Msc project

Dipole -Dipole interaction of Helium atoms in Highly excited states

Robin Philip ,Msci Theoretical Physics

PHASM201 Final Report

SupervisorsDr Gilllian PeachDr Stephen Hogan

1

Abstract

This project looks into the interaction of two helium atoms in highly excited states , upto n=52. We

are speci�cally intrested in the dipole dipole interaction that occurs between these two states in the absence

of any external �elds .To study the nature of this interaction ,I study the energy shifts for a range of

intermolecular distances "R" and principal quantum number "n" . The project then looks into how this

interaction varies for the same range of "R" and "n" in degenrate Energy levels and then goes on to make

a comparision between the non -degenrate and degnerate case.

2

Contents

1 Introduction 4

2 Methodology 52.1 Quantum defect theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.2 Calculation of Potential between two atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.3 Calculation of the Dipole -Dipole interaction term . . . . . . . . . . . . . . . . . . . . . . . . . . 82.4 Solving for Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.5 Application of theory to He-He Triplets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

3 FORTRAN -77 and its implementation 13

4 Results and Analysis 144.1 Non -Degenrate Energy levels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

4.1.1 Energy shifts with variation of intermolecular distance (R) . . . . . . . . . . . . . . . . . 154.1.2 Energy Shifts varying with principal quantum number (n) . . . . . . . . . . . . . . . . . . 17

4.2 Degenerate Energy Levels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184.2.1 Degenerate Energy shifts varying with intermolecular distance R . . . . . . . . . . . . . . 184.2.2 Degenerate Energy Shifts varying with principal quantum number . . . . . . . . . . . . . 19

4.3 Comparison of Non degenerate against Degenerate shifts . . . . . . . . . . . . . . . . . . . . . . . 20

5 Conclusion 22

6 Appendix 236.1 Appendix A: FORTRAN 77 Code . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236.2 Appendix B: Tables of Raw data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

6.2.1 Non-Degenrate Energy levels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276.2.2 Degenrate Energy Levels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

6.3 Appendix C : Numerical calculation of V(r1,r2,R) for l=l'=1 . . . . . . . . . . . . . . . . . . . . 29

7 References 33

3

1 Introduction

The study of Rydberg atoms �rst came into prominence in 1885 , in the balmer series of hydrogen atom forthe wavelengths of the visible series of the hydrogen atom . The signi�cance of these rydberg atoms were notapparent until Bohr's Theory of H atom was proposed in 1913. By de�nition, rydberg atom is an excited atomwith one or more electrons that have high principal quantm number[1].These atoms have a very particular setof properties[2], speci�cally:

1. The binding energy is small , and decrease with 1n∗2

2. The radiative life time is very long.

3. The dipole matrix elements provide a way of calculating the interaction between these atoms

The main attraction of these systems comes from the existence of large electric dipolar moments . And these canbe applied to manipulate rydberg particles with small gradients in Electric �eld [3], or even control interactionsbetween two particles at macrospic distances[4]. Hence , they provide a useful way of controlling the dynamicsof complex quantum systems in areas such as Solid State and Plasma Physics .

Another intresting phenomena that arises from rydberg atoms are that of the �Dipole blockade e�ect�.Thisarises due to the long range dipole interactions that causes the Rydberg energy levels to shift and therebypreventing laser excitation[5].

This has further applications in quantum computation , quantum cryptography , atomic clocks etc.

4

2 Methodology

We start by considering two atoms in a (np,np) and (ns,nd) state . We're intrested purely in how thesetwo atoms interact with each other , without the introduction of any external electric �eld or magnetic �eld. Mathematically , if one looks at 1

Rl+l′+1 in eqtn (7) then , both (l = 0, l = 2), (l = 1, 1) ⇒ 1R3 ,gives

us the dipole-dipole interaction term.However , if one were to look at experimentally , such dipole's onlyexist due to the exchange of excitation which corresponds to the existence of a transition dipole from a state{|nl >= |r >} =⇒ {|n′l′ >= |r′ >}[2]

The energy of a Rydberg atom [1]is given by :

Enl = −1

2× 1

(n− µ)2a.u (1)

where , µ is the quantum defect. Now compare this with that of a hydrogen [6]i.e

En = − 1

n2a.u (2)

we see , the quantum defect is essential a correction term from the energy levels of Hydrogen.To work out the quantum defect for the Helium atoms , we use the property that µ can be written in terms

of En , using the Rydberg Ritz coe�cients[7] we expand the quantum defect as :

µ = a+ bEn + cE2n + dE3

n . . . (3)

For He atoms , the coe�cients only up-to c are required , i.e we obtain a quadratic in En. The below tablelists the coe�cients obtained for di�erent values of l , which are calculated from experiments conducted atUniversity College London.

L values a b c

l=0 2.96749961*10−1 -3.373086437*10−2 8.20084060*10−3

l=1 6.83199414*10−2 1.78761373*10−2 -1.72931696*10−2

l=2 3.06013304*10−3 1.15510046*10−2 3.30544994*10−2

l=3 5.40573349*10−4 6.17605659*10−3 4.66550231*10−2

l=4 1.52541527*10−4 2.1074564*10−3 1.62811833*10−2

Table 1: Values of coe�cients a,b,c for di�erent l

Calculation of the quantum defect values from the tables and using equation (3) we �nd that for low �l�states the quantum defect is large , and does not necessarily follow the hydrogen energy levels . For higher lvalues , the quantum defect becomes increasingly smaller and begins to mimic that of the hydrogen.Furthermoreat higher n , µ→ constant , which is an essential property we exploit for Rydberg atoms .

Helium in its ground state , has a electronic con�guration of 1s2, however when an electron is excited , weeither have a singlet (parahelium) or triplet state(orthohelium) . For the purposes of this research we considerhelium atoms only in triplet states due to their lower energy and higher excitation times [8] .

5

2.1 Quantum defect theory

In general problem of solving for wavefunctions becomes complicated for non hydrogenic atoms due to theinseperability of the Schrödinger's equation, however in our case , quantum defects provide us with an alternativesolution. Quantum Defect theory (QDT) , �rst developed by Seaton [9] exploits the the fact that a Heliumatom in a Rydberg state behaves in a similar way to that of the hydrogen atom . In order to see why , let usconsider a helium atom in a rydberg state , i.e in our case the valence electron is at n=52 . The core electronsshield the outer electrons from the positive charge of the nucleus and thereby the outer electron experiences aelectric potential identical that to of a hydrogen atom. Furthermore , we assume the He core to be stationaryand spherically symmetric , therefore we expect the potential to be dependent only on r, i.e VHe(r) .

