[Attempt this problem only if you are familiar with vector calculus.] Define the (three-dimensional) probability current by generalization of Problem 1.14:

$\mathbf{J}\mathbf{=}\frac{\mathbf{i}\sout{\mathbf{h}}}{\mathbf{2}\mathbf{m}}\left(\psi \nabla {\psi }^{*}-{\psi }^{*}\nabla \psi \right)$

(a) Show that satisfies the continuity equation $\mathbf{\nabla }\mathbf{.}\mathbf{J}\mathbf{=}\mathbf{-}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{t}}{\left|\psi \right|}^{\mathbf{2}}$which expresses local conservation of probability. It follows (from the divergence theorem) that ${\mathbf{\int }}_{\mathbf{s}}\mathbf{J}\mathbf{.}\mathbf{da}\mathbf{=}\mathbf{-}\frac{\mathbf{d}}{\mathbf{dt}}{\mathbf{\int }}_{\mathbf{v}}{\left|\psi \right|}^{\mathbf{2}}{\mathbf{d}}^{\mathbf{3}}\mathbf{r}$where $\mathbf{V}$is a (fixed) volume and is its boundary surface. In words: The flow of probability out through the surface is equal to the decrease in probability of finding the particle in the volume.

(b) Find$\mathbf{J}$for hydrogen in the state$\mathbf{n}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{l}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{m}\mathbf{=}\mathbf{1}$ . Answer:

$\frac{\sout{\mathbf{h}}}{\mathbf{64}{\mathbf{ma}}^{\mathbf{5}}}{\mathbf{re}}^{\mathbf{-}\mathbf{r}\mathbf{/}\mathbf{a}}\mathbf{sin\theta }\hat{\mathbf{\varphi }}$

(c) If we interpret$\mathbf{mJ}$as the flow of mass, the angular momentum is

$\mathbf{L}\mathbf{=}\mathbf{m}\mathbf{\int }\left(r\times J\right){\mathbf{d}}^{\mathbf{3}}\mathbf{r}$

Use this to calculate ${\mathbf{L}}_{\mathbf{z}}$for the state${\mathbf{\psi }}_{\mathbf{211}}$, and comment on the result.

A mathematical quantity that describes the flow of probability is the probability current. If one imagines probability as a heterogeneous fluid, the probability current is the rate at which the fluid flows. It's a real-world vector that moves through space and time.

(a)

The probability current is,

$\mathrm{J}=\frac{\mathrm{i}\sout{\mathrm{h}}}{2\mathrm{m}}\left(\mathrm{\psi }\nabla {\mathrm{\psi }}^{*}-{\mathrm{\psi }}^{*}\nabla \mathrm{\psi }\right)$

Show that $J$satisfies the continuity equation

role="math" localid="1656046663375" $$\begin{gathered} \nabla .\mathrm{J}=\frac{\mathrm{i}\sout{\mathrm{h}}}{2\mathrm{m}}\left[\nabla \mathrm{\psi }.\nabla {\mathrm{\psi }}^{*}+\mathrm{\psi }\left({\nabla }^2{\mathrm{\psi }}^{*}\right)-\nabla {\mathrm{\psi }}^{*}.\nabla \mathrm{\psi }-{\mathrm{\psi }}^{*}\left({\nabla }^2\mathrm{\psi }\right)\right] \\ =\frac{\mathrm{i}\sout{\mathrm{h}}}{2\mathrm{m}}\left[\mathrm{\psi }\left({\nabla }^2{\mathrm{\psi }}^{*}\right)-{\mathrm{\psi }}^{*}\left({\nabla }^2\mathrm{\psi }\right)\right] \\ \mathrm{Consider}\mathrm{the}\mathrm{Schrodinger}’\mathrm{s}\mathrm{equation}, \\ \mathrm{i}\sout{\mathrm{h}}\frac{\partial \mathrm{\psi }}{\partial \mathrm{t}}=\frac{-\sout{\mathrm{h}}}{2\mathrm{m}}{\nabla }^2\mathrm{\psi }+\mathrm{V\psi } \\ \Rightarrow {\nabla }^2\mathrm{\psi }=\left(\mathrm{V\psi }-\mathrm{i}\sout{\mathrm{h}}\frac{\partial \mathrm{\psi }}{\partial \mathrm{t}}\right)\frac{2\mathrm{m}}{\sout{{\mathrm{h}}^2}} \\ \Rightarrow {\nabla }^2{\mathrm{\psi }}^{*}=\left({\mathrm{V\psi }}^{*}-\mathrm{i}\sout{\mathrm{h}}\frac{\partial {\mathrm{\psi }}^{*}}{\partial \mathrm{t}}\right)\frac{2\mathrm{m}}{\sout{{\mathrm{h}}^2}} \\ \Rightarrow \nabla .\mathrm{J}=\frac{\mathrm{i}\sout{\mathrm{h}}}{2\mathrm{m}}.\frac{2\mathrm{m}}{\sout{{\mathrm{h}}^2}}\left[\mathrm{\psi }\left({\mathrm{V\psi }}^{*}+\mathrm{i}\sout{\mathrm{h}}\frac{\partial {\mathrm{\psi }}^{*}}{\partial \mathrm{t}}\right)-{\mathrm{\psi }}^{*}\left(\mathrm{V\psi }-\mathrm{i}\sout{\mathrm{h}}\frac{\partial \mathrm{\psi }}{\partial \mathrm{t}}\right)\right] \\ =\left(\frac{\mathrm{i}}{\sout{\mathrm{h}}}\right)\mathrm{i}\sout{\mathrm{h}}\left(\mathrm{\psi }\frac{\partial {\mathrm{\psi }}^{*}}{\partial \mathrm{t}}+{\mathrm{\psi }}^{*}\frac{\partial \mathrm{\psi }}{\partial \mathrm{t}}\right) \\ =-\frac{\partial }{\partial \mathrm{t}}\left({\mathrm{\psi }}^{*}\mathrm{\psi }\right) \\ \Rightarrow \nabla .\mathrm{J}=-\frac{\partial }{\partial \mathrm{t}}{\left|\mathrm{\psi }\right|}^2 \end{gathered}$$

