\mathrm{\Psi }}_{\mathit{2}}dx=0$