The above assumptions , leads us to two consequences :

1. The Schrödinger equation becomes seperable .

2. The He wavefunction is anologus to the H wavefunction and di�ers only by the radial part.

Therefore one can write the wavefunction in the form, where Pnl are the solutions to the radiall equation. Onecan therefore , write the Schrödinger equation[10] as :

δ2Pnl(r)

δ2r−(l(l + 1)

r2− 2V (r) + Enl

)Pnl(r) = 0 (4)

which can be solved numerically as discussed in Section 4.

6

2.2 Calculation of Potential between two atoms

Figure 1: Calculation of potential for two atoms and their electrons seperated by a distance R

In order to �nd V(r) of the system , lets start by considering a two electron model for two atoms Atom1( A1)and Atom 2(A2) in Rydberg states as in Fig(1) . If we neglect the structure of the cores A1 and A2 and justtreat them as charged particles , the interaction between A1 and A2 can be written as :

V (r1, r2, R) =1

| ‖R− r1 + r2‖− 1

‖R− r1‖− 1

‖R+ r2‖+

1

R(5)

where R is the distance between A1 and A2 ,r1is the distance between atom A1 and electron 1 and r2is thedistance between atom A2 and electron 2.

We're intrested in the case where the atoms are well seperated assuming R≫r1, r2. From [11] , we get

V (r1,r2, R) =

∞∑l,l′

∞∑m,m′

(−1)l(4π) 32(l + 1)!

l!l′!×

[(2l + 1)(2l′ + 1)(2l + 2l′ + 1)]− 1

2

(l l′ l + l′

0 0 0

)(

l l′ l + l′

m m′ M

)rl1r

l′

2

Rl+l′+1Ylm(r1)Yl′m′(r2)Yl.+l′M (R) (6)

7

As stated previously , we're intrested in the dipole -dipole interaction between these two electrons . Whenl=l'=1 or l=0,l'=2 , we obtain the 1

Rl+l′+1 term gives as 1R3 ,which is known as the C3 coe�ecient representing

the dipole interaction we're intrested in .

Now Suppose we need the general matrix element of V (r1,r2, R) with respect to the atom-atom state(n1l1m1n2l2m2) and (n′1l

′1m′1n′2l′2m′2.) i.e

〈n1l1m1n2l2m2|V (r1,r2, R)|n′1l′1m′1n′2l′2m′2.〉 (7)

ignoring the the constants and the the matrix relations , we consider each spherical harmonic with itsrespective state. Therefore for electron 1 we get ,⟨

n′1l′1m′1|rl1Ylm(r1)|n1l1m1

⟩=

ˆ ∞0

Pn′1l′1(r1)rl1P (r1)dr1 ×

ˆY ∗l′1m′1(r1)Ylm(r1)Yl1m1

(r1)dr1 (8)

and similarly for electron 2 we get :⟨n′2l′2m′2|rl

′

2 Yl′m′(r2)|n2l2m2

⟩=

ˆ ∞0

Pn′2l′2(r2)rl′

2 P (r2)dr2 ×ˆY ∗l′2m′2(r2)Yl

′m′(r2)Yl2m2(r2)dr2 (9)

The radial part of the relations are normalised as

ˆ ∞0

P 2nl(r)dr = 1 (10)

and the angular part of the relation is given by 3jm Weigner symbol[12] :

ˆY ∗l′1m′1(r1)Ylm(r1)Yl1m1

(r1)dr1 =

[(2l′ + 1)(2l + 1)(2l1 + 1)

4π

] 12

×

(l′1 l l10 0 0

)(−1)m

′1

(l′1 l l1−m′1 m m1

)(11)

Hence ,we arrive at a general formula for the calculation of potential assuming the two cores are well seperated.

2.3 Calculation of the Dipole -Dipole interaction term

As mentioned earlier , we are solely intrested in the C3 coe�ecient . The states availablie in our transitions arel=l'=1 or l=0,l'=2 ,however for calculational simplicity we �rst consider the special case l=l'=1 .

Lets start by considering the z axis which is to be taken along R . We have then :

Yl+l′,M (R) = Yl+l′,0(R) =

[(2l + 2l′ + 1)

4π

] 12

Pl(1) =

[(2l + 2l′ + 1)

4π

] 12

(12)

Furthermore , the coe�ecient

(l l′ l + l′

m m′ M

)= 0 unless m+m′ +M = 0⇒ m′ = −m. If in addition ,

we consider our case of l = l′ = 1 then equation (6) becomes

V (r1,r2, R) =∑m

(−1)(4π)23

[(1 1 20 0 0

)−1(1 1 2m −m 0

)r1r2R3

Yl,m(r1)Yl,−m(r2)

](13)

8

and From Edmonds ,p127 (1 1 2m −m 0

)= (−1)1−m

(2(3m2 − 2)

(5!)12

)

⇒(

1 1 20 0 0

)−1(1 1 2m −m 0

)= (−1)m

(2(3m2 − 2)

2(−2)

)= (−1)m−1

((3m2 − 2)

2

)(14)

Substituing these relations into (6) , we obtain

V (r1,r2, R) =r1r2R3

∑m

(−1)(4π)23(−1)m−1 (3m

2 − 2)

2Yl,m(r1)Yl,−m(r2)

=r1r2R3

∑m

4π

3(3m2 − 2)Yl,m(r1)Y

∗l,m(r2)

Expanding the sum sign ,for m = −1, 0, 1

V (r1, r2, R) =r1r2R3

(4π

3

){3 (Y11(r1)Y

∗11(r2) + Y1,−1(r1)Y ∗1,−1 (r2))− 2

∑Yl,m(r1)Y

∗l,m(r2)

}similarly expanding for the term with coe�ecient 2 ,

V (r1, r2, R) =r1r2R3

(4π

3

){[Y11(r1)Y ∗11(r2) + Y10(r1)Y

∗10(r2) + Y1,−1(r1)Y ∗1,−1 (r2)]− 3 (Y10(r1)Y

∗10(r2))}

V (r1, r2, R) =r1r2R3

[r1.r2 − 3(r1.R)(r2.R)] (15)

Eq (15) is known as the Dipole-Dipole interaction term ,where

C3 = r1r2 [r1.r2 − 3(r1.R)(r2.R)] .

