The state that turned is not in the atom

Whether an alkali’s s and p levels first approach or first separate in a field is a question about two polarisabilities, and on potassium it was the one atom whose answer changed along its own Rydberg series. Summed over every level with measured Ritz defects, the ratio $\alpha(np_0)/\alpha(ns)$ passes one at $n=33.7$. That result came with a warning: every p level had been placed at the weighted average of its two fine-structure components, and placing them at $p_{1/2}$ or at $p_{3/2}$ instead sent the turn to the twenty-third or the forty-fifth shell. The essay said plainly that those two outer curves were bounds and not the polarisabilities of real states, since the angular factors had been left spinless, and it left the real calculation for later as one change of angular algebra.

That change has now been made. It does not narrow the bound, because the answer lies outside it.

No real state of potassium turns at the thirty-fourth shell. There, the three ratios are 1.067, 1.090 and 1.187. The dashed line that gave the earlier result falls through one with a confident slope, and none of the three solid lines does anything of the kind near it. $np_{1/2}$ is still at 1.017 on the hundredth shell. $np_{3/2}$ with $|m|=3/2$ rises as the series climbs, the opposite direction entirely. Only $np_{3/2}$ with $|m|=1/2$ crosses, between the seventy-seventh and seventy-eighth shells, more than forty shells from the published turn and well beyond even the outer bound.

Three levels and one of them missing #

A spinless p level in a field along $z$ has two kinds of state: $m_l=0$, its lobes along the field, and $m_l=\pm1$, its ring around it. The earlier sums followed the first, since it is the one with a route to the s level, and that route is why it can fall. In a real atom, spin–orbit coupling has already chosen the states before any field arrives. A weak field cannot split what the interval holds together, so the good labels are $j$ and $m$, and each state is a fixed blend of the two spinless kinds set by Clebsch–Gordan coefficients.

$np_{3/2}$ with $|m|=1/2$ is two thirds $m_l=0$, $np_{1/2}$ one third, and $np_{3/2}$ with $|m|=3/2$ none at all. Now the arithmetic that decided the earlier turn takes a different turn. With every level of a series at one energy, the two spinless ratios at the thirty-fourth shell are 0.998 and 1.173. They are nowhere near each other, for a reason that follows directly from the route to s. The $m_l=\pm1$ state cannot reach any s level, so it loses the term the s and p levels share with opposite sign, the term whose growth brings $m_l=0$ down along the series, and it answers only to the d levels. A blend containing it inherits a share of that surplus. So the ratio of 0.998 was the value of one component of a mixture, reported as though it were the mixture.

The same fact can be read off the angular factors, which are worked out here by expanding each $j$ state into uncoupled orbital and spin parts rather than quoted from a 6j table, so that they can be checked against the spinless ones.

Summed over $j'$ with one energy per series, they return the $1/3$, $4/15$ and $1/5$ of the spinless sum, and the $j$-resolved calculation then reproduces $\alpha(ns)$ and $\alpha(np_0)$ from the earlier essay at three shells to a part in a thousand million. So nothing about the spinless sum was wrong as a calculation; it is contained in this one. What was wrong was the name it went by.

Angles first, then positions #

The earlier outer curves moved only the positions of the levels. There are in fact two separate changes, and they can be made one at a time.

The angular factors do most of it, taking the ratio from 0.998 to 1.056, 1.115 and 1.173. Then the measured positions adjust those by up to five hundredths, downward on $np_{1/2}$ and upward on the others. That second step is the one the earlier curves tried, and it has the sign the earlier warning expected: the interval really is large for a cancellation this fine, and a thousandth in a defect really does move things. But moving a level’s position gives about as much as the cancellation has room for, and the surplus from the angular factors is larger than the cancellation. The fine-structure question was asked as one of where the levels are, and most of the answer was which levels exist.

That also disposes of the outer bound. Placing all the p levels at $p_{1/2}$'s defect with spinless factors gave a turn at $n=22.8$, and the natural reading was that $np_{1/2}$ would turn near there. On the twenty-third shell $np_{1/2}$'s ratio is 1.097. A bound computed for an object that is not in the atom has no reason to bracket the objects that are.

The thirty-fourth shell gets its meaning back #

A published number that survives all this can still be useful, and this one does. A state’s label is only a statement about weak fields: strong enough a field, compared with the interval, and it pulls the spin–orbit blend apart, so that the eigenstates end on $m_l=0$ and $m_l=\pm1$ after all. The fine-structure interval at the thirty-fourth shell is 0.0199 cm⁻¹, and the field needed is not large.

