Table Of ContentGeneral theory for Rydberg states of atoms: nonrelativistic case
Xiao-Feng Wang1,2, Zong-Chao Yan2,3,4
1 School of Physics and Technology, Wuhan University, Wuhan 430072, P. R. China
2 Department of Physics, University of New Brunswick,
7
Fredericton, New Brunswick, Canada E3B 5A3
1
0 3 State Key Laboratory of Magnetic Resonance and Atomic and Molecular Physics,
2
Wuhan Institute of Physics and Mathematics,
n
a Chinese Academy of Sciences, Wuhan 430071, P. R. China and
J
3 4 Center for Cold Atom Physics, Chinese Academy of Sciences, Wuhan 430071, P. R. China
2
(Dated: January 24, 2017)
]
h
We carry out a complete derivation on nonrelativistic energies of atomic Rydberg states,
p
-
m including finite nuclearmasscorrections. Severalmissingterms arefoundandadiscrepancy
o is confirmed in the works of Drachman [in Long Range Casimir Forces: Theory and Recent
t
a Experiments on AtomicSystems,editedbyF.S.LevinandD.A.Micha(Plenum,NewYork,
.
s
c 1993)]and Drake [Adv. At., Mol., Opt. Phys. 31, 1 (1993)]. As a benchmark, we present a
i
s detailed tabulation of different energy levels.
y
h
p PACS numbers: 31.15.ac
[
1
v
I. INTRODUCTION
9
3
3
6 The Rydberg states of few-electron atomic systems were investigated extensively from the mid-
0
. 1980s to 1990s [1–6]. According to the theory of Kelsey and Spruch [7, 8], experimental and
1
0
theoretical studies on high-(n,L) states can test the Casimir-Polder effect, where n and L are,
7
1
respectively, the principal and angular momentum quantum numbers of the Rydberg electron.
:
v
i The systems that have been studied include helium and lithium with one electron being excited to
X
r a high-(n,L) state. A series of precision measurements were performed by Hessels et al. [9–13] on
a
Rydberg states of helium using microwave spectroscopy. Hessels et al. [14, 15] also did the radio-
frequency measurements on lithium Rydberg states. On the theoretical side, a substantial work
on Rydberg states of helium was carried out independently by Drake [1–3] and by Drachman [5]
around the same period of time using the quantum mechanical perturbation method and the
optical potential method, including relativistic and quantum electrodynamic (QED) effects. These
methods are equivalent in nature and embody the picture of long-range interaction. A recent
extension to higher angular momentum states of helium was done by El-Wazni and Drake [16].
2
Bhatia and Drachman [17–19] also calculated relativistic and QED effects in the Rydberg states of
lithium. Later,WoodsandLundeen[20,21]extendedDrakeandDrachman’sworktomorecomplex
atoms, which allows for a high-L Rydberg atom to have nonzero core angular momentum, for the
purpose of modeling the effective potential and thus extracting core properties experimentally.
Very recently, a new exotic Rydberg atom H−+, which consists of a Rydberg positron e+ attached
to the ground state H−, was detected in the laboratory by Storry et al. [22]. Since these Rydberg
states are embedded in the Ps+H continuum, they are in fact resonant states [23]. It is therefore
interesting to do theoretical calculation on these states and explore the spectrum of H−+.
The main purpose of this paper is to present a complete calculation of nonrelativistic Rydberg
energy levels using the standard perturbationmethod upto the order of x−10 , where x stands for
h i
the distance of the Rydberg particle relative to the core, and to compare our results with the work
of Drake [2] and Drachman [5]. We find that there are several terms of order x−10 missing in the
h i
work of Drake [2] and Drachman [5]. We also confirm a discrepancy that exists between Drake [2]
andDrachman’s[5]calculations. Asabenchmarkforfuturereference,wetabulatenumericalvalues
for the nonrelativistic energy levels of helium in various Rydberg states.
