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# 8.3: Hydrogen Atom

A hydrogen atom consists of an electron, of charge $$-e$$ and mass $$m_e$$, and a proton, of charge $$+e$$ and mass $$m_p$$, moving in the Coulomb potential $V({\bf r}) = - \frac{e^{\,2}}{4\pi\,\epsilon_0\,|{\bf r}|},$ where $${\bf r}$$ is the position vector of the electron with respect to the proton. Now, according to the analysis in Section [stwo], this two-body problem can be converted into an equivalent one-body problem. In the latter problem, a particle of mass $\mu = \frac{m_e\,m_p}{m_e+m_p}$ moves in the central potential $V(r) = - \frac{e^{\,2}}{4\pi\,\epsilon_0\,r}.$ Note, however, that because $$m_e/m_p\simeq 1/1836$$ the difference between $$m_e$$ and $$\mu$$ is very small. Hence, in the following, we shall write neglect this difference entirely.

Writing the wavefunction in the usual form, $\psi(r,\theta,\phi) = R_{n,l}(r)\,Y_{l,m}(\theta,\phi),$ it follows from Section 1.2 that the radial function $$R_{n,l}(r)$$ satisfies $-\frac{\hbar^{\,2}}{2\,m_e}\left[\frac{d^{\,2}}{dr^{\,2}} + \frac{2}{r}\frac{d}{dr} -\frac{l\,(l+1)}{r^{\,2}}\right] R_{n,l} -\left(\frac{e^{\,2}}{4\pi\,\epsilon_0\,r}+E \right) R_{n,l}= 0.$ Let $$r = a\,z$$, with

$\label{e9.45} a = \sqrt{\frac{\hbar^{\,2}}{2\,m_e\,(-E)}}=\sqrt{\frac{E_0}{E}}\,a_0,$ where $$E_0$$ and $$a_0$$ are defined in Equations ([e9.56]) and ([e9.57]), respectively. Here, it is assumed that $$E<0$$, because we are only interested in bound-states of the hydrogen atom. The previous differential equation transforms to $\left[\frac{d^{\,2}}{dz^{\,2}} + \frac{2}{z}\frac{d}{dz}-\frac{l\,(l+1)}{z^{\,2}}+ \frac{\zeta}{z}-1\right] R_{n,l} = 0,$ where

$\label{e9.47} \zeta = \frac{2\,m_e\,a\,e^{\,2}}{4\pi\,\epsilon_0\,\hbar^{\,2}}=2\sqrt{\frac{E_0}{E}}.$ Suppose that $$R_{n,l}(r) = Z(r/a)\,\exp(-r/a)/(r/a)$$. It follows that

$\label{e9.48} \left[\frac{d^{\,2}}{dz^{\,2}} -2\,\frac{d}{dz} - \frac{l\,(l+1)}{z^{\,2}} + \frac{\zeta}{z}\right] Z = 0.$ We now need to solve the previous differential equation in the domain $$z=0$$ to $$z= \infty$$, subject to the constraint that $$R_{n,l}(r)$$ be square-integrable.

Let us look for a power-law solution of the form

$\label{e9.49} Z(z) = \sum_k c_k\,z^{\,k}.$ Substituting this solution into Equation ([e9.48]), we obtain $\sum_k c_k\left\{k\,(k-1)\,z^{\,k-2} - 2\,k\,z^{\,k-1} - l\,(l+1)\,z^{\,k-2} + \zeta\,z^{\,k-1}\right\} = 0.$ Equating the coefficients of $$z^{\,k-2}$$ gives the recursion relation

$\label{e9.51} c_k\,\left[k\,(k-1)-l\,(l+1)\right] = c_{k-1}\,\left[2\,(k-1) - \zeta\right].$ Now, the power series ([e9.49]) must terminate at small $$k$$, at some positive value of $$k$$, otherwise $$Z(z)$$ behaves unphysically as $$z\rightarrow 0$$ [i.e., it yields an $$R_{n,l}(r)$$ that is not square integrable as $$r\rightarrow 0$$]. From the previous recursion relation, this is only possible if $$[k_{\rm min}\,(k_{\rm min}-1)-l\,(l+1)]=0$$, where the first term in the series is $$c_{k_{\rm min}}\,z^{\,k_{\rm min}}$$. There are two possibilities: $$k_{\rm min}=-l$$ or $$k_{\rm min}=l+1$$. However, the former possibility predicts unphysical behavior of $$Z(z)$$ at $$z=0$$. Thus, we conclude that $$k_{\rm min}=l+1$$. Note that, because $$R_{n,l}(r)\simeq Z(r/a)/(r/a)\simeq (r/a)^{\,l}$$ at small $$r$$, there is a finite probability of finding the electron at the nucleus for an $$l=0$$ state, whereas there is zero probability of finding the electron at the nucleus for an $$l>0$$ state [i.e., $$|\psi|^{\,2}=0$$ at $$r=0$$, except when $$l=0$$].

