91影视

An equation that accounts for the electron's nature as a matter-wave inside of an atom describes the electron's energy and position in space and time.

$\frac{\mathbf{鈭�}{\mathbf{蠄}}^{\mathbf{*}}}{\mathbf{鈭�}\mathbf{t}}\mathbf{鈮�}\frac{\mathbf{-}\mathbf{i}\sout{\mathbf{h}}}{\mathbf{2}\mathbf{m}}\frac{{\mathbf{鈭�}}^{\mathbf{2}}{\mathbf{蠄}}^{\mathbf{*}}}{\mathbf{鈭�}{\mathbf{x}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{i}}{\sout{\mathbf{h}}}\frac{{\mathbf{鈭�}}^{\mathbf{2}}{\mathbf{蠄}}^{\mathbf{*}}}{\mathbf{鈭�}{\mathbf{x}}^{\mathbf{2}}}\mathit{V}{\mathbf{蠄}}^{\mathbf{*}}$ ....(1)

$\frac{\mathbf{鈭�}{\mathbf{蠄}}^{\mathbf{*}}}{\mathbf{鈭�}\mathbf{t}}\mathbf{鈮�}\frac{\mathbf{i}\sout{\mathbf{h}}}{\mathbf{2}\mathbf{m}}\frac{{\mathbf{鈭�}}^{\mathbf{2}}{\mathbf{蠄}}^{\mathbf{*}}}{\mathbf{鈭�}{\mathbf{x}}^{\mathbf{2}}}\mathbf{-}\frac{\mathbf{i}}{\sout{\mathbf{h}}}\mathit{V}{\mathbf{蠄}}^{\mathbf{*}}$ ....(2)

(a)

From Schrodinger equation

$\frac{鈭�\mathrm{蠄}}{鈭�\mathrm{t}}鈮�-\frac{h^2}{2\mathrm{m}}\frac{{鈭�}^2\mathrm{蠄}}{鈭�{\mathrm{x}}^2}+V^2\mathrm{蠄}$

Now,

$$\begin{gathered} P_{ab}={鈭�}_{-鈭�}^{鈭�}{\left|蠄\right|}^{2}dx \\ ={鈭�}_a^b{蠄}^{*}蠄dx \\ \frac{d{P}_{ab}}{dt}=\frac{d}{dt}{鈭�}_a^b{蠄}^{*}蠄dx \\ \frac{d{P}_{ab}}{dt}={鈭�}_a^b\frac{鈭�}{鈭�t}\left({蠄}^{*}蠄\right)dx \\ ={鈭�}_a^b\left(蠄\frac{鈭�{蠄}^{*}}{鈭�t}+{蠄}^{*}\frac{鈭�蠄}{鈭�t}\right)dx \end{gathered}$$

Substitute from Schrodinger equation and its conjugate,

$$\begin{gathered} \frac{d{P}_{ab}}{dt}={鈭�}_a^b\left\{蠄\left(\frac{-\mathrm{i}\sout{\mathrm{h}}}{2\mathrm{m}}\frac{{鈭�}^2{\mathrm{蠄}}^{*}}{鈭�{\mathrm{x}}^2}+\frac{\mathrm{i}}{\sout{\mathrm{h}}}{\mathrm{痴蠄}}^{*}\right)+{蠄}^{*}\left(\frac{\mathrm{i}\sout{\mathrm{h}}}{2\mathrm{m}}\frac{{鈭�}^2\mathrm{蠄}}{鈭�{\mathrm{x}}^2}+\frac{\mathrm{i}}{\sout{\mathrm{h}}}\mathrm{痴蠄}\right)\right\}dx \\ =-\frac{\mathrm{i}\sout{\mathrm{h}}}{2\mathrm{m}}{鈭�}_a^b\left(蠄\frac{{鈭�}^2{\mathrm{蠄}}^{*}}{鈭�{\mathrm{x}}^2}-{\mathrm{蠄}}^{*}\frac{{鈭�}^2\mathrm{蠄}}{鈭�{\mathrm{x}}^2}\right)dx \\ =-\frac{\mathrm{i}\sout{\mathrm{h}}}{2\mathrm{m}}{鈭�}_a^b\frac{鈭�}{鈭�x}\left(蠄\frac{鈭�{\mathrm{蠄}}^{*}}{鈭�\mathrm{x}}-{\mathrm{蠄}}^{*}\frac{鈭�\mathrm{蠄}}{鈭�\mathrm{x}}\right)dx \\ =-\frac{\mathrm{i}\sout{\mathrm{h}}}{2\mathrm{m}}{\left[蠄\frac{鈭�{\mathrm{蠄}}^{*}}{鈭�\mathrm{x}}-{\mathrm{蠄}}^{*}\frac{鈭�\mathrm{蠄}}{鈭�\mathrm{x}}\right]}_a^b \\ =-\frac{\mathrm{i}\sout{\mathrm{h}}}{2\mathrm{m}}\left[\left(蠄(b,t)\frac{鈭�{\mathrm{蠄}}^{*}(\mathrm{b},\mathrm{t})}{鈭�\mathrm{x}}-{\mathrm{蠄}}^{*}(\mathrm{b},\mathrm{t})\frac{鈭�\mathrm{蠄}(\mathrm{b},\mathrm{t})}{鈭�\mathrm{x}}\right)-\left(蠄(a,t)\frac{鈭�{\mathrm{蠄}}^{*}(a,\mathrm{t})}{鈭�\mathrm{x}}-{\mathrm{蠄}}^{*}(a,\mathrm{t})\frac{鈭�\mathrm{蠄}(a,\mathrm{t})}{鈭�\mathrm{x}}\right)\right] \\ =j(a,t)-j(b,t) \end{gathered}$$

Therefore, the unit of j ( x , t ) is j ( a , t ) -j ( b , t ) .

(b)

From problem 1.9 we have the wave-function

$蠄(x,t)=A{e}^{-a\left[\left(m{x}^2/\sout{h}\right)+it\right]}$

therefore,

$J(x,t)=\frac{i\sout{h}}{2m}{A}^2\left\{e^{-a\left[\left(m{x}^2/\sout{h}\right)+it\right]}\left(-2a\frac{mx}{\sout{h}}{e}^{-a\left[\left(m{x}^2/\sout{h}\right)+it\right]}\right)-e^{-a\left[\left(m{x}^2/\sout{h}\right)+it\right]}\left(-2a\frac{mx}{\sout{h}}{e}^{-a\left[\left(m{x}^2/\sout{h}\right)+it\right]}\right)\right\}$

Hence, the probability current of wave function is J ( x , t ) = 0