Residents of flatworld-a two-dimensional world far, far away-have it easy. Although quantum mechanics of course applies in their world, the equations they must solve to understand atomic energy levels involve only two dimensions. In particular, the Schrodinger equation for the one-electron flatrogen atom is

$\mathbf{-}\frac{\mathbf{h}}{\mathbf{2}\mathbf{m}}\frac{\mathbf{1}}{\mathbf{r}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{r}}\left(r\frac{\partial }{\partial r}\right)\mathit{\psi }\left(r,\theta \right)\mathbf{-}\frac{\mathbf{h}}{\mathbf{2}\mathbf{m}}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{{\mathbf{\partial }}^{\mathbf{2}}}{\mathbf{\partial }{\mathbf{\theta }}^{\mathbf{2}}}\mathit{\psi }\left(r,\theta \right)\mathbf{+}\mathit{U}\left(r\right)\mathit{\psi }\left(r,\theta \right)\mathbf{=}\mathit{E}\mathit{\psi }\left(r,\theta \right)$

(a) Separate variables by trying a solution of the form $\mathit{\psi }\left(r,\theta \right)\mathbf{=}\mathit{R}\left(r\right)\mathbf{⊝}\left(\theta \right)$, then dividing by$\mathit{R}\left(r\right)\mathbf{⊝}\left(\theta \right)$ . Show that the $\mathit{\theta }$equation can be written

$\frac{{\mathbf{d}}^{\mathbf{2}}}{\mathbf{d}{\mathbf{\theta }}^{\mathbf{2}}}\mathbf{⊝}\left(\theta \right)\mathbf{=}\mathit{C}\overline{)\mathbf{⊝}\left(\theta \right)}$

Here,$\left(C\right)$ is the separation constant.

(b) To be physically acceptable,$\mathbf{⊝}\left(\theta \right)$ must be continuous, which, since it involves rotation about an axis, means that it must be periodic. What must be the sign of C ?

(c) Show that a complex exponential is an acceptable solution for$\mathbf{⊝}\left(\theta \right)$ .

(d) Imposing the periodicity condition find allowed values of$\mathit{C}$ .

(e) What property is quantized according of C .

(f) Obtain the radial equation.

(g) Given that$\mathit{U}\left(r\right)\mathbf{=}\mathbf{-}\mathit{b}\mathbf{/}\mathit{r}$ , show that a function of the form$\mathit{R}\left(r\right)\mathbf{=}{\mathit{e}}^{\mathbf{r}\mathbf{/}\mathbf{a}}$ is a solution but only if C certain one of it, allowed values.

(h) Determine the value of a , and thus find the ground-state energy and wave function of flatrogen atom.

The Schrodinger equation for the one-electron flatrogen atom is

$\mathbf{-}\frac{\mathit{h}}{\mathbf{2}\mathbf{m}}\frac{\mathbf{1}}{\mathbf{r}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{r}}\mathbf{\left(}\mathbf{r}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{r}}\mathbf{\right)}\mathit{\psi }\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{\theta }\mathbf{)}\mathbf{-}\frac{\mathbf{h}}{\mathbf{2}\mathbf{m}}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{{\mathbf{\partial }}^{\mathbf{2}}}{\mathbf{\partial }{\mathbf{\theta }}^{\mathbf{2}}}\mathit{\psi }\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{\theta }\mathbf{)}\mathbf{+}\mathit{U}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathit{\psi }\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{\theta }\mathbf{)}\mathbf{=}\mathit{E}\mathit{\psi }\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{\theta }\mathbf{)}$

(a)

Separates the variables of the Schrodinger equation for the one-electron flatrogen atom.

