Last time, we did a lightning review of the hydrogen atom and first-order perturbation theory. We considered the corrections to the hydrogen spectrum due to the finite size of the nucleus, and found them to be utterly tiny (although potentially larger in atoms with large \( Z \) or muonic atoms.)
Let's keep going and dive into another example.
Example: relativistic corrections to kinetic energy
This time, we'll modify the kinetic term instead of the potential. The kinetic energy term
\[ \begin{aligned} T = \frac{p^2}{2m} \end{aligned} \]
is only correct non-relativistically; we know that the true kinetic energy of a relativistic particle is
\[ \begin{aligned} T = \sqrt{p^2 c^2 + m^2 c^4} - mc^2. \end{aligned} \]
Fortunately, we don't need to consider a fully relativistic version of quantum mechanics for this particular application. The speed of the electron is relatively small in the hydrogen atom, \( v/c \sim \alpha \), so we can expand the kinetic energy as a series in \( p \):
\[ \begin{aligned} T \approx \frac{p^2}{2m_e} - \frac{p^4}{8m_e^3c^2} + ... \end{aligned} \]
and treat the new term as a perturbation, so
\[ \begin{aligned} \hat{W} = -\frac{\hat{p}^4}{8m_e^3 c^2}. \end{aligned} \]
(Aside: this is, in principle, a degenerate-state perturbation theory problem, since the hydrogen states have degeneracy \( n^2 \). However, for this particular perturbation we can easily see that
\[ \begin{aligned} [\hat{\vec{L}}, \hat{W}] = 0 \end{aligned} \]
(and thus \( [\hat{L}^2, \hat{W}] = 0 \) as well), which means that as long as we work in the angular momentum eigenbasis \( \ket{nlm} \), the perturbation is diagonal and all ambiguities are removed.) So we can proceed to calculating the first-order energy shift as the usual expectation value,
\[ \begin{aligned} E_{nl}^{(1)} = \bra{nlm} \hat{W} \ket{nlm} = -\frac{1}{8m_e^3 c^2} \ev{(\hat{p}{}^2)^2}. \end{aligned} \]
It's possible to evaluate the expectation value by brute force, particularly if we're given a particular \( \ket{nlm} \) state, but some tricky rewriting will let us quickly find the right answer. Since the Hamiltonian contains two terms, the \( \hat{p}^2 \) kinetic term and the \( 1/r \) Coulomb potential, we can trade one for the other:
\[ \begin{aligned} \ev{\hat{W}} = -\frac{1}{2m_e c^2} \ev{ \left( \frac{\hat{p}^2}{2m_e}\right)^2} \\ = -\frac{1}{2m_e c^2} \ev{\left( \hat{H}_0 + \frac{Ze^2}{r} \right)^2} \\ = -\frac{1}{2m_e c^2} \left( (E_n^{(0)})^2 + 2E_n^{(0)} Ze^2 \ev{\frac{1}{r}} + Z^2 e^4 \ev{\frac{1}{r^2}} \right). \end{aligned} \]
The first expectation value \( \ev{1/r} \) is easy to calculate. Remember that the virial theorem relates the expectation values of kinetic and potential energy,
\[ \begin{aligned} 2\ev{T} = -\ev{V}. \end{aligned} \]
But we also know that \( \ev{T} = \ev{H} - \ev{V} = E_n^{(0)} - \ev{V} \), so \( \ev{V} = 2E_n^{(0)} \), i.e.
