Last time, we derived the quantum-mechanical version of the multipole expansion: for a potential \( V(\hat{\vec{r}}) \), we can rewrite it as the infinite sum
\[ \begin{aligned} V(\hat{\vec{r}}) = \sum_{l=0}^{\infty} \sum_{m=-l}^{l} c_{lm} \hat{Q}_m^{(l)}, \end{aligned} \]
where the multipole operators (multipole moments) \(\hat{Q}_m^{(l)}\) are defined by the relation
\[ \begin{aligned} \bra{\vec{r}'} \hat{Q}_m^{(l)} \ket{\vec{r}} = q \sqrt{\frac{4\pi}{2l+1}} r^l Y_l^m(\theta, \phi) \delta(\vec{r} - \vec{r}'). \end{aligned} \]
so that for example, the \( m=0 \) component of the \( l=2 \) quadrupole tensor is
\[ \begin{aligned} \hat{Q}_0^{(2)} = \frac{1}{2} q (3z^2 - r^2). \end{aligned} \]
Note that by construction, the multipole operator \( \hat{Q}^{(l)} \) is a rank-\( l \) spherical tensor. These multipole moments will often be particularly interesting for composite systems, where we have more than one particle to deal with. The modification turns out to be very simple: for a system of \( N \) particles, all of the multipole moments just add:
\[ \begin{aligned} \bra{\vec{r}'_1, \vec{r}'_2, ..., \vec{r}'_N} \hat{Q}_m^{(l)} \ket{\vec{r}_1, \vec{r}_2, ..., \vec{r}_N} = \sqrt{\frac{4\pi}{2l+1}} \sum_{n=1}^N q_n (r_n)^l Y_l^m(\theta_n, \phi_n). \end{aligned} \]
There exists a similar expansion for the magnetic multipole moments. We've been working in position space, but it's easy to see that since these things are all well-organized into spherical tensors, their matrix elements will be simpler in terms of the angular momentum eigenstates. By the Wigner-Eckart theorem,
\[ \begin{aligned} \bra{\alpha', l_2, m_2} \hat{Q}_m^{(l)} \ket{\alpha, l_1, m_2} = \sprod{l_1, l; m_1, m}{l_1, l; l_2, m_2} \frac{\bra{\alpha', l_2}|\hat{Q}^{(l)}|\ket{\alpha, l_1}}{\sqrt{2l_2+1}} \end{aligned} \]
If someone tries to tell you about "the dipole moment" or "the quadrupole moment" of a particle, what they really mean is the reduced matrix element here, or something proportional to it. The reduced matrix element can be computed if we know the radial solution for the wavefunction of our particle; a little bit of work with our series expansion above and formulas for integrals of three \( Y_l^m \)'s will lead you to the formula
\[ \begin{aligned} \bra{\alpha', l_2}|\hat{Q}^{(l)}|\ket{\alpha, l_1} = q\sqrt{2l_1+1} \sprod{l_1, l; 0, 0}{l_1, l; l_2, 0} \\ \ \times \int_0^{\infty} dr\ r^{l+2} R_{\alpha', l_2}^\star(r) R_{\alpha, l_1}(r). \end{aligned} \]
This leads to an interesting selection rule. Remember that the Clebsch-Gordan coefficients are symmetric under flipping the signs of all of the \( m \) values, up to an overall minus sign:
\[ \begin{aligned} \sprod{j_1, j_2; -m_1, -m_2}{j_1, j_2; j, -m} = (-1)^{j_1+j_2-j} \sprod{j_1, j_2; m_1, m_2}{j_1, j_2; j, m}. \end{aligned} \]
The C-G coefficient with all of the \( m \) numbers equal to zero therefore vanishes if \( l_1 + l_2 - l \) is an odd number. (The opposite condition happens to hold for magnetic multipole operators: they vanish if \( l_1 + l_2 - l \) is even.) This argument can also be made using the properties of the spherical harmonics under parity, as it turns out.
