Subsections


A.38 The relativistic hydrogen atom

The description of the hydrogen atom given earlier in chapter 4.3 is very accurate by engineering standards. However, it is not exact. This addendum examines various relativistic effects that were ignored in the analysis.

The approach will be to take the results of chapter 4.3 as the starting point. Then corrections are applied to them using perturbation theory as described in addendum {A.37}.


A.38.1 Introduction

According to the description of the hydrogen atom given in chapter 4.3, all energy eigenfunctions $\psi_{nlm}{\updownarrow}$ with the same value of $n$ have the same energy $E_n$. Therefore they should show up as a single line in an experimental line spectrum. But actually, when these spectra are examined very precisely, the $E_n$ energy levels for a given value of $n$ are found to consist of several closely spaced lines, rather than a single one. That is called the hydrogen atom fine structure. It means that eigenfunctions that all should have exactly the same energy, don’t.

To explain why, the solution of chapter 4.3 must be corrected for a variety of relativistic effects. Before doing so, it is helpful to express the nonrelativistic energy levels of that chapter in terms of the rest mass energy ${m_{\rm e}}c^2$ of the electron, as follows:

\begin{displaymath}
\fbox{$\displaystyle
E_n = - \frac{\alpha^2}{2 n^2} m_{\...
...frac{e^2}{4\pi\epsilon_0\hbar c}
\approx \frac1{137}
$} %
\end{displaymath} (A.247)

The constant $\alpha$ is called the “fine structure constant.” It combines the constants $e^2$$\raisebox{.5pt}{$/$}$$4\pi\epsilon_0$ from electromagnetism, $\hbar$ from quantum mechanics, and the speed of light $c$ from relativity into one nondi­men­sion­al number. It is without doubt the single most important number in all of physics, [19].

Nobody knows why it has the value that it has. Still, obviously it is a measurable value, so, following the stated ideas of quantum mechanics, maybe the universe measured this value during its early formation by a process that we may never understand, (since we do not have other measured values for $\alpha$ to deduce any properties of that process from.) If you have a demonstrably better explanation, Sweden awaits you.

In any case, for engineering purposes it is a small number, less than 1%. That makes the hydrogen energy levels really small compared to the rest mass energy of the electron, because they are proportional to the square of $\alpha$, which is as small as 0.005%. In simple terms, the electron in hydrogen stays well clear of the speed of light.

And that in turn means that the relativistic errors in the hydrogen energy levels are small. Still, even small errors can sometimes be very important. The required corrections are listed below in order of decreasing magnitude.

The following subsections discuss each correction in more detail.


A.38.2 Fine structure

From the Dirac equation, it can be seen that three terms need to be added to the nonrelativistic Hamiltonian of chapter 4.3 to correct the energy levels for relativistic effects. The three terms are worked out in derivation {D.82}. But that mathematics really provides very little insight. It is much more instructive to try to understand the corrections from a more physical point of view.

The first term is relatively easy to understand. Consider Einstein’s famous relation $E$ $\vphantom0\raisebox{1.5pt}{$=$}$ $mc^2$, where $E$ is energy, $m$ mass, and $c$ the speed of light. According to this relation, the kinetic energy of the electron is not ${\textstyle\frac{1}{2}}{m_{\rm e}}v^2$, with $v$ the velocity, as Newtonian physics says. Instead it is the difference between the energy $m_{{\rm {e}},v}c^2$ based on the mass $m_{{\rm {e}},v}$ of the electron in motion and the energy ${m_{\rm e}}c^2$ based on the mass $m_{\rm e}$ of the electron at rest. In terms of momentum $p$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m_{{\rm {e}},v}v$, chapter 1.1.2,

\begin{displaymath}
T = m_{\rm e}c^2 \sqrt{1 + \frac{p^2}{m_{\rm e}^2c^2}} - m_{\rm e}c^2
\end{displaymath} (A.248)

Since the speed of light is large compared to the typical speed of the electron, the square root can be expanded in a Taylor series, [40, 22.12], to give:

\begin{displaymath}
T \approx \frac{p^2}{2m_{\rm e}} - \frac{p^4}{8m_{\rm e}^3c^2} + \ldots
\end{displaymath}

