Solution of the Dirac Equation for Hydrogen

The standard Hydrogen atom problem can be solved exactly using relativistic quantum mechanics. The full solution is a bit long but short compared to the complete effort we made in non-relativistic QM. We have already seen that (even with no applied fields), while the total angular momentum operator commutes with the Dirac Hamiltonian, neither the orbital angular momentum operator nor the spin operators do commute with \bgroup\color{black}$H$\egroup. The addition of a spherically symmetric potential does not change these facts.

We have shown in the section on conserved quantities that the operator

\begin{displaymath}\bgroup\color{black} K=\gamma_4\vec{\Sigma}\cdot\vec{J}-\gamma_4{\hbar\over 2} \egroup\end{displaymath}

also commutes with the Hamiltonian and with $\vec{J}$. \bgroup\color{black}$K$\egroup is a measure of the component of spin along the total angular momentum direction. We will use \bgroup\color{black}$K$\egroup to help solve problems with spherical symmetry and ultimately the problem of hydrogen. We therefore have four mutually commuting operators the eigenvalues of which can completely label the eigenstates:

\begin{displaymath}\bgroup\color{black} H, J^2, J_z, K\rightarrow n_r, j, m_j, \kappa .\egroup\end{displaymath}

The operator $K$ may be written in several ways.

K&=&\gamma_4\vec{\Sigma}\cdot\vec{J}-\gamma_4{\hbar\over 2}
...}\cdot\vec{L}+\hbar & 0 \cr 0 & -\vec{\sigma}\cdot\vec{L}-\hbar}

Assume that the eigenvalues of $K$ are given by

\begin{displaymath}\bgroup\color{black} K\psi=-\kappa\hbar\psi . \egroup\end{displaymath}

We now compare the \bgroup\color{black}$K^2$\egroup and \bgroup\color{black}$J^2$\egroup operators.

...\vec{L}\times\vec{L})+2\hbar\vec{\Sigma}\cdot\vec{L}+\hbar^2 \\

...\kappa&=&\pm(j+{1\over 2}) \\
K\psi&=&\pm(j+{1\over 2})\psi \\

The eigenvalues of \bgroup\color{black}$K$\egroup are

\bgroup\color{black}$\displaystyle \kappa=\pm\left(j+{1\over 2}\right)\hbar.$\egroup
We may explicitly write out the eigenvalue equation for \bgroup\color{black}$K$\egroup for \bgroup\color{black}$\kappa=\pm\left(j+{1\over 2}\right)\hbar$\egroup.

...\left(j+{1\over 2}\right)\hbar\pmatrix{\psi_A\cr \psi_B}\egroup\end{displaymath}

The difference between \bgroup\color{black}$J^2$\egroup and \bgroup\color{black}$L^2$\egroup is related to \bgroup\color{black}$\vec{\sigma}\cdot\vec{L}$\egroup.

\begin{displaymath}\bgroup\color{black} L^2=J^2-\hbar\vec{\sigma}\cdot\vec{L}-{3\over 4}\hbar \egroup\end{displaymath}

We may solve for the effect of \bgroup\color{black}$\vec{\sigma}\cdot\vec{L}$\egroup on the spinor \bgroup\color{black}$\psi$\egroup, then, solve for the effect of \bgroup\color{black}$L^2$\egroup. Note that since \bgroup\color{black}$\psi_A$\egroup and \bgroup\color{black}$\psi_B$\egroup are eigenstates of \bgroup\color{black}$J^2$\egroup and \bgroup\color{black}$\vec{\sigma}\cdot\vec{L}$\egroup, they are eigenstates of \bgroup\color{black}$L^2$\egroup but have different eigenvalues.

\pmatrix{\vec{\sigma}\cdot\vec{L}+\hbar & 0 \cr 0 & -\vec{\sig...
...m(j+{1\over 2}\mp 1)\psi_A\cr (\pm(j+{1\over 2}\pm 1)\psi_B} \\

\bgroup\color{black}$\displaystyle \vec{\sigma}\cdot\vec{L}\pmatrix{\psi_A\cr \p...
...\pmatrix{(\pm(j+{1\over 2}\mp 1)\psi_A\cr (\mp(j+{1\over 2}\pm 1)\psi_B}$\egroup

L^2\pmatrix{\psi_A\cr \psi_B}
...ver 4})\psi_A\cr (j^2+j\pm j\pm{1\over 2}+{1\over 4})\psi_B} \\

Note that the eigenvalues for the upper and lower components have the same possible values, but are opposite for energy eigenstates. We already know the relation \bgroup\color{black}$\ell=j\pm{1\over 2}$\egroup from NR QM. We simply check that it is the same here.

