Hydrogen Atom

Illustration of the hydrogen atom.

We now discuss the solution of the Schrödinger equation for the hydrogen atom. We are especially interested in this system because it is of direct physical interest and because it is possible to solve it exactly.

The effective one-dimensional problem for this system is

$\left[ -\frac{\hbar^2}{2\mu}\frac{d^2}{dr^2}+\frac{\hbar^2}{2\mu}\frac{l(l+1)}{r^2}-\frac{Ze^2}{r}\right] u_{nl}=Eu_{nl},$

where $Z=1\!$ represents the hydrogen atom, $Z=2\!$ represents the helium ion, $He^+,\!$ and so on. Here, $\mu\!$ is the effective mass of the atom.

We will focus on the bound states here. To solve the above equation, let us begin by writing down its solutions in the limits of small and large $r\!.$

In the limit of small $r;\!$ i.e., $r \to 0,$ the equation becomes

$\left[ -\frac{\hbar^2}{2\mu}\frac{d^2}{dr^2}+\frac{\hbar^2}{2\mu}\frac{l(l+1)}{r^2}\right] u_{nl}=0.$

The only solution to this equation that is not divergent as $r\to 0$ is $u_{nl}(r)\sim r^{l+1}.\!$

In the opposite limit, $r \to \infty,$ we obtain

$\frac{d^2 u_{nl}}{dr^2}+\frac{2\mu E}{\hbar^2}u_{nl}=0.$

The only solution to this equation that does not diverge as $r\to\infty$ is $u_{nl}(r)\sim e^{-r/a},\!$ where $a=\sqrt{\frac{-\hbar^2}{2\mu E}}.$

Let us now define $\kappa=a^{-1}.\!$ The above asymptotic limit for large $r\!$ can now be written as

$u_{nl}(r)\sim e^{-\kappa r}.\!$

We now assume that the full expression for $u_{nl}(r)\!$ has the form, $u_{nl}(r)=(\kappa r)^{l+1}e^{-\kappa r}W(\kappa r).\!$

To simplify the equation, let us introduce the dimensionless radial coordinate, $\rho=\kappa r,\!$ so that $u_{nl}(r)\!$ may now be written as $u_{nl}(\rho)=\rho^{l+1}e^{-\rho}W(\rho).\!$

Substituting this into the effective Schrödinger equation and simplifying, it becomes

$\frac{d^2W}{d\rho^2}+2\left(\frac{l+1}{\rho}-1\right)\frac{dW}{d\rho}+\left(\frac{\rho_0}{\rho}-\frac{2(l+1)}{\rho}\right)W=0,$

where $\rho_{0}=Ze^{2}\sqrt{\frac{2\mu }{\hbar^{2}E}}.$

Let us now try a series solution for $W(\rho):\!$

$W(\rho)=a_0+a_1 \rho+a_2\rho^2+...=\sum_{k=0}^\infty a_k \rho^k$

Our equation now becomes

$\sum_{k=0}^\infty (a_{k}k(k-1)\rho^{k-2}+2(l+1)k\rho^{k-2}a_k-2\rho^{k-1}a_k k)+\sum_{k=0}^\infty(\rho_0 a_k\rho^{k-1}-2(l+1)a_k\rho^{k-1})=0,$

or, upon simplification,

$\sum_{k=0}^\infty (a_{k+1}(k+1)k\rho^{k-1}+2(l+1)(k+1)\rho^{k-1}a_{k+1}-2\rho^{k-1}a_k k+(\rho_0-2(l+1))a_k\rho^{k-1})=0.$

Setting the coefficients of all powers of $\rho\!$ to zero, we obtain

$k(k+1)a_{k+1}+2(l+1)(k+1)a_{k+1}-2ka_k+(\rho_0-2(l+1))a_k=0,\!$

or

$\frac{a_{k+1}}{a_k}=\frac{2(k+l+1)-\rho_0}{(k+1)(k+2l+2)}.$

In the limit of large $k,\!$ the recursion relation becomes

$\frac{a_{k+1}}{a_k}=\frac{2}{k},$

or

$a_{k+1}=\frac{2}{k}a_k=\frac{2^k}{k!}.$

These are the coefficients of the series for $e^{2\rho}.\!$ We therefore see that, unless we terminate the series at a finite order (i.e., we make $W(\rho)\!$ a polynomial), the wave function will diverge as $r\to\infty.$ If we wish to terminate the series at $N^{\text{th}}\!$ order, we set $a_{N+1}=0.\!$ This yields the condition,

