5.73 Lecture 32: Configurations and Resultant L–S–J “Terms” (States)

5.73 Lecture #32 32 - 1 Configuration and Resultant L-S-J “Terms” (States) Last time: Matrix elements of Slater determinantal wavefunctions Normalization: (N!)–1/2 F(i): selection rule (∆s-o ≤ 1), sign depending on order G(i,j): selection rule (∆s-o ≤ 2), two additive terms with opposite signs TODAY: Configuration → which L-S terms? → L-S basis states → matrix elements ⎡Method of crossing out M L , M S boxes ⎢ ⎢ Ladders plus orthogonality ⎢⎣ Many worked out examples that will not be covered in lecture. KEY IDEAS: * 1/rij destroys spin-orbital labels as good quantum numbers. * Configuration splits into widely spaced L-S-J “terms.” * ∑ 1/rij is a scalar operator with respect to L, S, and J, thus matrix elements i>j are independent of ML, MS, and MJ. * Configuration generates all possible ML, MS components of each L-S term. * It can’t matter which ML, MS component we use to evaluate the 1/rij matrix elements * Method of microstates and boxes: Book-keeping for which L-S states are present, organizes the algebra to find eigenstates of L2 and S2, as basis for “sum rule” method (Lecture #33). Longer term goals: represent “electronic structure” in terms of properties of atomic orbitals 1. Configuration → L,S terms 2. Derive correct linear combination of Slater determinants for each L,S term: several methods 3. 1/rij matrix elements → Fk, Gk Slater-Condon parameters, Slater sum rule trick 4. HSO Spin-Orbit * �(NLS) — coupling constant for each L-S term in an electronic configuration * �(NLS) ↔ �nℓ a single spin-orbit orbital integral for the entire configuration * full HSO matrix in terms of �nℓ" 5. Stark effect, Zeeman effect, optical transitions 6. transition strengths nℓ r n′ℓ +1 ( matrix elements of r!, magnetic g-values ) There are a vastly smaller number of orbital parameters than the number of electronic states. The periodic table provides a basis for rationalization of orbital parameters (dependence on atomic number and on number of electrons.) Intuition vs. numerical description. 1 5.73 Lecture #32 32 - 2 Which L-S terms belong to (nf)2 * shorthand notation for spin-orbitals nℓm ℓα/β e.g. 4f3α , could suppress 4 and f ( main diagonal represents Slater determinant, …represents simple product of spin-orbitals) * standard order (to get signs internally consistent): for f spin-orbitals 3α3β2α2β … -3α -3β is my standard order for f (ℓ = 3) (2ℓ +1)(2s +1) = (7)(2) = 14 spin-orbitals * which Slater determinants are nonzero and distinct (i.e., not identical when spin-orbitals are permuted to a different ordering)? f2 - take any 2 s-o’s and list in standard order ||2α0α|| is OK, but ||0α2α|| is not in standard order, and ||2β2β|| = 0. How many nonzero and distinct Slater determinants are there for f2? 