$$\begin{gathered} -\frac{\sout{\mathrm{h}}}{2\mathrm{m}}\frac{1}{\mathrm{r}}\frac{\partial }{\partial \mathrm{r}}\left(\mathrm{r}\frac{\partial }{\partial \mathrm{r}}\right)\psi (\mathrm{r},\mathrm{\theta })-\frac{\sout{\mathrm{h}}}{2\mathrm{m}}\frac{1}{{\mathrm{r}}^2}\frac{{\partial }^2}{\partial {\mathrm{\theta }}^2}\psi (\mathrm{r},\mathrm{\theta })+U\left(\mathrm{r}\right)\psi (\mathrm{r},\mathrm{\theta })=E\psi (\mathrm{r},\mathrm{\theta }) \\ \frac{\sout{\mathrm{h}}}{2\mathrm{m}}\frac{1}{{\mathrm{r}}^2}\frac{{\partial }^2}{\partial {\mathrm{\theta }}^2}\psi (\mathrm{r},\mathrm{\theta })=\left[-\frac{\sout{\mathrm{h}}}{2\mathrm{m}}\frac{1}{\mathrm{r}}\frac{\partial }{\partial \mathrm{r}}\left(\mathrm{r}\frac{\partial }{\partial \mathrm{r}}\right)+U\left(r\right)-E\right]\psi (\mathrm{r},\mathrm{\theta }) \\ \frac{{\partial }^2}{\partial {\mathrm{\theta }}^2}\psi (\mathrm{r},\mathrm{\theta })=\frac{2m{r}^2}{\sout{\mathrm{h}}}\left[-\frac{\sout{\mathrm{h}}}{2\mathrm{m}}\frac{1}{\mathrm{r}}\frac{\partial }{\partial \mathrm{r}}\left(\mathrm{r}\frac{\partial }{\partial \mathrm{r}}\right)+U\left(r\right)-E\right]\psi (\mathrm{r},\mathrm{\theta }) \\ \frac{{\partial }^2}{\partial {\mathrm{\theta }}^2}\psi (\mathrm{r},\mathrm{\theta })=\left[-r\frac{\partial }{\partial \mathrm{r}}\left(r\frac{\partial }{\partial \mathrm{r}}\right)-\frac{2m{r}^2}{\sout{\mathrm{h}}}\left(E-U\left(r\right)\right)\right]\psi (\mathrm{r},\mathrm{\theta }) \end{gathered}$$

Substitute $R\left(r\right)⊝\left(\theta \right)$for $\psi \left(r,\theta \right)$into above equation.

$$\begin{gathered} \frac{{\partial }^2}{\partial {\theta }^2}R\left(r\right)⊝\left(\theta \right)=\left[-r\frac{\partial }{\partial r}\left(r\frac{\partial }{\partial r}\right)-\frac{2m{r}^2}{\sout{h}}\left(E-U\left(r\right)\right)\right]R\left(r\right)⊝\left(\theta \right) \\ \frac{{\partial }^2}{\partial {\theta }^2}R\left(r\right)⊝\left(\theta \right)=-r\frac{\partial }{\partial r}\left(r\frac{\partial }{\partial r}\right)R\left(r\right)⊝\left(\theta \right)-\frac{2m{r}^2}{\sout{h}}\left(E-U\left(r\right)\right)R\left(r\right)⊝\left(\theta \right) \\ R\left(r\right)\frac{{\partial }^2}{\partial {\theta }^2}⊝\left(\theta \right)=-r⊝\left(\theta \right)\frac{\partial }{\partial r}\left(r\frac{\partial }{\partial r}\right)R\left(r\right)-\frac{2m{r}^2}{\sout{h}}\left(E-U\left(r\right)\right)R\left(r\right)⊝\left(\theta \right) \end{gathered}$$

Divide the above equation by.

role="math" localid="1660042716100" $$\begin{gathered} \frac{R\left(r\right)}{R\left(r\right)⊝\left(\theta \right)}\frac{{\partial }^2}{\partial {\theta }^2}⊝\left(\theta \right)=-\frac{r⊝\left(\theta \right)}{R\left(r\right)⊝\left(\theta \right)}\frac{\partial }{\partial r}\left(r\frac{\partial }{\partial r}\right)R\left(r\right)-\frac{2m{r}^2}{\sout{h}}\left(E-U\left(r\right)\right)\frac{R\left(r\right)⊝\left(\theta \right)}{R\left(r\right)⊝\left(\theta \right)} \\ \frac{1}{⊝\left(\theta \right)}\frac{{\partial }^2}{\partial {\theta }^2}⊝\left(\theta \right)=-\frac{r}{R\left(r\right)}\frac{\partial }{\partial r}\left(r\frac{\partial }{\partial r}\right)R\left(r\right)-\frac{2m{r}^2}{\sout{h}}\left(E-U\left(r\right)\right)............\left(i\right) \end{gathered}$$

From the above equation, variables r and$\theta$ is separated on either side of the equation alone, leading deduction that both the sides must equal to constant, which is C ,

$$\begin{gathered} \frac{1}{⊝\left(\theta \right)}\frac{d^2}{d{\theta }^2}⊝\left(\theta \right)=C \\ \frac{d^2}{d{\theta }^2}⊝\left(\theta \right)=C⊝\left(\theta \right).....\left(ii\right) \end{gathered}$$ ……. (ii)

Hence the required equation$\frac{d^2}{d{\theta }^2}⊝\left(\theta \right)=C⊝\left(\theta \right)$ is proved.