For large values of $$z$$, the ratio of successive coefficients in the power series ([e9.49]) is $\frac{c_k}{c_{k-1}} = \frac{2}{k},$ according to Equation ([e9.51]). This is the same as the ratio of successive coefficients in the power series $\sum_k \frac{(2\,z)^{\,k}}{k!},$ which converges to $$\exp(2\,z)$$. We conclude that $$Z(z)\rightarrow \exp(2\,z)$$ as $$z\rightarrow\infty$$. It thus follows that $$R_{n,l}(r)\sim Z(r/a)\,\exp(-r/a)/(r/a)\rightarrow \exp(r/a)/(r/a)$$ as $$r\rightarrow\infty$$. This does not correspond to physically acceptable behavior of the wavefunction, because $$\int|\psi|^2\,dV$$ must be finite. The only way in which we can avoid this unphysical behavior is if the power series ([e9.49]) terminates at some maximum value of $$k$$. According to the recursion relation ([e9.51]), this is only possible if

$\label{e9.54} \frac{\zeta}{2} = n,$ where $$n$$ is an integer, and the last term in the series is $$c_n\,z^{\,n}$$. Because the first term in the series is $$c_{l+1}\,z^{\,l+1}$$, it follows that $$n$$ must be greater than $$l$$, otherwise there are no terms in the series at all. Finally, it is clear from Equations ([e9.45]), ([e9.47]), and ([e9.54]) that

$\label{e9.55} E = \frac{E_0}{n^{\,2}}$ and $a = n\,a_0,$ where

$\label{e9.56} E_0 = -\frac{m_e\,e^{\,4}}{2\,(4\pi\,\epsilon_0)^2\,\hbar^{\,2}} = - \frac{e^{\,2}}{8\pi\,\epsilon_0\,a_0} = -13.6\,{\rm eV},$ and

$\label{e9.57} a_0 = \frac{4\pi\,\epsilon_0\,\hbar^{\,2}}{m_e\,e^{\,2}} = 5.3\times 10^{-11}\,{\rm m}.$ Here, $$E_0$$ is the energy of so-called ground-state (or lowest energy state) of the hydrogen atom, and the length $$a_0$$ is known as the Bohr radius. Note that $$|E_0|\sim \alpha^{\,2}\,m_e\,c^{\,2}$$, where $$\alpha = e^{\,2}/ (4\pi\,\epsilon_0\,\hbar\,c)\simeq 1/137$$ is the dimensionless fine-structure constant. The fact that $$|E_0|\ll m_e\,c^{\,2}$$ is the ultimate justification for our non-relativistic treatment of the hydrogen atom.

We conclude that the wavefunction of a hydrogen atom takes the form

$\label{e9.59} \psi_{n,l,m}(r,\theta,\phi) = R_{n,l}(r)\,Y_{l,m}(\theta,\phi).$ Here, the $$Y_{l,m}(\theta,\phi)$$ are the spherical harmonics (see Section [sharm]), and $$R_{n,l}(z=r/a)$$ is the solution of $\left[\frac{1}{z^{\,2}}\,\frac{d}{dz}\,z^{\,2}\,\frac{d}{dz}- \frac{l\,(l+1)}{z^{\,2}} + \frac{2\,n}{z}-1\right] R_{n,l} = 0$ which varies as $$z^{\,l}$$ at small $$z$$. Furthermore, the quantum numbers $$n$$, $$l$$, and $$m$$ can only take values that satisfy the inequality

$\label{e9.61} |m| \leq l < n,$ where $$n$$ is a positive integer, $$l$$ a non-negative integer, and $$m$$ an integer.