14 spin - orbitals ⎤ 14 ⋅13 = 91 ⎥ 2 identical electrons ⎦ 2 Slater determinants! ⎡⎣ 2 ( 2ℓ +1)⎤⎦! 1 general ( nℓ) : ∏ nℓ ⎡ ⎣ 2 ( 2ℓ +1) − p⎤⎦! p! p put p indistinguishable e– and 2(2ℓ+1) –p indistinguishable “holes” into 2(2ℓ+1) boxes subshell : one such factor for each subshell How to generate all 91 linear combinations of Slater determinants that correspond to the 91 possible ⏐LMLSMS〉 basis states that arise from f2? Next lecture. all of these are labor intensive you know how to do this * ladders plus orthogonality * construct and diagonalize L2 and S2 matrices * projection operators you should begin to know this * 3-j, 6-j, 9-j coefficients 2 5.73 Lecture #32 32 - 3 Sometimes all we want to know is “which L-S terms”? [WHY? 1/rij is scalar with respect to L,S, and J, thus eigenenergies are independent of ML, MS and MJ.] L z =L∑=ℓ iz and S z =S∑=s iz directly EASY read from thethe spin -orbital EASY because because can we can read from spin-orbital ∑ ℓ 1z and ∑ s1zdirectly z z i labels. labels. i i i 2 L z 2α1β = ∑ ℓ iz 2α1β = " ⎡⎣2+1⎤⎦ 2α1β i=1 ML = 3 ML is sum of mℓ’s MS is sum of mS’s NONLECTURE What about L2 ? Can do this in either of two ways: * as below (very cumbersome) L2 = L2z + (1/ 2)(L + L − + L − L + ) [separately apply each * 1e – operator rather than treat entire operator as a 2e – operator.] very laborious because L2 = ∑ ℓ i ⋅ ℓ j = ∑ ℓ 2i + 2∑ ℓ i ℓ j i, j i one e– i> j two–e– L2 2α1β ≠ ∑ ! 2 ℓ i ( ℓ i + 1) 2α1β ℓ i = 3 for f i WORK OUT L2 matrix for M L = 3, M S = 0 block of f 2 for future reference ( ) ⎡ ⎤ 1 L2 = ∑ ℓ i ⋅ ℓ j = ∑ ⎡⎣ ℓ 2i ⎤⎦ + 2∑ ⎢ ℓ iz ℓ jz + ℓ i+ ℓ j− + ℓ i− ℓ j+ ⎥ 2 i, j i i> j ⎣ ⎦ ∆ℓ = 0, ∆Mℓ = 0 all are ∆ M S = ∆ ms = ∆ ms = 0 1 2 3 ∆ℓ = 0, ∆Mℓ = 0 ∆mℓ1 = –∆mℓ2 = ± 1 5.73 Lecture #32 32 - 4 mℓ1mℓ2 ℓ(ℓ + 1) ⎡(12 + 12 ) 2α1β + 2 ( 2 ⋅1) 2α1β + ⎢ 1/2 1/2 2 2⎢ L 2α1β = ! ⎡⎣3⋅ 4 − 2 ⋅1⎤⎦ ⎡⎣3⋅ 4 − 1⋅ 2⎤⎦ 1α2β ⎢ ⎢ ⎡3 ⋅ 4 − 2 ⋅3⎤1/2 ⎡3⋅ 4 − 1⋅0 ⎤1/2 3α0β ⎦ ⎣ ⎦ ⎣⎣ Many steps skipped … ℓ ℓ 1− 2+ 2 −1/2 ℓ1+ ℓ 2− = ! ⎡⎣ 28 2α1β − 10 2β1α + 6 ⋅ 2 ⎤ ⎥ ⎥ + ⎥ ⎥ ⎦ 3α0β ⎤⎦ nonstandard order All of the 12, 21 exchange-type matrix elements are 0 because of ms mismatch. e.g. 2α(1)1β(1) space only 1β(1)2α(2) = 0 operator Recall ± (〈12|G|12〉 – 〈12|G|21〉) for 2e– operator. We get: L2 2β1α = ! 2 ⎡⎣ 28 2β1α − 10 2α1β + (12 ⋅ 2 −1/2 ) 3β0α ⎤⎦ L2 3α0β = ! 2 ⎡⎣(24 + 3⋅ 0) 3α0β + (12 ⋅ 2 −1/2 ) 2α1β ⎤⎦ L2 3β0α = ! 