We also have the usual triangle selection rule, which requires that \( |l_1 - l_2| \leq l \leq l_1 + l_2 \). This has interesting implications when we ask about the expectation values of multipole operators, i.e. set \( l_2 = l_1 \). Then we must have \( 0 \leq l \leq 2l_1 \). In other words, for a state with angular momentum \( l_1 \), _all electric multipole moments with \( l > 2l1 \) must vanish. Moreover, since \( 2l_1 - l \) must now be an even number, we see that \( l \) must be even for any non-vanishing electric multipole moment. Writing this out in equation form, the selection rules are:
\[ \begin{aligned} \bra{l_1} \hat{Q}^{(l)} \ket{l_1} = 0\ \textrm{unless}\ l = 0, 2, ..., 2l_1. \end{aligned} \]
Here are a few simple examples of how this can be applied to atomic and nuclear systems:
Our last stop this semester will be putting together many of the things we've learned to study the fine and hyperfine structure of the hydrogen atom. This exploration will rely heavily on perturbation theory, which you will see in detail next semester - but I should say a few words about it first, at least.
The basic idea here should be very familiar: perturbation theory simply means finding solutions to an otherwise intractable system by systematically expanding in some small parameter. In quantum mechanics, there are large differences in how perturbations are handled depending on whether they are time-dependent or not. For our current purposes, we consider only time-independent perturbation theory.
Suppose that we are interested in a physical system described by the Hamiltonian
\[ \begin{aligned} \hat{H} = \hat{H}_0 + \lambda \hat{W}, \end{aligned} \]
where \( \lambda \) is a small and continuous parameter, and we assume that we have a full analytic solution in the \( \lambda \rightarrow 0 \) limit, i.e. we have solved the eigenvalue problem
\[ \begin{aligned} \hat{H}_0 \ket{n^{(0)}} = E_n^{(0)} \ket{n^{(0)}}. \end{aligned} \]
The question then is to find the energy eigenvalues and eigenstates of the full Hamiltonian,
\[ \begin{aligned} (\hat{H}_0 + \lambda \hat{W}) \ket{n} = E_n \ket{n}. \end{aligned} \]
As I said above, the way forward will be systematic expansion: we write
\[ \begin{aligned} E_n = E_n^{(0)} + \lambda E_n^{(1)} + ... \\ \ket{n} = \ket{n^{(0)}} + \lambda \ket{n^{(1)}} + ... \end{aligned} \]
and work order by order in \( \lambda \). There are a lot of subtleties that start to crop up when we consider how the eigenstates themselves are shifted, even at leading order in the expansion. Fortunately, if we just want to know first-order energy corrections, we don't care about the states at all! The eigenvalue equation in this expansion can be written out as
\[ \begin{aligned} (\hat{H}_0 + \lambda \hat{W}) (\ket{n^{(0)}} + \lambda \ket{n^{(1)}} + ...) = (E_n^{(0)} + \lambda E_n^{(1)} + ...) (\ket{n^{(0)}} + \lambda \ket{n^{(1)}} + ...) \end{aligned} \]
At order \( \lambda \), this gives the equation
\[ \begin{aligned} (\hat{H}_0 - E_n^{(0)}) \ket{n^{(1)}} = (E_n^{(1)} - \hat{W}) \ket{n^{(0)}} \end{aligned} \]
or dotting in \( \bra{n^{(0)}} \) on the left,
\[ \begin{aligned} E_n^{(1)} = \bra{n^{(0)}} \hat{W} \ket{n^{(0)}}. \end{aligned} \]
This proves the result that we've already borrowed a couple of times: if we just care about leading-order perturbative energy corrections, all we have to do is calculate the expectation value of the perturbation \( \hat{W} \) against the exact analytic eigenstates \( \ket{n^{(0)}} \) that we already have. (Intuitively, this occurs because the shifts in the states at order \( \lambda \) can only affect expectation values at order \( \lambda^2 \).)
One word of caution: the derivation above assumes that we have a one-to-one map between states and energies. For systems with degenerate states, i.e. energy eigenstates that share an energy eigenvalue, some assumptions will generally break and we have to use a more elaborate approach (known as "degenerate-state perturbation theory".) The good news is that we can sidestep this problem for calculation of first-order energy corrections as long as we have a CSCO that includes \( \hat{W} \) to uniquely label all the states with. The hydrogen atom has lots of degeneracy, but we can ignore the problem of degeneracy for basically all of the fine and hyperfine structure, which we'll turn to now.