\begin{displaymath}\bgroup\color{black} \ell(\ell+1)=(j\pm{1\over 2})(j+1\pm{1\o...
...1\over 2}+{1\over 4}=j^2+j\pm j\pm{1\over 2}+{1\over 4} \egroup\end{displaymath}

It is correct. So \bgroup\color{black}$\psi_A$\egroup and \bgroup\color{black}$\psi_B$\egroup are eigenstates of $L^2$ but with different eigenvalues.
\bgroup\color{black}$\displaystyle L^2\pmatrix{\psi_A\cr \psi_B}=\hbar^2\pmatrix{\ell_\mp(\ell_\mp+1)\psi_A\cr \ell_\pm(\ell_\pm+1)\psi_B}$\egroup
\bgroup\color{black}$\displaystyle \ell_\pm=j\pm{1\over 2}$\egroup
Now we apply the Dirac equation and try to use our operators to help solve the problem.

\left(\gamma_\mu{\partial\over\partial x_\mu}+\gamma_\mu{ie\ov...
...-V(r)-mc^2 & 0\cr 0 & E-V(r)+mc^2}\pmatrix{\psi_A\cr \psi_B} \\

The Dirac Equation then is.
\bgroup\color{black}$\displaystyle c\vec{\sigma}\cdot\vec{p}\pmatrix{\psi_B\cr \...
...}=\pmatrix{E-V(r)-mc^2 & 0\cr 0 & E-V(r)+mc^2}\pmatrix{\psi_A\cr \psi_B}$\egroup

We can use commutation and anticommutation relations to write $\vec{\sigma}\cdot\vec{p}$ in terms of separate angular and radial operators.

... r{\partial\over\partial r}+i\vec{\sigma}\cdot\vec{L}\right) \\

\bgroup\color{black}$\displaystyle \vec{\sigma}\cdot\vec{p}={1\over r}{\vec{\sig...
...left(-i\hbar r{\partial\over\partial r}+i\vec{\sigma}\cdot\vec{L}\right)$\egroup

Note that the operators \bgroup\color{black}${\vec{\sigma}\cdot\vec{x}\over r}$\egroup and \bgroup\color{black}$i\vec{\sigma}\cdot\vec{L}$\egroup act only on the angular momentum parts of the state. There are no radial derivatives so they commute with \bgroup\color{black}$-i\hbar r{\partial\over\partial r}$\egroup. Lets pick a shorthand notation for the angular momentum eigenstates we must use. These have quantum numbers \bgroup\color{black}$j$\egroup, \bgroup\color{black}$m_j$\egroup, and \bgroup\color{black}$\ell$\egroup. \bgroup\color{black}$\psi_A$\egroup will have \bgroup\color{black}$\ell=\ell_A$\egroup and \bgroup\color{black}$\psi_B$\egroup must have the other possible value of \bgroup\color{black}$\ell$\egroup which we label \bgroup\color{black}$\ell_B$\egroup. Following the notation of Sakurai, we will call the state \bgroup\color{black}$\vert jm_j\ell_A\rangle\equiv {\cal Y}^{m_j}_{j\ell_A}=\alpha Y_{\ell_A,m_j-{1\over 2}}\chi_++\beta Y_{\ell_A,m_j+{1\over 2}}\chi_-$\egroup. (Note that our previous functions made use of \bgroup\color{black}$m=m_\ell$\egroup particularly in the calculation of \bgroup\color{black}$\alpha$\egroup and \bgroup\color{black}$\beta$\egroup.)

c{1\over r}{\vec{\sigma}\cdot\vec{x}\over r}\left(-i\hbar r{\p...
...r){\cal Y}^{m_j}_{j\ell_A}\cr if(r){\cal Y}^{m_j}_{j\ell_B}} \\

The effect of the two operators related to angular momentum can be deduced. First, \bgroup\color{black}$\vec{\sigma}\cdot\vec{L}$\egroup is related to \bgroup\color{black}$K$\egroup. For positive \bgroup\color{black}$\kappa$\egroup, \bgroup\color{black}$\psi_A$\egroup has \bgroup\color{black}$\ell=j+{1\over 2}$\egroup. For negative \bgroup\color{black}$\kappa$\egroup, \bgroup\color{black}$\psi_A$\egroup has \bgroup\color{black}$\ell=j-{1\over 2}$\egroup. For either, \bgroup\color{black}$\psi_B$\egroup has the opposite relation for \bgroup\color{black}$\ell$\egroup, indicating why the full spinor is not an eigenstate of \bgroup\color{black}$L^2$\egroup.