2.4 Solving for Hamiltonian

The Hamiltonian of the two electron system is therefore given by :

H = −1

2∇2

1 −1

2∇2

2 +r1r2R3

[r1.r2 − 3(r1.R)(r2.R)] (16)

One has to now use the tensor algebra notation to obtain , we can write

V (r1, r2,R) = Vp(r1, r2,R)T (2).C(2) (17)

where T(2) and C(2) are two second rank irreducible tensors written by :

T (2, q) = [C1(1)× C2(2)]2q =

∑µ

C1,1,2µ,q−µ,qC1(1, q)C2(1, q − µ) (18)

Furthermore C1(1), C2(2) are �rst -rank tensors and are de�ned by :

C1(1, q) =

√4π

3Y1q(θ1, φ1);C2(1, q) =

√4π

3Y1q(θ2, φ2);C(2, q) =

√4π

5Y2q(θ, φ) (19)

9

where the spherical polar angles (θ1, φ1) and (θ2, φ2) specify the directions of r1and r2and (θ, φ) specifythe direction of R relative to the quantisization axis z.If there is no external �eld applied to the system , thenwithout the loss of generality R may be taken to coincide with z and we can therfore write :

C(2, q) = δq0 (20)

Further more

Vp(r1, r2, R) = −√6r1r2R3

, Cl1,l2,Lm1,m2,M

= (−1)l1−l2+M (2L+ 1)12

(l1 l2 Lm1 m2 −M

)(21)

We can now use these relations to obtain the reduced matrix elements of the potential, .First, in an uncoupledrepresentation for the atom-atom basis function is given by

however in our case , using a coupled representation gives us :

|γl1l2LM >=∑m1m2

Cl1l2Lm1m2M|n1l1m1 > |n2l2m2 > (22)

where γdenotes all other quantum numbers specifying the atom-atom pair.From Edmonds ,equation (5.2.4) :

T (2).C(2) =∑q

(−1)qT (2, q)C(2,−q) (23)

Therefore equation (17) becomes

V (r1, r2,R) = Vp(r1, r2,R)∑q

(−1)qT (2, q)C(2,−q) (24)

If we now consider the matrix elements of the T(2,q),from Edmonds ,equation (5.2.4) we obtain⟨γ′l′

1l′2L′M ′|T (2, q)|γl1l2LM

⟩= (−1)L

′−M ′(

L′ 2 L−M ′ q M

)⟨γ′l′

1l′2L′|T (2)|γl1l2L

⟩(25)

i.e we're able to reduce the 2nd rank tensor to a 1st rank one and thereby eliminating the dependence on�M� in the matrix elements .

One has to further make use of the below relations , which are not proved here , but obtained from�Edmonds�[12]:

From Edmonds (7.1.5)

⟨γ′l′

1l′2L′|T (2)|γl1l2L

⟩=⟨l′

1|C1(1)|l1⟩⟨

l′′

2 |C2(1)|l2⟩[5(2L′ + 1)(2L+ 1)]

12 ×

l′1 l1 1l′2 l2 1L′ L 2

(26)

and ⟨l′

1|C1(1)|l1⟩= (−1)l

′1 [[(2l′1 + 1)] (2l1 + 1)]

12

(l′1 1 l10 0 0

)(27)

⟨l′′

2 |C2(1)|l2⟩= (−1)l

′2 [[(2l′2 + 1)] (2l2 + 1)]

12

(l′2 1 l20 0 0

)(28)

10

substituting (26,27,28) into (25) we obtain :

⟨γ′l′

1l′2L′M ′|T (2, q)|γl1l2LM

⟩= (−1)L

′−M ′(

L′ 2 L−M ′ q M

)[5(2L′ + 1)(2L+ 1)]

12 ×

l′1 l1 1l′2 l2 1L′ L 2

×(−1)l

′1 [[(2l′1 + 1)] (2l1 + 1)]

12

(l′1 1 l10 0 0

)× (−1)l

′2 [[(2l′2 + 1)] (2l2 + 1)]

12

(l′2 1 l20 0 0

)(29)

Eqtn (29) essentially provides us with a route to calculating the matrix elements of our intersted quantity ,simply by a few matrix relations . These relations can be implemented in Fortran routine , which essentiallysaves a lot of computational time.

11

2.5 Application of theory to He-He Triplets

We consider the interaction that links the atom-atom states (np ,np) and (ns,nd).We therefore have ,l′1 = l′2 =1for the (np,np) and l1 = 0, l2 = 2 for the (ns,nd) state, with n values upto 52. We consider the simplest case inwhich R = z so that eqtn (20) is valid and from eqtn (25) it follows that M'=M and so is a constant of motion.So the cases of M=0,1,2 can be considered seperately and taking the matrix elements of equation we obtain :

⟨γ′l′

1l′2L′M ′| − 1

2∇2

1 −1

2∇2

2 +r1r2R3

[r1.r2 − 3(r1.R)(r2.R)] |γl1l2LM⟩

= δl′1l1δl′2l′2δL′L (Enl1 + Enl2) (30)

where Enlis the energy in atomic units of the nl level of He .We're intrested in the o� -diagonal terms of theabove Hamiltonian , which gives us the coupling terms we're looking for given by :⟨

γ′l′

1l′2L′M ′|V (r1, r2,R)|γl1l2LM

⟩= −√6

R3〈l′1|r1|l1〉〈l′2|r2|l2〉 ×

⟨γ′l′

1l′2L′M ′|T (2, 0)|γl1l2LM

⟩(31)

where 〈l′1|r1|l1〉 , 〈l′2|r2|l2〉 are radial equations to be calculated using the Fortran routine.As in our case [l1 = 0, l2 = 2⇒ L = 2], and [l′1 = 1, l′2 = 1⇒ L′ = {0, 1, 2}] and subbing (29) into (31) we

obtain

⟨γ′l′

1l′2L′M |V (r1, r2,R)|γl1l2LM

⟩= −√6

R3〈l′1|r1|l1〉〈l′2|r2|l2〉×

(−1)L′−M

(L′ 2 2−M 0 M

)[5(2L′ + 1)(5)]

12 ×

1 0 11 2 1L′ 2 2

×(

1 1 00 0 0

)× [[(3)] (5)]

12

(1 1 20 0 0

)(32)

However , we can diagonalise the original Hamilitonian matrix to obtain the values in eqtn (32) as

U+HU = E (33)

where E is the diagonal matrix containing the eigenvalues i.e the interaction terms we're intrested in andthe Unitary matrix containing the corresponding eigenvectors.

12

3 FORTRAN -77 and its implementation

The above theory that has been discussed in Chapter 2 is implemented in the fortran programme . From eqtn(33) , the programme is used to implement the matrix relations by their recursion relations given in Appendixsection. However in order to calculate the radial matrix elements〈l′1|r1|l1〉 , 〈l′2|r2|l2〉 , the Numerov method[13]is used .An alternative approach is to search for solutions analytically[14] , however was discarded due tothe size of the data in this particular project.

