Properties:

• consists of a single electron in a bound state around a single proton held together by Coulomb force
• We will start by assuming the proton is infinitely heavy (relative to the electron) and stationary at the origin

• Start with the TDSE

  $$ \begin{aligned} \left(-\frac{\hbar^2}{2m}\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\right)+V(x,y,z,t)\right)\Psi(x,y,z,t)=i\hbar\,\frac{\partial}{\partial t}\Psi(x,y,z,t)

  \end{aligned} $$

  Convert it to spherical polar coordinates

  $$ \begin{aligned} &\left(\frac{-\hbar^2}{2m}\left(\frac{\partial^2}{\partial r^2}+\frac2r\,\frac{\partial}{\partial r}+\frac1{r^2}\left[\frac{\partial^2}{\partial\theta^2}+\cot\theta\frac{\partial}{\partial\theta}+\frac1{\sin^2\theta}\,\frac{\partial^2}{\partial\phi^2}\right]\right)+V(r)\right) \hookleftarrow \\ &\times\psi(r,\theta,\phi)=E\,\psi(r,\theta,\phi)

  \end{aligned} $$

• We recognise that

  $$ \widehat{L}^2=-\hbar^2\left[\frac{\partial^2}{\partial\theta^2}+\cot\theta\frac{\partial}{\partial\theta}+\frac1{\sin^2\theta}\,\frac{\partial^2}{\partial\phi^2}\right] $$

  so that

  $$ \begin{aligned} \left(\frac{-\hbar^2}{2m}\underbrace{\left[\frac{\partial^2}{\partial r^2}+\frac2r\,\frac{\partial}{\partial r}\right]}_{\frac1{r^2}\frac{\partial}{\partial r}r^2\frac{\partial}{\partial r}}+\frac{\widehat{L}^2}{2mr^2}+V(r)\right)\psi(r,\theta,\phi)=E\,\psi(r,\theta,\phi) \end{aligned} $$

• The Coulomb potential between the positively charge nucleus (proton) and electron is

  $$ V(r)=\frac{-e^2}{4\pi \epsilon_0 r} $$

  its negative since it is an attractive force with energies lower than infinity thus we expect bound state.

• We will try separation of solutions

  $$ \psi(r,\theta,\phi) = R(r)\,Y^m_{\ell}(\theta,\phi) $$

  therefore

  $$ \left(\frac{-\hbar^2}{2m}\,\frac1{r^2}\,\frac{\partial}{\partial r}r^2\frac{\partial}{\partial r}+\frac{\ell(\ell+1)\hbar^2}{2mr^2}-\frac{e^2}{4\pi\varepsilon_0r}\right)R(r)\,Y^m_{\ell}(\theta,\phi)=E\,R(r)\,Y^m_{\ell}(\theta,\phi) $$

  $Y^m_l$ cancels. We will use the substitution $R(r)=U(r)/r$

  $$ \frac{-\hbar^2}{2m}\,\frac{\mathrm{d}^2U}{\mathrm{d} r^2}+\frac{\ell(\ell+1)\hbar^2}{2mr^2}U-\frac{e^2}{4\pi\varepsilon_0r}U=E\,U $$

  💎 Conclusion: we have just derived the radial Schrodinger equation for the hydrogen atom. It describes the allowed energy levels and the wavefunctions of the electron in the hydrogen atom

Dimensionless Variables

🧠 Remember: Bohr model of the atom had energy levels

$$ E_n=\frac{-E_R}{n^2} \qquad \text{with}\; E_R=\frac{m_ee^4}{2(4\pi\varepsilon_0)^2\hbar^2}=13.6\,\mathrm{eV} $$

with radii

$$ a_n = a_0n^2 \quad \text{with}\; a_0=\frac{4\pi\varepsilon_0\hbar^2}{m_ee^2}=0.53\,\r{\text{A}} $$

