D.82 Dirac fine structure Hamiltonian

This note derives the fine structure Hamiltonian of the hydrogen atom. This Hamiltonian fixes up the main relativistic errors in the classical solution of chapter 4.3. The derivation is based on the relativistic Dirac equation from chapter 12.12 and uses nontrivial linear algebra.

According to the Dirac equation, the relativistic Hamiltonian and wave function take the form

\begin{displaymath}
H_D = m_{\rm e}c^2 \left(\begin{array}{cr} 1&0\\ 0&-1 \end...
...{array}{c} \vec\psi^{\,p}\\ \vec\psi^{\,n} \end{array}\right)
\end{displaymath}

where $m_{\rm e}$ is the mass of the electron when at rest, $c$ the speed of light, and the $\sigma_i$ are the 2 $\times$ 2 Pauli spin matrices of chapter 12.10. Similarly the ones and zeros in the shown matrices are 2 $\times$ 2 unit and zero matrices. The wave function is a four-di­men­sion­al vector whose components depend on spatial position. It can be subdivided into the two-di­men­sion­al vectors $\vec\psi^{\,p}$ and $\vec\psi^{\,n}$. The two components of $\vec\psi^{\,p}$ correspond to the spin up and spin down components of the normal classical electron wave function; as noted in chapter 5.5.1, this can be thought of as a vector if you want. The two components of the other vector $\vec\psi^{\,n}$ are very small for the solutions of interest. These components would be dominant for states that would have negative rest mass. They are associated with the anti-particle of the electron, the positron.

The Dirac equation is solvable in closed form, but that solution is not something you want to contemplate if you can avoid it. And there is really no need for it, since the Dirac equation is not exact anyway. To the accuracy it has, it can easily be solved using perturbation theory in essentially the same way as in derivation {D.80}. In this case, the small parameter is 1/$c$: if the speed of light is infinite, the nonrelativistic solution is exact. And if you ballpark a typical velocity for the electron in a hydrogen atom, it is only about one percent or so of the speed of light.

So, following derivation {D.80}, take the Hamiltonian apart into successive powers of 1$\raisebox{.5pt}{$/$}$$c$ as $H_D$ $\vphantom0\raisebox{1.5pt}{$=$}$ $H_{D,0}+H_{D,1}+H_{D,2}$ with

\begin{displaymath}
H_{D,0} = \left(\begin{array}{cc} m_{\rm e}c^2&0\\ 0&-m_{\...
..._{D,2} = \left(\begin{array}{cc} V&0\\ 0&V \end{array}\right)
\end{displaymath}

and similarly for the wave function vector:

\begin{displaymath}
\vec\psi_D =
\left(\begin{array}{c}\vec\psi^{\,p}_0\\ \v...
...c\psi^{\,p}_4\\ \vec\psi^{\,n}_4\end{array}\right) +
\ldots
\end{displaymath}

and the energy:

\begin{displaymath}
E_D = E_{D,0} + E_{D,1}+ E_{D,2} + E_{D,3} + E_{D,4} + \ldots
\end{displaymath}

Substitution into the Hamiltonian eigenvalue problem $H_D\vec\psi_D$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E_D\vec\psi_D$ and then collecting equal powers of 1$\raisebox{.5pt}{$/$}$$c$ together produces again a system of successive equations, just like in derivation {D.80}:

\begin{eqnarray*}
c^2:&&
\left[
\left(\begin{array}{cc} m_{\rm e}c^2&0\\ 0...
...\ \vec\psi^{\,n}_0\end{array}\right)
= 0
\hspace{.45truein}
\end{eqnarray*}


\begin{eqnarray*}
c^1:&&
\left[
\left(\begin{array}{cc} m_{\rm e}c^2&0\\ 0...
...{array}{c}\vec\psi^{\,p}_0\\ \vec\psi^{\,n}_0\end{array}\right)
\end{eqnarray*}


\begin{eqnarray*}
c^0:&&
\left[
\left(\begin{array}{cc} m_{\rm e}c^2&0\\ 0...
...{array}{c}\vec\psi^{\,p}_0\\ \vec\psi^{\,n}_0\end{array}\right)
\end{eqnarray*}


\begin{eqnarray*}
c^{-1}:&&
\left[
\left(\begin{array}{cc} m_{\rm e}c^2&0\...
...{array}{c}\vec\psi^{\,p}_0\\ \vec\psi^{\,n}_0\end{array}\right)
\end{eqnarray*}


