Energy levels in the jj coupling scheme - Journal of Chemical

These authors show how to determine "Energy levels in the jj coupling scheme" so they can be compared with the energy levels from the LS scheme to fin...
0 downloads 0 Views 3MB Size
Energy Levels in the jjcoupling Scheme J. Rubio Dpto. de Quimica Fisica, Facultad de Quimica, Tarragona, Spain J. J. Perez Dpto. de Quimica, ETS d'Enginyers Industrials, Av. Diagonal, 647, 08028 Barcelona, Spain The theory of atomic structure is dominated by. angular .. momentum, and the coupling between angular momenta ia a subject dealt with in most undergraduate quantum mechanics courses. The Hamiltonian operator for an N-electron atom can he written in dimensionless form as

[. . . slpl] Configuration Consider the individual quantum numbers for each electron: for the s electron 11 = 0 and sl = 112; for the p electron lz = 1and sz = 112. In general, the possible values for j; are 1; si, 1; + q - 1,.. . ,ili - sd (9);in thiscase, for the first electron jl = 112; and for the second electron j z = 312 and j z = 112. Now by combining jl and jz wecan obtain the possihle values for J (i.e., jl jz,jl+ jz - 1,...,1j1- jd). In this case, when we comhine jl = 112 and jz = 312, we obtain the values 2 and 1 for J; and if we combine jl = 112 and jz = 112 we obtain the values 1and 0 for J. This is summarized in Table 1. After these considerations, the energy levels can be written

+

+

where the first sum is the electron kinetic energy term, the second the electron nuclear attraction, the third the electronic repulsion, and the last the spin-orbit coupling. For light atoms (up to Z = 30), the latter term is small in comparison to the electrostatic repulsion and can he treated as a perturbation. In such cases the total angular momentum, L, and the total soin anmlar momentum. S. are said to be "good quantum- numb&" for the classification of energy levels and one speaks of the Russell-Saunders or L S couoline scheme. When the spin-orbit term becomes large, (as the case of heavy atoms), L and S are no longer good quantum numhers, hut the total angular momentum J, defined as the sum of individual total angular momenta, has to be worked with. This case is termed the jj coupling scheme. Which of these two schemes gives a better match with experimental results for a given atom usually has to be tested. Most of the textbooks deal with theLScouplingscheme in some detail. Of these, the more sophisticated tackle the problems raised by orbital configurations such as [. . . pz] where some of the possihle comhinations of one-electron angular momentum quantum numbers violate the Pauli Principle. Indeed, a great deal of literature has been concerned with the development of algorithms for evaluating the different terms and energy levels in theLSscheme (1-4). Very little attention, however, is usually paid t o the jj scheme; thus, of well-known texts, Johnson and Pedersen (5) deal only with LS while JBrgensen and Oddershede (6) do make a brief mention of jj hut without offering real explanations. Pilar (7) appears to be the exception, hut he only addresses the [. . . s'p'] case, where none of the possible comhinations of quantum numbers violate the Pauli Principle. The only publication known to the present authors concerned with the evaluation of i i levels without the use of group theory is that of Tuttle (8j. We propose here a simple alternative treatment based on the usual method emoloved in the L S case.

nj,

The Method For the sake of simplicity we first give the trivial example [. . . slpl] where the electrons are nonequivalent. We then develop the mechanism for writing the levels for equivalent electrons by considering explicitly the [. . . pz] and [. . . pa] orhital configurations. 476

Journal of Chemical Education

[. . . p2] Configuration First of all, i t is necessary to know the number of states of the configuration. For an N-electron configuration, this is the binomial coefficient (3)

$2, i.e., the total number of different ways in which N identical particles can he distributed in 2(21+ 1) possible states, 1 being the orhital angular momentum quantum number. In the case of a [. . .pz] configuration this value is 15. As the electrons are equivalent, both have the same quantum numhers: 11 = lz = 1 and SI = 82 = 112. The possible values for jl and jz are: j~= j p = 312,112. Now in order to avoid Pauli-Principle-violating levels we proceed as follows: 1) Write down a matrix whose rows are lahelled with all possible combinations Gl)Gz) and whose columns are Labelled with all pmsiblevalues ofMj= Zlrn,l, Elrnjl - 1,... ,-EJm,l. In this case the maximum value for M~willbe 3; GI + jz = 312 + 312). This is shown in Table 2.

