Question

Question: Let ${\mathit{p}}_{\mathbf{a}\mathbf{b}}\mathbf{\left(}\mathit{t}\mathbf{\right)}$be the probability of finding a particle in the range $\mathbf{(}\mathit{a}\mathbf{<}\mathit{x}\mathbf{<}\mathit{b}\mathbf{)}$,at time t.

(a)Show that

$\frac{\mathbf{d}{\mathbf{p}}_{\mathbf{a}\mathbf{b}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{j}\mathbf{(}\mathit{a}\mathbf{.}\mathit{t}\mathbf{)}\mathbf{-}\mathit{j}\mathbf{(}\mathit{b}\mathbf{,}\mathit{t}\mathbf{)}\mathbf{,}$

Where

$\mathit{j}\mathbf{(}\mathit{x}\mathbf{,}\mathit{t}\mathbf{)}\mathbf{\equiv }\frac{\mathbf{i}\sout{\mathbf{h}}}{\mathbf{2}\mathbf{m}}\left(\psi \frac{\partial {\psi }^{*}}{\partial x}-{\psi }^{*}\frac{\partial \psi }{\partial x}\right)$

What are the units of j(x,t)?

Comment: j is called the probability current, because it tells you the rate at which probability is "flowing" past the point x. If${\mathit{p}}_{\mathbf{a}\mathbf{b}}\mathbf{\left(}\mathit{t}\mathbf{\right)}$ is increasing, then more probability is flowing into the region at one end than flows out at the other.

(b) Find the probability current for the wave function in Problem 1.9. (This is not a very pithy example, I'm afraid; we'll encounter more substantial ones in due course.)

Short answer

(a)The unit of $j(x,t)$is $j(a,t)-j(b,t)$.

(b)The probability current of wave function is$j(x,t)=0$

Step-by-step solution

Define the Schrödinger equation

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.

$\frac{\mathbf{\partial }{\mathbf{\psi }}^{\mathbf{*}}}{\mathbf{\partial }\mathbf{t}}\mathbf{\equiv }\frac{\mathbf{-}\mathbf{i}\sout{\mathbf{h}}}{\mathbf{2}\mathbf{m}}\frac{{\mathbf{\partial }}^{\mathbf{2}}{\mathbf{\psi }}^{\mathbf{*}}}{\mathbf{\partial }{\mathbf{x}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{i}}{\sout{\mathbf{h}}}\frac{{\mathbf{\partial }}^{\mathbf{2}}{\mathbf{\psi }}^{\mathbf{*}}}{\mathbf{\partial }{\mathbf{x}}^{\mathbf{2}}}\mathit{V}{\mathbf{\psi }}^{\mathbf{*}}$ ....(1)

$\frac{\mathbf{\partial }{\mathbf{\psi }}^{\mathbf{*}}}{\mathbf{\partial }\mathbf{t}}\mathbf{\equiv }\frac{\mathbf{i}\sout{\mathbf{h}}}{\mathbf{2}\mathbf{m}}\frac{{\mathbf{\partial }}^{\mathbf{2}}{\mathbf{\psi }}^{\mathbf{*}}}{\mathbf{\partial }{\mathbf{x}}^{\mathbf{2}}}\mathbf{-}\frac{\mathbf{i}}{\sout{\mathbf{h}}}\mathit{V}{\mathbf{\psi }}^{\mathbf{*}}$ ....(2)

Determine the units of  j ( x , t )

(a)

From Schrodinger equation

$\frac{\partial \mathrm{\psi }}{\partial \mathrm{t}}\equiv -\frac{h^2}{2\mathrm{m}}\frac{{\partial }^2\mathrm{\psi }}{\partial {\mathrm{x}}^2}+V^2\mathrm{\psi }$

Now,

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

Substitute from Schrodinger equation and its conjugate,

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

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

Determine the probability current of the wave function

(b)

From problem 1.9 we have the wave-function

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

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