The first term corresponds to the kinetic energy operator used in the nonrelativistic quantum solution of chapter 4.3. (It may be noted that the relativistic momentum ${\skew0\vec p}$ is based on the moving mass of the electron, not its rest mass. It is this relativistic momentum that corresponds to the operator ${\skew 4\widehat{\skew{-.5}\vec p}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar\nabla$$\raisebox{.5pt}{$/$}$${\rm i}$. So the Hamiltonian used in chapter 4.3 was a bit relativistic already, because in replacing ${\skew0\vec p}$ by $\hbar\nabla$$\raisebox{.5pt}{$/$}$${\rm i}$, it used the relativistic expression.) The second term in the Taylor series expansion above is the first of the corrections needed to fix up the hydrogen energy levels for relativity. Rewritten in terms of the square of the classical kinetic energy operator, the Bohr ground state energy $E_1$ and the fine structure constant $\alpha$, it is
\begin{displaymath}
H_{1,{\rm Einstein}}= - \frac{\alpha^{\,2}}{4\vert E_1\vert}
\left(\frac{{\widehat p}^2}{2m_{\rm e}}\right)^2 %
\end{displaymath} (A.249)

The second correction that must be added to the nonrelativistic Hamiltonian is the so-called spin-orbit interaction. In classical terms, it is due to the spin of the electron, which makes it into a magnetic dipole. Think of it as a magnet of infinitesimally small size, but with infinitely strong north and south poles to make up for it. The product of the infinitesimal vector from south to north pole times the infinite strength of the poles is finite, and defines the magnetic dipole moment $\vec\mu$. By itself, it is quite inconsequential since the magnetic dipole does not interact directly with the electric field of the nucleus. However, moving magnetic poles create an electric field just like the moving electric charges in an electromagnet create a magnetic field. The electric fields generated by the moving magnetic poles of the electron are opposite in strength, but not quite centered at the same position. Therefore they correspond to a motion-induced electric dipole. And an electric dipole does interact with the electric field of the nucleus; it wants to align itself with it. That is just like the magnetic dipole wanted to align itself with the external magnetic field in the Zeeman effect.

So how big is this effect? Well, the energy of an electric dipole $\vec\wp$ in an electric field $\skew3\vec{\cal E}$ is

\begin{displaymath}
E_{1,\mbox{\scriptsize spin-orbit}} = -\vec\wp \cdot \skew3\vec{\cal E}
\end{displaymath}

As you might guess, the electric dipole generated by the magnetic poles of the moving electron is proportional to the speed of the electron $\vec{v}$ and its magnetic dipole moment $\vec\mu$. More precisely, the electric dipole moment $\vec\wp$ will be proportional to $\vec{v}$ $\times$ $\vec\mu$ because if the vector connecting the south and north poles is parallel to the motion, you do not have two neighboring currents of magnetic poles, but a single current of both negative and positive poles that completely cancel each other out. Also, the electric field $\skew3\vec{\cal E}$ of the nucleus is minus the gradient of its potential $e$$\raisebox{.5pt}{$/$}$$4\pi\epsilon_0r$, so

\begin{displaymath}
E_{1,\mbox{\scriptsize spin-orbit}} \propto
(\vec v\times\vec\mu) \cdot \frac{e}{4\pi\epsilon_0 r^3}{\skew0\vec r}
\end{displaymath}

Now the order of the vectors in this triple product can be changed, and the dipole strength $\vec\mu$ of the electron equals its spin $\vec{S}$ times the charge per unit mass $\vphantom0\raisebox{1.5pt}{$-$}$$e$$\raisebox{.5pt}{$/$}$$m_{\rm e}$, so

\begin{displaymath}
E_{1,\mbox{\scriptsize spin-orbit}} \propto
\frac{e^2}{m...
...r^3}
\left({\skew0\vec r}\times\vec{v}\right) \cdot \vec{S}
\end{displaymath}

The expression between the parentheses is the angular momentum $\vec{L}$ save for the electron mass. The constant of proportionality is worked out in derivation {D.83}, giving the spin-orbit Hamiltonian as
\begin{displaymath}
H_{1,\mbox{\scriptsize spin-orbit}} =
\alpha^2 \vert E_1...
...^2} {\skew 4\widehat{\vec L}}\cdot{\skew 6\widehat{\vec S}} %
\end{displaymath} (A.250)

The final correction that must be added to the nonrelativistic Hamiltonian is the so-called “Darwin term:”

\begin{displaymath}
H_{1,{\rm Darwin}} = \alpha^2 \vert E_1\vert\; \pi a_0^3\delta^3({\skew0\vec r}) %
\end{displaymath} (A.251)

According to its derivation in {D.82}, it is a crude fix-up for an interaction with a virtual positron that simply cannot be included correctly in a nonrelativistic analysis.