Table 1. Energy Levels for an [. . . s'p'] Conflguratlon

Table 2.

Matrlx for a [. . .dl Conflguratlon MJ

3

2

1

0

-1

-2

-3

Table 3.

,"

Posslble JVahes for the j, = 112: h = 312 Case

Matrix lor a [. . .$1 conflguratlon

m.

"7.

"L

Table 4.

4) Finally the (312P case can he thought of as the addition of a (312) electron to the (3/212pair. This gives rise to only four states because many of the possihle combinations appear twice.

2) The table will he filled by placing a mark for each of the 15 states of the configuration. This is done by evaluating the possihle states for each row. In order to do this we hore in mind the following picture: if j = 312 (i.e., mj = 312,112, -112, -312) the electron can be placed either in the ml = 1or ml = -1 orhital, hut, if j = 112 (i.e., mj = 112, -1/2), the electron can he placed only in the ml = 0 orbital. a) When jl = jz = 112 the electrons are placed in m, = 0, so they must have opposite spins and MJ = 0.For this case we have only one state, which we mark on the row (112)2,column MJ = 0 in Table 2. h) When j, = 112 and jz= 312 one of the electrons is placed in the ml = 0 orhital and the other either in the ml = 1or ml = -1 orbitals. This means they can he arranged in all possihle combinations as there are no restrictions imposed by the Pauli Principle. We have outlined the ~rocedurefor countine - them all in Tahle 3. e ) Finally, when,, = j~ = 312, the electrons are placed either in the rnl = 1 or ma = -1 orbital$. If they are in the same orbital. the onlv ~oosiblevalues for M,are 2. -2. Ifthevare indifferent orbitals, the-possiblevalues for ~ i a r 1;0, e -1,O. With the latter states we can complete Tahle 2. Now we proceed to calculate the possible values for J.In case (a),since MJ = 0, the only possible value for J i s 0. In case ( ~ ) , M=J 2,1,0, -1, -2 (J= 2) and MJ = 1,0, -1 ( J = 1) are possihle. Finally, in case (4 J = 2.0. Now we can write the levels

The fallowing comments may he useful: When we have a combination of the type &)(iz) with jl # j z all possihle values for Jare allowed. In the example we are considerine. rise to the J values 2 ". the combination (112)(3/2) . . . . eives a n d l o , + j z , ...,I ] ~ - i d ) . In general, for any arrangement of the type ti)2 the allowed values for J are (8):2j - 1,2j - 3,2j - 5,. . . ,0, where each of the values occurs once. So, in this example, for (1/2)2the only possibility is J = 0, and in the (312P case J = 2, 0 as we have previously derived. .

. .

We now write the levels for the case of a [. . .p3] configuration. Although it can be done without any previous information. for the sake of sim~licitv . we are eoine to make use of the previous results. Let us construct the matrix of the Dossible combinations of the type 0.1)0.~)0.~) and the possibc values for MJ (= 912 as maximum in this case). This is shown in Table 4. In this case the number of states is 20. Now we proceed to analyze the possible states for each row.

-

- -

1) The comhination (1/2)3 can be thought of as the addition of a (112) electron to the (1/2)2pair. As the only possibility for (1/2)2was MJ = 0, there is no roam for the third electron and no Levels will he possible. 2) The (1/2)2(3/2)case can he thought of as the addition of a (312) electron to the (1/2)2 pair. As the former is placed in either the orhital mn = 1 or mn = -1 and the oair (1/2)2is laced in the ml = 0 orbital, ail the combinations are iossihle. As tde (312) electron has four possibilities, in this case four states will arise. 3) The (1/2)(3/2)zcase can be thought of as the addition of a (112) electron to the (3/2)2pair. In this case there are two possibilities for each of the (312)zpairs, so this will give rise to 12 states.

Now, with aid of Table 4, i t is easy t o write the energy levels:

The procedure developed above is not only applicable t o

[. . .p"] configurations hut also to any desired configuration.