(b)

The solution of equation (i) depends on the sign of the constant C . If the sign of the constant C is positive then the solution of an equation must be non-periodic and real exponential. But if the sign of the constant is negative, then the solution of the equation must be periodic of the complex exponential. Therefore, the sign of C should be negative.

Hence the sign of C is negative.

(c)

It is proved that the sign of the constant is negative in part (b).

Rewrite the equation (1).

$\frac{d^2}{d{\theta }^2}⊝\left(\theta \right)=-\left|C\right|⊝\left(\theta \right)$ …… (iii)

Let assume the angular solution has the complex exponential that therefore,$⊝\left(\theta \right)=e^{i\alpha \theta }$ for the real scalar$\alpha$ .

$$\begin{gathered} {\alpha }^2⊝\left(\theta \right)=-\left|C\right|⊝\left(\theta \right) \\ {\alpha }^2=-\left|C\right| \\ \alpha =\sqrt{\left|C\right|} \end{gathered}$$

Hence the complex exponential of the form$e^{i\sqrt{\left|C\right|}\theta }$ is acceptable solution for$⊝\left(\theta \right)$ .

(d)

The function$⊝\left(\theta \right)$ is said to be periodic when full rotation about the polar axis brings the same value of$⊝\left(\theta \right)$ .

Therefore, the periodic condition is$⊝\left(\theta \right)=⊝\left(\theta +2\mathrm{\pi }\right)$ .

$$\begin{gathered} e^{i\sqrt{\left|C\right|}\left(\theta +2\mathrm{\pi }\right)}=e^{i\sqrt{\left|C\right|}\theta } \\ 2\mathrm{\pi }\sqrt{\left|\mathrm{C}\right|}=2\mathrm{\pi m}\left(\mathrm{m}=0\pm 1,\pm 2,.....\right) \end{gathered}$$

From the above, the allowed values of C are$m=0\pm 1,\pm 2,....$ that is$C=0,-1,-4,....$ .

Hence, the allowed values of$C$ are $C=0,-1,-4,.....$.

(e)

The z component of the angular momentum is in the cylindrical coordinates is given by,

$L_z=-i\sout{h}\frac{\partial }{\partial \theta }$

Thus, the solution is given by,

$$\begin{gathered} L_z⊝\left(\theta \right)=-i\sout{h}\frac{\partial }{\partial \theta }{e}^{i\sqrt{\left|C\right|}\theta } \\ {L}_z⊝\left(\theta \right)=-\sout{h}\sqrt{\left|C\right|}⊝\left(\theta \right) \end{gathered}$$

The angular momentum$L_z=\sout{h}\sqrt{\left|C\right|}$, with$\sqrt{\left|C\right|}=0,\pm 1,\pm 2,......$ .

The above discussion tells us that property quantized according to the value C is component of angular momentum.

(f)

Determine the radial equation by substituting the left side of the equation (i) to C .

$$\begin{gathered} -\frac{r}{R\left(r\right)}\frac{\partial }{\partial r}\left(r\frac{\partial }{\partial r}\right)R\left(r\right)-\frac{2m{r}^2}{\sout{h}}\left(E-U\left(r\right)\right)=C \\ r\frac{\partial }{\partial r}\left(r\frac{\partial }{\partial r}\right)R\left(r\right)+\left(\frac{2m{r}^2}{\sout{h}}\left(E-U\left(r\right)\right)+C\right)R\left(r\right)=0........\left(iv\right) \end{gathered}$$

Hence the radial equation is$r\frac{\partial }{\partial r}\left(r\frac{\partial }{\partial r}\right)R\left(r\right)+\left(\frac{2m{r}^2}{\sout{h}}\left(E-U\left(r\right)\right)+C\right)R\left(r\right)=0$ .

(g)

Given that $U\left(r\right)=-b/r$

Recall the equation (iv),

$r\frac{\partial }{\partial r}\left(r\frac{\partial }{\partial r}\right)R\left(r\right)+\left(\frac{2m{r}^2}{\sout{h}}\left(E-U\left(r\right)\right)+C\right)R\left(r\right)\overline{)=0}$

Substitute $-b/r$for $U\left(r\right)$into above equation.

$$\begin{gathered} r\frac{\partial }{\partial r}\left(r\frac{\partial }{\partial r}\right)R\left(r\right)+\left(\frac{2m{r}^2}{\sout{h}}\left(E-\left(-\frac{b}{r}\right)\right)+C\right)R\left(r\right)=0 \\ r\frac{\partial }{\partial r}\left(r\frac{\partial }{\partial r}\right)R\left(r\right)+\left(\frac{2m{r}^2}{\sout{h}}\left(E+\frac{b}{r}\right)+C\right)R\left(r\right)=0.......\left(v\right) \end{gathered}$$