2 ⎡⎣ 24 3β0α + (12 ⋅ 2 −1/2 ) 2β1α ⎤⎦ L2 = !2 3α0β ⎛ 24 6⋅21/2 0 0 ⎜ 2α1β ⎜ 6⋅21/2 28 −10 0 − 1α2β ⎜ 6⋅21/2 0 −10 28 − 0α3β ⎜⎝ 6⋅21/2 0 0 24 ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ [the bottom two Slater determinants are intentionally out of standard order to display effects of decreasing values of mℓ(1) and increasing values of mℓ(2).] 4 5.73 Lecture #32 32 - 5 LSM L M S = 5130 Example: found eigenvalues and eigenvectors of this block ML = 3, MS = 0 of f2 L2 !2 L2 !2 L2 !2 L2 !2 ⎡⎣3 −1/2 3α 0 β + 3 −1/2 3β 0α + 6 −1/2 2α 1β + 6 −1/2 L=5 2 β1α ⎤⎦ = 30 [ ] L=3 ⎡⎣ 6 −1/2 3α 0 β + 6 −1/2 3β 0α − 3−1/2 2α 1β − 3−1/2 2 β1α ⎤⎦ = 12 [ ] L=6 ⎡⎣11 −1/2 ⎡⎣3⋅ 22 3α 0 β −11 −1/2 −1/2 3β 0α + 3⋅ 22 3α 0 β − 3⋅ 22 −1/2 2α 1β − 3⋅ 22 −1/2 3β 0α −11 −1/2 −1/2 2α 1β +11 −1/2 2 β1α ⎤⎦ = 42 [ ] L=4 2 β1α ⎤⎦ = 20 [ ] (Note how easy it is to see that normalization is correct.) Look at the sum of the squares of each coefficient! a lot of algebra is not presented here! (especially the derivation of the 4 eigenvectors) * each Slater basis state gets “used up” [sum of squares of that basis set is 1] * the first 2 eigenfunctions are in the form: αβ + βα → S = 1 * the second 2 eigenfunctions are in the form: αβ – βα → S = 0 You could prove these S = 1 and S = 0 results by applying S2 to above eigenfunctions of L2. We have obtained LSM L M S = 5130 , 3130 , 6030 , and 4030 eigenstates. END OF NON-LECTURE 5 5.73 Lecture #32 32 - 6 Non-lecture pages were intended to show that applying L2 and S2 to Slater determinants is laborious — much more so than applying Lz and Sz. This is one reason why we use the “crossing out ML, MS microstates” method to figure out which L,S states must be considered. Often this is sufficient — and it can be the basis for some shortcut tricks! The ML, MS method works because: * each configuration generates the full (2L + 1) (2S + 1) manifold of ML, MS states associated with each L,S term. Why? If you have one |MLMS〉 member of |LMLSMS〉 you can generate all of the others for that L,S using L± and S± operators. * This must be true because, starting with ML = L, MS = S, L– and S– can be used to generate all ML, MS components of the full L,S term without the need to go outside the specific configuration. ML, MS method ML= LMAX MS = SMAX LMAX – 1 LMAX – 2 … 0 list all Slater determinants S–1 0 SMAX = (# of e–)/2. No need to include negative values of MS or ML. Why? They are accessed by L– + S– and contain no new information. 