The method applies to equations of the type :

d2y

d2x+ f(x)y = 0 (34)

which in our case is equation (4).Consider now the function , which is tabulated at a number of points seperated by equal intervals of length

h.We use the fact that

δ2yn ' h2(1 +

1

12δ2)y′′n (35)

and then substituting into eqtn (35) we get :

h2(1 +

1

12δ2)(y′′n + fnyn) = 0; [fn ≡ f(xn), yn ≡ y(xn)] (36)

so that:

yn+1 − 2yn + yn−1 + h2fnyn +h2

12(fn+1yn+1 − 2fnyn + fn−1yn−1) = 0 (37)

or. . . (

1 +h2

12× fn+1

)yn+1 −

(2− 10

12h2fn

)yn +

(1 +

h2

12fn−1

)yn−1 = 0 (38)

The above is a step by step method , we start out with at a small x , with two values of reasonable guessesof y(x) say y1, y2,then eqtn(39) gives y3.

Therefore y2, y3 → y4 . . . etc.Applying this to equation (4) we are faced with the problem of choosing boundary conditions .Quantum

mechanically wavefuctions are de�ned for −∞ < x < ∞, fo r≥ 0, therfore the boundary conditons need tobe chosen carefully at the two end points say [−xm, xm] to be solved numerically. We therfore need to �nd{Enl, Pnl}which not only satis�es the Schrodinger equation but the boundary conditions.To get around thisproblem , we start in a region −xm and integrate towards the origin , and similarly we start from the regionxmand integrate towards the origin . The solutions of these two routes are matched at a suitable point say, xc.To �nd a suitable matching point , the lograthmic derivative of both the solutions is compared.A boundstate can be obtained by setting these two derivatives equal at xc.

13

4 Results and Analysis

4.1 Non -Degenrate Energy levels

In order to create a meaningful sense of the anlaysed data . A few de�nitions need to be implementeed .Asmentioned previoulsy , we're intrested in the dipole interaction of the subsuquent states . To do this ,wecalculate the Energy shifts that ocurs betweeen these where 4Eshift is calculated as :

4Eshift = (Ens + End)− (Enp + Enp) (39)

In our case , we take our initial state to be (np,np) , and the energies of these states are given by (1). Theenergies of (ns,nd) state when interaction occurs can be calculated from the values using equation (32) &(33) .

we de�ne for (ns,nd) state[l1 = 0, l2 = 2⇒ L = 2]→ State1

Similarly for (np,np),[l′1 = 1, l′2 = 1⇒ |L′| = 0]→ State 2

-[l′1 = 1, l′2 = 1⇒ |L′| = 1]→ State 3

[l′1 = 1, l′2 = 1⇒ |L′| = 2]→ State 4

where we have ignored the negative values due to the symmetry of the situation.We further, need to consider the possiblities of M , ranging from {M = 0, 1, 2} for each state as above ,

again ignoring the negative values due to symmetry.

14

4.1.1 Energy shifts with variation of intermolecular distance (R)

We start by plotting Energy shifts as a function of R for the various states .

Figure 2: Energy Shifts for M=0 ,State 1-4 against R

NB : The graph in the above and consequent �gures do not intersect the x axis . As per Table 1 , at largedistances the interaction is of non-apperciable quantity to be displayed by Excel. For a closer look at the data, please look at appendix B.

15

From Fig2 , we notice the energy shifts , decrease as the distance between the atoms increases . This due tothe relation V α 1

R3 , from eqtn(15) , the dipole intreaction strength e�ectively decreases as R3.However , states2 and 3 are missing from the above graph, if we take a closer look at the values involved in Table (2):

R/a.u Energy Shifts for State 1/MHz Energy shifts State2/Mhz Energy Shifts State 3/Mhz Energy Shifts State 4/Mhz

108 −5.25817 ∗ 10−19 0 0 5.25817 ∗ 10−19

107 −5.25817 ∗ 10−13 −3.09332855 ∗ 10−28 0 5.25817 ∗ 10−13

106 −5.25817 ∗ 10−7 0 0 5.25817 ∗ 10−7

105 −5.25817 ∗ 10−1 0 0 5.25817 ∗ 10−1

104 −6.01012725 ∗ 104 0 0 6.01012725 ∗ 104

Table 2: Energy Shifts for M=0 ,State 1-4 for a range of R

We notice state 2 & 3 essentialy produce no shifts ,due to these states coupling with each other .Furthermore, the shifts between 1 and 4 are symmetric in nature , therfore considering a single state ( state 1 in our case)would provide the information of the whole system. This is attributed to the fact , we have considered onlytriplet He states , which have symmetric wavefunctions . These wavefunctions inturn strongly interact with thestates 1 &4 symmetric to produce a symmetric energy shift .The Range of R ,chosen is such that for R > 108 ,thedipole-dipole interaction becomes negligible and crosses over to van-der waals interaction C6coe�ecient takesup importance known as the "van der waals radius "[2].Hence at small distances , dipole -dipole interactionproduces the largest interaction between two Rydberg atoms.

Now by plotting these states 1-4 , for M=0,1,2 as in Fig (3) we can see how �M� which is a constant ofmotion in our problem e�ects the energy shifts .

Figure 3: Energy Shifts for M=0,1,2 ,State 1-4 for a range of R

We see ,the symmetry throughout the di�erent M's for the states 1&4 and furthermore one notices ,increasingM decreases eneryshifts.A possible reason for this , is by increasing M , the size of the matrix elements is reducedand therby the reducing the interaction terms , creating a smaller energy shift.�M� desciribed here is simply theprojection of internuclear axis of the two atoms , in our case the �z� axis.

16

4.1.2 Energy Shifts varying with principal quantum number (n)

Another intresting variation ,is to see how the shifts vary with principal quantum number n . We choose aupper limit of n=52 , and lower limit of n=4 essentially maintaining the characterstics of Rydberg atoms.Weknow , as the dipole varies as n2 , the C3coe�cient goes to n∗4.Therefore one would expect a rapid rise in ,theEnergy shift for higher principal quantum i.e the Rydberg states. Plotting the Energy shifts against n we get :

Figure 4: Energy Shifts (state 1 )with varying principal quantum number at R=104

It is clear from the above graph , at lower n's the dipole dipole interaction is of negligible magnitude .

17

4.2 Degenerate Energy Levels

As a further extension , we look at setting energy levels close-by to each other , which can lead to interestingquantum phenomena between the two Rydberg atoms

We calculate the initial energies for the (np,np) and (ns,nd) for He using QDT as beforeFor the (np,np) state ,

Enp+np =−1

(n− µ)2a.u (40)

And similarly for the (ns ,nd ) state :

Ens+nd = −1

2×

[1

(n− µ1)2 +

1

(n− µ2)2

]a.u (41)

whereµ, µ1, µ2are quantum defects of {np,ns,nd} respectively.Now , we're intrested in what happens when these energy levels are close together , therefore we de�ne a

new level as :

E[np+np,ns+nd] =1

2[Enp+np + Ens+nd] (42)

Once these new energy levels are de�ned , we look at how the introduction of the perturbation 1R3 term

a�ects the levels with increasing R.