II. THEORY AND METHOD
A. The Hamiltonian
Consideranatomicormolecularsystemthatconsistsofn+2chargedparticles. TheHamiltonian
of the system (in a.u.) is
n n+1
1 1 1 q q
H = 2R 2R 2R + i j , (1)
−2m ∇ 0 − 2m ∇ i − 2m ∇ n+1 R R
0 i n+1 i j
i=1 i>j≥0| − |
X X
where R is the position vector of the ith particle relative to the origin of a laboratory frame, with
i
0 i n+1, m its mass, and q its charge. We assume that the (n+1)th particle is far away
i i
≤ ≤
from the ”core”, which is made up of the remaining n+1 particles. We also take the 0th particle
as a reference one. In reality, it could be the nucleus. In order to eliminate the center of mass
degree of freedom for the whole system, we make the following coordinate transformations [24]:
n+1
1
X = m R (2)
j j
M
T
j=0
X
r = R R , i= 1,2,...,n (3)
i i 0
−
n
1
r = R m R , (4)
n+1 n+1 j j
− M
C
j=0
X
3
where M = n+1m is the total mass of the whole system, and M = n m the total mass
T j=0 j C j=0 j
of the core. FProm the above expressions, we can see that X represents thPe position vector of the
center of mass of the whole system, r is the position vector of ithe particle in the core relative
i
to the reference particle, and r is the position vector of the Rydberg particle relative to the
n+1
center of mass of the core. Thus, we have established a one to one transformation between the
set (R ,R ,R ,...,R ,R ) and the set (X,r ,r ,...,r ,r ). The corresponding differential
0 1 2 n n+1 1 2 n n+1
operators transform according to
n
m m
0 0
∇R0 = − ∇i− M ∇n+1+ M ∇X (5)
C T
i=1
X
m m
i i
∇Ri = ∇i− M ∇n+1+ M ∇X (6)
C T
m
n+1
∇Rn+1 = ∇n+1+ M ∇X, (7)
T
where ∇i ≡ ∇ri and ∇n+1 ≡ ∇rn+1. After some simplification, the Hamiltonian (1) can be
rewritten in the form
n n n n
1 1 1 1 q q q q
H = 2 2 2 + i 0 + i j
− 2µ ∇i − 2µ ∇n+1− 2M ∇X− m ∇i·∇j r r
i x T 0 i ij
i=1 i>j≥1 i=1 i>j≥1
X X X X
n
q q q q
i n+1 0 n+1
+ + , (8)
n n
i=1 r r 1 m r r + 1 m r
X i− n+1− MC j j n+1 MC j j
(cid:12) j=1 (cid:12) (cid:12) j=1 (cid:12)
(cid:12) P (cid:12) (cid:12) P (cid:12)
wherer = r r i(cid:12)stherelativepositionbetw(cid:12)een(cid:12)twocoreparticlesia(cid:12)ndj,µ = mim0 (1 i n)
ij i− j (cid:12) (cid:12) (cid:12) (cid:12) i mi+m0 ≤ ≤
is the reduced mass of ith electron in the core with the reference particle 0, and µ = mn+1MC
x mn+1+MC
is the reduce mass of the Rydberg particle relative to the core. Since H does not contain X, X
is a cyclic coordinate and thus can be ignored. Furthermore, the last two terms of (8) may be
combined by introducing
ǫ = δ m /M , 0 i n, 1 j n (9)
ij ij j C
− ≤ ≤ ≤ ≤
i.e.,
n
q q q q
i n+1 0 n+1
+
n n
i=1 r r 1 m r r + 1 m r
X i− n+1− MC j j n+1 MC j j