Hence, J satisfies the continuity equation.

(b)

At time $t$, the wave function for the hydrogen atom is $$\begin{gathered} {\mathrm{\psi }}_{\mathrm{mim}}={\mathrm{R}}_{\mathrm{nl}}\left(\mathrm{r}\right){\mathrm{Y}}_{\mathrm{l}}^{\mathrm{m}}\left(\mathrm{\theta },\mathrm{\varphi }\right){\mathrm{e}}^{\frac{{\mathrm{iE}}_{\mathrm{n}}\mathrm{t}}{\sout{\mathrm{h}}}} \\ \Rightarrow {\mathrm{\psi }}_{211}={\mathrm{R}}_{21}\left(\mathrm{r}\right){\mathrm{Y}}_1^1\left(\mathrm{\theta },\mathrm{\varphi }\right){\mathrm{e}}^{\frac{{\mathrm{iE}}_{\mathrm{n}}\mathrm{t}}{\sout{\mathrm{h}}}} \\ \mathrm{Now},\mathrm{apply}, \\ {\mathrm{R}}_{21}\left(\mathrm{r}\right)=\frac{1}{\sqrt{24}}{\mathrm{a}}^{\frac{3}{2}}\frac{\mathrm{r}}{\mathrm{a}}\exp \left(\frac{-\mathrm{r}}{2\mathrm{a}}\right) \\ \mathrm{Also}, \\ {\mathrm{Y}}_1^1\left(\mathrm{\theta },\mathrm{\varphi }\right)=-\sqrt{\frac{3}{8\mathrm{\pi }}}{\mathrm{sin\theta e}}^{\mathrm{i\varphi }} \\ \Rightarrow {\mathrm{\psi }}_{211}=-\frac{1}{\sqrt{\mathrm{\pi a}}}.\frac{1}{8{\mathrm{a}}^2}{\mathrm{re}}^{\frac{-\mathrm{r}}{2\mathrm{a}}}{\mathrm{sin\theta e}}^{\mathrm{i\varphi }}{\mathrm{e}}^{\frac{-{\mathrm{iE}}_2\mathrm{t}}{\sout{\mathrm{h}}}} \\ \mathrm{For}\mathrm{spherical}\mathrm{coordinates}, \\ \nabla \mathrm{\psi }=\frac{\partial \mathrm{\psi }}{\partial \mathrm{r}}\hat{\mathrm{r}}+\frac{1}{\mathrm{r}}\frac{\partial \mathrm{\psi }}{\partial \mathrm{\theta }}\hat{\mathrm{\theta }}+\frac{1}{\mathrm{r}\mathrm{sin\theta }}.\frac{\partial \mathrm{\psi }}{\partial \mathrm{\varphi }}\hat{\mathrm{\varphi }} \\ \nabla {\mathrm{\psi }}_{211}=\frac{-1}{\sqrt{\mathrm{\pi a}}}\frac{1}{8{\mathrm{a}}^2}\left[\begin{array}{c}\left(1-\frac{\mathrm{r}}{2\mathrm{a}}\right){\mathrm{e}}^{\frac{-\mathrm{r}}{2\mathrm{a}}}{\mathrm{sin\theta e}}^{\mathrm{i\varphi }}{\mathrm{e}}^{\frac{-{\mathrm{iE}}_2\mathrm{t}}{\sout{\mathrm{h}}}}\hat{\mathrm{r}}+\frac{1}{\mathrm{r}}{\mathrm{re}}^{\frac{-\mathrm{r}}{2\mathrm{a}}}{\mathrm{cos\theta e}}^{\mathrm{i\varphi }}{\mathrm{e}}^{\frac{-{\mathrm{iE}}_2\mathrm{t}}{\sout{\mathrm{h}}}\hat{\mathrm{\theta }}}\\ +\frac{1}{\mathrm{r}\mathrm{sin\theta }}{\mathrm{re}}^{\frac{-\mathrm{r}}{2\mathrm{a}}}{\mathrm{sin\theta ie}}^{\mathrm{i\varphi }}.{\mathrm{e}}^{\frac{-{\mathrm{iE}}_2\mathrm{t}}{\mathrm{h}}}\mathrm{\varphi }\end{array}\right] \\ =\left[\left(1-\frac{\mathrm{r}}{2\mathrm{a}}\right)\hat{\mathrm{r}}+\mathrm{cos\theta }\hat{\mathrm{\theta }}+\frac{\mathrm{i}}{\mathrm{sin\theta }}\mathrm{\varphi }\right]\frac{1}{\mathrm{r}}{\mathrm{\psi }}_{211} \\ \mathrm{Replace}\mathrm{i}\mathrm{by}-\mathrm{i}\mathrm{to}\mathrm{get}\nabla {{\mathrm{\psi }}^{*}}_{211}, \\ \Rightarrow \nabla {\mathrm{\psi }}_{211}^{*}=\left[\left(1-\frac{\mathrm{r}}{2\mathrm{a}}\right)\hat{\mathrm{r}}+\mathrm{cos\theta }\hat{\mathrm{\theta }}-\frac{\mathrm{i}}{\mathrm{sin\theta }}\mathrm{\varphi }\right]\frac{1}{\mathrm{r}}{\mathrm{\psi }}_{211}^{*} \\ \mathrm{Substitute}\mathrm{the}\mathrm{two}\mathrm{expressions}\mathrm{into}\mathrm{the}\mathrm{equation}\mathrm{for}\mathrm{J}, \end{gathered}$$