K&=&\pmatrix{\vec{\sigma}\cdot\vec{L}+\hbar & 0 \cr 0 & -\vec{...
...B \\
\vec{\sigma}\cdot\vec{L}\psi_B&=&(\kappa-1)\hbar\psi_B \\

Second, \bgroup\color{black}${\vec{\sigma}\cdot\vec{x}\over r}$\egroup is a pseudoscalar operator. It therefore changes parity and the parity of the state is given by \bgroup\color{black}$(-1)^\ell$\egroup; so it must change \bgroup\color{black}$\ell$\egroup.

\begin{displaymath}\bgroup\color{black} {\vec{\sigma}\cdot\vec{x}\over r}{\cal Y}^{m_j}_{j\ell_A}=C{\cal Y}^{m_j}_{j\ell_B} \egroup\end{displaymath}

The square of the operator \bgroup\color{black}$\left({\vec{\sigma}\cdot\vec{x}\over r}\right)^2$\egroup is one, as is clear from the derivation above, so we know the effect of this operator up to a phase factor.

\begin{displaymath}\bgroup\color{black} {\vec{\sigma}\cdot\vec{x}\over r}{\cal Y}^{m_j}_{j\ell_A}=e^{i\delta}{\cal Y}^{m_j}_{j\ell_B} \egroup\end{displaymath}

The phase factor depends on the conventions we choose for the states \bgroup\color{black}${\cal Y}^{m_j}_{j\ell}$\egroup. For our conventions, the factor is \bgroup\color{black}$-1$\egroup.

\begin{displaymath}\bgroup\color{black} {\vec{\sigma}\cdot\vec{x}\over r}{\cal Y}^{m_j}_{j\ell_A}=-{\cal Y}^{m_j}_{j\ell_B} \egroup\end{displaymath}

We now have everything we need to get to the radial equations.

c{1\over r}{\vec{\sigma}\cdot\vec{x}\over r}\left(-i\hbar r{\p...
...j\ell_B}} \\
\hbar c{1\over r}{\vec{\sigma}\cdot\vec{x}\over r}

\pmatrix{\left( r{\partial\over\partial r}-(\kappa-1)\right)f(...\over r}g\right)}
&=&\pmatrix{(E-V-mc^2)g\cr (E-V+mc^2)f} \\

This is now a set of two coupled radial equations. We can simplify them a bit by making the substitutions \bgroup\color{black}$F=rf$\egroup and \bgroup\color{black}$G=rg$\egroup. The extra term from the derivative cancels the 1's that are with \bgroup\color{black}$\kappa$\egroups.

\hbar c\pmatrix{\left(-{1\over r}{\partial F\over\partial r}+{... G\over r}\right)}
&=&\pmatrix{(E-V-mc^2)G\cr (E-V+mc^2)F} \\

\bgroup\color{black}$\displaystyle \pmatrix{\left({\partial F\over\partial r}-{\...
=\pmatrix{{mc^2-E+V\over\hbar c}G\cr {mc^2+E-V\over\hbar c}F}$\egroup

These equations are true for any spherically symmetric potential. Now it is time to specialize to the hydrogen atom for which \bgroup\color{black}${V\over\hbar c}=-{Z\alpha\over r}$\egroup. We define \bgroup\color{black}$k_1={mc^2+E\over\hbar c}$\egroup and \bgroup\color{black}$k_2={mc^2-E\over\hbar c}$\egroup and the dimensionless \bgroup\color{black}$\rho=\sqrt{k_1k_2}r$\egroup. The equations then become.

\pmatrix{\left({\partial F\over\partial r}-{\kappa F\over r}\r...
...t)G-\left(\sqrt{k_1\over k_2}+{Z\alpha\over \rho}\right)F}=0 \\

With the guidance of the non-relativistic solutions, we will postulate a solution of the form

F=e^{-\rho}\rho^s\sum\limits_{m=0}^\infty a_m\rho^m=e^{-\rho}\...
...y b_m\rho^m=e^{-\rho}\sum\limits_{m=0}^\infty b_m\rho^{s+m}. \\

The exponential will make everything go to zero for large \bgroup\color{black}$\rho$\egroup if the power series terminates. We need to verify that this is a solution near \bgroup\color{black}$\rho=0$\egroup if we pick the right \bgroup\color{black}$a_0$\egroup, \bgroup\color{black}$b_0$\egroup, and \bgroup\color{black}$s$\egroup. We now substitute these postulated solutions into the equations to obtain recursion relations.