(cid:12) j=1 (cid:12) (cid:12) j=1 (cid:12)
n (cid:12) P (cid:12) (cid:12) P (cid:12)
(cid:12) q q (cid:12)q q (cid:12) (cid:12)
= (cid:12) i n+1 + (cid:12) 0 n(cid:12)+1 (cid:12)
n n
i=1 r ǫ r r ǫ r
X n+1 ij j n+1 0j j
− −
(cid:12) j=1 (cid:12) (cid:12) j=1 (cid:12)
n (cid:12) P (cid:12) (cid:12) P (cid:12)
(cid:12) q q (cid:12) (cid:12) (cid:12)
= (cid:12) i n+1 (cid:12). (cid:12) (cid:12) (10)
n
i=0 r ǫ r
X n+1 ij j
−
(cid:12) j=1 (cid:12)
(cid:12) P (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
4
The Hamiltonian can thus be partitioned into the form
H = H +H +V , in 2R , (11)
c x cx ∞
where
n n n n
1 1 q q q q
H = 2 + 0 i + i j (12)
c − 2µ ∇i − m ∇i·∇j r r
i 0 i ij
i=1 i>j≥1 i=1 r>j≥1
X X X X
1 q q
H = 2 + x c (13)
x −2µ ∇x x
x
n
q q q q
i x c x
V = (14)
cx n − x
i=0 x ǫ r
X ij j
−
(cid:12) j=1 (cid:12)
(cid:12) P (cid:12)
(cid:12) (cid:12)
with qx ≡ qn+1, x ≡ rn+1, and(cid:12) qc ≡ nj=0(cid:12)qj being the total charge of the core. In (11), R∞ is the
Rydberg constant and 2R∞ represePnts the atomic units of energy expressed in cm−1. It is clear
that H is the Hamiltonian of the core [24], H the Hamiltonian of the Rydberg particle in the
c x
field of point charge q , and V the interaction potential energy between the core and the Rydberg
c cx
particle.
For a highly excited Rydberg particle, we may assume that x > n ǫ r for 0 i n.
| | | j=1 ij j| ≤ ≤
Under this condition, we have P
1 ∞ ℓ 4π dℓ
= Y∗ (xˆ)Y (dˆ), (15)
x d 2ℓ+1xℓ+1 ℓm ℓm
| − | ℓ=0m=−ℓ
X X
with d = n ǫ r . Using the formula [24]
j=1 ij j
P
ℓ−1
3 2s+3
Y (ˆr) = (ˆr ˆr ˆr)(ℓ) (16)
ℓm 4π s+1 ⊗ ⊗··· m
r (cid:18)sY=1r (cid:19) ℓ
| {z }
with the understanding that ℓ−1 2s+3 = 1 when ℓ = 1, we obtain
s=1 s+1
q
Q
ℓ−1
3 2s+3
dℓY (dˆ) = (d d d)(ℓ)
ℓm 4π s+1 ⊗ ⊗··· m
r (cid:18)sY=1r (cid:19) ℓ
ℓ−1 n
3 2s+3 | {z }
= (ǫ ǫ ǫ )(r r r )(ℓ). (17)
4π s+1 ij1 ij2··· ijℓ j1 ⊗ j2 ⊗··· jℓ m
r (cid:18)sY=1r (cid:19)j1j2X···jℓ≥1
Thus we have
n ∞ ℓ
q q 4π
i x = [q x−ℓ−1Y∗ (xˆ)]T (r ,r ,...,r ), (18)
n 2ℓ+1 x ℓm ℓm 1 2 n
i=0 x ǫ r ℓ=0m=−ℓ
X ij j X X
−
(cid:12) j=1 (cid:12)
(cid:12) P (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
5
where
ℓ−1
3 2s+3
T (r ,r ,...,r ) =
ℓm 1 2 n
4π s+1
r (cid:18)s=1r (cid:19)
Y
n n
q ǫ ǫ ǫ (r r r )(ℓ). (19)
× i ij1 ij2··· ijℓ j1 ⊗ j2 ⊗··· jℓ m
j1j2X···jℓ≥1(cid:18)Xi=0 (cid:19)
It is easy to see from (15) that the term with ℓ = 0 is 1/x and its corresponding term in V is
cx
q q /x, which cancels exactly with the second term in V . In other words, there is no monopole
c x cx
contribution to the interaction potential. Finally we obtain the following multipole expansion for
the interaction potential energy V , where in each term the degree of freedom of the Rydberg