$$\begin{gathered} J=\frac{\mathrm{i}\sout{\mathrm{h}}}{2\mathrm{m}}\left\{{\mathrm{\psi }}_{211}\left[\left(1-\frac{\mathrm{r}}{2\mathrm{a}}\right)\hat{\mathrm{r}}+\mathrm{cos\theta }\hat{\mathrm{\theta }}-\frac{\mathrm{i}}{\mathrm{sin\theta }}\hat{\mathrm{\varphi }}\right]\frac{1}{\mathrm{r}}{\mathrm{\psi }}_{211}^{*}\right]-{\mathrm{\psi }}_{211}^{*}\left[\left(1-\frac{\mathrm{r}}{2\mathrm{a}}\right)\hat{\mathrm{r}}+\mathrm{cos\theta }\hat{\mathrm{\theta }}\right]\frac{1}{\mathrm{r}}{\mathrm{\psi }}_{211} \\ =\frac{\mathrm{i}\sout{\mathrm{h}}}{2\mathrm{m}}\left[\left(1-\frac{\mathrm{r}}{2\mathrm{a}}\right)\hat{\mathrm{r}}+\mathrm{cos\theta }\hat{\mathrm{\theta }}-\frac{\mathrm{i}}{\mathrm{sin\theta }}\mathrm{\varphi }-\left(1-\frac{\mathrm{r}}{2\mathrm{a}}\right)\hat{\mathrm{r}}-\mathrm{cos\theta }\hat{\mathrm{\theta }}\frac{1}{\mathrm{sin\theta }}\hat{\mathrm{\varphi }}\right]\frac{1}{\mathrm{r}}{\left|{\mathrm{\psi }}_{211}\right|}^2 \\ =\frac{\mathrm{i}\sout{\mathrm{h}}}{2\mathrm{m}}\frac{\left(-2\mathrm{i}\right)}{\mathrm{r}\mathrm{sin\theta }}{\left|{\mathrm{\psi }}_{211}\right|}^2\hat{\mathrm{\varphi }} \\ =\frac{\sout{\mathrm{h}}}{\mathrm{m}}\frac{1}{\mathrm{\pi a}}\frac{1}{64{\mathrm{a}}^4}\frac{{\mathrm{r}}^2{\mathrm{e}}^{{\displaystyle \frac{-\mathrm{r}}{\mathrm{a}}}}{\sin }^2\mathrm{\theta }}{\mathrm{r}\sin \mathrm{\theta }}\mathrm{\varphi } \\ =\frac{\sout{\mathrm{h}}}{64{\mathrm{\pi ma}}^5}{\mathrm{re}}^{\frac{-\mathrm{r}}{\mathrm{a}}}\mathrm{sin\theta }\hat{\mathrm{\varphi }} \\ \mathrm{Therefore},\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{J}\mathrm{is}\frac{\sout{\mathrm{h}}}{64{\mathrm{\pi ma}}^5}{\mathrm{re}}^{-\frac{\mathrm{r}}{\mathrm{a}}}\mathrm{sin\theta }\widehat{\mathrm{\varphi }.} \end{gathered}$$