6 5.73 Lecture #32 f ML MS 1 6(I) 3α 3α 5(H) 3α2α exclusively S = 1 states 0 Both S = 1 and S = 0 states 3α 3β 3α2β 3β2α 4(G) 3α1α 2α2α 3α1β 3β1α 2α2β 32 - 7 2 3(F) 3α0α 2(D) 2α0α 1(P) 3α − 2α 0(S) 3α − 3α 2α1α 3α − 1α 1α1α 2α − 1α 1α0α 2α − 2α 1α − 1α 0α0α 3α0β 3β0α 2α1β 2β1α 2α0β 2β0α 3α − 1β 3β − 1α 1α1β 3α − 2β 3β − 2α 2α − 1β 2β − 1α 1α0β 1β0α 3α − 3β 3β − 3α 2α − 2β 2β − 2α 1α − 1β 1β − 1α 0α0β Need not include MS < 0 or ML < 0 because these are identical to the ML > 0 and MS > 0 quadrant. Notice that as you go down by 1 in ML, the number of Slater determinants in each ML, MS box increases only by 1. This is a prerequisite for using the L– plus orthogonality method! This useful simplicity does not occur as you go down a column in MS. This convenient situation does not occur for d3 or f3. Why? Because there can be more than one L-S term of a specified symmetry. For example, for d2 there are 1S, 3P, 1D, 3F, 1G terms, but for d3 there are 2P, 4P, two 2D, 2F, 4F, 2G and 2H terms. 7 Slaters for f2 5.73 Lecture #32 32 - 8 No J by convention S P D F G H I K L = 0, 1, 2, 3, 4, 5, 6, 7 Start in extreme ML, MS corner — This generally contains only one Slater determinant L = MLMAX , S = MS MAX so we have one of the L - S terms –L ≤ ML ≤ L –S ≤ M S ≤ S This L-S term includes one of each ML, MS in the range This means this L-S term will “use up” the equivalent of one Slater determinant in each ML,MS box. Bookkeeping: cross out one Slater determinant, any one, from each relevant ML,MS box. Now repeat, again starting at the remaining extreme ML,MS corner etc. * 1I * 3H * 1G * 3F * 1D * 3P * 1S 1 × 13 3 × 11 1×9 3×7 1×5 3×3 1×1 = 13 = 33 =9 = 21 =5 =9 =1 91 as required! [It is a good idea to use this total degeneracy of the configuration as a check.] Since there is only one Slater determinant in the ML = 5, MS = 1 box, generate all triplets by repeated application of L– to ||3α2α|| (plus orthogonality) and generate all singlets by L– on ||3α3β||. Many orthogonalization steps are needed! Especially for singlets. Need to use S– also. 8 5.73 Lecture #32 32 - 9 Before illustrating the ladders plus orthogonality method, it is useful to show some patterns and list some valuable tricks. The most difficult cases are (nℓ)m where m = 2, 3,… 2ℓ. Easy to combine nℓ with n′ℓ′ because no need for special bookkeeping. ℓ s p d f (nℓ)2 (nℓ)3 1 S — 1 3 1 4 2 D, P, S S, D, 2P 1 2 G, 3F, 1D, 3P, 1S H, 2G, 2F, 4F, 2D(2), 4P, 2P 1 3 I, H, 1G, 3F, 1D, 3P, 1S A simple, memorable pattern Rather complicated Get the same L-S states for 2 and 3 “holes” (e.g. p4⟷p2, d3⟷d7) instead of electrons. ( ) 2 Also nℓ n′ℓ′ → ⎡⎣ nℓ 2 2S +1 L ⎤⎦ ⊗ ( ℓ′) = ( 2 2S +2, and 2S ) ( L + ℓ′, L + ℓ′ −1,! L − ℓ′ ) Simple vector coupling of the n'ℓ' electron to the two-electron nℓ2 2S+1L term. No Pauli exclusion because n'ℓ' is distinguished from nℓ. When the e– are in distinct subshells (different values of ℓ and n), there is no need to be as careful about the exclusion principle. 