We expect the stationary states of the hydrogen atom to be orthonormal: that is, $\int \psi^\ast_{n',l',m'}\,\psi_{n,l,m}\,dV = \delta_{nn'}\,\delta_{ll'}\,\delta_{mm'},$ where $$dV$$ is a volume element, and the integral is over all space. Of course, $$dV = r^{\,2}\,dr\,d{\mit\Omega}$$, where $$d{\mit\Omega}$$ is an element of solid angle. Moreover, we already know that the spherical harmonics are orthonormal [see Equation ([spho])]: that is, $\oint Y_{l',m'}^{\,\ast}\,Y_{l,m}\,d{\mit\Omega} = \delta_{ll'}\,\delta_{mm'}.$ It, thus, follows that the radial wavefunction satisfies the orthonormality constraint $\int_0^{\infty} R_{n',l}^\ast\,R_{n,l}\,r^{\,2}\,dr = \delta_{nn'}.$ The first few radial wavefunctions for the hydrogen atom are listed below: \begin{aligned} R_{1,0}(r)&= \frac{2}{a_0^{\,3/2}}\,\exp\left(-\frac{r}{a_0}\right),\\[0.5ex] R_{2,0}(r) &= \frac{2}{(2\,a_0)^{3/2}}\left(1-\frac{r}{2\,a_0}\right) \exp\left(-\frac{r}{2\,a_0}\right),\\[0.5ex] R_{2,1}(r)&= \frac{1}{\sqrt{3}\,(2\,a_0)^{3/2}}\,\frac{r}{a_0}\, \exp\left(-\frac{r}{2\,a_0}\right),\\[0.5ex] R_{3,0}(r)&= \frac{2}{(3\,a_0)^{3/2}}\left(1- \frac{2\,r}{3\,a_0} + \frac{2\,r^{\,2}}{27\,a_0^{\,2}}\right)\exp\left(-\frac{r}{3\,a_0}\right),\\[0.5ex] R_{3,1}(r) &= \frac{4\sqrt{2}}{9\,(3\,a_0)^{3/2}}\,\frac{r}{a_0} \left(1-\frac{r}{6\,a_0}\right)\,\exp\left(-\frac{r}{3\,a_0}\right),\\[0.5ex] R_{3,2}(r)&= \frac{2\sqrt{2}}{27\sqrt{5}\,(3\,a_0)^{3/2}} \left(\frac{r}{a_0}\right)^2 \exp\left(-\frac{r}{3\,a_0}\right).\end{aligned} These functions are illustrated in Figures [coul1] and [coul2].

Figure 21: The $$$$a_{0} r^{2}\left|R_{n, l}(r)\right|^{2}$$$$ plotted as a functions of  $$$$r / a_{0}$$$$. The solid, short-dashed, and long-dashed curves correspond to $$$$n, l=1,0, \text { and } 2,0, \text { and } 2,1,$$$$ respectively.

Figure 22: The $$$$a_{0} r^{2}\left|R_{n, l}(r)\right|^{2}$$$$ plotted as a functions of  $$$$r / a_{0}$$$$.The solid, short-dashed, and long-dashed curves correspond to $$$$n, l=3,0, \text { and } 3,1, \text { and } 3,2,$$$$ respectively.

Given the (properly normalized) hydrogen wavefunction ([e9.59]), plus our interpretation of $$|\psi|^{\,2}$$ as a probability density, we can calculate $\langle r^{\,k}\rangle = \int_0^\infty r^{\,2+k}\,|R_{n,l}(r)|^{\,2}\,dr,$ where the angle-brackets denote an expectation value. For instance, it can be demonstrated (after much tedious algebra) that

\begin{aligned} \langle r^{\,2}\rangle &= \frac{a_0^{\,2}\,n^{\,2}}{2}\,[5\,n^{\,2}+1-3\,l\,(l+1)],\label{e9.73}\\[0.5ex] \langle r\rangle &= \frac{a_0}{2}\,[3\,n^{\,2}-l\,(l+1)],\\[0.5ex] \left\langle \frac{1}{r}\right\rangle &= \frac{1}{n^{\,2}\,a_0},\label{e9.74}\\[0.5ex] \left\langle\frac{1}{r^{\,2}}\right\rangle &= \frac{1}{(l+1/2)\,n^{\,3}\,a_0^{\,2}},\label{e9.75}\\[0.5ex] \left\langle\frac{1}{r^{\,3}}\right\rangle &=\frac{1}{l\,(l+1/2)\,(l+1)\,n^{\,3}\,a_0^{\,3}}.\label{e9.75a}\end{aligned}

According to Equation ([e9.55]), the energy levels of the bound-states of a hydrogen atom only depend on the radial quantum number $$n$$. It turns out that this is a special property of a $$1/r$$ potential. For a general central potential, $$V(r)$$, the quantized energy levels of a bound-state depend on both $$n$$ and $$l$$. (See Section 1.3.)

The fact that the energy levels of a hydrogen atom only depend on $$n$$, and not on $$l$$ and $$m$$, implies that the energy spectrum of a hydrogen atom is highly degenerate: that is, there are many different states which possess the same energy. According to the inequality ([e9.61]) (and the fact that $$n$$, $$l$$, and $$m$$ are integers), for a given value of $$l$$, there are $$2\,l+1$$ different allowed values of $$m$$ (i.e., $$-l,-l+1, \cdots, l-1, l$$). Likewise, for a given value of $$n$$, there are $$n$$ different allowed values of $$l$$ (i.e., $$0,1,\cdots, n-1$$). Now, all states possessing the same value of $$n$$ have the same energy (i.e., they are degenerate). Hence, the total number of degenerate states corresponding to a given value of $$n$$ is $1 + 3 + 5 + \cdots +2\,(n-1)+1 = n^{\,2}.$ Thus, the ground-state ($$n=1$$) is not degenerate, the first excited state ($$n=2$$) is four-fold degenerate, the second excited state ($$n=3$$) is nine-fold degenerate, et cetera (Actually, when we take into account the two spin states of an electron, the degeneracy of the $$n$$th energy level becomes $$2\,n^{\,2}$$.)