\[ \begin{aligned} \ev{\frac{1}{r}} = \frac{-2E_n^{(0)}}{Ze^2}. \end{aligned} \]
Let's take a quick aside to derive a very useful litle theorem, called the Feynman-Hellmann theorem. The theorem allows us to calculate expectation values by taking derivatives of energy eigenvalues:
\[ \begin{aligned} \frac{dE_\lambda}{d\lambda} = \bra{E_{\lambda}} \frac{d\hat{H}}{d\lambda} \ket{E_{\lambda}}. \end{aligned} \]
Here \( \lambda \) can be any parameter that appears in the Hamiltonian; the eigenstates and eigenenergies then depend implicitly on \( \lambda \) as well. To derive this result, we start by observing that
\[ \begin{aligned} E_\lambda = \bra{E_{\lambda}} \hat{H}_{\lambda} \ket{E_{\lambda}} \end{aligned} \]
and then differentiating both sides:
\[ \begin{aligned} \frac{dE_{\lambda}}{d\lambda} = \left( \frac{d}{d\lambda} \bra{E_{\lambda}}\right) \hat{H}_{\lambda} \ket{E_{\lambda}} + \bra{E_{\lambda}} \frac{d\hat{H}_{\lambda}}{d\lambda} \ket{E_{\lambda}} + \bra{E_{\lambda}} \hat{H}_{\lambda} \left( \frac{d}{d\lambda} \ket{E_{\lambda}} \right). \end{aligned} \]
In the first and third terms, we can act \( \hat{H}{\lambda} \) on the energy eigenstates and pull out \( E{\lambda} \):
\[ \begin{aligned} \frac{dE_{\lambda}}{d\lambda} = \bra{E_{\lambda}} \frac{d\hat{H}_{\lambda}}{d\lambda} \ket{E_{\lambda}} + E_{\lambda} \left[ \left( \frac{d}{d\lambda} \bra{E_{\lambda}} \right) \ket{E_{\lambda}} + \bra{E_{\lambda}} \left( \frac{d}{d\lambda} \ket{E_{\lambda}} \right) \right] \\ = \bra{E_{\lambda}} \frac{d\hat{H}_{\lambda}}{d\lambda} \ket{E_{\lambda}} + E_{\lambda} \frac{d}{d\lambda} (\sprod{E_{\lambda}}{E_{\lambda}}). \end{aligned} \]
But the inner product of the state with itself \( \sprod{E_\lambda}{E_\lambda} = 1 \), so the last term vanishes, and we have proven the theorem.
The Feynman-Hellmann theorem is a simple but powerful result for evaluating certain expectation values. To apply it to the hydrogen atom, let's write out the Hamiltonian in position space:
\[ \begin{aligned} \hat{H} = \frac{\hat{p}^2}{2m_e} - \frac{Ze^2}{\hat{r}} \\ \Rightarrow \bra{\hat{\vec{r}}} \hat{H} \ket{nlm} = -\frac{\hbar^2}{2m_er^2} \left[ \frac{d}{dr} \left( r^2 \frac{d}{dr} \right) - l(l+1) \right] \psi_{nlm}(r) - \frac{Ze^2}{r} \psi_{nlm}(r). \end{aligned} \]
Now, here's the tricky part. The Feynman-Hellmann theorem is a purely mathematical statement about the expectation values; it essentially doesn't "know about" the physics. In particular, it is perfectly valid to apply it to the angular momentum eigenvalue \( l \):
\[ \begin{aligned} \frac{dE_{nlm}}{dl} = \bra{nlm} \frac{d\hat{H}}{dl} \ket{nlm} \\ = \frac{\hbar^2 (2l+1)}{2m_e} \ev{\frac{1}{r^2}}. \end{aligned} \]
Again, we can differentiate like this even though we know that only integer values of \( l \) are physical; we're making a purely mathematical statement.