4.2.1 Degenerate Energy shifts varying with intermolecular distance R

As before , we start by plotting the energy shifts with varying R , for M=0,1,2

Figure 5: Degenerate Energy Shifts for state 1 against R,M=0,1,2

18

We notice , a similar behavior to the non-degenerate energy levels , a larger energy shift is observed asexpected . However at M=2 , the symmetry of the graph is broken. The energy shift changes sign at M=2 .It is possible at M=2 , this is a result of the mixing of the states 1 and 4 , causing the states to change sign .Furthermore , at M=2 , the size of the matrix is reduced to a 2×2 from a 3× 3matrix , attributed to the factthat much less interaction between states occur.

One can also see from Table 3 .the symmetry between the states 1& 4 is preserved as before .Hence , themethod of labeling "states" is a approach that can be extended to further cases ,especially when external �eldsare applied .

R/a.u Energy Shifts for State 1/MHz Energy shifts State2/MHz Energy Shifts State 3/Mhz Energy Shifts State 4/Mhz

108 6.38617× 10−80 0 -6.38617× 10−8

107 6.38617× 10−5 3.09332855 ∗ 10−280 −6.38617× 10−5

106 0.063861709 0 0 -0.063861709

105 63.8617087 3.09332855 ∗ 10−280 -63.8617087

104 63861.7087 3.09332855 ∗ 10−280 -63861.7087

Table 3: Degenerate Energy Shifts for M=0 ,State 1-4 for a range of R

4.2.2 Degenerate Energy Shifts varying with principal quantum number

As before , we proceed to look the behavior of the shifts , for varying principal quantum number . As we haveimposed degeneracy on the levels one would expect a certain quantum mechanical behavior at lower levels of nand to be smoothed out as we progress to higher values of n.

Figure 6: Degenerate Energy Shifts for state 1 Vs n at R=104,M=0,1,2

We notice the curve do not essentially follow a smooth curve as before.I ,hypothesize this is due to the e�ectsof �rst order perturbation causing resonance at n≈28 . Also as previously , the curves are not symmetric innature , due to the mixing that occurs between state 1 and state 4 and hence the change of sign in the shifts

19

4.3 Comparison of Non degenerate against Degenerate shifts

Finally , if we compare the two cases for varying R , one can notice some interesting peculiarities .

Figure 7: Non Degenerate(LHS) against Degenrate Energies(RHS) for R=104, 106, 108a.u

Initially for R = 104range, the magnitude of the energy shifts for the non degenerate and degenerate occursin a similar range ,i.e 105MHz . However as we increases R to 106,we see a rapid increase in the shift for thedegenerate levels , almost an increase by a factor of 105MHz .Now if we continue , to R = 108a.u,we see thisfactor to almost double to 1011MHz . Clearly , by bringing the energy levels closer together , and increasingthe intermolecular distance creates a large energy shift . This demonstrates , how a small change in the spacingof Energy levels can cause such large variations in the shifts . And hence , one has to be accurate with thequantum defects supplied initially to obtain a good result for the dipole-dipole interaction.

20

Further more , for a given R (say 108) , looking at the non-degenerate case (Fig 7), we see a smooth behaviorfor the shifts through n , i.e a preferred sign (-ve) (See Table 4,5,6).If we now consider the degenerate case forthe same R (Table 7,8,9) , we see at n=28 , the energy shift changes sign to positive (for m=2) and then switchessign back or vice versa for (m=0,1). I imagine this is due to signi�cant mixing occurs between states 1 and 4 (states 2&3 produce no shifts ) creating a kind of resonance at n=28. However , to investigate why such mixingoccurs at n=28 , the speci�c region needs to be investigated further. A possible approach , would be to obtainshifts for either side of n ( say 26 and 30) , and see if the sign change is local only to n=28 or throughout (26-30). If the latter , it would point to a possible quantum behavior rather than a bug/error in the program . Due totime constraints , I was unable to look further into this , and would be the starting line for further work .

21

5 Conclusion

As a result of this project , I have gained a deeper and thorough understanding of the mathematics involvedwith Rydberg atoms By studying Rydberg atoms and exploiting their properties , I have been able to gatherquanti�able data, and explore the interaction between the two atoms.In order to do so ,I started of withcalculating relatively simple energies using QDT , and then derived the energies of the atoms due to theperturbation of the dipole interaction in analytical form . then proceeded onto looking into how these interactionvaried for di�erent R ,n in both degenerate and non degenerate cases. By making such a comparison , it becameclear that by arti�cially imposing the energy levels at such close proximity very large increase in energy shift'sare created . A much better approach , would have been to gradually bring the levels closer together and henceproduce a more ,meaningful and accurate comparison between the non-degenerate and degenerate case.We alsocame across how labeling and symmetry in these states plays a crucial role in our calculations .

Naturally ,the next step in the project ,would be to investigate, if the symmetry of the states 1-4 hold whenan external �eld is applied .And if so, use this symmetry to study the variation of the 1

R3 interaction and theresulting quantum phenomena .Further more , if we consider the excitation np 3

2+ np 3

2→ ns+ (n+ 1)s , then

under stark shifting , resonance occurs if the np 32is located midway between ns (n+1)s states . The resulting

phenomena due to the dipole-dipole coupling of the atoms known as "Förster resonance " can be studied ingreater detail [5].

As a �nal word, I feel privileged to have the opportunity to work with Dr G.peach and would like to thankher , for the endless guidance and insight she provided throughout the project. Thank you Dr Peach.

22

6 Appendix

6.1 Appendix A: FORTRAN 77 Code

IMPLICIT DOUBLE PRECISION (A-H,O-Z)REAL*16 BB,CC,DDCOMMON/MAR/ RI(502001),WI(502001),YI(502001),RX,JI,JF0COMMON/PARM/ GZ,DA,DB,DC,DD,ZNZ,GZD,ZND,BTZ,BTQ,BTQ1,BTQ2,1 BTD,ALZ,ALQ,BLZ,GMZ,GEC COMMON/ANGL/ L1,L1P,L2,L2P,L,LP,MCOMMON/COEF/ AUDIMENSION YH(502001,3),BN(3),S(2),B(4,4),C(4,4),E(4,4),D(4)DIMENSION BB(4,4),CC(4,4),DD(4),EE(4),WW(384),LL(3),IJ(4)WRITE(6,96)WRITE(6,*) ' COUPLED ATOMS IN RYDBERG STATES'CALL INITIALIC=0 NG0=5 N0=20BN0=DBLE(N0)BN2=BN0*BN0HB0=1.0D-2 IF(N0.GE.49) RX=BN2IF(N0.LE.48.AND.N0.GE.43) RX=1.5D0*BN2IF(N0.LE.42.AND.N0.GE.3) RX=1.75D0*BN2IDEL=1BN(1)=BN0-0.29703D0BN(2)=BN0-6.8192D-2BN(3)=BN0-3.526D-3LL(1)=0 LL(2)=1LL(3)=2N=1R=1.0D8WRITE(6,96)WRITE(6,98) N0,RDO 1 K=1,3 BNU=BN(K) L=LL(K)CALL RADICL(BNU,L,HB0,NG0,NODE,IDEL,IC) BN(K)=BNUDO 2 J=1,JF 2 YH(J,K)=YI(J)1 CONTINUEWRITE(6,96)WRITE(6,*) ' N1* L1 N2* L2 <N1* L1 |r| N2* L2 >'0 DO 3 K=1,2 K1=K+1S(K)=0.0D0DO 4 J=2,JI 4S(K)=S(K)+WI(J)*YH(J,K)*YH(J,K1)*RI(J)**N DO 5 J=JI,JFDO 5 J=JI,JF5 S(K)=S(K)+WI(J)*YH(J,K)*YH(J,K1)*RI(J)**N3 WRITE(6,92)BN(K),LL(K),BN(K1),LL(K1),S(K)WRITE(6,96) T=-S(1)*S(2)*SQRT(6.0D0)/R**3L1=LL(1)L2=LL(3)L1P=LL(2)