\begin{eqnarray*}
c^{-2}:&&
\left[
\left(\begin{array}{cc} m_{\rm e}c^2&0\...
...{array}{c}\vec\psi^{\,p}_0\\ \vec\psi^{\,n}_0\end{array}\right)
\end{eqnarray*}


\begin{eqnarray*}
c^{-3}:&& \smash{\cdots}\hspace{4.13truein}
\end{eqnarray*}

The first, order $c^2$, eigenvalue problem has energy eigenvalues $\pm{m}_ec^2$, in other words, plus or minus the rest mass energy of the electron. The solution of interest is the physical one with a positive rest mass, so the desired solution is

\begin{displaymath}
E_{D,0}=m_{\rm e}c^2
\qquad
\vec\psi^{\,p}_0 = \mbox{still arbitrary}
\qquad
\vec\psi^{\,n}_0 = 0
\end{displaymath}

Plug that into the order $c^1$ equation to give, for top and bottom subvectors

\begin{displaymath}
0 = E_{D,1} \vec\psi^{\,p}_0
\qquad
- 2 m_{\rm e}c^2 \...
...si^{\,n}_1 = -\sum_i c{\widehat p}_i\sigma_i \vec\psi^{\,p}_0
\end{displaymath}

It follows from the first of those that the first order energy change must be zero because $\vec\psi^{\,p}_0$ cannot be zero; otherwise there would be nothing left. The second equation gives the leading order values of the secondary components, so in total

\begin{displaymath}
E_{D,1} = 0
\qquad
\vec\psi^{\,p}_1 = \mbox{still arbi...
... \frac{1}{2m_{\rm e}c}{\widehat p}_j\sigma_j \vec\psi^{\,p}_0
\end{displaymath}

where the summation index $i$ was renamed to $j$ to avoid ambiguity later.

Plug all that in the order $c^0$ equation to give

\begin{displaymath}
0 = - \frac{1}{2m_{\rm e}}\sum_i\sum_j{\widehat p}_i{\wide...
... \frac{1}{2m_{\rm e}c}{\widehat p}_j\sigma_j \vec\psi^{\,p}_1
\end{displaymath}

The first of these two equations is the nonrelativistic Hamiltonian eigenvalue problem of chapter 4.3. To see that, note that in the double sum the terms with $j$ $\raisebox{.2pt}{$\ne$}$ $i$ pairwise cancel since for the Pauli matrices, $\sigma_i\sigma_j+\sigma_j\sigma_i$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 when $j$ $\raisebox{.2pt}{$\ne$}$ $i$. For the remaining terms in which $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $i$, the relevant property of the Pauli matrices is that $\sigma_i\sigma_i$ is one (or the 2 $\times$ 2 unit matrix, really,) giving

\begin{displaymath}
\frac{1}{2m_{\rm e}}\sum_i\sum_j{\widehat p}_i{\widehat p}...
...= \frac{1}{2m_{\rm e}}\sum_i{\widehat p}_i^2 + V
\equiv H_0
\end{displaymath}

where $H_0$ is the nonrelativistic hydrogen Hamiltonian of chapter 4.3.

So the first part of the order $c^0$ equation takes the form

\begin{displaymath}
H_0\vec\psi^{\,p}_0=E_{D,2}\vec\psi^{\,p}_0
\end{displaymath}

The energy $E_{D,2}$ will therefore have to be a Bohr energy level $E_n$ and each component of $\vec\psi^{\,p}_0$ will have to be a nonrelativistic energy eigenfunction with that energy:

\begin{displaymath}
E_{D,2} = E_n
\qquad
\vec\psi^{\,p}_0 =
\sum_l\sum_m...
...nlm}{\uparrow}+ \sum_l\sum_m c_{lm{-}} \psi_{nlm}{\downarrow}
\end{displaymath}

The sum multiplying ${\uparrow}$ is the first component of vector $\vec\psi^{\,p}_0$ and the sum multiplying ${\downarrow}$ the second. The nonrelativistic analysis in chapter 4.3 was indeed correct as long as the speed of light is so large compared to the relevant velocities that 1$\raisebox{.5pt}{$/$}$$c$ can be ignored.