cx
particle is separated from the core coordinates
∞ ℓ
4π
V = [q x−ℓ−1Y∗ (xˆ)] T (r ,r ,...,r ) . (20)
cx 2ℓ+1 x ℓm ℓm 1 2 n
ℓ=1m=−ℓ
X X Rydberg core
If we make the scaling transformation x| µ {xz, we ob}ta|in the H{zamiltoni}an
x
→
H = h +h +v , in 2R , (21)
c x cx ∞
where
n n n n
1 1 q q q q
h = 2 + 0 i + i j (22)
c − 2µ ∇i − m ∇i·∇j r r
i 0 i ij
i=1 i>j≥1 i=1 r>j≥1
X X X X
1 q q
h = µ 2 + x c (23)
x x − 2∇x x
(cid:18) (cid:19)
n
q q q q
i x c x
v = µ
cx 1 n − x x
i=0 x ǫ r
X µx − ij j
(cid:12) j=1 (cid:12)
(cid:12) P (cid:12)
∞ (cid:12) ℓ 4π (cid:12)
= (cid:12) µℓ+1(cid:12)[q x−ℓ−1Y∗ (xˆ)]T (r ,r ,...,r ). (24)
2ℓ+1 x x ℓm ℓm 1 2 n
ℓ=1m=−ℓ
X X
The above formulation is general for any system containing n + 2 charged particles. If the
system under consideration is an atomic system with n+1 electrons and one nucleus, we assume
that the 0th particle (the reference particle) is the nucleus with its mass M and its nuclear charge
Z. The Hamiltonian of the system becomes
n n n n n
1 1 Z 1 1 q q
H = 2 + 2 + x i ,(25)
−2µ ∇i − M ∇i·∇j − r r − 2µ ∇x n
i=1 i>j≥1 i=1 i i>j≥1 ij x i=0 x ǫ r
X X X X X ij j
−
(cid:12) j=1 (cid:12)
(cid:12) P (cid:12)
(cid:12) (cid:12)
where q0 = Z, qi = 1 (1 i n), qx = 1, µ is the reduced mass of th(cid:12)e electron r(cid:12)elative
− ≤ ≤ −
to the nucleus, and µ is the reduced mass of the Rydberg electron relative to the core mass
x
6
M+nm . In order to see the finite nuclear mass effect more clearly, we make the following scaling
e
transformations:
r µr , i= 1,2,...n (26)
i i
→
x µ x. (27)
x
→
The Hamiltonian (25) can thus be transformed to
H = h +h +v , in 2R , (28)
c x cx M
where R = µ R and
M me ∞
n n n n
1 µ Z 1
h = 2 + (29)
c −2 ∇i − M ∇i·∇j − r r
i ij
i=1 i>j≥1 i=1 r>j≥1
X X X X
µ 1 Z n
h = x 2 − (30)
x µ − 2∇x− x
(cid:18) (cid:19)
n
q µ (Z n)
i x
v = + −
cx − n µ x
i=0 µ x ǫ r
X µx − ij j
(cid:12) j=1 (cid:12)
(cid:12) P (cid:12)
∞ ℓ(cid:12) 4π µ (cid:12)ℓ+1
= (cid:12) x (cid:12) [q x−ℓ−1Y∗ (xˆ)]T (r ,r ,...,r ). (31)
2ℓ+1 µ x ℓm ℓm 1 2 n
ℓ=1m=−ℓ (cid:18) (cid:19)
X X
From now on, we use the following unified expressions for h and v
x cx
1 Z
h = a 2 1 (32)
x − 2∇x− x
(cid:18) (cid:19)
∞ ℓ
v = C u∗ (x)T (r ,r ,...,r ), (33)
cx ℓ ℓm ℓm 1 2 n
ℓ=1m=−ℓ
X X
where a = µ or µ /µ, Z = q q with Z > 0 in order to form a bound or quasi-bound Rydberg
x x 1 x c 1
−
state,
4π
C aℓ+1q , (34)
ℓ x
≡ 2ℓ+1
and
u (x) x−ℓ−1Y (xˆ) (35)
ℓm ℓm
≡
denotes the irregular solid harmonics satisfying the Laplace equation 2u (x) = 0. It should be
ℓm
∇
mentioned that the Rydberg particle could be either an electron or positron, or any other charged
particle.