What about the left-hand side? As we know, the energies depend solely on the value of principal quantum number \( n \):
\[ \begin{aligned} E_{nlm} = -\frac{Z^2}{n^2}\ \textrm{Ry} = -\frac{Z^2 e^2}{2a_0 n^2}. \end{aligned} \]
This looks independent of \( l \), and in fact we've emphasized that for a given value of \( n \), the corresponding angular momentum levels are degenerate. However, mathematically we remember that \( n \) is a just a convenient rewriting of the true radial and angular quantum numbers,
\[ \begin{aligned} n = q + l + 1. \end{aligned} \]
So
\[ \begin{aligned} \frac{dE_{nlm}}{dl} = \frac{dE_{nlm}}{dn} = \frac{Z^2 e^2}{a_0 n^3}. \end{aligned} \]
Substituting back in, we find that
\[ \begin{aligned} \ev{\frac{1}{r^2}} = \frac{2m_e Z^2 e^2}{a_0 \hbar^2 n^3 (2l+1)} \\ = \frac{1}{a_0^2 n^3 (l+1/2)}, \end{aligned} \]
recalling the definition of the Bohr radius \( a_0 = \hbar^2 / m_e e^2 \). Notice that we could also have found \( \ev{1/r} \) through the Feynman-Hellmann theorem, by applying it to the electric charge; it's worth going through the exercise at home if you're still uncertain about this derivation.
Finally, back to the relativistic correction: we now have
\[ \begin{aligned} \ev{\hat{W}} = -\frac{1}{2m_e c^2} \left( -3 (E_n^{(0)})^2 + Z^4 e^4 \left( \frac{1}{a_0^2 n^3 (l+1/2)}\right) \right) \\ = -\frac{Z^4 e^4}{2m_e c^2 n^3} \left( \frac{-3}{4 n a_0^2} + \frac{1}{a_0^2 (l+1/2)} \right) \\ = \frac{Z^4 e^2}{m_e c^2 n^3 a_0} \left( \frac{3}{4n} - \frac{1}{l+1/2} \right)\ \textrm{Ry} \end{aligned} \]
Let's try to combine some of the parameters out front, with a preference for dimensionless numbers like \( \alpha \). We can see that
\[ \begin{aligned} \frac{e^2}{m_e c^2 a_0} = \frac{e^4}{\hbar^2 c^2} = \alpha^2, \end{aligned} \]
so finally, the relativistic energy shift at first order is
\[ \begin{aligned} \Delta_n^{(1)} = \frac{Z^4 \alpha^2}{n^3} \left[ \frac{3}{4n} - \frac{1}{l+1/2} \right]\ \textrm{Ry}. \end{aligned} \]
Since the ground-state energy itself scales as \( Z^2 \), we see that the relative correction here is on the order of \( Z^2 \alpha^2 \). For hydrogen itself this is on the order of a part per ten thousand, which is quite small; it is one of the many terms that make up the so-called "hyperfine corrections" to the spectrum of hydrogen. Clearly for larger nuclei, the relativistic shift will be more significant; the stronger Coulomb potential for larger \( Z \) pulls in the electron orbits and gives them a higher average speed.
Fine and hyperfine structure of hydrogen
Now let's move on to a more organized and complete discussion of corrections to the hydrogen atom's energy levels.
There are a number of physical effects which give small corrections to the exact quantum mechanical solution for hydrogen; these corrections are known as the fine (small) and hyperfine (really small) structure of the hydrogen atom. (A brief aside: actually, there is an exact solution which incorporates all of these effects, but it requires a relativistic formulation of quantum mechanics using the Dirac equation. I'll come back to this again at the end of our study.)
I will from here on out set \( Z=1 \), specializing to the hydrogen atom and not just single-electron atoms. For atoms with larger \( Z \) there are other corrections which can change the relative magnitude of these terms or introduce new effects to deal with. We will have to include the spin of the electron in our calculations now, so we label the hydrogen eigenstates as \( \ket{n, l, m, m_s} \).
We begin with the fine structure. There are three terms we must consider, which give rise to corrections of roughly the same magnitude:
\[ \begin{aligned} \hat{H}_{fine} = \hat{H}_0 + \hat{W}_{rel} + \hat{W}_{SO} + \hat{W}_D. \end{aligned} \]
\( \hat{H}0 \) here is the kinetic term and the Coulomb interaction. The next term, \( \hat{W}{rel} \), contains the leading relativistic correction to the kinetic energy. We looked at this perturbation just above, and found it to give energy corrections of order \( \alpha^2 \).