We'll be dealing with many examples based on the hydrogen atom, so it's worth going through a short review. The full solution is rather involved; Sakurai presents it very briefly, but chapter 13 of Shankar offers a much more detailed treatment if you're interested.
Strictly speaking, our solution is for a "hydrogenic" atom, i.e. a bound state of a single electron in the electric field generated by a nucleus with total charge \( Z \):
\[ \begin{aligned} V(r) = -\frac{Ze^2}{r}. \end{aligned} \]
Clearly to be a bound state, the electron energy must be less than zero. Since the Coulomb potential is a central potential (spherically symmetric), the solution to the Schrödinger equation splits into radial and angular components,
\[ \begin{aligned} \psi_{nlm}(r, \theta, \phi) = R_{nl}(r) Y_l^m(\theta, \phi). \end{aligned} \]
Solving the radial equation for the hydrogen atom is a difficult task, and as we've seen in other examples the solutions are written only in terms of special functions, either the associated Laguerre polynomials \( L_p^q(\rho) \) or the confluent hypergeometric functions, which are written as \( F(a;c;\rho) \). Here is the general formula in Sakurai's notation, and conveniently in Mathematica code:
R[n_, l_] := 1/(2*l + 1)!*((2*Z*r)/(n*a0))^l*Exp[(-Z*r)/(n*a0)]*
Sqrt[((2*Z)/(n*a0))^3*(n + l)!/(2 n*(n - l - 1)!)]*
Hypergeometric1F1[-n + l + 1, 2 l + 2, (2*Z*r)/(n*a0)];
The constant length scale \( a_0 \) is known as the Bohr radius,
\[ \begin{aligned} a_0 = \frac{\hbar^2}{m_e e^2}. \end{aligned} \]
As we observed generally for a particle in a central potential, the wavefunctions will be peaked away from the origin except for \( l=0 \); in particular only \( l=0 \) gives \( R_{n0}(0) \neq 0 \). The energy eigenstates are labelled by the principal quantum number, \( n \), which is a linear combination of a radial quantum number \( q \) and the angular momentum quantum number \( l \):
\[ \begin{aligned} n = q + l + 1. \end{aligned} \]
This imposes the restrictions that \( l \leq n-1 \) and \( n \geq 1 \); for the \( n=1 \) state, only \( l=0 \) is allowed. The energy eigenvalues are
\[ \begin{aligned} E_n = -\frac{1}{2} mc^2 \frac{Z^2 \alpha^2}{n^2} \end{aligned} \]
where \( \alpha \) is the fine-structure constant,
\[ \begin{aligned} \alpha = \frac{e^2}{\hbar c}. \end{aligned} \]
The value of the ground-state \( n=1 \) energy for the hydrogen atom (\( Z=1 \)) is sometimes used as a unit of energy, known as the Rydberg: we see that
\[ \begin{aligned} \textrm{Ry} = \frac{me^4}{2\hbar^2} = \frac{e^2}{2a_0}, \end{aligned} \]
and so
\[ \begin{aligned} E_n = -\frac{Z^2}{n^2}\ \textrm{Ry}. \end{aligned} \]
I'll emphasize that the energy eigenvalues of this solution depend only on \( n \), not \( l \) or \( m \). The \( E_n \) energy level is thus made up of \( \sum_l (2l+1) = n^2 \) degenerate states.
The ground state radial wavefunction is relatively simple, and we should have it handy for some of the examples we're about to work: it is equal to
\[ \begin{aligned} R_{10}(r) = 2 \sqrt{\frac{Z^3}{a_0^3}} e^{-Zr/a_0}. \end{aligned} \]
Although the "hydrogen atom" as you learn it in class is a nice, simple, exactly solvable quantum system, it has one drawback; the simple model doesn't completely describe the real hydrogen atom. Let's start to think about possible corrections that might appear.
