Fine Structure of Hydrogen Energy Levels

For the case of a hydrogen atom,

$$\phi(r) = \frac{e}{4\pi\,\epsilon_0\,r}.$$ (11.170)

Hence, Equations (11.168) and (11.169) yield

$$\left(\frac{1}{a_1} - \frac{\alpha}{y}\right)\psi_a - \left(\frac{d}{dy} + \frac{k+1}{y}\right)\psi_b =0,$$ (11.171)

$$\left(\frac{1}{a_2} +\frac{\alpha}{y}\right)\psi_b - \left(\frac{d}{dy} - \frac{k-1}{y}\right)\psi_a =0,$$ (11.172)

where $y=r/a_0$ , and

$$a_1 = \frac{\alpha}{1-{\cal E}},$$ (11.173)

$$a_2 = \frac{\alpha}{1+{\cal E}},$$ (11.174)

with ${\cal E} = E/(m_e\,c^{\,2})$ . Here, $a_0= 4\pi\,\epsilon_0\,\hbar^{\,2}/(m_e\,e^{\,2})$ is the Bohr radius, and $\alpha=e^{\,2}/(4\pi\,\epsilon_0\,\hbar\,c)$ the fine structure constant. Writing

$$\psi_a(y) = \frac{{\rm e}^{-y/a}}{y}\,f(y),$$ (11.175)

$$\psi_b(y) = \frac{{\rm e}^{-y/a}}{y}\,g(y),$$ (11.176)

where

$$a = (a_1\,a_2)^{1/2} = \frac{\alpha}{\sqrt{1-{\cal E}^{\,2}}},$$ (11.177)

we obtain

$$\left(\frac{1}{a_1}-\frac{\alpha}{y}\right)f - \left(\frac{d}{d y}- \frac{1}{a}+\frac{k}{y}\right)g =0,$$ (11.178)

$$\left(\frac{1}{a_2}+\frac{\alpha}{y}\right)g - \left(\frac{d}{d y}- \frac{1}{a}-\frac{k}{y}\right)f = 0.$$ (11.179)

Let us search for power-law solutions of the form

$$f(y) = \sum_{s} c_s\,y^{\,s},$$ (11.180)

$$g(y) = \sum_s c_s'\,y^{\,s},$$ (11.181)

where successive values of $s$ differ by unity. Substitution of these solutions into Equations (11.178) and (11.179) leads to the recursion relations

$$\frac{c_{s-1}}{a_1} -\alpha\,c_s - (s+k)\,c_s' + \frac{c_{s-1}'}{a} =0,$$ (11.182)

$$\frac{c_{s-1}'}{a_2}+\alpha\,c_s' -(s-k)\,c_s + \frac{c_{s-1}}{a} = 0.$$ (11.183)

Multiplying the first of these equations by $a$ , and the second by $a_2$ , and then subtracting, we eliminate both $c_{s-1}$ and $c_{s-1}'$ , because $a/a_1=a_2/a$ . We are left with

$$[a\,\alpha-a_2\,(s-k)]\,c_s + [a_2\,\alpha+a\,(s+k)]\,c_{s}' = 0.$$ (11.184)

The physical boundary conditions at $y=0$ require that $y\,\psi_a\rightarrow 0$ and $y\,\psi_b\rightarrow 0$ as $y\rightarrow 0$ . Thus, it follows from Equations (11.175) and (11.176) that $f\rightarrow 0$ and $g\rightarrow 0$ as $y\rightarrow 0$ . Consequently, the series (11.180) and (11.181) must terminate at small positive $s$ . If $s_0$ is the minimum value of $s$ for which $c_s$ and $c_s'$ do not both vanish then it follows from Equations (11.182) and (11.183), putting $s=s_0$ and $c_{s_0-1}=c_{s_0-1}'=0$ , that

$$\alpha\,c_{s_0}+(s_0+k)\,c_{s_0}' =0,$$ (11.185)

$$\alpha\,c_{s_0}' - (s_0-k)\,c_{s_0} =0,$$ (11.186)

which implies that

$$\alpha^{\,2} = - s_0^{\,2} + k^{\,2}.$$ (11.187)