If that is not very satisfactory, the following much more detailed derivation can be found on the web. It does succeed in explaining the Darwin term fully within the nonrelativistic picture alone. First assume that the electric potential of the nucleus does not really become infinite as 1$\raisebox{.5pt}{$/$}$$r$ at $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, but is smoothed out over some finite nuclear size. Also assume that the electron does not see this potential sharply, but perceives of its features a bit vaguely, as diffused out symmetrically over a typical distance equal to the so-called Compton wave length $\hbar$$\raisebox{.5pt}{$/$}$${m_{\rm e}}c$. There are several plausible reasons why it might: (1) the electron has illegally picked up a chunk of a negative rest mass state, and it is trembling with fear that the uncertainty in energy will be noted, moving rapidly back and forwards over a Compton wave length in a so-called Zitterbewegung; (2) the electron has decided to move at the speed of light, which is quite possible nonrelativistically, so its uncertainty in position is of the order of the Compton wave length, and it just cannot figure out where the right potential is with all that uncertainty in position and light that fails to reach it; (3) the electron needs glasses. Further assume that the Compton wave length is much smaller than the size over which the nuclear potential is smoothed out. In that case, the potential within a Compton wave length can be approximated by a second order Taylor series, and the diffusion of it over the Compton wave length will produce an error proportional to the Laplacian of the potential (the only fully symmetric combination of derivatives in the second order Taylor series.). Now if the potential is smoothed over the nuclear region, its Laplacian, giving the charge density, is known to produce a nonzero spike only within that smoothed nuclear region, figure 13.7 or (13.30). Since the nuclear size is small compared to the electron wave functions, that spike can then be approximated as a delta function. Tell all your friends you heard it here first.

The key question is now what are the changes in the hydrogen energy levels due to the three perturbations discussed above. That can be answered by perturbation theory as soon as the good eigenfunctions have been identified. Recall that the usual hydrogen energy eigenfunctions $\psi_{nlm}{\updownarrow}$ are made unique by the square angular momentum operator $\L ^2$, giving $l$, the $z$ angular momentum operator $\L _z$, giving $m$, and the spin angular momentum operator ${\widehat S}_z$ giving the spin quantum number $m_s$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\pm\frac12$ for spin up, respectively down. The decisive term whether these are good or not is the spin-orbit interaction. If the inner product in it is written out, it is

\begin{displaymath}
H_{1,\mbox{\scriptsize spin-orbit}} =
\alpha^2 \vert E_1...
...{\widehat S}_x+\L _y{\widehat S}_y+\L _z{\widehat S}_z\right)
\end{displaymath}

The radial factor is no problem; it commutes with every orbital angular momentum component, since these are purely angular derivatives, chapter 4.2.2. It also commutes with every component of spin because all spatial functions and operators do, chapter 5.5.3. As far as the dot product is concerned, it commutes with $\L ^2$ since all the components of ${\skew 4\widehat{\vec L}}$ do, chapter 4.5.4, and since all the components of ${\skew 6\widehat{\vec S}}$ commute with any spatial operator. But unfortunately, $\L _x$ and $\L _y$ do not commute with $\L _z$, and ${\widehat S}_x$ and ${\widehat S}_y$ do not commute with ${\widehat S}_z$ (chapters 4.5.4 and 5.5.3):

\begin{displaymath}[\L _x,\L _z]= - {\rm i}\hbar\L _y \quad
[\L _y,\L _z] = {\...
... [{\widehat S}_y,{\widehat S}_z] = {\rm i}\hbar{\widehat S}_x
\end{displaymath}

The quantum numbers $m$ and $m_s$ are bad.

Fortunately, ${\skew 4\widehat{\vec L}}\cdot{\skew 6\widehat{\vec S}}$ does commute with the net $z$ angular momentum ${\widehat J}_z$, defined as $\L _z+{\widehat S}_z$. Indeed, using the commutators above and the rules of chapter 4.5.4 to take apart commutators:

\begin{eqnarray*}
& & [\L _x{\widehat S}_x,\L _z+{\widehat S}_z] = [\L _x,\L _...
...\L _z]{\widehat S}_z + \L _z[{\widehat S}_z,{\widehat S}_z] = 0
\end{eqnarray*}

and adding it all up, you get $[{\skew 4\widehat{\vec L}}\cdot{\skew 6\widehat{\vec S}},{\widehat J}_z]$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. The same way of course ${\skew 4\widehat{\vec L}}\cdot{\skew 6\widehat{\vec S}}$ commutes with the other components of net angular momentum ${\skew 6\widehat{\vec J}}$, since the $z$-​axis is arbitrary. And if ${\skew 4\widehat{\vec L}}\cdot{\skew 6\widehat{\vec S}}$ commutes with every component of ${\skew 6\widehat{\vec J}}$, then it commutes with their sum of squares ${\widehat J}^2$. So, eigenfunctions of $\L ^2$, ${\widehat J}^2$, and ${\widehat J}_z$ are good eigenfunctions.