First, for a hydrogenic atom (one electron), we assume that the electron moves subject to the Coulomb potential; this is valid at large \( r \), but if we look at very short distances, we will realize that the nucleus is not a point charge, but rather has some finite size associated with it. If we treat the nucleus as a uniform ball of charge with finite radius \( r_0 \), then the potential is corrected to
\[ \begin{aligned} V(r) = \begin{cases} -\frac{Ze^2}{r}, & r \geq r_0; \\ \frac{Ze^2}{2r_0} \left[ \left( \frac{r}{r_0}\right)^2 - 3 \right], & r \leq r_0. \end{cases} \end{aligned} \]
We can thus rewrite our Hamiltonian as \( \hat{H} = \hat{H}_0 + \lambda \hat{W} \), where
\[ \begin{aligned} \hat{W} = \begin{cases} 0, & r \geq r_0; \\ \frac{Ze^2}{2r_0} \left[ \left( \frac{r}{r_0}\right)^2 - 3 + \frac{2r_0}{r} \right], & r \leq r_0. \end{cases} \end{aligned} \]
where this time I am using \( \lambda \) purely as an auxiliary constant, which we'll set to \( 1 \) at the end.
Why is it valid to treat this correction perturbatively? We know that the nucleus is very small, and so we expect that the overlap of the electron wavefunction with the region we're changing the potential in is quite small. We will find that the matrix elements of \( \hat{W} \) depend on the ratio \( r_0 / a_0 \), where \( a_0 \) is the Bohr radius. The small size of this ratio validates our perturbative expansion; for hydrogen, \( a_0 \sim 10^{-10} \) m, while the charge radius of the proton is closer to \( 10^{-15} \) m.
To compute the energy shift due to this finite-size effect, we have to integrate using hydrogenic wavefunctions. Let's just look at the ground state \( \ket{n,l,m} = \ket{1,0,0} \). Here the angular dependence vanishes, and we simply have
\[ \begin{aligned} \psi_{100}(r, \theta, \phi) = 2 \sqrt{\frac{Z^3}{4\pi a_0^3}} e^{-Zr/a_0}. \end{aligned} \]
The first-order energy correction is then
\[ \begin{aligned} E_{100}^{(1)} = \bra{100} \hat{W} \ket{100} \\ = \int_0^{r_0} dr\ r^2 \frac{2Z^4 e^2}{r_0 a_0^3} \left[ \left( \frac{r}{r_0} \right)^2 - 3 + \frac{2r_0}{r} \right] e^{-2Zr/a_0}. \end{aligned} \]
(the \( 4\pi \) vanished due to the angular integral we skipped over.) This is a straightforward integral, which you can do in Mathematica or with integral tables.
The result, as promised, depends on the small ratio \( r_0 / a_0 \). Although taken at face value, the energy shift will contain higher powers of \( r_0 / a_0 \), we know that we're only working at first order here, so we should really only keep the first term in an expansion of this quantity. (The second-order perturbative term in \( \lambda \) is expected to give us new corrections of order \( \ev{W}^2 \).)
Carrying out the expansion, to first order in \( r_0 / a_0 \) we find
\[ \begin{aligned} E_{100}^{(1)} = \frac{2Z^4 e^2 r_0^2}{5 a_0^3} = \frac{4}{5} \frac{Z^4 r_0^2}{a_0^2} \left( \frac{e^2}{2a_0} \right) = \frac{4}{5} Z^2 E_1^{(0)} \left( \frac{r_0}{a_0}\right)^2. \end{aligned} \]
The correction is therefore of order \( (r_0 / a_0)^2 \), i.e. a \( 10^{-8} \) percent correction to the ground-state energy of the hydrogen atom; very small indeed!
This is, in fact, the largest correction factor out of all of the states of hydrogen; we know that for higher \( n \) or higher \( l \), the wavefunction of the electron will be localized further away from the origin, and they will turn out to be suppressed by further powers of \( (r_0 / a_0) \).
The finite-size energy shift is clearly is a tiny effect in hydrogen, but can be relevant in very precise atomic physics experiments for some systems. In particular, for higher-\( Z \) nuclei the nuclear radius increases, while the effective Bohr radius decreases, and so the size of the perturbation can be substantially larger. Replacing the electron with a muon, the electron's heavier cousin, can also lead to appreciable effects.