9 5.73 Lecture #32 32 - 10 Laddersplus plusOrthogonality OrthogonalityMethod Method Ladder 2 f example Start with 2 extreme UNIQUE states 1. 3 H M L = 5, M s = 1 = 3α2α Use this to generate all triplet states by applying L– repeatedly and using orthogonality when necessary. Note that # of determinants in each ML,MS=1 box increases no faster than in steps of 1. 3 To get to P, must not only apply orthogonality several times, but must follow each L state down to the ML = 1 box! 1 2. To I Munique = 6, M|s1I= M 0 L = 6, Ms = 0⟩ state. Toget getsinglets, singlets,start startwith with the L Again, as L– takes us to successively lower-ML boxes, # of determinants increases in steps of 1. But some of these steps are due to triplets with MS = 0. Need to step triplets down into MS = 0 territory using S– once. Lots more orthogonality steps, lots more trails being followed. AWFUL, but do-able. Nonlecture 3 H ML MS = L − 3 H 51 = ∑ ℓ i− 3α2α 0 i 1/2 3 " ⎡⎣5⋅6 − 5⋅ 4⎤⎦ 3 1/2 H 41 = " ⎣⎡3⋅ 4 − 3⋅ 2⎤⎦ H 41 = 3α1α 2α2α + " ( 3⋅ 4 − 2 ⋅1) 1/2 big surprise! L − 3 H 41 = Σℓ i− 3α1α 3 orthogonality: H 31 = (1/ 3) 1/2 3 ⎛ 2⎞ F 31 = ⎜ ⎟ ⎝ 3⎠ ML 1/2 2α1α + ( 2 / 3) 1/2 ⎛ 1⎞ 2α1α − ⎜ ⎟ ⎝ 3⎠ 3α0α 1/2 3α0α MS and so on, to get all | 3L L 1〉 many-electron functions 10 3α1α 5.73 Lecture #32 32 - 11 MS = 0 Try a detour into singlet territory, and then check for self-consistency. 1/2 ⎤ (by orthogonality ⎡⎛ 2⎞ 1/2 ⎛ 1⎞ 3 S − F 31 = ∑ s i− ⎢⎜ ⎟ 2α1α − ⎜ ⎟ 3α0α ⎥ 3 ⎝ 3⎠ ⎥⎦ with ) H 31 ⎢⎣⎝ 3⎠ i ⎡⎛ 2⎞ 1/2 ⎡ 1 3 1 ⎛ 1⎞ ⎤1/2 1/2 3 ! ⎡⎣1⋅ 2 − 1⋅0⎤⎦ F 30 = ! ⎢⎜ ⎟ ⎢ ⋅ − ⎜ − ⎟ ⎥ 2β1α + 2α1β ⎢⎝ 3⎠ ⎣ 2 2 2 ⎝ 2⎠ ⎦ ⎣ 1/2 ⎤ this factor is 1/2 ⎛ 1⎞ 3β0α + 3α0β ⎥ − ⎜ ⎟ ⎡⎣1⎤⎦ always 1 or 0 ⎝ 3⎠ ⎥⎦ ( ( 3 ⎛ 1⎞ F 30 = ⎜ ⎟ ⎝ 3⎠ S− 3 3 1/2 ( ⎛ 1⎞ 2β1α + 2α1β − ⎜ ⎟ ⎝ 6⎠ ) 1/2 ) ) ( 3β0α + 3α0β ) 1/2 ⎡⎛ ⎞ 1/2 ⎤ ⎛ 2⎞ 1 ⎢ H 31 = ∑ si− ⎜ ⎟ 2α1α + ⎜ ⎟ 3α0α ⎥ ⎝ 3⎠ ⎢⎣⎝ 3 ⎠ ⎥⎦ i ⎛ 1⎞ H30 = ⎜ ⎟ ⎝ 6⎠ 1/2 ( ⎛ 1⎞ 2β1α + 2α1β + ⎜ ⎟ ⎝ 3⎠ ) 1/2 ( 3β0α + 3α0β ) There are 4 Slater determinants in the ML = 3, MS = 0 box. We can’t find the other two singlet linear combinations uniquely without using L– on the extreme singlets. L− 1 I 60 " ⎡⎣6⋅7 − 6⋅5⎤⎦ = Σℓ i − 3α3β 1/2 1 I 50 = " ⎡⎣3⋅4 − 3⋅2⎤⎦ 1/2 ( 2α3β + 3α2β ) wrong order: must change sign to reverse order 1 I 50 ⎛ 1⎞ =⎜ ⎟ ⎝ 2⎠ 1/2 L− ⎡⎣ 3α2β − 3β2α ⎤⎦ orthogonality 1 I 50 ⎛ 1⎞ = Σℓ i − ⎜ ⎟ ⎝ 2⎠ 11 3 H 50 ⎛ 1⎞ =⎜ ⎟ ⎝ 2⎠ 1/2 ⎡⎣ 3α2β + 3β2α ⎤⎦ 1/2 ⎡⎣ 3α2β − 3β2α ⎤⎦ 5.73 Lecture #32 ⎛ 1⎞ I 40 = ⎜ ⎟ ⎝ 44 ⎠ 1 1 1/2 ⎛ 5⎞ I 40 = ⎜ ⎟ ⎝ 22 ⎠ 1/2 3 ⎛ 1⎞ H 40 = ⎜ ⎟ ⎝ 20 ⎠ 3 ⎛ 1⎞ H 40 = ⎜ ⎟ ⎝ 2⎠ ( ) ( ) ⎡(10)1/2 3α1β − 3β1α + 61/2 2α2β − 2β2α ⎤ ⎦ ⎣ ⎡ ⎤ ⎛ 6⎞ ⎢ 3α1β − 3β1α + ⎜⎝ ⎟⎠ 2α2β ⎥ 11 ⎣ ⎦ 1/2 1/2 32 - 12 wrong order ( ) ( ) ( ) ⎡(6)1/2 2α2β + 2β2α +101/2 3α1β + 3β1α ⎤ ⎣ ⎦ wrong order ( 3α1β + 3β1α ) orthogonality 1/2 ⎡ ⎤ ⎛ 5⎞ ⎢ 3α1β − 3β1α − ⎜ ⎟ 2α2β ⎥ ⎝ 11 ⎠ ⎢⎣ ⎥⎦ At last we are ready to enter the ML = 3, MS = 0 block! It is clear that if we apply L– to |3H 40⟩, we will get the same form that we already derived starting from |3H 51⟩ Instead, let's lower |1I 40⟩. 