23

L2P=LL(2) L=2WRITE(6,*) ' COUPLED STATES, SEE BOTTOM OF PAGE 4'WRITE(6,96)WRITE(6,*) ' L1 L2 L'WRITE(6,90) L1,L2,LDO 6 LP=0,26 WRITE(6,90) L1P,L2P,LPWRITE(6,96)DO 7M=0,2WRITE(6,97) MDO 8 I=1,4DO 9 J=1,49 E(I,J)=0.0D08 CONTINUE E(1,1)=1.0D0DO 10 K=2,4 LP=K-2CALL ANGLE(Y)CALL ANGLM(Z)E(1,K)=Y*Z*T 10E(K,1)=E(1,K) II=0DO 11 I=1,4 Z=E(I,1) IJ(I)=0IF(Z.EQ.0.0D0) GO TO 11II=II+1 IJ(I)=II11 CONTINUE E(1,1)=0.0D0DO 12 I=1,4 II=IJ(I)IF(II.EQ.0)GO TO 12DO 13 J=1,4 JJ=IJ(J)IF(JJ.EQ.0) GO TO 13BB(II,JJ)=E(I,J)B(II,JJ)=E(I,J)13 CONTINUE12 CONTINUENI=IIEE(1)=-0.5D0/BN(1)**2-0.5D0/BN(3)**2DO 14 I=2,414 EE(I)=-1.0D0/BN(2)**2 DD(1)=0.5D0*(EE(1)+EE(2))DO 20 I=1,420 EE(I)=DD(1)DO 15 I=1,4 II=IJ(I)IF(II.EQ.0) GO TO 15BB(II,II)=EE(I) B(II,II)=EE(I)15 CONTINUEDO 16 I=1,NI 16WRITE(6,93) (B(I,J),J=1,NI)WRITE(6,96)CALL DMATRIX(BB,4,NI,CC,4,DD)CALL SORT(DD,NI)DO 17 I=1,4 II=IJ(I)IF(II.EQ.0) GO TO 17DD(II)=AU*(DD(II)-EE(I)) D(II)=DD(II)

24

DO 18 J=1,4 JJ=IJ(J)IF(JJ.EQ.0) GO TO 18C(II,JJ)=CC(II,JJ)18 CONTINUE17 CONTINUEWRITE(6,*) ' NUMBER OF COUPLED STATES'WRITE(6,90) (I,I=1,4)WRITE(6,90) (IJ(I),I=1,4)WRITE(6,96)WRITE(6,*) ' ENERGY SHIFTS IN MHz'WRITE(6,93) (D(J),J=1,NI)WRITE(6,96)DO 19 I=1,NI19 WRITE(6,93) (C(I,J),J=1,NI)WRITE(6,96) 7CONTINUE 90FORMAT(10I5)92 FORMAT(0P,F12.6,I3,0P,F12.6,I3,1P,D18.9)93 FORMAT(1P,4D16.8)96 FORMAT(/)97 FORMAT(' M =',I3/)98 FORMAT(' PRINCIPAL QUANTUM NO: n=',I3,' ','R = ',1P,D7.1)STOPENDSUBROUTINEANGLE(Y) C COMPUTES <...||T(2)||...>, SEE EQUATION(27) IN NOTESIMPLICIT DOUBLE PRECISION (A-H,O-Z)COMMON/ANGL/L1,L1P,L2,L2P,L,LP,MJ=(2*L1P+1)*(2*L1+1) A=SQRT(DBLE(J))B=C3JL(L1P,1,L1)C1=A*B*(-1.0D0)**L1P J=(2*L2P+1)*(2*L2+1)A=SQRT(DBLE(J))B=C3JL(L2P,1,L2)C2=A*B*(-1.0D0)**L2PJ=5*(2*LP+1)*(2*L+1)A=SQRT(DBLE(J))B=C9JL(L1P,L1,1,L2P,L2,1,LP,L,2)Y=C1*C2*A*BRETURNENDSUBROUTINE ANGLM(Z) C ANGULAR COEFFICIENT, SEE EQUATION (29) IN NOTESIMPLICIT DOUBLE PRECISION (A-H,O-Z)COMMON/ANGL/ L1,L1P,L2,L2P,L,LP,M A=(-1.0D0)**(LP-M) B=S3JL(LP,2,L,-M,0,M) Z=A*BIMPLICIT DOUBLE PRECISION (A-H,O-Z) COMMON/ANGL/ L1,L1P,L2,L2P,L,LP,M A=(-1.0D0)**(LP-M) B=S3JL(LP,2,L,-M,0,M) Z=A*BRETURNEND

25

SUBROUTINE SORT(DD,NI)IMPLICIT REAL*16 (A-H,O-Z)DIMENSION DD(4),AA(4)IF(NI.EQ.2)RETURN AA(2)=DD(2) AA(1)=DD(3) AA(3)=DD(1)DO 1 I=1,NI 1DD(I)=AA(I)RETURNENDSUBROUTINEINITIALIMPLICITDOUBLE PRECISION (A-H,O-Z)0COMMON/PARM/ GZ,DA,DB,DC,DD,ZNZ,GZD,ZND,BTZ,BTQ,BTQ1,BTQ2,1 BTD,ALZ,ALQ,BLZ,GMZ,GECCOMMON/COEF/ AUCOMMON/FACTL/FCL(2000)FCL(1)=0.0D0DO 1 I=1,19991 FCL(I+1)=LOG(DBLE(I))+FCL(I) RYC=3.28984196036D9BMEC=5.4857990945D-4BM=4.00260325415D0RYH=RYC/(1.0D0+BMEC/BM)AU=2.0D0*RYHGZ=3.514113522451D+00DA=3.295929341351D+00DB=-3.245921168509D-01DC=3.907841427636D+00DD=-2.919710630508D-01ZNZ=1.000000000000D+00BTZ=2.967345502362D+00BTQ=2.317501491612D+00BTQ1=2.037020420889D+00BTQ2=1.560583333745D+00ALZ=2.812500000000D-01ALQ=2.343750000000D-01BLZ=8.398437500000D-02GMZ=2.596028646000D-02GEC=2.540849230640D+00 BTD=1.0D0GZD=1.0D0ZND=0.0D0RETURNEND