The calculation takes the two $m=1/2$ states as a pair with their interval between them, adds the matrix of polarisabilities in that basis, and reads off how each eigenvalue falls with the square of the field. At weak field the two curves are 1.067 and 1.090. Near 33 V/cm, where the difference in field shift reaches the interval, they cross into each other’s characters, and they settle at 0.995 and 1.161. The lower of those is within a per cent of the spinless ratio from the thirtieth shell upward, and it is the same number within the error of treating positions that way. So the turn at $n=33.7$ is the strong-field result: exact as far as it goes, about a state the atom has only once the field has undone the spin–orbit blend.

And this produces something the spinless calculation could not have reached, an answer to the question the last essay deferred to the next order of perturbation theory. Where a state’s separation from $ns_{1/2}$ first shrinks and then grows, it has a minimum, and until now the minimum needed a hyperpolarisability. Here the local polarisability itself passes that of the s level as the field rises. The separation stops falling without any fourth-order term.

A minimum of this kind exists only inside a window, and the window’s two ends are the two turns. Below the thirty-fourth shell the strong-field limit stays above one, so nothing passes through: 1.002 on the thirty-third and 0.995 on the thirty-fourth. Above the seventy-seventh, the weak-field ratio is below one already, so the separation grows from the start and there is no fall to reverse. Between those ends the field drops fast, 57 V/cm at the thirty-fourth, 11.9 at the fortieth, 2.2 at the fiftieth and 0.56 at the sixtieth. It is below the crossover’s middle from about the thirty-eighth shell, since the lower ratio has less far to fall.

That gives the two numbers in the argument a joint meaning. Thirty-four is where a field first makes a minimum possible, and seventy-seven is where it stops being needed. They come out of two opposite limits of one two-state problem. Earlier in this line of essays a limit was mistaken for something it was not, and here a limit was in plain sight and it was its regime that went unstated.

The field that matters is a differential one, and it differs from the figure in the earlier essay for that reason. That essay compared $\tfrac12\alpha F^2$ for the s level itself with the interval, and got 14 V/cm on the thirty-fourth shell. But a shared shift does nothing to a pair: what pulls the blend apart is the difference between the two spinless polarisabilities, which is about a sixth of each. That puts the crossover at 33 V/cm, and the corresponding minimum well above 14, which answers that essay’s closing question in the direction it hoped against: at the thirty-fourth shell the minimum does not arrive while the fine structure still holds. It cannot. It exists because the fine structure stopped holding.

Why the lines go where they go #

The three curves have reasons, and the table of factors gives them. The term that makes a p state fall along a series is its coupling to the s level of its own shell, which pushes the s level down and the p level up: a term the s and p states share with opposite signs, whose relative size is why a ratio that started near 2.7 at the bottom of the series comes down at all. How much of it a state carries is its angular factor to s, and the factors are $2/9$, $1/9$ and $0$.

So $np_{3/2}$ with $|m|=3/2$ has no such term, and nothing brings it down. It answers only to the d levels, and the ratio rises slowly from 1.16 near the twentieth shell to 1.23 at the hundredth. $np_{3/2}$ with $|m|=1/2$ carries twice as much of the term as $np_{1/2}$, so at high shells it falls faster, and the two curves change places near the forty-eighth shell: 1.0534 against 1.0495 at the forty-fifth, 1.0414 against 1.0440 at the fiftieth. That ordering, with the weakest-coupled state left standing, is why $np_{1/2}$ has not crossed even by the hundred-and-fiftieth shell, where it reads 1.007 while the other has reached 0.958. Whether it crosses far above that is a question for a series no measured defect has been tested at, and this essay does not claim that it never does; a state whose ratio is within a per cent of one at that height is outside what the integrals can promise.

This also gives a measurement something specific to look for. Three states at one shell have three different ratios, and the differences are not small: at the thirty-fourth shell they span twelve per cent, much more than the half a per cent or so the earlier argument was resolving. A resolved Stark map at that shell would show the states separately without any need to fix the sign of a near-cancellation.

A label is a claim about a range of fields #

There is an instructive shape in how this was missed. The essay that made the shell the wrong unit was right to discard the single-shell threshold. The earlier threshold essay was right that a verdict should belong to the atom. And the four alkalis the three-level model could not hold were real failures. Each step improved a calculation of $\alpha(np_0)$, which is why $np_0$ went on being the object: a quantity that keeps being refined starts to feel like a property of the atom. The $m_l$ label was chosen in the first essay of the sequence, with no spin in the model, when it was a fair choice. Nothing later forced it to be asked again, because adding precision to a number does not test whether the number names something.