To find out the error in it, the relativistic expansion must be taken to higher order. To order $c^{-1}$, you get for the top vector

\begin{displaymath}
0 = - (H_0 - E_n) \vec\psi^{\,p}_1 + E_{D,3} \vec\psi^{\,p}_0
\end{displaymath}

Now if $\vec\psi^{\,p}_1$ is written as a sum of the eigenfunctions of $H_0$, including $\vec\psi^{\,p}_0$, the first term will produce zero times $\vec\psi^{\,p}_0$ since $(H_0-E_n)\vec\psi^{\,p}_0$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. That means that $E_{D,3}$ must be zero. The expansion must be taken one step further to identify the relativistic energy change. The bottom vector gives

\begin{displaymath}
\vec\psi^{\,n}_3 = \sum_j \frac{1}{2m_{\rm e}c}{\widehat p...
... \frac{1}{2m_{\rm e}c}{\widehat p}_j\sigma_j \vec\psi^{\,p}_0
\end{displaymath}

To order $c^{-2}$, you get for the top vector

\begin{displaymath}
0 = - (H_0 - E_n) \vec\psi^{\,p}_2
- \sum_i\sum_j
{\wi...
...at p}_j\sigma_j \vec\psi^{\,p}_0
+ E_{D,4} \vec\psi^{\,p}_0
\end{displaymath}

and that determines the approximate relativistic energy correction.

Now recall from derivation {D.80} that if you do a nonrelativistic expansion of an eigenvalue problem $(H_0+H_1)\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E\psi$, the equations to solve are (D.55) and (D.56);

\begin{displaymath}
(H_0-E_{{\vec n},0})\psi_{{\vec n},0} = 0
\qquad
(H_0-...
...\psi_{{\vec n},1}
= - (H_1-E_{{\vec n},1})\psi_{{\vec n},0}
\end{displaymath}

The first equation was satisfied by the solution for $\vec\psi^{\,p}_0$ obtained above. However, the second equation presents a problem. Comparison with the final Dirac result suggests that the fine structure Hamiltonian correction $H_1$ should be identified as

\begin{displaymath}
H_1 \stackrel{\mbox{?}}{=}
\sum_i\sum_j {\widehat p}_i\sigma_i \frac{V-E_n}{4m_{\rm e}^2c^2}{\widehat p}_j\sigma_j
\end{displaymath}

but that is not right, since $E_n$ is not a physical operator, but an energy eigenvalue for the selected eigenfunction. So mapping the Dirac expansion straightforwardly onto a classical one has run into a snag.

It is maybe not that surprising that a two-di­men­sion­al wave function cannot correctly represent a truly four-di­men­sion­al one. But clearly, whatever is selected for the fine structure Hamiltonian $H_1$ must at least get the energy eigenvalues right. To see how this can be done, the operator obtained from the Dirac equation will have to be simplified. Now for any given $i$, the sum over $j$ includes a term $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $i$, a term $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\overline{\imath}}$, where ${\overline{\imath}}$ is the number following $i$ in the cyclic sequence $\ldots123123\ldots$, and it involves a term $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\overline{\overline{\imath}}}$ where ${\overline{\overline{\imath}}}$ precedes $i$ in the sequence. So the Dirac operator falls apart into three pieces:

\begin{displaymath}
H_1 \stackrel{\mbox{?}}{=}
\sum_i {\widehat p}_i\sigma_i...
...e{\overline{\imath}}}}\sigma_{{\overline{\overline{\imath}}}}
\end{displaymath}

or using the properties of the Pauli matrices that $\sigma_i\sigma_i$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, $\sigma_i\sigma_{{\overline{\imath}}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\rm i}\sigma_{{\overline{\overline{\imath}}}}$, and $\sigma_i\sigma_{{\overline{\overline{\imath}}}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-{\rm i}\sigma_{{\overline{\overline{\imath}}}}$ for any $i$,
\begin{displaymath}
H_1 \stackrel{\mbox{?}}{=}
\sum_i {\widehat p}_i \frac{V...
...{\overline{\overline{\imath}}}}\sigma_{{\overline{\imath}}} %
\end{displaymath} (D.59)

The approach will now be to show first that the final two terms are the spin-orbit interaction in the fine structure Hamiltonian. After that, the much more tricky first term will be discussed. Renotate the indices in the last two terms as follows:

\begin{displaymath}
H_{1,\mbox{\scriptsize spin-orbit}} =
{\rm i}\sum_i {\wi...
...{4m_{\rm e}^2c^2}{\widehat p}_{{\overline{\imath}}}\sigma_{i}
\end{displaymath}