Because the boundary condition requires that the minimum value of $s_0$ be greater than zero, we must take

$$s_0 = (k^{\,2}-\alpha^{\,2})^{1/2}.$$ (11.188)

To investigate the convergence of the series (11.180) and (11.181) at large $y$ , we shall determine the ratio $c_s/c_{s-1}$ for large $s$ . In the limit of large $s$ , Equations (11.183) and (11.184) yield

$$s\,c_s \simeq \frac{c_{s-1}}{a}+\frac{c_{s-1}'}{a_2},$$ (11.189)

$$a_2\,c_s \simeq a\,c_s',$$ (11.190)

because $\alpha\simeq 1/137 \ll 1$ . Thus,

$$\frac{c_s}{c_{s-1}}\simeq \frac{2}{a\,s}.$$ (11.191)

However, this is the ratio of coefficients in the series expansion of $\exp(2\,y/a)$ . Hence, we deduce that the series (11.180) and (11.181) diverge unphysically at large $y$ unless they terminate at large $s$ .

Suppose that the series (11.180) and (11.181) terminate with the terms $c_s$ and $c_s'$ , so that $c_{s+1}=c_{s+1}'=0$ . It follows from Equations (11.182) and (11.183), with $s+1$ substituted for $s$ , that

$$\frac{c_s}{a_1} + \frac{c_s'}{a} =0,$$ (11.192)

$$\frac{c_s'}{a_2} + \frac{c_s}{a} = 0.$$ (11.193)

These two expressions are equivalent, because $a^{\,2}=a_1\,a_2$ . When combined with Equation (11.184) they give

$$a_1\left[a\,\alpha-a_2\,(s-k)\right] = a\left[a_2\,\alpha+a\,(s+k)\right],$$ (11.194)

which reduces to

$$2\,a_1\,a_2\,s = a\,(a_1-a_2)\,\alpha,$$ (11.195)

or

$${\cal E} = \left(1+\frac{\alpha^{\,2}}{s^{\,2}}\right)^{-1/2}.$$ (11.196)

Here, $s$ , which specifies the last term in the series, must be greater than $s_0$ by some non-negative integer $i$ . Thus,

$$s = i+ (k^{\,2}-\alpha^{\,2})^{1/2}= i+\left[(j+1/2)^{\,2}-\alpha^{\,2}\right]^{1/2}.$$ (11.197)

where $j\,(j+1)\,\hbar^{\,2}$ is the eigenvalue of $J^{\,2}$ . Hence, the energy eigenvalues of the hydrogen atom become

$$\frac{E}{m_e\,c^{\,2}} =\left\{1 + \frac{\alpha^{\,2}}{\left(i+[(j+1/2)^{\,2}-\alpha^{\,2}]^{1/2}\right)^2}\right\}^{-1/2}$$ (11.198)

[24,57]. Given that $\alpha\simeq 1/137$ , we can expand the previous expression in $\alpha^{\,2}$ to give

$$\frac{E}{m_e\,c^{\,2}} = 1 - \frac{\alpha^{\,2}}{2\,n^{\,2}}- \fr... ...4}}{2\,n^{\,4}}\left(\frac{n}{j+1/2}-\frac{3}{4}\right)+{\cal O}(\alpha^{\,6}),$$ (11.199)

where $n= i + j+1/2$ is a positive integer. Of course, the first term in the previous expression corresponds to the electron's rest mass energy. The second term corresponds to the standard non-relativistic expression for the hydrogen energy levels, with $n$ playing the role of the radial quantum number. (See Section 4.6.) Finally, the third term corresponds to the fine structure correction to these energy levels. (See Exercise 14). Note that this correction only depends on the quantum numbers $n$ and $j$ . Now, we showed in Exercise 14 that the fine structure correction to the energy levels of the hydrogen atom is a combined effect of spin-orbit coupling, the electron's relativistic mass increase, and the Darwin term. Hence, it is evident that all of these effects are automatically taken into account in the Dirac equation.