26

6.2 Appendix B: Tables of Raw data

6.2.1 Non-Degenrate Energy levels

n 4Eshift,R=104(Mhz) 4Eshift,R=105(Mhz) 4Eshift,R=106(Mhz) 4Eshift,R=107(Mhz) 4Eshift,R=108(Mhz)

52 -60101.2725 -0.525781217 -5.25817E-07 -5.25817E-13 -5.25817E-19

44 -26914.8818 -0.083350907 -8.33514E-08 -8.33514E-14 -8.33514E-20

36 -7016.21212 -0.009108854 -9.10886E-09 -9.10886E-15 -9.10886E-21

28 -561.571631 -0.000567842 -5.67842E-10 -5.67842E-16 -5.67842E-22

20 -13.728748 -1.37301E-05 -1.37301E-11 -1.37301E-17 -1.37294E-23

12 -0.046991824 -4.69918E-08 -4.69918E-14 -4.69918E-20 -4.45439E-26

4 -1.5805E-07 -1.5805E-13 -1.5805E-19 -1.58378E-25 0

Table 4: Energy shifts for M=0 , state 1 for various R at various N


52 -54548.2659 -0.438155964 -4.38181E-07 -4.38181E-13 -4.38181E-19

44 -24123.2218 -0.069459165 -6.94595E-08 -6.94595E-14 -6.94595E-20

36 -6039.92395 -0.007590712 -7.59071E-09 -7.59071E-15 -7.59071E-21

28 -468.831287 -0.000473201 -4.73201E-10 -4.73201E-16 -4.73201E-22

20 -11.4408109 -1.14417E-05 -1.14417E-11 -1.14417E-17 -1.14428E-23

12 -0.039159854 -3.91599E-08 -3.91599E-14 -3.91599E-20 -3.95946E-26

4 -1.31708E-07 -1.31708E-13 -1.31708E-19 -1.58378E-25 0



52 -33195.8853 -0.175268326 -1.75272E-07 -1.75272E-13 -1.75272E-19

44 -13527.4641 -0.027783756 -2.77838E-08 -2.77838E-14 -2.77838E-20

36 -2721.44594 -0.003036285 -3.03629E-09 -3.03629E-15 -3.03629E-21

28 -188.573572 -0.000189281 -1.89281E-10 -1.89281E-16 -1.89281E-22

20 -4.5765495 -4.5767E-06 -4.5767E-12 -4.5767E-18 -4.57565E-24

12 -0.015663942 -1.56639E-08 -1.56639E-14 -1.56639E-20 -1.4848E-26

4 -5.26832E-08 -5.26832E-14 -5.26832E-20 -7.91892E-26 0


27

6.2.2 Degenrate Energy Levels


52 63861.70898 63.8617089 0.063861709 6.38617E-05 6.38617E-08

44 32709.1124 32.7091124 0.032709112 3.27091E-05 3.27091E-08

36 14638.1733 14.6381733 0.014638173 1.46382E-05 1.46382E-08

28 -5344.1755 -5.3441755 -0.005344176 -5.34418E-06 -5.34418E-09

20 1384.07954 1.38407954 0.00138408 1.38408E-06 1.38408E-09

12 176.521555 0.176521555 0.000176522 1.76522E-07 1.76522E-10

4 1.81248401 0.001812484 1.81248E-06 1.81248E-09 1.81248E-12



52 58297.4973 58.2974973 0.058297497 5.82975E-05 5.82975E-08

44 29859.1979 29.8591979 0.029859198 2.98592E-05 2.98592E-08

36 13362.7628 13.3627628 0.013362763 1.33628E-05 1.33628E-08

28 -4878.54245 -4.87854245 -0.004878542 -4.87854E-06 -4.87854E-09

20 1263.48597 1.26348597 0.001263486 1.26349E-06 1.26349E-09

12 161.141396 0.161141396 0.000161141 1.61141E-07 1.61141E-10

4 1.65456397 0.001654564 1.65456E-06 1.65456E-09 1.65456E-12

Table 8: Energy shifts for M=1, state 1 for various R at various N


52 -36870.5747 -36.8705747 -0.036870575 -3.68706E-05 -3.68706E-08

44 -18884.6149 -18.8846149 -0.018884615 -1.88846E-05 -1.88846E-08

36 -8451.35327 -8.45135327 -0.008451353 -8.45135E-06 -8.45135E-09

28 3085.46116 3.08546116 0.003085461 3.08546E-06 3.08546E-09

20 -799.098693 -0.799098693 -0.000799099 -7.99099E-07 -7.99099E-10

12 -101.914767 -0.101914767 -0.000101915 -1.01915E-07 -1.01915E-10

4 -1.0464381 -0.001046438 -1.04644E-06 -1.04644E-09 -1.04644E-12

Table 9: Energy shifts for M=2, state 1 for various R at various N

28

6.3 Appendix C : Numerical calculation of V(r1,r2,R) for l=l'=1

We're intrested in the equation (6) , for the case l = l′ = 1 , i.e 1R3 case.

Equation (6) becomes:2

V (r1, r2, R) = (4π)32 × (−1)× (2!) [3× 3× 5]

12

×(

1 1 20 0 0

)−1(1 1 2m m′ M

)

×r1r2R3× Y1m (r1)Y1m′ (r2)Y2M (R) (43)

We know from equation (12)

Y2M (R) =

[(2l′ + l + 1)

12Pl(1)

]4π

=

[(2l′ + l + 1)

12

]4π

setting l=l'=1

Y20 =

[5

4π

] 12

(44)

Further more , From Edmonds,(1 1 2m −m 0

)=

(−1)1−m × 2×(3m2 − 2

)(5!)

