Rydberg States of ThIII and UV W. R. Johnson Department of Physics, Notre Dame University June 25, 2003 Abstract We examine the Rydberg spectrum of the two actinide ions Th 2+ and U 5+. The states of interest consist of a wealy bound electron interacting with the 5F 5/2 ground states of the respective one electron ions. First- and second-order correlation corrections are wored out. Basic Assumptions We consider an N+ electron with atom with one electron in a high Rydberg state. We neglect the exchange interaction between the Rydberg electron and the N-electron ion. The Hamiltonian of the atom is written H = H N ( r r N ) + h( r) + U( r, r r N ), () where H N is the Hamiltonian of the N-electron ion, h is the Rydberg electron: and and U is the interaction Hamiltonian, h( r) = T + V nuc (r) + N r, (2) U( r, r r N ) = = N i= r r i N r L L= M= L C LM (ˆr) r L+ (3) N ri L CLM(ˆr i ). (4) i= In the expansion of the interaction Hamiltonian, we assume that the Rydberg electron wave-function has no overlap with the ionic core.
where In lowest-order, the atomic wave function may be written Ψ 0 = Φ 0 ( r r N ) φ ( r), (5) H N ( r r N ) Φ n ( r r N ) = E n Φ n ( r r N ) (6) The lowest-order energy of a Rydberg state is E 0 + ɛ. 2 First-Order Energy h( r) φ l ( r) = ɛ l φ l ( r). (7) The ground state of both of the N = 87 electron ions, Th 3+ and U 5+, is a J 0 = 5/2 state with a 5f 5/2 valence electron outside a Rn-lie core. The two valence electron systems of interest here have a Rydberg electron with angular momentum (j m ) coupled to the (J 0 M 0 ) ground-state to give a coupled two-electron state (F M F ). We use the notation of hyperfine structure here, because of the similarity of the present analysis to the analysis of hyperfine interactions. The first-order energy is E () F = (J 0 M 0, m )F M F U (J 0 M 0, m )F M F = J 0 M 0, m F M F J 0 M 0, m F M F ( ) M m C LM(ˆr) m r L+ J 0 M 0 C L M (ˆr i ) r L i J 0 M 0. (9) LM i Carrying out the sums over magnetic quantum numbers, we find E () F = { } ( ) J 0++F J0 F J L=2,4 0 L C L r L+ J 0 C L (ˆr i ) ri L J 0. (0) i Only two terms L=2 and L=4 contribute for the cases of interest where J 0 = 5/2. In the HF approximation, the core parameters F 5/2 i rl i C L(ˆr i ) F 5/2 are found to be: Ion F 5/2 r 2 C 2 F 5/2 F 5/2 r 4 C 4 F 5/2 Th 3+ -3.755 5.97 U 5+ -2.207 4.78 (8) 2
3 Second-Order Energy The second-order energy is = (nl) (0) = n 0, all l l 0 U nlnl U 0 + ɛ ɛ l () 0 U nlnl U 0 + ɛ ɛ l We expand the denominator in the first term of Eq. (2) as to find + ɛ l ɛ = + = n 0 l 3. Polarizability correction n 0 0 U 0l0l U 0 ɛ ɛ l. (2) ɛ ɛ l ( ) 2 + (ɛ ɛ l ) 2 ( ) 3 + 0 U nn U 0 0 [h, U] nn U 0 ( ) 2 + 0 U 0l0l U 0 ɛ ɛ l. (3) In this subsection, we examine the first term in Eq. (3), which leads to the correction to the energy of the Rydberg state from the core polarizability. We have pol = J 0M 0, m F M F J 0 M 0, m F M F m C L M L (ˆr)C LM L (ˆr) m L M L LM L n 0 J 0 M 0 i rl i CL M L J n M n J n M n j rl j C LM L J 0 M 0 (4) We expand the product of two spherical tensors as C L M L (ˆr) C LM L (ˆr) = JM J A JMJ C JMJ (ˆr), (5) 3
and find A JMJ = [J] L M [L][L L C JMJ LM L. (6) ] Therefore, C L M L (ˆr) C LM L (ˆr) = JM J [J] L M [L][L L C JMJ LM L CJMJ (ˆr). (7) ] It follows that m C L M L (ˆr) C LM L (ˆr) m = JM J [J] [L][L ] L M L C JMJ LM L m C JMJ m Carrying out the sums over magnetic quantum numbers, we find pol = { } ( ) J 0++F J0 F [J] C J J 0 J J [L][L ] L C J L LL { L L J ( ) L J 0 J 0 J n n 0, (8) } J 0 i rl i C L J n J n j rl j C L J 0. (9) For the 5F 5/2 state of interest here, J 0 =5/2 and J < 2J 0 = 5; moreover, J is even. It follows that J = 0, 2, 4. 3.. Contribution of J = 0 In the following, we include only those terms with L + L 4. If we set J = 0 in Eq. (9), then we find 0 = ᾱ 2 r 4 ᾱ2 2 r 6, (20) where ᾱ and ᾱ 2 are the scalar parts of the dipole and quadrupole polarizabilities, respectively. The scalar polarizabilities are defined by ᾱ L = 2 [L][J 0 ] n J n j rl j C 2 L J 0. (2) 4
3..2 Contribution from J = 2, 4 The J = 2 and 4 terms in Eq. (9) consists of partial contributions from all L >. The corresponding values of L have the same parity as L and satisfy max (, L J ) L L + J. Terms with L = L = : We consider the L = L = term first. This term arises only for J = 0 and 2. We have already evaluated the J = 0 part. The J = 2, L = L = term can be rewritten in terms of the L = tensor polarizability α t = 4 5 6 ( J0 2 J 0 J 0 0 J 0 ) { ( ) J 0+J n 2 J J 0 J 0 J n n } J n j r j C J 0 2. (22) We note in passing that ( J0 2 J 0 J 0 0 J 0 ) J 0 (2J 0 ) = (J 0 + )(2J 0 + )(2J 0 + 3). (23) Substituting into Eq. (9), we find that the L = L = contribution to the J = 2 term is: 2 = 2 ( )J 0++F { J0 F J 0 2 } C 2 ( J0 2 J 0 There are no further second-order corrections falling off as /r 4. J 0 0 J 0 ) αt Terms with L = L = 2: The J > 0 terms with L = L = 2 can be summarized as 2 = { } ( ) J 0++F J0 F C 22 J J=2,4 0 J J r 6 } J n j r2 j C 2 2 J 0 2 C J 2 Jn For the record, we note that ( ) J 0+J n { 2 J0 J n J 0 2 J r 4. (24). (25) 70 2 C 2 2 = 7 and 2 C 4 2 = 70 7. 5
Terms with L =, 3 and L = 3, : The sum of the terms with L =, 3 and L = 3, can be written 2 = 2 { } ( ) J 0++F J0 F C 3 J 0 J J r 6 We note that J=2,4 [J] 3 C J { 3 J0 J n 2 J Jn 0 J 3 C 2 = 3 5 5 } J0 i r3 i C 3 J n J n j r j C J 0 and 3 C 4 = 2 3 3, and further that C J 3 = 3 C J, for J = 2, 4. energy contributions that fall off as /r 6. 3..3 Summary. (26) This completes the second-order The contributions to the second-order polarization energy that fall off as /r 4 and /r 6 are given by Eqs. (20), (24), (25), and (26), above. The contributions to the first-order energy fall off as /r 3 and /r 5. These first-order contributions are given given in Eq. (0). 6