The degeneracy group-theory argument is the general form of what should have raised the question. Which labels are good is set by what commutes with the Hamiltonian, and in an atom with spin–orbit coupling at weak field, $m_l$ does not. A symmetry label names a regime of the Hamiltonian, as well as a state. Every potassium result in this line of argument was correct within the regime nobody had stated.

What was computed, and how #

The polarisability of each state is the sum over every s, p and d fine-structure component from the bottom of each series to twice the state’s own principal quantum number, using Coulomb-approximation radial integrals from the measured Ritz series of Peper and co-workers (2019), component by component, with the measured term values at the bottom of each series. The angular factor is $|\langle l\,j\,m|\cos\theta|l'\,j'\,m\rangle|^2$, assembled from the expansion of each $j$ state in uncoupled states. Ratios are taken to $ns_{1/2}$ at the same $n$, and a crossing is located by linear interpolation between neighbouring shells.

The crossover takes the two $m=1/2$ states with the measured interval, diagonal polarisabilities at their real positions, and an off-diagonal of $(\sqrt2/3)(P_0-P_1)$ from the two spinless ones, and differentiates each eigenvalue numerically in $F^2$. The test of what is claimed: the spinless sum recovered with one energy per series; no crossing of $np_{1/2}$ or of $|m|=3/2$ to the hundredth shell; exactly one crossing of $|m|=1/2$; the strong-field ratio within one per cent of the spinless one from the thirtieth shell up; and a minimum existing on the thirty-fourth shell and not on the thirty-third or the seventy-eighth. The refusal is the pair of selection rules, which any error in the expansion would break: no s term for $|m|=3/2$, no $d_{5/2}$ term for $np_{1/2}$.

Where it stops #

The weak-field ratios are the strong result, and the minimum is not. The three solid curves are second-order perturbation theory on measured levels, with one approximate input, the integrals. The minimum’s field sits against the field where $ns$ and $np$ couple as strongly as they are separated: 57 against 106 V/cm at the thirty-fourth shell. Since $\alpha_p$ and $\alpha_s$ cancel to a few per cent, and the next order scales as the square of that field ratio, a hyperpolarisability is not safely negligible at the low end of the window. The minimum’s existence at $n=34$ and its field there should be taken as second-order statements. By the fiftieth shell the ratio of fields is six and the case is much firmer.

The window’s lower edge rests on a difference of under one per cent. On the thirty-third shell the strong-field ratio is 1.0018, and on the thirty-fourth 0.9948, and treating real positions in a two-state matrix is itself approximate at order interval over gap. The edge could be a shell lower or higher.

The low levels carry no fine structure. The bottom of each series uses spinless measured terms. At these shells the lowest levels contribute almost nothing and the choice cannot matter much, but the calculation says nothing reliable about $4p_{1/2}$ and $4p_{3/2}$ themselves, where it does matter, and they were not compared with measured values.

The f series is missing. The d levels of an alkali couple strongly to f, which does not enter any p level’s sum at second order but will enter at the next one. That is a further reason to treat the exact field of a minimum with care.

Nothing here has been observed. The series would be testable in cold-atom Rydberg experiments, which do resolve fine structure and $m$ at these shells, but no measured differential polarisability is used or compared.

Who worked this out #

The $j$-resolved scalar and tensor polarisability formulae have been textbook material since the middle of the twentieth century, and precision calculations for alkali Rydberg states routinely use them. Nothing about the angular algebra is new. A field competing with spin–orbit coupling is older still, and one of the first things a course in atomic physics treats. Peper, Merkt and co-workers supplied the defects.

What was computed here is the three real states’ ratios along potassium’s series against the spinless turn, the identification of that turn as a strong-field limit, and the window of shells over which a minimum of separation arises from the crossover alone. The essay that located the crossing nothing couples and the argument about two levels and minima both read differently once the regime is stated.

Still open: the next order at the low edge, and other alkalis #

The obvious open question is the hyperpolarisability, now that the minimum’s field is known. On the lower part of the window the second-order argument is close to its limit, and a direct diagonalisation of the $m=1/2$ Stark problem in a truncated basis of s, p, d and f components, at the thirty-fourth and fortieth shells, would say whether the minimum sits where this puts it. It needs measured or extrapolated f defects for potassium, which are small and available.

The nearer question is sodium, rubidium and caesium. Their spinless verdict was that the p level wins at every shell, and that verdict now needs reading state by state like potassium’s. For rubidium and caesium, with intervals far larger, the crossover occupies a much higher field, and the $j$-resolved ratios are what a laboratory would ever see. The same sum with their measured series should say whether any of them has a state that crosses at all.