...+m+1+\kappa)b_{m+1}-\sqrt{k_1\over k_2}a_m-Z\alpha a_{m+1}=0 \\

\bgroup\color{black}$\displaystyle -a_m+(s+m+1-\kappa)a_{m+1}-\sqrt{k_2\over k_1}b_m+Z\alpha b_{m+1}=0$\egroup
\bgroup\color{black}$\displaystyle -\sqrt{k_1\over k_2}a_m-Z\alpha a_{m+1}-b_m+(s+m+1+\kappa)b_{m+1}=0$\egroup

For the lowest order term \bgroup\color{black}$\rho^s$\egroup, we need to have a solution without lower powers. This means that we look at the $m=-1$ recursion relations with \bgroup\color{black}$a_m=b_m=0$\egroup and solve the equations.

(s-\kappa)a_{0}+Z\alpha b_{0}=0 \\
-Z\alpha a_{0}+(s+\kappa)b...
...^2=\kappa^2-Z^2\alpha^2 \\
s=\pm\sqrt{\kappa^2-Z^2\alpha^2} \\

Note that while \bgroup\color{black}$\kappa$\egroup is a non-zero integer, \bgroup\color{black}$Z^2\alpha^2$\egroup is a small non-integer number. We need to take the positive root in order to keep the state normalized.

\begin{displaymath}\bgroup\color{black} s=+\sqrt{\kappa^2-Z^2\alpha^2} \egroup\end{displaymath}

As usual, the series must terminate at some $m=n_r$ for the state to normalizable. This can be seen approximately by assuming either the \bgroup\color{black}$a$\egroup's or the \bgroup\color{black}$b$\egroup's are small and noting that the series is that of a positive exponential.

Assume the series for \bgroup\color{black}$F$\egroup and \bgroup\color{black}$G$\egroup terminate at the same $n_r$. We can then take the equations in the coefficients and set \bgroup\color{black}$a_{n_r+1}=b_{n_r+1}=0$\egroup to get relationships between \bgroup\color{black}$a_{n_r}$\egroup and \bgroup\color{black}$b_{n_r}$\egroup.

a_{n_r}=-\sqrt{k_2\over k_1}b_{n_r} \\
b_{n_r}=-\sqrt{k_1\over k_2}a_{n_r} \\

These are the same equation, which is consistent with our assumption.

The final step is to use this result in the recursion relations for $m=n_r-1$ to find a condition on $E$ which must be satisfied for the series to terminate. Note that this choice of \bgroup\color{black}$m$\egroup connects \bgroup\color{black}$a_{n_r}$\egroup and \bgroup\color{black}$b_{n_r}$\egroup to the rest of the series giving nontrivial conditions on \bgroup\color{black}$E$\egroup. We already have the information from the next step in the recursion which gives \bgroup\color{black}$a_{n_r+1}=b_{n_r+1}=0$\egroup.

-a_m+(s+m+1-\kappa)a_{m+1}-\sqrt{k_2\over k_1}b_m+Z\alpha b_{m...
..._{n_r-1} -Z\alpha a_{n_r}-b_{n_r-1}+ (s+n_r+\kappa)b_{n_r}=0 \\

At this point we take the difference between the two equations to get one condition.

\left((s+n_r-\kappa)\sqrt{k_1\over k_2}+Z\alpha\right)a_{n_r}+...
...r+\sqrt{\left(j+{1\over 2}\right)^2-Z^2\alpha^2}\right)^2}}} \\

Using the quantum numbers from four mutually commuting operators, we have solved the radial equation in a similar way as for the non-relativistic case yielding the exact energy relation for relativistic Quantum Mechanics.

\bgroup\color{black}$\displaystyle E={mc^2\over\sqrt{1+{Z^2\alpha^2\over\left(n_r+\sqrt{\left(j+{1\over 2}\right)^2-Z^2\alpha^2}\right)^2}}}$\egroup

We can identify the standard principle quantum number in this case as \bgroup\color{black}$n=n_r+j+{1\over 2}$\egroup. This result gives the same answer as our non-relativistic calculation to order \bgroup\color{black}$\alpha^4$\egroup but is also correct to higher order. It is an exact solution to the quantum mechanics problem posed but does not include the effects of field theory, such as the Lamb shift and the anomalous magnetic moment of the electron.

Relativistic corrections become quite important for high \bgroup\color{black}$Z$\egroup atoms in which the typical velocity of electrons in the most inner shells is of order \bgroup\color{black}$Z\alpha c$\egroup.

Jim Branson 2013-04-22