12

(45)

And multiplying the constant with the above we have :(1 1 2m −m 0

)×

((1 1 20 0 0

)−1)=

(−1)m−1(3m2 − 2)

2(46)

where , m+m'+M=0 ⇒m=m' , for M=0Now for eqtn (46) , we consider the possible values of m possible [m = −1, 0, 1]For m=-1 , (

1 1 2−1 1 0

)=

1

2(47)

For m=1 , (1 1 21 −1 0

)= −1

2(48)

and m=0 , (1 1 20 0 0

)= 1 (49)

29

And since we're summing up the values in equation 6 ,we obtain:∑m=−1,0,1

(1 1 2m −m 0

)= 1 (50)

If we now plug this back into our orginal equation (6), we get it simpli�ed to:

V (r1, r2, R) =∑m

(4π)×−(2

3

)× r1r2

R3Y1m (r1)Y1−m (r2) (51)

We can now consider the general state of the two atoms :

〈n1l1m1n2l2m2|V (r1,r2, R)|n′1l′1m′1n′2l′2m′2.〉 =∑m

(4π)×−(2

3

)

×⟨n′1l′1m′1|rl1Ylm (r1) |n1l1m1

⟩ ⟨n′2l′2m′2|rl2Yl′m′ (r2) |n2l2m2

⟩× 1

R3(52)

Each of the matrix elements can be expanded further using the relation (8) and (11)Considering atom 1 , we have :

⟨n′1l′1m′1|rl1Ylm (r1) |n1l1m1

⟩=

ˆ ∞0

Pn′1l′1(r1)rl1P (r1)dr1

×[(2l′ + 1)(2l + 1)(2l1 + 1)

4π

] 12

×(l′1 l l10 0 0

)(−1)m

′1

(l′1 l l1−m′1 m m1

)(53)

One obtains a similar relation for atom 2.Taking a closer look at atom1 , ⇒ l′1 = 0, l1 = 1Using the program we obtain

ˆ ∞0

Pn′1,0 (r1) r1Pn11 (r1) dr1 = −3.80273586× 103 (54)

Similarly fo atom 2,⇒ l′2 = 2, l2 = 1

ˆ ∞0

Pn′2,0 (r2) r2Pn11 (r1) dr1 = −4.036134889× 103 (55)

for convinence de�nea1 = −3.80273586× 103 ×−4.036134889× 103

30

Now considering the angular part of the atom-atom state i.e the matrix relation in equation (53)noting [l = l′ = 1] , [l′1 = 0, l1 = 1] , [l′2 = 2, l2 = 1] , equation (52) becomes := ∑

m1m2

∑mm′

∑m′1m

′2

(−2)× a1 × 3√5

×(

0 1 10 0 0

)(−1)m

′1 ×

(0 1 1−m′1 m m1

)

×(

2 1 10 0 0

)(−1)m

′2 ×

(2 1 1−m′2 m′ m2

)× 1

R3(56)

where the relations are directly obtained from the program.(0 1 10 0 0

)= −0.57735⇒ A1 (57)

(2 1 10 0 0

)= 0.36515⇒ A2 (58)

Now one can evaluate the matrix relations in the summation sign ,Consider each coe�ecient seperately :

B1 = (−1)m′1 ×

(0 1 1−m′1 m m1

)where m′1 = 0, (m = −1, 0, 1) , (m1 = 1, 0,−1)also using the condition m+m1 + 0 = 0⇒ (m = −1,m1 = 1) or (m = 1,m1 = −1) or (m = 0,m1 = 1)This gives us for B1:

−(1)0(

0 1 10 −1 1

)+ (−1)0

(0 1 10 1 −1

)+ (−1)0

(0 1 10 0 0

)

= 2A3 +A1,where A3 =

(0 1 10 −1 1

)= 0.57735 (59)

Similary for the other relation:

B2 = (−1)m′2 ×

(2 1 1−m′2 m′ m2

)(60)

where [m′2 = −2,−1, 0, 1, 2] , [m′ = −1, 0, 1] , [m2 = −1, 0, 1]Expanding (60) and simplifying :

2×(

2 1 1−2 1 1

)− 2×

(2 1 11 −1 0

)− 2×

(2 1 11 0 −1

)(

2 1 10 0 0

)+ 2×

(2 1 10 −1 1

)(61)

Labelling the matrices as before :

31

A4 =

(2 1 1−2 1 1

)= −0.31623

A5 =

(2 1 11 −1 0

)= −0.31623 =

(2 1 11 0 −1

)

A7 =

(2 1 10 0 0

)= 0.36515

A8 =

(2 1 10 −1 1

)= 0.18257

Thefore our original equation (52) becomes :

〈n1l1m1n2l2m2|V (r1,r2, R)|n′1l′1m′1n′2l′2m′2.〉 =(a1 ×

(−6√5)A1A2 ×

1

R3

)

× (2A4 − 4A5 +A7 + 2A8) (62)

Now inorder to sub a suitable R , we take the number density to be 109cm−3

We get

R =1

N13

(63)

and converting into a.u109 × 5.29177× 10−9 = (5.29177)

3 × 10−18a.u (64)

Subbign this into equation (62) and simplifying we get

〈n1l1m1n2l2m2|V (r1,r2, R)|n′1l′1m′1n′2l′2m′2.〉 = 1.07328× 10−8a.u (65)

32

7 References

References

[1] Thomas F .Gallagher. Rydberg Atoms. Cambridge University Press, 2005.

[2] Daniel Comparat and Pierre Pillet. Dipole blockade in a cold rydberg atomic sample. J.Opt.Soc.Am.B,27(6), June 2010.

[3] S.D.Hogan and F Merkt. Demonstration of three dimensional electrostatic trapping of state -selected rydergatoms. Phys.Rev.Lett, (043001), 2008.

[4] Th.F.Gallagher and P.Pillet. Dipole dipole iinteractions of rydberg atoms. Adv.At.,Mol.,Opt.Phys., 56(161),2008.

[5] Amodsen Chotia Jiamming Zhao Daniel Comparat Thibault Vogt, Matthieu and Pierre. Electric-�eldinduced dipole blockade with rydberg atoms. Physical Review Letters, 99(7), 08 2007.

[6] C.J.Joachain B.H Bransden. Physics of Atoms and Molecules. Pearson, 2003.

[7] Robert Jastrow. On the rydberg-ritz formula in quantum mechanics. Phys.Rev., 73:60, January 1948.

[8] James William Rohlf. Modern Physics from a to z. Wiley, 1st edition edition, March 1994.

[9] M.J.Seaton. Quantum defect theory. Reports on Progress in Physics, 46(2):167, February 1983.

[10] C.J.Joachain B.H Bransden. Quantum Mechanics. Prentice Hall, 2000.

[11] P.G.Burke. Atoms in AstroPhysics. Plenum Press, 1983.

[12] A.R.Edmonds. Angular Momentum in Quantum Mechanics. Princeton University Press, January 1996.

[13] National Physical Laboratory Sta� of Mathematics Division. Modern Computing Methods. HMSO, 1961.

[14] J D Hey. On the determination of radial matrix elements for high-n transitions in hydrogenic atoms andions. J.Phys.B.At.Mol.Opt.Phys, 39:2641�2664, 2006.

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