Let us see how it works for the [. . . d2] case. The total number of states for this configuration is 45. Since 11 = 12 = 2 and sl = s2= 112, the possible values for j will he 512 and 312. We proceed t o write the matrix with the possible comhinations of pairs ti1)(&) (i.e., (312)2; (3/2)(512) and (5/2)2), and the possible values for MJ = 5,4, . . ,0, . . , -5. Now, in order to mark the states, we should bear in mind that, when j = 312 (mj = 312,112, -112, -3/2), the electron can be considered to be placed in the ml = 1or m l = -1 orbitals, and when j = 512 (m, = 512, 312, 112, -112, -312, -512) the electron can be considered t o be in either ml = 2,0, -2 orbitals. By writing out the tahle, i t is easy t o see that the levels are [(312)q2; [(312)10; [(312)(512)1,; [(3l2)(5l2)l3; [(312)(512)1,; [(3/2)(5I2)l1; [(5/2)1,; [(5/2)zl,; [(5/2)q0 In order to assign the relative energies once we have the levels. it is useful to correlate them with those arisine from the L;S coupling scheme where Hund rules are applkahle. This can be done bv means of J. since i t is a nood quantum number for level classification in both schem&. For a I.. .u21confieuration Hund rules ~ r e d i cthe3P t to be the low& term and& order of increasing energy the levels are 3P0,3P,, and 3P2. We expect these levels to he close in energy in comparison with IDz and 'So. Of the levels arising from the j j coupling the lowest must be [(1/2)2]0, since we would expect [(3/2)2]0and [(312)2]2 to he close t o each other. The next level will be that of J = 1 [(1/2)(3/2)]1, which correlates t o 3 P ~and the next that of J = 2 1(112)(3/2)1~. . .. . ,- close in energy to the latter, which correlates with theV2. Asaumine the ID., level to be the next. it will correlate with 1(3/2r11~. a& finall; the 'So will correlate with the [(3/2)2]0level. is displayed in Figure 1. The correlation diagram for the [. . . p3] configuration can he drawn u p in a similar way. The levels for the LS scheme are: 2P112, 2P312. 2D312,2D512, and 4S3n The lowest, 4S31z, will be correlated with [(1/2)2(312)]312.For the intermediate levels 2D512with [(112)2(3/2)]512;2D31zwith [(112)(3/2)2]3/2and zPllzwith [(112)(3/2)2]1/2.Finally the2Pll~levelwill correlate with [(3/2)3]112. This is displayed in Figure 2.

.

Match wlth the Experimental Results and Flnal Remarks Let us see for the configurations mentioned above how the ex~erimentalvalues match best with the different s ~ e c t r u m paiteru schemes. For the [. p 2 ] configuration we can comparethe values (in cm-I) for carbon 0, 16, 43, 10193, 21648, with those of lead. 0. 7819. 10650. 21458. 29467 (10). where the nround state total eiergy is set equal to zero. 1t can he cleariy seen with the aid of Figure 1that carbon values match fairly well

..

Volume 63

Number 6

June 1986

477

an? J=r J=O

Is ,..&.-----.

L S COUPLING

-

,, I

jj

COUPLING

.

Figure I.Level conelation diagram for a [. . p2] conflguratlon. LS coupling scheme levels (left) and jjcoupllng scheme levels (rlght).

with the LS scheme pattern and lead values matchbetter with the j j scheme pattern. In the case of the [. . . p3] configmation we can Compare the experimental energy values (in cm-') of nitrogen, 0, 19224.19233.28839.28840. with those of bismuth. 0.11419. e ' good 15438; 21661, 33165 (10).w i t h the aid of ~ i ~ u r2'a match for nitrogen with the L S scheme Dattern and a better match for bismith with the j j scheme p k t e r n can be seen. Finally we should comment that the number of therms in the LS scheme is usually greater than that in the j j scheme. Thus for the [. . . dZ] case we have five terms in the L S scheme, 3F, 3P, ID, lG, IS, and just three in the j j scheme, (3/2)2, (3/2)(5/2), (5/2)2. Exceptions t o this are the I... P"] configurations, where the number of terms is the same. ~ e v :

478

Journal o f C h e m i c a l Education

LS

COUPLING

,i,i COUPLING ~-

Flgure 2. Level correlation diagram far a [. . . ps] configuration. LScoupllng scheme levels (left) and jjcoupllng scheme levels (rlght).

Acknowledgment

We would like to thank Alan Hinchliffe (U.M.I.S.T.) for h e l ~ f usueeestions l "" in readine the manuscriot.

...

.

-

Lnerature cnea

(71 P ~ . FL ~ E ~ ~ ~ . ~ ~ ~ ~ Q ~ ~ ~ ~ C ~ ~ ~ ~ ~ ~ ~ ~ : M ~ G ~