The second term is the spin-orbit coupling,
\[ \begin{aligned} \hat{W}_{SO} = \frac{e^2}{2m_e^2 c^2 r^3} \hat{\vec{L}} \cdot \hat{\vec{S}} \end{aligned} \]
Physically, this term arises from the interaction of the electron's magnetic moment with the magnetic field which the electron "sees" due to its motion through the proton's static electric field. (As you know from E&M, this is again a relativistic correction.)
We have studied this interaction before, when considering addition of angular momentum. We rewrote this in terms of the total angular momentum \( \hat{\vec{J}} = \hat{\vec{L}} + \hat{\vec{S}} \),
\[ \begin{aligned} \hat{\vec{L}} \cdot \hat{\vec{S}} = \frac{1}{2} \left( \hat{J}^2 - \hat{L}^2 - \hat{S}^2 \right). \end{aligned} \]
We advertised this rewriting as a convenient change of basis before, but it's actually necessary in the context of perturbation theory, to avoid running into the degenerate state problem; working in the \( \hat{\vec{J}} \) eigenbasis diagonalizes the spin-orbit term.
Relative to the Coulomb potential \( e^2 / r \), the size of the spin-orbit term can be written
\[ \begin{aligned} \frac{W_{SO}}{V_0} = \frac{\hbar^2}{m_e^2 c^2 r^2} \sim \frac{\hbar^2}{m_e^2 c^2 a_0^2} = \alpha^2, \end{aligned} \]
where we're making a rough approximation by assuming that \( r \) is most probably the Bohr radius \( a_0 \). This leads to another result proportional to \( \alpha^2 \), like the relativistic correction. (You can probably guess now why \( \alpha \) is the "fine-structure constant".)
The third term is known as the Darwin term,
\[ \begin{aligned} \hat{W}_D = -\frac{\hbar^2}{8m_e^2 c^2} \nabla^2 V(\hat{\vec{r}}) = \frac{\pi e^2 \hbar^2}{2m_e^2 c^2} \delta(\hat{\vec{r}}), \end{aligned} \]
since the Laplacian of the Coulomb \( 1/r \) potential is zero everywhere except the origin. This is a very mysterious-looking term, and it doesn't really have a nice physical story unlike the previous two interactions. In fact, this term was first written down by Sir Charles Darwin (grandson of the other Darwin) after he solved the Dirac equation in full, and then discovered it in the non-relativistic expansion.
The Darwin term comes from the motion of the electron itself; there is really no good non-relativistic justification for it, in my opinion. It is a purely relativistic effect, which arises from the fact that a relativistic quantum particle with spin does not conserve orbital angular momentum when it moves; the result of this is that if we study the evolution of the position operator \( \hat{\vec{r}}(t) \), we will find a "jittering" of the motion about the classical path.
(Like many hard-to-understand physics concepts, this one has a fancy German word associated with it; the phenomenon is called zitterbewegung, or "trembling motion".) This instrinsic uncertainty in the motion of the electron causes it to be sensitive to the potential \( V(r) \) non-locally when we expand non-relativistically, with the Darwin term proportional to \( \nabla^2 V(r) \) being the leading effect.
You can find some slightly more rigorous attempts to justify the Darwin term non-relativistically if you look through the literature; in my opinion, you won't gain much from any of them. Understanding the Dirac equation itself is probably the best way you can feel better about the Darwin term, but we won't get to that this semester.