Such good eigenfunctions can be constructed from the $\psi_{nlm}{\updownarrow}$ by forming linear combinations of them that combine different $m$ and $m_s$ values. The coefficients of these good combinations are called Clebsch-Gordan coefficients and are shown for $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 and $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 in figure 12.5. Note from this figure that the quantum number $j$ of net square momentum can only equal $l+\frac12$ or $l-\frac12$. The half unit of electron spin is not big enough to change the quantum number of square orbital momentum by more than half a unit. For the rest, however, the detailed form of the good eigenfunctions is of no interest here. They will just be indicated in ket notation as $\big\vert nljm_j\big\rangle $, indicating that they have unperturbed energy $E_n$, square orbital angular momentum $l(l+1)\hbar^2$, square net (orbital plus spin) angular momentum $j(j+1)\hbar^2$, and net $z$ angular momentum $m_j\hbar$.

As far as the other two contributions to the fine structure are concerned, according to chapter 4.3.1 ${\skew 4\widehat{\skew{-.5}\vec p}}^{\,2}$ in the Einstein term consists of radial functions and radial derivatives plus $\L ^2$. These commute with the angular derivatives that make up the components of ${\skew 4\widehat{\vec L}}$, and as spatial functions and operators, they commute with the components of spin. So the Einstein Hamiltonian commutes with all components of ${\skew 4\widehat{\vec L}}$ and ${\skew 6\widehat{\vec J}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\skew 4\widehat{\vec L}}+{\skew 6\widehat{\vec S}}$, hence with $\L ^2$, ${\widehat J}^2$, and ${\widehat J}_z$. And the delta function in the Darwin term can be assumed to be the limit of a purely radial function and commutes in the same way. The eigenfunctions $\big\vert nljm_j\big\rangle $ with given values of $l$, $j$, and $m_j$ are good ones for the entire fine structure Hamiltonian.

To get the energy changes, the Hamiltonian perturbation coefficients

\begin{displaymath}
\langle m_jjln \vert
H_{1,{\rm Einstein}}+H_{1,\mbox{\scriptsize spin-orbit}}+H_{1,{\rm Darwin}}
\vert nljm_j \rangle
\end{displaymath}

must be found. Starting with the Einstein term, it is

\begin{displaymath}
\langle m_jjln\vert H_{1,{\rm Einstein}}\vert nljm_j\rangl...
...ert\frac{{\widehat p}^{\,4}}{4m_{\rm e}^2}\vert nljm_j\rangle
\end{displaymath}

Unlike what you may have read elsewhere, ${\widehat p}^{\,4}$ is indeed a Hermitian operator, but ${\widehat p}^{\,4}\vert nljm_j\rangle$ may have a delta function at the origin, (13.30), so watch it with blindly applying mathematical manipulations to it. The trick is to take half of it to the other side of the inner product, and then use the fact that the eigenfunctions satisfy the nonrelativistic energy eigenvalue problem:

\begin{eqnarray*}
\langle m_jjln \vert \frac{{\widehat p}^2}{2m_{\rm e}} \bigg...
...
& = & \langle m_jjln\vert E_n^2-2VE_n +V^2\vert nljm_j\rangle
\end{eqnarray*}

Noting from chapter 4.3 that $E_n$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E_1$$\raisebox{.5pt}{$/$}$$n^2$, $V$ $\vphantom0\raisebox{1.5pt}{$=$}$ $2E_1a_0$$\raisebox{.5pt}{$/$}$$r$ and that the expectation values of $a_0$$\raisebox{.5pt}{$/$}$$r$ and $(a_0/r)^2$ are given in derivation {D.84}, you find that

\begin{displaymath}
\langle m_jjln\vert H_{1,{\rm Einstein}}\vert nljm_j\rangl...
...pha^2}{4n^2}\left(\frac{4n}{l+\frac12}-3\right)\vert E_n\vert
\end{displaymath}

The spin-orbit energy correction is

\begin{displaymath}
\langle m_jjln\vert H_{1,\mbox{\scriptsize spin-orbit}}\ve...
...hat{\vec L}}\cdot{\skew 6\widehat{\vec S}}\vert nljm_j\rangle
\end{displaymath}