$\rho_0=2(N+l+1).\!$

If we now solve this for the energy, we obtain the energy eigenvalues of the system,

$E_{Nl}=-\frac{Z^2\mu e^4}{2\hbar^2(N+l+1)^2}.$

The recursion relation now becomes

$\frac{a_{k+1}}{a_k}=\frac{2(k-N)}{(k+1)(k+2l+2)}.$

The function, $W(\rho),\!$ can be expressed in terms of the confluent hypergeometric function,

$_1F_1(a,c,z)=1+\frac{a}{c}\frac{z}{1!}+\frac{a(a+1)}{c(c+1)}\frac{z^2}{2!}+...=\sum_{k=0}^\infty a_k z^k,$

which is a solution of Kummer's equation,

$z\frac{d^2w}{dz^2} + (b-z)\frac{dw}{dz} - aw = 0.\,\!$

If we assume a power series solution as before, then the recursion relations are

$\frac{a_{k+1}}{a_k}=\frac{a+k}{(c+k)(k+1)}.$

Comparing this form to the recursion relations for our solution for $W(\rho),\!$ we see that

$c=2(l+1),\!$

$a=-N,\!$

and

$z=2\rho.\!$

Therefore, we may write $W(\rho)\!$ as

$W(\rho)=C _1F_1(-N,2(l+1),2\rho),\!$

where $\rho=\kappa r=\sqrt{\frac{-2\mu E}{\hbar^2}}r.$

The full normalized wave function is given by

$\psi_{n,l,m}(r,\theta,\phi)=\frac{e^{-\kappa r}(2\kappa r)^l}{(2l+1)!}\sqrt{\frac{(2\kappa)^3(n+l)!}{2n(n-l+1)}}\ _1F_1(-n+l+1,2(l+1),2\kappa r)Y_{l}^m(\theta,\phi),$

where we have introduced the more commonly-used principal quantum number $n=N+l+1.\!$

The first few normalized wave functions for the hydrogen atom are as follows

$\psi_{100}= \dfrac{e^{-r/a}}{\sqrt{\pi a^3}}$

$\psi_{200}= \dfrac{e^{-r/2a}}{\sqrt{32\pi a^3}} \left( 2-\dfrac{r}{a} \right)$

$\psi_{210}= \dfrac{e^{-r/2a}}{\sqrt{32\pi a^3}} \left( \dfrac{r}{a} \right) \cos(\theta)$

$\psi_{2 \pm 10}= \dfrac{e^{-r/2a}}{ \sqrt{64\pi a^3}} \left( \dfrac{r}{a} \right) \sin(\theta) e^{\pm i \phi}$

The energy may be written as

$E=-\frac{Z^2\mu e^4}{2\hbar^2 n^2}=-\frac{Z^2}{n^2}\,\text{Ry},$

where the Rydberg $\text{Ry}=13.6\,\text{eV}\!$ for the hydrogen atom and $n=1,2,3,...\!$ The degeneracy of each energy level is $n^2.\!$

To the side is a chart that depicts the energy levels for the hydrogen atom graphically for $n=1,2,3\!$ in units of $\text{Ry}\!$. The parenthesis indicates the degeneracy due to possibile values of the magnetic quantum number m from l to + l.

Problems

(1) (N. Zettili, Quantum Mechanics: Concepts and Applications, Exercise 6.3)

An electron in a hydrogen atom is in the energy eigenstate, $\psi_{2,1,-1} \left(r, \theta, \phi \right) = Nre^{-r/2a}Y_1^{-1}\left(\theta, \phi \right).$

(a) Find the normalization constant, $N.\!$

(b) What is the probability per unit volume of finding the electron at $r = a,\!$ $\theta = 45^{\circ},\!$ and $\phi = 60^{\circ}?\!$

(c) What is the probability per unit radial interval $dr\!$ of finding the electron at r = a?

(d) What are the expectation values of $\hat{\mathbf{L}}^2$ and $\hat{L}_{z}?$

(2) (Griffiths, Quantum Mechanics, Problems 4.13 and 4.14)

(a) Find $\langle r\rangle$ and $\langle r^2\rangle$ for an electron in the ground state of hydrogen. Express your answers in terms of the Bohr radius $a.\!$

(b) What is the most probable value of $r\!$ in the ground state of hydrogen?