The term $\frac{\partial }{\partial r}\left(r\frac{\partial }{\partial r}\right)R\left(r\right)$can be calculated as,

$$\begin{gathered} \frac{\partial }{\partial r}\left(r\frac{\partial }{\partial r}\right)R\left(r\right)=\frac{\partial }{\partial r}\left(r\frac{\partial }{\partial r}{e}^{-r/a}\right) \\ \frac{\partial }{\partial r}\left(r\frac{\partial }{\partial r}\right)R\left(r\right)=-\frac{\partial }{\partial r}\left(\frac{r}{a}{e}^{-r/a}\right) \\ \frac{\partial }{\partial r}\left(r\frac{\partial }{\partial r}\right)R\left(r\right)=\left(-\frac{1}{a}+\frac{r}{a^2}\right)R\left(r\right) \end{gathered}$$

Substitute $\left(-\frac{1}{a}+\frac{r}{a^2}\right)R\left(r\right)$for $\frac{\partial }{\partial r}\left(r\frac{\partial }{\partial r}\right)R\left(r\right)$into equation (v).

$$\begin{gathered} r\left(-\frac{1}{a}+\frac{r}{a^2}\right)R\left(r\right)+\left(\frac{2m{r}^2}{\sout{h}}\left(E+\frac{b}{r}\right)+C\right)R\left(r\right)=0 \\ \left[\left(\frac{1}{a^2}+\frac{2mE}{\sout{h}}\right)r^2+\left(-\frac{1}{a}+\frac{2mr}{{\sout{h}}^2}\right)r+C\right]R\left(r\right)=0......\left(vi\right) \end{gathered}$$

From the above discussion, R(r) is the solution under some requirement and c = 0 is the one of them.

(h)

Since the C = 0 is the required condition for R(r) to be solution.

Substitute 0 for C into the equation (vi).

$$\begin{gathered} \left[\left(\frac{1}{a^2}+\frac{2mE}{{\sout{h}}^2}\right)r^2+\left(-\frac{1}{a}+\frac{2mb}{{\sout{h}}^2}\right)r+0\right]R\left(r\right)=0 \\ \left[\left(\frac{1}{a^2}+\frac{2mE}{{\sout{h}}^2}\right)r^2+\left(-\frac{1}{a}+\frac{2mb}{{\sout{h}}^2}\right)r\right]R\left(r\right)=0 \end{gathered}$$

The other requirement to satisfy the R(r) into above equation are $\frac{1}{a}=\frac{2mb}{{\sout{h}}^2}$and $-\frac{1}{a^2}=\frac{2mE}{{\sout{h}}^2}$.

Now the value of the is given by,

$$\begin{gathered} \frac{1}{a}=\frac{2mb}{{\sout{h}}^2} \\ a=\frac{{\sout{h}}^2}{2mb}......\left(vii\right) \end{gathered}$$

The allowed energy is given by,

localid="1660046666625" $$\begin{gathered} -\frac{1}{a^2}=\frac{2mE}{{\sout{h}}^2} \\ E=\frac{{\sout{h}}^2}{2m{a}^2} \end{gathered}$$

Substitute localid="1660046671256" $\frac{{\sout{h}}^2}{2mb}$for into above equation.

localid="1660046660072" $$\begin{gathered} E=\frac{{\sout{h}}^2}{2m{\left({\displaystyle \frac{{\sout{h}}^2}{2mb}}\right)}^2} \\ E=\frac{4m^2{b}^2\times {\sout{h}}^2}{2m\times {\sout{h}}^2} \\ E=-\frac{2m{b}^2}{{\sout{h}}^2}........\left(viii\right) \end{gathered}$$

From the equation (vii) and (viii), the required value of C is 0 . It follows the ground state wave function,

localid="1660046678252" $$\begin{gathered} {\psi }_0\left(r,\theta \right)=R_0{⊝}_0\left(\theta \right) \\ {\psi }_0\left(r,\theta \right)=A{e}^{-2mbr/{\sout{h}}^2} \end{gathered}$$

Here, A is the normalization constant.

Hence the value of a is localid="1660046691263" $\frac{{\sout{h}}^2}{2mb}$, the ground state energy islocalid="1660046684817" $-\frac{2mb}{{\sout{h}}^2}$and eave wave function of flatrogen atom is$A{e}^{-2mbr/{\sout{h}}^2}$.