For states with no orbital angular momentum, all components of ${\skew 4\widehat{\vec L}}$ produce zero, so there is no contribution. Otherwise, the dot product ${\skew 4\widehat{\vec L}}\cdot{\skew 6\widehat{\vec S}}$ can be rewritten by expanding

\begin{displaymath}
{\widehat J}^2 = ({\skew 4\widehat{\vec L}}+{\skew 6\wideh...
...S}^2+2{\skew 4\widehat{\vec L}}\cdot{\skew 6\widehat{\vec S}}
\end{displaymath}

to give

\begin{eqnarray*}
{\skew 4\widehat{\vec L}}\cdot{\skew 6\widehat{\vec S}}\vert...
...ac{1}{2}} (1+{\textstyle\frac{1}{2}}) \Big) \vert nljm_j\rangle
\end{eqnarray*}

That leaves only the expectation value of $(a_0/r)^3$ to be determined, and that can be found in derivation {D.84}. The net result is

\begin{displaymath}
\langle m_jjln\vert H_{1,\mbox{\scriptsize spin-orbit}}\ve...
...})}{l(l+\frac12)(l+1)}\vert E_n\vert
\quad\mbox{if }l \ne 0
\end{displaymath}

or zero if $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0.

Finally the Darwin term,

\begin{displaymath}
\langle m_jjln\vert H_{1,{\rm Darwin}}\vert nljm_j\rangle ...
...langle m_jjln\vert\delta^3({\skew0\vec r})\vert nljm_j\rangle
\end{displaymath}

Now a delta function at the origin has the property to pick out the value at the origin of whatever function it is in an integral with, compare chapter 7.9.1. Derivation {D.15}, (D.9), implies that the value of the wave functions at the origin is zero unless $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, and then the value is given in (D.10). So the Darwin contribution becomes

\begin{displaymath}
\langle m_jjln\vert H_{1,{\rm Darwin}}\vert nljm_j\rangle ...
...frac{\alpha^2}{4n^2} 4n \vert E_n\vert \qquad \mbox{if $l=0$}
\end{displaymath}

To get the total energy change due to fine structure, the three contributions must be added together. For $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, add the Einstein and Darwin terms. For $l$ $\raisebox{.2pt}{$\ne$}$ 0, add the Einstein and spin-orbit terms; you will need to do the two possibilities that $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l+{\textstyle\frac{1}{2}}$ and $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l-{\textstyle\frac{1}{2}}$ separately. All three produce the same final result, anyway:

\begin{displaymath}
\fbox{$\displaystyle
E_{nljm_j,1}= - \left(\frac{1}{n(j+...
...ac{3}{4}\frac{1}{n^2}\right)
\alpha^2 \vert E_n\vert
$} %
\end{displaymath} (A.252)

Since $j+{\textstyle\frac{1}{2}}$ is at most $n$, the energy change due to fine structure is always negative. And it is the biggest fraction of $E_n$ for $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\textstyle\frac{1}{2}}$ and $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2, where it is $-\frac{5}{16}\alpha^2\vert E_n\vert$, still no more than a sixth of a percent of a percent change in energy.

In the ground state $j$ can only be one half, (the electron spin), so the ground state energy does not split into two due to fine structure. You would of course not expect so, because in empty space, both spin directions are equivalent. The ground state does show the largest absolute change in energy.

Woof.


A.38.3 Weak and intermediate Zeeman effect

The weak Zeeman effect is the effect of a magnetic field that is sufficiently weak that it leaves the fine structure energy eigenfunctions almost unchanged. The Zeeman effect is then a small perturbation on a problem in which the unperturbed (by the Zeeman effect) eigenfunctions $\big\vert nljm_j\big\rangle $ derived in the previous subsection are degenerate with respect to $l$ and $m_j$.

The Zeeman Hamiltonian

\begin{displaymath}
H_1 = \frac{e}{2m_{\rm e}}{\cal B}_{\rm ext}\left(\L _z + 2 {\widehat S}_z\right)
\end{displaymath}

commutes with both $\L ^2$ and ${\widehat J}_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\widehat S}_z+\L _z$, so the eigenfunctions $\big\vert nljm_j\big\rangle $ are good. Therefore, the energy perturbations can be found as

\begin{displaymath}
\frac{e}{2m_{\rm e}} {\cal B}_{\rm ext} \langle m_jjln\vert \L _z+2{\widehat S}_z \vert nljm_j\rangle
\end{displaymath}

To evaluate this rigorously would require that the $\big\vert nljm_j\big\rangle $ state be converted into the one or two $\psi_{nlm}{\updownarrow}$ states with $\vphantom0\raisebox{1.5pt}{$-$}$$l$ $\raisebox{-.3pt}{$\leqslant$}$ $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m_j\pm{\textstyle\frac{1}{2}}$ $\raisebox{-.3pt}{$\leqslant$}$ $l$ and $m_s$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\mp\frac12$ using the appropriate Clebsch-Gordan coefficients from figure 12.5.