7
B. Perturbation expansion
1. Second-Order Energy: General Expression
In(28), wecan treat v as a perturbationtotheunperturbedHamiltonian H = h +h , which
cx 0 c x
is uncoupled. The eigenvalue equations for h and h are respectively
c x
h φ = ε (L )φ (36)
c ncLcMc nc c ncLcMc
h χ = e χ , (37)
x nxLxMx nx nxLxMx
where the eigenvalue e only depends on the principal quantum number n because of the hydro-
nx x
genic nature of h . The initial eigenstates for h and h are assumed to be
x c x
h φ = ε φ (38)
c 0 0 0
h χ = e χ . (39)
x n0L0M0 n0 n0L0M0
Thus
H Ψ = E Ψ (40)
0 0 0 0
where
Ψ = φ χ (41)
0 0 n0L0M0
E = ε +e . (42)
0 0 n0
Inthiswork,weonlyconsiderthecasewhereφ isinanS state,whichresultsintheconsequence
0
that the first-order energy correction due to v is zero, i.e.,
cx
∆E = Ψ v Ψ = 0, (43)
1 0 cx 0
h | | i
The reason why (43) is valid is that there is no monopole term in the multipole expansion of v
cx
in (33).
The second-order energy correction can be calculated according to
∆E = Ψ v Ψ , (44)
2 0 cx 1
h | | i
where
Ψ v Ψ
n cx 0
Ψ = h | | i Ψ (45)
1 n
| i E E | i
0 n
n −
X
8
and n represents a set of quantum numbers describing an intermediate eigenstate of H , i.e.,
0
H Ψ = E Ψ (46)
0 n n n
where
Ψ = φ χ (47)
n ncLcMc nxLxMx
E = ε (L )+e . (48)
n nc c nx
We first denote the excitation energies for the core and the Rydberg electron by
δε (L ) = ε (L ) ε (49)
nc c nc c − 0
δe = e e . (50)
nx nx − n0
ConsideringtheRydbergparticleisinahighlyexcited state, wemakethefollowingkeyassumption
that [2]
δe < δε (L ) . (51)
| nx| | nc c |
In the above we have implicitly assumed that δε (L ) = 0. Now we can perform the following
nc c 6
expansion
∞
1 1 1 (δe )i
= = ( 1)i+1 nx (52)
E0−En −δεnc(Lc)1+ δεδnecn(Lxc) Xi=0 − (δεnc(Lc))i+1
Substituting (52) into (45) yields
∞
Ψ = ( 1)i+1 φ χ v φ χ
| 1i − h ncLcMc nxLxMx| cx| 0 n0L0M0i
Xi=0 ncXLcMcnxXLxMx
hi
s φ χ , (53)
× (δε (L ))i+1| ncLcMc nxLxMxi
nc c
where we have applied the eigenvalue equation (37) of h
x
(δe )i χ = hi χ , (54)
nx | nxLxMxi s| nxLxMxi
with the definition of h h e operating on the Rydberg electron. Now the second-order
s ≡ x − n0
energy correction (44) becomes
∞
∆E = ( 1)i+1 φ χ v φ χ
2 − h ncLcMc nxLxMx| cx| 0 n0L0M0i
Xi=0 ncXLcMcnxXLxMx
1
φ χ v hi φ χ . (55)
× (δε (L ))i+1h 0 n0L0M0| cx s| ncLcMc nxLxMxi
nc c
9
Substituting (33) into (55) and using the Wigner-Eckart theorem for the matrix element T
ℓm
L ℓ 0