Since the relative order of the subscripts in the cycle was maintained in the renotation, the sums still contain the exact same three terms, just in a different order. Take out the common factors;

\begin{displaymath}
H_{1,\mbox{\scriptsize spin-orbit}} =
\frac{{\rm i}}{4m_...
...}}}(V-E_n){\widehat p}_{{\overline{\imath}}}\right]\sigma_{i}
\end{displaymath}

Now according to the generalized canonical commutator of chapter 4.5.4:

\begin{displaymath}
{\widehat p}_i (V-E_n) = (V-E_n) {\widehat p}_i - {\rm i}\hbar\frac{\partial (V-E_n)}{\partial r_i}
\end{displaymath}

where $E_n$ is a constant that produces a zero derivative. So ${\widehat p}_{{\overline{\imath}}}$, respectively ${\widehat p}_{{\overline{\overline{\imath}}}}$ can be taken to the other side of $V-E_n$ as long as the appropriate derivatives of $V$ are added. If that is done, $(V-E_n){\widehat p}_{{\overline{\imath}}}{\widehat p}_{{\overline{\overline{\imath}}}}$ and $-(V-E_n){\widehat p}_{{\overline{\overline{\imath}}}}{\widehat p}_{{\overline{\imath}}}$ cancel since linear momentum operators commute. What is left are just the added derivative terms:

\begin{displaymath}
H_{1,\mbox{\scriptsize spin-orbit}} =
\frac{\hbar}{4m_{\...
...th}}}}}{\widehat p}_{{\overline{\imath}}}
\right]\sigma_{i}
\end{displaymath}

Note that the errant eigenvalue $E_n$ mercifully dropped out. Now the hydrogen potential $V$ only depends on the distance $r$ from the origin, as 1/$r$, so

\begin{displaymath}
\frac{\partial V}{\partial r_i} = - \frac{V}{r^2}r_i
\end{displaymath}

and plugging that into the operator, you get

\begin{displaymath}
H_{1,\mbox{\scriptsize spin-orbit}} =
- \frac{\hbar V}{4...
...\imath}}}}{\widehat p}_{{\overline{\imath}}}\right]\sigma_{i}
\end{displaymath}

The term between the square brackets can be recognized as the $i$th component of the angular momentum operator; also the Pauli spin matrix $\sigma_i$ is defined as ${\widehat S}_i$$\raisebox{.5pt}{$/$}$${\textstyle\frac{1}{2}}\hbar$, so

\begin{displaymath}
H_{1,\mbox{\scriptsize spin-orbit}}
= - \frac{V}{2m_{\rm e}^2c^2r^2} \sum_i \L _i{\widehat S}_i
\end{displaymath}

Get rid of $c^2$ using $\vert E_1\vert$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\textstyle\frac{1}{2}}\alpha^2{m_{\rm e}}c^2$, of $V$ using $V$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-2\vert E_1\vert a_0$$\raisebox{.5pt}{$/$}$$r$, and $m_{\rm e}$ using $\vert E_1\vert$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar^2$$\raisebox{.5pt}{$/$}$$2{m_{\rm e}}a_0^2$ to get the spin-orbit interaction as claimed in the section on fine structure.

That leaves the term

\begin{displaymath}
\sum_i {\widehat p}_i \frac{V-E_n}{4m_{\rm e}^2c^2}{\widehat p}_i
\end{displaymath}

in (D.59). Since $V$ $\vphantom0\raisebox{1.5pt}{$=$}$ $H_0-{\widehat p}^2/2m_{\rm e}$, it can be written as

\begin{displaymath}
\sum_i {\widehat p}_i \frac{H_0-E_n}{4m_{\rm e}^2c^2}{\wid...
...
- \frac{\left({\widehat p}^{\,2}\right)^2}{8m_{\rm e}^3c^2}
\end{displaymath}

The final term is the claimed Einstein correction in the fine structure Hamiltonian, using $\vert E_1\vert$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\textstyle\frac{1}{2}}\alpha^2{m_{\rm e}}c^2$ to get rid of $c^2$.