However, the following simplistic derivation is usually given instead, including in this book. First get rid of $L_z$ by replacing it by ${\widehat J}_z-{\widehat S}_z$. The inner product with ${\widehat J}_z$ can then be evaluated as being $m_j\hbar$, giving the energy change as

\begin{displaymath}
\frac{e}{2m_{\rm e}} {\cal B}_{\rm ext}
\left[m_j\hbar + \langle m_jjln\vert{\widehat S}_z \vert nljm_j\rangle\right]
\end{displaymath}

For the final inner product, make a semi-classical argument that only the component of ${\skew 6\widehat{\vec S}}$ in the direction of $\vec{J}$ gives a contribution. Don’t worry that $\vec{J}$ does not exist. Just note that the component in the direction of $\vec{J}$ is constrained by the requirement that ${\skew 4\widehat{\vec L}}$ and ${\skew 6\widehat{\vec S}}$ must add up to ${\skew 6\widehat{\vec J}}$, but the component normal to $\vec{J}$ can be in any direction and presumably averages out to zero. Dismissing this component, the component in the direction of $\vec{J}$ is

\begin{displaymath}
{\skew 6\widehat{\vec S}}_J = \frac{1}{J^2}({\skew 6\wideh...
...c S}}\cdot{\skew 6\widehat{\vec J}}){\skew 6\widehat{\vec J}}
\end{displaymath}

and the dot product in it can be found from expanding

\begin{displaymath}
\L ^2 = {\skew 4\widehat{\vec L}}\cdot{\skew 4\widehat{\ve...
... {\skew 6\widehat{\vec J}}\cdot{\skew 6\widehat{\vec S}}+ S^2
\end{displaymath}

to give

\begin{displaymath}
{\skew 6\widehat{\vec S}}_J = \frac{J^2 - \L ^2 + S^2}{2J^2}{\skew 6\widehat{\vec J}}
\end{displaymath}

For a given eigenfunction $\big\vert nljm_j\big\rangle $, $J^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar^2j(j+1)$, $\L ^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar^2l(l+1)$, and $S^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar^2s(s+1)$ with $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12$.

If the $z$-​component of ${\skew 6\widehat{\vec S}}_J$ is substituted for ${\widehat S}_z$ in the expression for the Hamiltonian perturbation coefficients, the energy changes are

\begin{displaymath}
\left[1+\frac{j(j+1)-l(l+1)+s(s+1)}{2j(j+1)}\right]
\frac{e\hbar}{2m_{\rm e}} {\cal B}_{\rm ext} m_j %
\end{displaymath} (A.253)

(Rigorous analysis using figure 12.5, or more generally item 2 in chapter 12.8, produces the same results.) The factor within the brackets is called the “Landé $g$-​factor.” It is the factor by which the magnetic moment of the electron in the atom is larger than for a classical particle with the same charge and total angular momentum. It generalizes the $g$-​factor of the electron in isolation to include the effect of orbital angular momentum. Note that it equals 2, the Dirac $g$-​factor, if there is no orbital momentum, and 1, the classical value, if the orbital momentum is so large that the half unit of spin can be ignored.

In the intermediate Zeeman effect, the fine structure and Zeeman effects are comparable in size. The dominant perturbation Hamiltonian is now the combination of the fine structure and Zeeman ones. Since the Zeeman part does not commute with ${\widehat J}^2$, the eigenfunctions $\big\vert nljm_j\big\rangle $ are no longer good. Eigenfunctions with the same values of $l$ and $m_j$, but different values of $j$ must be combined into good combinations. For example, if you look at $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2, the eigenfunctions $\big\vert 21{\textstyle\frac{3}{2}}{\textstyle\frac{1}{2}}\big\rangle $ and $\big\vert 21{\textstyle\frac{1}{2}}{\textstyle\frac{1}{2}}\big\rangle $ have the same unperturbed energy and good quantum numbers $l$ and $m_j$. You will have to write a two by two matrix of Hamiltonian perturbation coefficients for them, as in addendum {A.37.3}, to find the good combinations and their energy changes. And the same for the $\big\vert 21{\textstyle\frac{3}{2}}\rule[2.5pt]{5pt}{.5pt}{\textstyle\frac{1}{2}}\big\rangle $ and $\big\vert 21{\textstyle\frac{1}{2}}\rule[2.5pt]{5pt}{.5pt}{\textstyle\frac{1}{2}}\big\rangle $ eigenfunctions. To obtain the matrix coefficients, use the Clebsch-Gordan coefficients from figure 12.5 to evaluate the effect of the Zeeman part. The fine structure contributions to the matrices are given by (A.252) when the $j$ values are equal, and zero otherwise. This can be seen from the fact that the energy changes must be the fine structure ones when there is no magnetic field; note that $j$ is a good quantum number for the fine structure part, so its perturbation coefficients involving different $j$ values are zero.