1 ⎛ 3⎞ G 40 = ⎜ ⎟ ⎝ 11 ⎠ 1/2 ( ) 1/2 ⎧⎛ 5 ⎞ 1/2 ⎫ ⎛ 6⎞ LL– I 40 = ∑ ℓ i− ⎨⎜ ⎟ ⎡⎣ 3α1β − 1β3α ⎤⎦ + ⎜ ⎟ 2α2β ⎬ ⎝ 11⎠ ⎪⎩⎝ 22 ⎠ ⎪⎭ i 1/2 1/2 ⎧ ⎛ 5⎞ 1 1/2 ⎛ 5 ⎞ 1/2 I 30 = (30) ⎨⎜ ⎟ (6) 2α1β − 2β1α + ⎜ ⎟ (12)1/2 3α0β − 3β0α ⎝ 22 ⎠ ⎩⎪⎝ 22 ⎠ 1 ( ⎛ 6⎞ +⎜ ⎟ ⎝ 11⎠ 1 1/2 ) ( ⎫ (10)1/2 2α1β − 2β1α ⎬ ⎭⎪ ( ) 1/2 ⎡⎛ 1 ⎞ 1/2 ⎛ 4 ⎞ 1/2 ⎤ ⎛ 2⎞ I 30 = ⎢⎜ ⎟ + ⎜ ⎟ ⎥ 2α1β − 2β1α + ⎜ ⎟ 3α0β − 3β0α ⎝ 22 ⎠ ⎦ ⎝ 22 ⎠ ⎣⎝ 22 ⎠ 1 ( ⎛ 9⎞ I 30 = ⎜ ⎟ ⎝ 22 ⎠ ) 1/2 ( ) ) ⎛ 2⎞ 2α1β − 2β1α + ⎜ ⎟ ⎝ 22 ⎠ 1/2 ( ) ( 3α0β − 3β0α ) Finally, by orthogonality: ⎯⎯⎯⎯ ⎯→ IMPORTANT ⎛ 1⎞ 1 G 30 = − ⎜ ⎟ ⎝ 11 ⎠ 1/2 ( ) ⎛ 9⎞ 2α1β − 2β1α + ⎜ ⎟ ⎝ 22 ⎠ Does this match what one would get from L– |1G 40〉? 12 1/2 ( 3α0β − 3β0α ) 5.73 Lecture #32 32 - 13 1/2 ⎧⎛ 3 ⎞ 1/2 ⎫ ⎛ 5⎞ L − G 40 = ∑ ℓ i− ⎨⎜ ⎟ ⎡⎣ 3α1β − 1β3α ⎤⎦ − ⎜ ⎟ 2α2β ⎬ ⎝ 11⎠ ⎪⎭ ⎩⎪⎝ 11⎠ i 1/2 1/2 ⎧ ⎛ 3⎞ 1 1/2 ⎛ 5 ⎞ 1/2 ⎨ G 30 = (8) ⎜ ⎟ (6) 2α1β − 2β1α + ⎜ ⎟ (12)1/2 3α0β − 3β0α ⎝ 11⎠ ⎝ ⎠ ⎪⎩ 11 1/2 ⎫ ⎛ 5⎞ − ⎜ ⎟ (10)1/2 2α1β − 2β1α ⎬ ⎝ 11⎠ ⎭⎪ 1 ( ) ( ⎯⎯⎯⎯ ⎯→ IMPORTANT 1 ⎛ 1⎞ G 30 = − ⎜ ⎟ ⎝ 11 ⎠ 1/2 ( ( ) ) ⎛ 9⎞ 2α1β − 2β1α + ⎜ ⎟ ⎝ 22 ⎠ ) 1/2 ( 3α0β − 3β0α ) checks! End of Non-Lecture As you see, this ladders plus orthogonality method is extremely laborious. There is a much better way! ⎡There are several patterns: singlets for M = 0 always have the S ** ⎢ αβ − βα and M S = 0 triplets always αβ + βα . ⎣⎢form This can be generalized for any value of S (page 151 of Hélène LefebvreBrion-Robert Field Perturbations 2004 book) [Also M. Yamazaki, Sci. Rep. Kanezawa Univ. 8, 371 (1963).] ( ) ( ) 2. Failure and Inconvenience of ladder method The ladder method is OK when you have a single target LML SMS state, especially when it is near an edge of the ML,MS box diagram. Essential that # of Slater determinants in each MLMS box increases in steps of 1 as you step down in ML or MS. Fails when there are two L-S terms of same L and S in a given configuration. Then we must set up a 2 × 2 secular equation anyway. e.g. (nd)3 2 H,2 G,2 F,4 F,2 D(2),4 P,2 P 3. L2 and S2 Matrix Method Another method is based on constructing L2 and S2 matrices in the Slater determinantal basis set. This is no cakewalk either (but this is easier)! Since usually SMAX ≪ LMAX for a configuration when using L2 + S2 matrices method, it is best to start with the S2 matrix because it is simpler. 13