5.1 The Hydrogen Hamiltonian

We will now look at the major early success story of quantum mechanics, solving for the motion of an electron in a hydrogen atom. Our assumptions are that the nucleus (the proton) is a particle with mass $m_{p}$ and charge $+e$ , and the electron is a particle with mass $m_{e}$ and charge $-e$ . Reality, or rather our latest understanding of the hydrogen atom, is more complex and there are things that we are neglecting here such as coupling between spin and angular momentum, special relativity, and the quantum vacuum.

5.1.1 Setting up the Problem

The electron is bound to the proton by the Coulomb force, which is

\boldsymbol{F}=-\frac{e^{2}}{4\pi\varepsilon_{0}r^{2}}\hat{\boldsymbol{r}},

(5.1)

where $\varepsilon_{0}$ is the permittivity of free space and $r$ is the distance between the two particles.

To find the potential energy, we integrate from $\infty$ to $r$ (we are taking the potential energy to be zero at infinity) to calculate the work done to bring the electron from infinity to $r$ , so we get

V(r)=-\int_{\infty}^{r}F(r^{\prime})\operatorname{\mathrm{d}}\!r^{\prime}=\int% _{\infty}^{r}\frac{e^{2}}{4\pi\varepsilon_{0}r^{\prime 2}}\operatorname{% \mathrm{d}}\!r^{\prime}=-\frac{e^{2}}{4\pi\varepsilon_{0}r}.

(5.2)

The Coulomb force does positive work to bring the electron closer to the proton, so the potential energy is negative for any finite separation. This means that the energy eigenvalues will also be negative.

Taking the reduced mass of the electron as

\mu=\frac{m_{p}m_{e}}{m_{p}+m_{e}},

(5.3)

the kinetic energy operator is

\hat{T}=\frac{\hat{p}^{2}}{2\mu}=-\frac{\hbar^{2}}{2\mu}\nabla^{2}.

(5.4)

Combining these, the time-independent Schrodinger equation for the hydrogen atom electron can be written as

\left(-\frac{\hbar^{2}}{2\mu}\nabla^{2}-\frac{e^{2}}{4\pi\varepsilon_{0}r}% \right)\psi=E\psi.

(5.5)

This potential has spherical symmetry, so we will solve the Schrodinger equation in spherical coordinates. Expanding the Laplacian, the full equation is

-\frac{\hbar^{2}}{2\mu}\left[\frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r% ^{2}\frac{\partial}{\partial r}\right)+\frac{1}{r^{2}\sin\theta}\frac{\partial% }{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{% 1}{r^{2}\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}\right]\psi-\frac{% e^{2}}{4\pi\varepsilon_{0}r}\psi=E\psi

(5.6)

5.1.2 Separating and Solving

Equation 5.6 is separable, so we look for solutions of the form

\psi(r,\theta,\phi)=R(r)\Theta(\theta)\Phi(\phi).

(5.7)

This gives us the derivatives of the wavefunction:

\frac{\partial\psi}{\partial r}=\Theta\Phi\frac{\operatorname{\mathrm{d}}\!R}{% \operatorname{\mathrm{d}}\!r},\quad\frac{\partial\psi}{\partial\theta}=R\Phi% \frac{\operatorname{\mathrm{d}}\!\Theta}{\operatorname{\mathrm{d}}\!\theta},% \quad\frac{\partial\psi}{\partial\phi}=R\Theta\frac{\operatorname{\mathrm{d}}% \!\Phi}{\operatorname{\mathrm{d}}\!\phi}.

(5.8)

Now, if we substitute these into equation 5.6 and rearrange so that we have all terms containing $r$ on one side and all terms containing $\theta$ and $\phi$ on the other, we get

\frac{1}{\Theta\sin\theta}\frac{\operatorname{\mathrm{d}}\!}{\operatorname{% \mathrm{d}}\!\theta}\left(\sin\theta\frac{\operatorname{\mathrm{d}}\!\Theta}{% \operatorname{\mathrm{d}}\!\theta}\right)+\frac{1}{\Phi\sin^{2}\theta}\frac{% \operatorname{\mathrm{d}}\!^{2}\Phi}{\operatorname{\mathrm{d}}\!\phi^{2}}=-% \frac{1}{R}\frac{\operatorname{\mathrm{d}}\!}{\operatorname{\mathrm{d}}\!r}% \left(r^{2}\frac{\operatorname{\mathrm{d}}\!R}{\operatorname{\mathrm{d}}\!r}% \right)-\frac{2\mu r^{2}}{\hbar^{2}}\left(\frac{e^{2}}{4\pi\varepsilon_{0}r}+E% \right).

(5.9)

Using the method of separation of variables, both sides must be equal to the same constant, which we will set to be $-\lambda^{2}$ . Thus we have the radial equation (rearranging slightly and cancelling off the minus sign from all terms):

\frac{\operatorname{\mathrm{d}}\!}{\operatorname{\mathrm{d}}\!r}\left(r^{2}% \frac{\operatorname{\mathrm{d}}\!R(r)}{\operatorname{\mathrm{d}}\!r}\right)+% \frac{2\mu r^{2}}{\hbar^{2}}\left(\frac{e^{2}}{4\pi\varepsilon_{0}r}+E\right)R% (r)=\lambda^{2}R(r),

(5.10)

and the angular equation:

\frac{1}{\Theta(\theta)\sin\theta}\frac{\operatorname{\mathrm{d}}\!}{% \operatorname{\mathrm{d}}\!\theta}\left(\sin\theta\frac{\operatorname{\mathrm{% d}}\!\Theta(\theta)}{\operatorname{\mathrm{d}}\!\theta}\right)+\frac{1}{\Phi(% \phi)\sin^{2}\theta}\frac{\operatorname{\mathrm{d}}\!^{2}\Phi(\phi)}{% \operatorname{\mathrm{d}}\!\phi^{2}}=-\lambda^{2}.

(5.11)

Notice that equation 5.11 is completely independent of the potential energy, and it is also separable. Moving all terms with $\theta$ onto one side and all terms with $\phi$ onto the other, we get

-\lambda^{2}\sin^{2}\theta-\frac{\sin\theta}{\Theta(\theta)}\frac{% \operatorname{\mathrm{d}}\!}{\operatorname{\mathrm{d}}\!\theta}\left(\sin% \theta\frac{\operatorname{\mathrm{d}}\!\Theta(\theta)}{\operatorname{\mathrm{d% }}\!\theta}\right)=\frac{1}{\Phi(\phi)}\frac{\operatorname{\mathrm{d}}\!^{2}% \Phi(\phi)}{\operatorname{\mathrm{d}}\!\phi^{2}}.

(5.12)

Now setting both sides equal to a separation constant $-m^{2}$ , we get the polar equation:

\sin\theta\frac{\operatorname{\mathrm{d}}\!}{\operatorname{\mathrm{d}}\!\theta% }\left(\sin\theta\frac{\operatorname{\mathrm{d}}\!\Theta(\theta)}{% \operatorname{\mathrm{d}}\!\theta}\right)=(m^{2}-\lambda^{2}\sin^{2}\theta)% \Theta(\theta),

(5.13)

and the azimuthal equation:

\frac{\operatorname{\mathrm{d}}\!^{2}\Phi(\phi)}{\operatorname{\mathrm{d}}\!% \phi^{2}}=-m^{2}\Phi(\phi).

(5.14)