(c) The value of J from part (b),

$$\begin{gathered} \overline{\mathrm{J}}=\frac{\sout{\mathrm{h}}}{64{\mathrm{\pi ma}}^5}{\mathrm{re}}^{\frac{-\mathrm{r}}{\mathrm{a}}}\mathrm{sin\theta }\hat{\mathrm{\varphi }} \\ \mathrm{Now}, \\ \overline{\mathrm{r}}\times \overline{\mathrm{J}}=\frac{\sout{\mathrm{h}}}{64{\mathrm{\pi ma}}^5}{\mathrm{re}}^{\frac{-\mathrm{r}}{\mathrm{a}}}\mathrm{sin\theta }\left(\hat{\mathrm{r}}\times \hat{\mathrm{\varphi }}\right) \\ \mathrm{Consider},\hat{\mathrm{r}}\times \hat{\mathrm{\varphi }}=-\hat{\mathrm{\theta }} \\ \mathrm{And}\hat{\mathrm{z}}.\hat{\mathrm{\theta }}=-\sin \mathrm{\theta } \\ \mathrm{So},{\left(\overline{\mathrm{r}}\times \overline{\mathrm{J}}\right)}_{\mathrm{z}}=\frac{\sout{\mathrm{h}}}{64{\mathrm{\pi ma}}^5}{\mathrm{r}}^2{\mathrm{e}}^{\frac{-\mathrm{r}}{\mathrm{a}}}{\sin }^2\mathrm{\theta } \\ \mathrm{As}, \\ {\mathrm{L}}_{\mathrm{z}}=\mathrm{m}\int {\left(\overline{\mathrm{r}}\times \overline{\mathrm{J}}\right)}_{\mathrm{z}}{\mathrm{d}}^3\mathrm{r} \\ {\mathrm{L}}_{\mathrm{z}}=\mathrm{m}\frac{\sout{\mathrm{h}}}{64{\mathrm{\pi ma}}^5}\int {\mathrm{r}}^2{\mathrm{e}}^{\frac{-\mathrm{r}}{\mathrm{a}}}{\sin }^2{\mathrm{\theta r}}^2\mathrm{sin\theta drd\theta d\varphi } \\ =\frac{\sout{\mathrm{h}}}{64{\mathrm{\pi ma}}^5}{\int }_0^{\mathrm{a}}{\mathrm{r}}^4{\mathrm{e}}^{\frac{-\mathrm{r}}{\mathrm{a}}}\mathrm{dr}{\int }_0^{\mathrm{\pi }}{\sin }^3\mathrm{\theta d\theta }{\int }_0^{2\mathrm{\pi }}\mathrm{d\varphi } \\ =\frac{\sout{\mathrm{h}}}{64{\mathrm{\pi ma}}^5}\left(4!{\mathrm{a}}^5\right)\left(\frac{4}{3}\right)\left(2\mathrm{\pi }\right) \\ \mathrm{Hence},\mathrm{the}\mathrm{value}{\mathrm{L}}_{\mathrm{z}}\mathrm{is}\frac{\sout{\mathrm{h}}}{64{\mathrm{\pi a}}^5}\left(4!{\mathrm{a}}^5\right)\left(\frac{4}{3}\right)\left(2\mathrm{\pi }\right). \end{gathered}$$