• Therefore we define dimensionless variables

  $$ \rho\equiv\frac{r}{a_0} \quad \widetilde{E}\equiv\frac{E}{E_R} $$

  which gives

  $$ -\frac{\mathrm{d}^2U}{\mathrm{d}\rho^2}+\frac{\ell(\ell+1)}{\rho^2}U-\frac2\rho U=\widetilde{E}\,U $$

  where:

  $U$ is the radial wave function

  $\rho$ is the dimensionless radial coordinate (depends on the mass and charge of the electron and nucleus)

  $\widetilde E$ is the dimensionless energy

Large $\rho$

• For large $\rho$ the $1/\rho^2$ and $1/\rho$ terms must both be tiny relative to the constant $\widetilde{E}$ so

  $$ -\frac{\mathrm{d}^2U}{\mathrm{d}\rho^2}\approx\widetilde{E}\,U $$

  Since we are seeking bound states $\widetilde{E}<0$ so

  $$ \widetilde{E}=-b^2 $$

• For some real constant $b$ we get

  $$ U(\rho)=\mathrm{e}^{\pm b\rho} $$

  But the wave function must vanish for large $\rho$ so it must be that for large $\rho$

  $$ U(\rho)\to\mathrm{e}^{-b\rho} $$

Small $\rho$

• For small $\rho$ the $\ell(\ell+1)/\rho^2$ term dominates (when $\ell\not=0$ )

  $$ -\frac{\mathrm{d}^2U}{\mathrm{d}\rho^2}+\frac{\ell(\ell+1)}{\rho^2}U\approx0 $$

  Since $U(0)$ has to be zero we try $U(\rho)\sim\rho^a$ for some constant $a$

  $$ \to -a(a-1)+\ell(\ell+1)=0 \quad \Rightarrow a=\ell+1 \;\text{or}\; a=-\ell $$

• However $U(0)=0$ only for $a>0$ so we get

  $$ U(\rho)\to\rho^{\ell+1} $$

  🗒️ Note: For $\ell=0$ we have to consider the $-2/\rho$ term in the same way but come to the same conclusion that $U\sim \rho^1$ in this case

General solution

• We can propose a general solution to be

  $$ U(\rho)=\rho^{\ell+1}\,\mathrm{e}^{-b\rho}\,f(\rho) \quad \text{with}\; \widetilde{E}=-b^2 $$

  where $f(\rho)$ is a smooth function with $f(0)=\text{const.}$

• Subbing in we get

  $$ \rho\frac{\mathrm{d}^2f}{\mathrm{d}\rho^2}+2(\ell+1-b\rho)\frac{\mathrm{d}f}{\mathrm{d}\rho}+2(1-b(\ell+1))f=0 $$

  The systematic solution shows that $f(\rho)$ mut be a polynomial, let us call its degree $p$

• For $p=0$ so when $f(\rho)= \text{cons.}$ we get

  $$ 1-b(\ell+1)=0 \quad\Rightarrow\quad b=\frac1{\ell+1} \quad\Rightarrow\quad \widetilde{E}=\frac{-1}{(\ell+1)^2}\, $$

• For $p=1$ so when $f(\rho)=c_0+c_1\rho$ we get

  $$ \begin{aligned} 2(\ell+1-b\rho)c_1+2(1-b(\ell+1))(c_0+c_1\rho)&=0 \\ \to \Bigl[(\ell+1)c_1+(1-b(\ell+1))c_0\Bigr]-\Bigl[b-(1-b(\ell+1))\Bigr]c_1\rho&=0

  \end{aligned} $$

  this is true for all $\rho$ so we get

  $$ \Rightarrow c_1=\frac{-c_0}{(\ell+1)(\ell+2)} \qquad\Rightarrow\quad b=\frac1{\ell+2} \quad\Rightarrow\quad \widetilde{E}=\frac{-1}{(\ell+2)^2} $$

• For $p=2$ so when $f(\rho)=c_0+c_1\rho+c_2\rho^2$ we obtain three equations which we solve to get

  $$ b=\frac1{\ell+3} \quad\Rightarrow\quad \widetilde{E}=\frac{-1}{(\ell+3)^2} $$

  and $c_1,c_2 \propto c_0$