The first term,

\begin{displaymath}
H_{1,{\rm Darwin}} \stackrel{\mbox{?}}{=}
\sum_i {\widehat p}_i \frac{H_0-E_n}{4m_{\rm e}^2c^2}{\widehat p}_i
\end{displaymath}

is the sole remaining problem. It cannot be transformed into a decent physical operator. The objective is just to get the energy correction right. And to achieve that requires only that the Hamiltonian perturbation coefficients are evaluated correctly at the $E_n$ energy level. Specifically, what is needed is that

\begin{displaymath}
H_{\underline{\vec n}{\vec n},1,{\rm Darwin}} \equiv
\la...
...{\widehat p}_i(H_0-E_n){\widehat p}_i\psi_{{\vec n},0}\rangle
\end{displaymath}

for any arbitrary pair of unperturbed hydrogen energy eigenfunctions $\psi_{\underline{\vec n},0}$ and $\psi_{{\vec n},0}$ with energy $E_n$. To see what that means, the leading Hermitian operator ${\widehat p}_i$ can be taken to the other side of the inner product, and in half of that result, $H_0-E_n$ will also be taken to the other side:

\begin{displaymath}
H_{\underline{\vec n}{\vec n},1,{\rm Darwin}} = \frac{1}{8...
...ec n},0}\vert{\widehat p}_i\psi_{{\vec n},0}\rangle
\right)
\end{displaymath}

Now if you simply swap the order of the factors in $(H_0-E_n){\widehat p}_i$ in this expression, you get zero, because both eigenfunctions have energy $E_n$. However, swapping the order of $(H_0-E_n){\widehat p}_i$ brings in the generalized canonical commutator $[V,{\widehat p}_i]$ that equals ${\rm i}\hbar\partial{V}$$\raisebox{.5pt}{$/$}$$\partial{r}_i$. Therefore, writing out the remaining inner product you get

\begin{displaymath}
H_{\underline{\vec n}{\vec n},1,{\rm Darwin}} = \frac{-\hb...
...*\psi_{{\vec n},0}}{\partial r_i}
{\,\rm d}^3{\skew0\vec r}
\end{displaymath}

Now, the potential $V$ becomes infinite at $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, and that makes mathematical manipulation difficult. Therefore, assume for now that the nuclear charge $e$ is not a point charge, but spread out over a very small region around the origin. In that case, the inner product can be rewritten as

\begin{displaymath}
H_{\underline{\vec n}{\vec n},1,{\rm Darwin}} = \frac{-\hb...
... n},0}^*\psi_{{\vec n},0}
\right] {\,\rm d}^3{\skew0\vec r}
\end{displaymath}

and the first term integrates away since $\psi_{\underline{\vec n},0}^*\psi_{{\vec n},0}$ vanishes at infinity. In the final term, use the fact that the derivatives of the potential energy $V$ give $e$ times the electric field of the nucleus, and therefore the second order derivatives give $e$ times the divergence of the electric field. Maxwell’s first equation (13.5) says that that is $e$$\raisebox{.5pt}{$/$}$$\epsilon_0$ times the nuclear charge density. Now if the region of nuclear charge is allowed to contract back to a point, the charge density must still integrate to the net proton charge $e$, so the charge density becomes $e\delta^3({\skew0\vec r})$ where $\delta^3({\skew0\vec r})$ is the three-di­men­sion­al delta function. Therefore the Darwin term produces Hamiltonian perturbation coefficients as if its Hamiltonian is

\begin{displaymath}
H_{1,{\rm Darwin}} = \frac{\hbar^2e^2}{8m_{\rm e}^2c^2\epsilon_0} \delta^3({\skew0\vec r})
\end{displaymath}

Get rid of $c^2$ using $\vert E_1\vert$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12\alpha^2{m_{\rm e}}c^2$, of $e^2$$\raisebox{.5pt}{$/$}$$\epsilon_0$ using $e^2$$\raisebox{.5pt}{$/$}$$4\pi\epsilon_0$ $\vphantom0\raisebox{1.5pt}{$=$}$ $2\vert E_1\vert a_0$, and $m_{\rm e}$ using $\vert E_1\vert$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar^2$$\raisebox{.5pt}{$/$}$$2{m_{\rm e}}a_0^2$ to get the Darwin term as claimed in the section on fine structure. It will give the right energy correction for the nonrelativistic solution. But you may rightly wonder what to make of the implied wave function.