A.38.4 Lamb shift

A famous experiment by Lamb & Retherford in 1947 showed that the hydrogen atom state $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2, $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12$, also called the 2S$_{1/2}$ state, has a somewhat different energy than the state $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2, $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12$, also called the 2P$_{1/2}$ state. That was unexpected, because even allowing for the relativistic fine structure correction, states with the same principal quantum number $n$ and same total angular momentum quantum number $j$ should have the same energy. The difference in orbital angular momentum quantum number $l$ should not affect the energy.

The cause of the unexpected energy difference is called Lamb shift. To explain why it occurs would require quantum electrodynamics, and that is well beyond the scope of this book. Roughly speaking, the effect is due to a variety of interactions with virtual photons and electron/positron pairs. A good qualitative discussion on a nontechnical level is given by Feynman [19].

Here it must suffice to list the approximate energy corrections involved. For states with zero orbital angular momentum, the energy change due to Lamb shift is

\begin{displaymath}
E_{{\vec n},1,{\rm Lamb}} = -\frac{\alpha^3}{2n} k(n,0) E_n
\qquad \mbox{if } l = 0 %
\end{displaymath} (A.254)

where $k(n,0)$ is a numerical factor that varies a bit with $n$ from about 12.7 to 13.2. For states with nonzero orbital angular momentum,
\begin{displaymath}
E_{{\vec n},1,{\rm Lamb}} = -\frac{\alpha^3}{2n}
\left[k...
...} l \ne 0
\mbox { and } j = l \pm {\textstyle\frac{1}{2}} %
\end{displaymath} (A.255)

where $k(n,l)$ is less than 0.05 and varies somewhat with $n$ and $l$.

It follows that the energy change is really small for states with nonzero orbital angular momentum, which includes the 2P$_{1/2}$ state. The change is biggest for the 2S$_{1/2}$ state, the other state in the Lamb & Retherford experiment. (True, the correction would be bigger still for the ground state $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, but since there are no states with nonzero angular momentum in the ground state, there is no splitting of spectral lines involved there.)

Qualitatively, the reason that the Lamb shift is small for states with nonzero angular momentum has to do with distance from the nucleus. The nontrivial effects of the cloud of virtual particles around the electron are most pronounced in the strong electric field very close to the nucleus. In states of nonzero angular momentum, the wave function is zero at the nucleus, (D.9). So in those states the electron is unlikely to be found very close to the nucleus. In states of zero angular momentum, the square magnitude of the wave function is 1$\raisebox{.5pt}{$/$}$$n^3{\pi}a_0^3$ at the nucleus, reflected in both the much larger Lamb shift as well as its approximate 1$\raisebox{.5pt}{$/$}$$n^3$ dependence on the principal quantum number $n$.


A.38.5 Hyperfine splitting

Hyperfine splitting of the hydrogen atom energy levels is due to the fact that the nucleus acts as a little magnet just like the electron. The single-proton nucleus and electron have magnetic dipole moments due to their spin equal to

\begin{displaymath}
\vec \mu_{\rm {p}} = \frac{g_pe}{2m_{\rm p}} {\skew 6\wide...
...- \frac{g_ee}{2m_{\rm e}} {\skew 6\widehat{\vec S}}_{\rm {e}}
\end{displaymath}

in which the $g$-​factor of the proton is about 5.59 and that of the electron 2. The magnetic moment of the nucleus is much less than the one of the electron, since the much greater proton mass appears in the denominator. That makes the energy changes associated with hyperfine splitting really small compared to other effects such as fine structure.