φ T φ = ( 1)Lc−Mc c φ T φ
h ncLcMc| ℓm| 0i − h ncLck ℓk 0i
M m 0
c
−
1
= δ δ φ T φ , (56)
√2L +1 ℓLc mMch ncLck ℓk 0i
c
we arrive at
φ χ v φ χ
h ncLcMc nxLxMx| cx| 0 n0L0M0i
1
= C φ T φ χ u∗ (x)χ , (57)
Lc√2L +1h ncLck Lck 0ih nxLxMx| LcMc | n0L0M0i
c
where C is defined in (34). It is noted here that L 1 in (57), as indicated in (33). Similarly,
Lc c ≥
φ χ v hi φ χ
h 0 n0L0M0| cx s| ncLcMc nxLxMxi
1
= C ( 1)Lc φ T φ χ u (x)hi χ . (58)
Lc − √2L +1h 0k Lck ncLcih n0L0M0| LcMc s| nxLxMxi
c
Substituting (57) and (58) into (55) leads to the final expression for ∆E
2
∞ 1 φ T φ 2
∆E = ( 1)i+1 C2 |h 0k Lck ncLci| w(2)(L ). (59)
2 − Lc2L +1 (δε (L ))i+1 i c
Xi=0 nXcLc c nc c
(2)
In the above w (L ) is the quantity that describes the Rydberg particle and is given by
i c
w(2)(L ) = χ ˆ(L )χ , (60)
i c h n0L0M0|Ui c | n0L0M0i
where the operator ˆ(ℓ) is defined by
i
U
ˆ(ℓ) u hi u∗ . (61)
Ui ≡ ℓm s ℓm
m
X
It is seen that ˆ(ℓ) is an Hermitian operator. In obtaining (59), the following two relations have
i
U
been used, namely, the closure relation
χ χ = I (62)
| nxLxMxih nxLxMx|
nxXLxMx
and
φ T φ = ( 1)Lc φ T φ ∗. (63)
h ncLck Lck 0i − h 0k Lck ncLci
It would be convenient to define the 2Lc-pole “generalized polarizability” for the state of the S-
symmetric core
23−iπ φ T φ 2
α(i,L ) |h 0k Lck ncLci| . (64)
c ≡ (2L +1)2 (δε (L ))i+1
c Xnc nc c
10
In fact for the first few values of i, we have
α(0,L ) = α (65)
c Lc
α(1,L ) = β (66)
c Lc
α(2,L ) = γ (67)
c Lc
α(3,L ) = δ (68)
c Lc
α(4,L ) = ς (69)
c Lc
α(5,L ) = η (70)
c Lc
α(6,L ) = θ (71)
c Lc
α(7,L ) = ι (72)
c Lc
as defined by Drake [2] up to i = 3. We therefore have the final expression for the second-order
energy correction
∞ ∞
∆E = δe (i,L ), (73)
2 2 c
Xi=0LXc=1
where
2i+1π
δe (i,L ) = q2 ( 1)i+1 a2Lc+2α(i,L )w(2)(L ). (74)
2 c x − 2L +1 c i c
c
2. Second-Order Energy: Calculations
(2)
Consider w (L ) first. Using the formula
0 c
ℓ
2ℓ+1
Y (xˆ)Y∗ (xˆ) = , (75)
ℓm ℓm 4π
m=−ℓ
X
we have
2L +1 2L +1
w(2)(L ) = c χ x−2Lc−2 χ = c x−2Lc−2 (76)
0 c 4π h n0L0M0| | n0L0M0i 4π h in0L0
with χ . It should be pointed out that x−s diverges unless s 2L +2. The
| in0L0 ≡ | n0L0M0i h in0L0 ≤ 0
analytical expressionsfor x−s withsupto16aregiven explicitlybyDrakeandSwainson[25].
h in0L0
Thus the result for i = 0 is
1
δe (0,L ) = q2a2Lc+2α x−2Lc−2 . (77)
2 c −2 x Lch in0L0
For the case of i = 1, we first notice that
h χ = 0. (78)
s| n0L0M0i