We can see that the expectation value of the Darwin term will be zero unless we have \( \psi(0) \neq 0 \), which only holds for \( s \) orbitals; this correction is zero for higher values of \( l \). For the \( s \) orbitals, we can see that
\[ \begin{aligned} \ev{\hat{W}_D} = \frac{\pi e^2 \hbar^2}{2m_e^2 c^2} |\psi(0)|^2. \end{aligned} \]
If you go back to the general formula for the hydrogen radial wavefunctions, you'll find that they simplify drastically if \( l=0 \) and \( r=0 \); we find some terms depending on \( n \), and then a factor of \( 1/\sqrt{\pi a0^{3}} \). (We could have guessed this from dimensional analysis, since we must have \( \int dr r^2 |R{nl}(r)|^2 = 1 \).) Substituting in and simplifying, we find that since \( a_0^3 = \hbar^6 / m_e^3 e^6 \),
\[ \begin{aligned} \ev{\hat{W}_D} = m_e c^2 \frac{e^8}{\hbar^4 c^4} = m_e c^2 \alpha^4. \end{aligned} \]
Expressing the leading-order energy in the same units, we see
\[ \begin{aligned} \ev{H_0} \sim \frac{e^2}{a_0} = \frac{m_e e^4}{\hbar^2} = m_e c^2 \alpha^2, \end{aligned} \]
so once again the correction is of order \( \alpha^2 \) times the Rydberg.
Return to \( n=2 \) hydrogen: fine structure
Now let's return to an explicit study of the \( n=2 \) energy level of hydrogen, and work out all of the fine-structure corrections. Most of these corrections are not difficult to find for general \( n \), but the number of states to deal with proliferates, so we'll focus on \( n=2 \) to be concrete; our results here are easily extended to higher \( n \).
We should begin by noting that none of the fine structure perturbations connect states with different \( l \); this is an obvious statement for any energy level, not just \( n=2 \), since it's easy to see that \( [\hat{L}^2, \hat{W}_{fine}] = 0 \).
Better yet, if we use the appropriate angular-momentum basis, the perturbation \( \hat{W}_{fine} \) is in fact automatically diagonal. For the kinetic energy correction and the Darwin term, this statement is fairly trivial; these operators are both scalars (rank-0 spherical tensors), so they only connect matrix elements with equal \( l \), \( m \), and \( m_s \).
Let's go through and calculate the explicit corrections.
Corrections to the \( 2s \) energy
We found above a general formula for the first-order energy correction due to the relativistic \( \hat{p}^4 \) kinetic term:
\[ \begin{aligned} \ev{\hat{W}_{rel}} = \frac{\alpha^2}{n^3} \left[ \frac{3}{4n} - \frac{1}{l+1/2} \right]\ \textrm{Ry} \end{aligned} \]
So for the \( 2s \) level, we have
\[ \begin{aligned} \ev{\hat{W}_{rel}}_{2s} = -\frac{13}{64}\ \alpha^2 \textrm{Ry} \end{aligned} \]
The Darwin term, which involves a delta function in position space, is non-zero here since \( \psi(0) \neq 0 \) in the \( s \)-wave. There's no good trick to evaluate it, we simply have to plug in the explicit form of the wavefunction; we find after simplifying that
\[ \begin{aligned} \ev{\hat{W}_D}_{2s} = \frac{1}{8}\ \alpha^2 \textrm{Ry} \end{aligned} \]
Finally, the spin-orbit term is zero in this state. Since \( l=0 \), the expectation value of \( \hat{\vec{L}} \cdot \hat{\vec{S}} \) is proportional to the matrix elements
\[ \begin{aligned} \bra{l=0,m=0} \hat{L}_{x,y,z} \ket{l=0,m=0} \end{aligned} \]
which uniformly vanish. Thus,
\[ \begin{aligned} \ev{\hat{W}_{SO}}_{2s} = 0. \end{aligned} \]
Corrections to the \( 2p \) energy
This time, we'll start with the Darwin term: since for any \( l>0 \) hydrogen state the wavefunction is zero at the origin, we have immediately
\[ \begin{aligned} \ev{\hat{W}_D}_{2p} = 0. \end{aligned} \]
The kinetic energy correction is easy, we just have to plug in to our formula above once more:
\[ \begin{aligned} \ev{\hat{W}_{rel}}_{2p} = -\frac{7}{192}\ \alpha^2 \textrm{Ry} \end{aligned} \]
Here the spin-orbit coupling is non-zero, and requires slightly more effort to calculate. As we know, to deal with this perturbation we must work in the \( \ket{nljm} \) eigenbasis, in which case we can write out the desired expectation value as
\[ \begin{aligned} \ev{\hat{\vec{L}} \cdot \hat{\vec{S}} } \propto \frac{1}{2} \hbar^2 \left[ j(j+1) - l(l+1) - s(s+1) \right] \end{aligned} \]
However, this isn't the whole story; the spin-orbit coupling itself contains not just this operator, but also a factor of \( 1/r^3 \), so we need to evaluate a radial integral. Note that we can't use the Feynman-Hellmann theorem to obtain \( \ev{1/r^3} \). There does exist a recursion relation between expectation values of various powers of \( r \) in hydrogen, known as the Kramers relation, which can be used to obtain all such expectation values based on the ones we found with Feynman-Hellmann; its derivation is tedious and we're focused on the \( n=2 \) state anyway, so I'll just integrate explicitly.