This discussion will restrict itself to the ground state, which is by far the most important case. For the ground state, there is no orbital contribution to the magnetic field of the electron. There is only a spin-spin coupling between the magnetic moments of the electron and proton, The energy involved can be thought of most simply as the energy $-\vec\mu_{\rm {e}}\cdot\skew2\vec{\cal B}_p$ of the electron in the magnetic field $\skew2\vec{\cal B}_{\rm {p}}$ of the nucleus. If the nucleus is modelled as an infinitesimally small electromagnet, its magnetic field is that of an ideal current dipole as given in table 13.2. The perturbation Hamiltonian then becomes

\begin{displaymath}
H_{1,\mbox{\scriptsize spin-spin}} =
\frac{g_{\rm {p}}g_...
...hat{\vec S}}_{\rm {e}})}{3}\delta^3({\skew0\vec r})
\right]
\end{displaymath}

The good states are not immediately self-evident, so the four unperturbed ground states will just be taken to be the ones which the electron and proton spins combine into the triplet or singlet states of chapter 5.5.6:

\begin{displaymath}
\mbox{triplet:}\quad\psi_{100}\big\vert 1\:1\big\rangle \q...
... \qquad\mbox{singlet:}\quad\psi_{100}\big\vert\:0\big\rangle
\end{displaymath}

or $\psi_{100}\big\vert s_{\rm {net}}m_{\rm {net}}\big\rangle $ for short, where $s_{\rm {net}}$ and $m_{\rm {net}}$ are the quantum numbers of net spin and its $z$-​component. The next step is to evaluate the four by four matrix of Hamiltonian perturbation coefficients

\begin{displaymath}
\langle {\underline m}_{\rm net}{\underline s}_{\rm net}\v...
... spin-spin}}
\psi_{100} \vert s_{\rm net}m_{\rm net}\rangle
\end{displaymath}

using these states.

Now the first term in the spin-spin Hamiltonian does not produce a contribution to the perturbation coefficients. The reason is that the inner product of the perturbation coefficients written in spherical coordinates involves an integration over the surfaces of constant $r$. The ground state eigenfunction $\psi_{100}$ is constant on these surfaces. So there will be terms like $3{\widehat S}_{{\rm {p}},x}{\widehat S}_{{\rm {e}},y}xy$ in the integration, and those are zero because $x$ is just as much negative as positive on these spherical surfaces, (as is $y$). There will also be terms like $3{\widehat S}_{{\rm {p}},x}{\widehat S}_{{\rm {e}},x}x^2-{\widehat S}_{{\rm {p}},x}{\widehat S}_{{\rm {e}},x}r^2$ in the integration. These will be zero too because by symmetry the averages of $x^2$, $y^2$, and $z^2$ are equal on the spherical surfaces, each equal to one third the average of $r^2$.

So only the second term in the Hamiltonian survives, and the Hamiltonian perturbation coefficients become

\begin{displaymath}
\frac{g_{\rm {p}}g_{\rm {e}}e^2}{6m_{\rm e}m_{\rm p}\epsil...
...kew0\vec r})
\psi_{100} \vert s_{\rm net}m_{\rm net}\rangle
\end{displaymath}

The spatial integration in this inner product merely picks out the value $\psi_{100}^2(0)$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1$\raisebox{.5pt}{$/$}$${\pi}a_0^3$ at the origin, as delta functions do. That leaves the sum over the spin states. According to addendum {A.10},

\begin{displaymath}
\mbox{triplet:}\quad {\skew 6\widehat{\vec S}}_{\rm {p}}\c...
...
-{\textstyle\frac{3}{4}} \hbar^2 \big\vert\,0\big\rangle
\end{displaymath}

Since the triplet and singlet spin states are orthonormal, only the Hamiltonian perturbation coefficients for which ${\underline s}_{\rm {net}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $s_{\rm {net}}$ and ${\underline m}_{\rm {net}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m_{\rm {net}}$ survive, and these then give the leading order changes in the energy.

Plugging it all in and rewriting in terms of the Bohr energy and fine structure constant, the energy changes are:

\begin{displaymath}
\mbox{triplet: } E_{1,\mbox{\scriptsize spin-spin}} =
{\...
...g_{\rm {e}} \frac{m_{\rm e}}{m_{\rm p}}\alpha^2\vert E_1\vert
\end{displaymath} (A.256)

The energy of the triplet states is raised and that of the singlet state is lowered. Therefore, in the true ground state, the electron and proton spins combine into the singlet state. If they somehow get kicked into a triplet state, they will eventually transition back to the ground state, say after 10 million years or so, and release a photon. Since the difference between the two energies is so tiny on account of the very small values of both $\alpha^2$ and $m_{\rm e}$$\raisebox{.5pt}{$/$}$$m_{\rm p}$, this will be a very low energy photon. Its wave length is as long as 0.21 m, producing the 21 cm hydrogen line.