You can verify that taking the \( R_{21}(r) \) function and integrating with the \( 1/r^3 \) from the spin-orbit term, after the dust settles you find that
\[ \begin{aligned} \ev{\hat{W}_{SO}} = \frac{\alpha^2}{48 \hbar^2} \ev{\hat{\vec{L}} \cdot \hat{\vec{S}}}\ \textrm{Ry}, \end{aligned} \]
so plugging in the angular momentum values, we see that
\[ \begin{aligned} \ev{\hat{W}_{SO}}_{2p,j=1/2} = -\frac{1}{24} \alpha^2\ \textrm{Ry} \\ \ev{\hat{W}_{SO}}_{2p,j=3/2} = \frac{1}{48} \alpha^2\ \textrm{Ry} \end{aligned} \]
Energy levels and the Lamb shift
We combine all of the above results to find the energy levels of \( n=2 \) hydrogen at leading order in \( \alpha \):
\[ \begin{aligned} \Delta E_{2s} = \left( -\frac{13}{64} + \frac{8}{64} \right) \alpha^2\ \textrm{Ry} \\ = -\frac{5}{64}\alpha^2 \ \textrm{Ry}; \\ \Delta E_{2p,j=1/2} = \left( -\frac{7}{192} - \frac{8}{192} \right) \alpha^2\ \textrm{Ry} \\ = -\frac{5}{64}\alpha^2 \ \textrm{Ry}; \\ \Delta E_{2p,j=3/2} = \left( -\frac{7}{192} + \frac{4}{192} \right) \alpha^2\ \textrm{Ry} \\ = -\frac{1}{64} \alpha^2\ \textrm{Ry}. \end{aligned} \]
Notice first of all that the splitting between the \( 2p \) energy levels is entirely determined by the spin-orbit coupling; the other effects don't depend on the value of \( j \).
A striking fact about these energy levels is that the \( 2s \) and \( 2p_{1/2} \) levels are degenerate, despite the fact that their energy corrections include contributions from seemingly different sources (the Darwin term vs. spin-orbit coupling)! This is sometimes called an "accidental" degeneracy, since there's no apparent symmetry that forces the two levels to be equal.
However, the degeneracy isn't really accidental. Remember that all of this fine structure comes from a non-relativistic expansion, and underlying it all is an exact relativistic solution using the Dirac equation. We won't get to the equation itself, but the exact energy level is given by
\[ \begin{aligned} E_{\textrm{Dirac}} = m_e c^2 \left[ 1 + \alpha^2 \left(n - j - \frac{1}{2} + \sqrt{(j+1/2)^2 - \alpha^2} \right)^{-2} \right]^{-1/2} \end{aligned} \]
This reveals the reason for the degeneracy; energies in hydrogen from the Dirac equation depend on \( n \) and \( j \), and the two levels which are degenerate here both have \( j = 1/2 \).