5.73 Lecture 25: HSO + HZeeman: Coupled vs. Uncoupled Basis

5.73 Lecture #25 25 - 1 HSO + HZeeman Coupled vs. Uncoupled Basis Sets Last time: matrices for J2, J+, J–, Jz, Jx, Jy in |jmj⟩ basis for J = 0, ½, 1 Pauli spin ½ matrices ! ! arbitrary 2 × 2 matrix M = a0 I + a1 ⋅ σ decomposed as scalar plus vector. When M is ! → visualization via fictitious vector in fictitious B-field. When M is a term in H ® idea that arbitrary operator can be decomposed as a sum of the terms that behave like components of J = 0, J = 1, J = 2… This leads to spherical tensor algebra. types of operators aJ ⎤ ! q ⎥⎥ J1 ⋅ J 2 ⎥⎦ e.g. magnetic moment (a is a known constant or a function of r) how to evaluate matrix elements (e.g. Stark Effect) e.g. Spin-Orbit Special simplification of Trace (AH) For example ⎛ 0 1 0 ⎜ A=⎜ 0 0 0 ⎝ 0 0 0 ⎛ H 21 ⎜ AH = ⎜ 0 ⎜⎝ 0 ⎛ H ⎞ 11 ⎜ ⎟ ⎟ , H = ⎜ H 21 ⎜ ⎠ ⎝ H 31 H 22 0 0 Trace ( AH ) = H 21 H12 H 22 H 32 H 23 ⎞ ⎟ 0 ⎟ 0 ⎟⎠ H13 ⎞ ⎟ H 23 ⎟ ⎟ H 33 ⎠ simpler extreme simplification! A12 picks out only H21, A21 picks out only H12. Extreme labor saving trick! 1 5.73 Lecture #25 25 - 2 TODAY: 1. HSO + HZeeman as illustrative 2. Dimension of two basis sets, ⎜JLSMJ〉 and ⎜LMLSMS〉, is the same 3. matrix elements of HSO in both basis sets 4. matrix elements of HZeeman in both basis sets 5. ladder operators and orthogonality for transformation between basis sets. Necessary to be able to evaluate matrix elements of HZeeman in “coupled basis”. Why? Because coupled basis set does not explicitly reveal the effects of Lz or Sz. Nos. 3, 4 and 5 will be repeated in Lecture #26. Suppose we have 2 kinds of angular momenta, which can be coupled to each other to form a total angular momentum. ! L orbital ⎫⎪ ! ⎬ operate on different coordinates or in different vector spaces S spin ⎪⎭ ! ! ! J=L+S total The components of L,S, and J each follow the standard angular momentum definition commutation rule, but, in addition ! ! ⎡⎣ L, S⎤⎦ = 0 ⎡⎣ J i , L j ⎤⎦ = i! Σ ε ijk L k , k ⎡⎣ J i , S j ⎤⎦ = i! Σ ε ijk Sk . k These commutation rules specify that L and S act like vectors with respect to J but as scalars with respect to each other. ! J → jm j ! L → ℓmℓ ! S → sms Coupled ∣jℓsmj〉 vs. uncoupled ∣ℓmℓ〉∣sms〉 representations. * * * matrix elements of certain operators are more convenient in one basis set than the other a unitary transformation between basis sets must exist limiting cases for energy level patterns (and Zeeman tuning rates assignment and intensities for transitions into eigenstates) determination of key parameters, structure, and dynamics 2 5.73 Lecture #25 25 - 3 matrix elements of ℓ and s will each give a factor of "# ⎡ SO ζ nℓ ℓ ⋅s ⎢ H = ξ(r)ℓ ⋅s ≡ " ⎢ each will give a factor of ! anomalous g - value of e 1. ⎢ ⎢ Zeeman = −γBz ( ℓ z + 2s z ) ≡ − ( ω 0 )( ℓ z + 2s z ) ⎢H ⎢⎣ (ζ and ω are in units of rad/s) − nℓ 0 * evaluate matrix elements in both basis sets * look at energy levels and their Zeeman tuning rate in high field γBz ≫ ζnℓ limit * and in low field γBz ≪ ζnℓ limit ⎧⎪lower case for 1e − atom angular momenta Notation: ⎨ ⎪⎩upper case for many - e − angular momenta two different CSCOs a) H elect , J 2 , J z ,L2 ,S2 nJLSM J b) H elect , L2 ,L z ,S2 ,Sz nLM L SM S ⎫ ⎪ (can't be factored) ⎪ recall tensor product ⎬ uncoupled basis ⎪ states and “entanglement” (explicitly factored) ⎪⎭ coupled basis 3 5.73 Lecture #25 25 - 4 2. Coupled and Uncoupled Basis Sets have the same dimension COUPLED ! ! ! J = L +S L −S ≤ J ≤ L +S each J has 2J + 1 MJ ’s J S L J= every term in this sum has 2L + 1 and there are 2S + 1 of them. The second term shows that the S,S–1,…–S terms in sum all cancel. Dimension 2(L + S) + 1 L+S 2 ( L + S − 1) + 1 L + S −1 L+S−2 2 ( L + S − 2) + 1 …… !!!! 2( L − S ) + 1 L−S Every allowed value of J contributes 2L + 1 to sum. How many allowed values of J are there? If L > S, there are 2S + 1 terms in sum. (2S + 1)(2L + 1) + 2[S + (S − 1) + ! (−S)] = (2S + 1)(2L + 1) =0 ⇑� total dimension of basis set for specified L and S UNCOUPLED LM L SM S !"#! "# 2L +1 2 S+ 1 total dimension (2L + 1)( 2S + 1) again There is a term by term correspondence between the 2 basis sets ∴ a transformation must exist: Coupled basis state in terms of uncoupled basis states: JLSM J = Σ aM LML S M S = M J − ML L ML !# #"## $ constraint Trade J, MJ for ML, MS, but MJ = ML + MS. 4 5.73 Lecture #25 25 - 5 Going in the opposite direction: express uncoupled basis state in terms of coupled basis states: OR LML SM S = L+S ∑ b J JLS M J = ML + M S !# #"## $ J= L−S 3. Matrix elements of H = SO ζnℓ " constraint ℓ ⋅s A. Coupled Representation ! ! ! J = L + S J2 = L2 + S2 + 2L ⋅ S L and S commute because they operate in different vector spaces J2 − L2 − S2 L⋅S = (useful trick!) 2 J′L′S′M J′ L ⋅S JLSM J = ! 2 2 [ J(J + 1) − L(L + 1) − S(S + 1)] δ J ′J δ L′L δ S′S δ M J′ M J ( ) an entirely diagonal matrix. B. Uncoupled Representation: work out all of the matrix elements. L · S = LzSz + ½(L+S– + L–S+): because L+S– + L–S+ = (Lx +iLy)(Sx – iSy) diagonal + (Lx – iLy)(Sx + iSy) = 2(LxSx + LySy) off-diagonal L′M L′S′M S′ L ⋅S LM L SM S = ! 2δ L′Lδ S ′S × can’t change L can’t change S 1/2 1 ⎧⎡ ⎨ ⎣ M L M Sδ M L′ M L δ M S′ M S ⎤⎦ + ⎡⎣ L ( L +1) − M L′ M L ⎤⎦ × 2 ⎩ ⎡⎣ S ( S +1) − M S′M S ⎤⎦ δ M L′ M L ±1 × δ M S′ M S ∓1 1/2 } Non-Lecture notes for evaluated matrices S = 1 / 2, L = 0,1,2 2 5 S, 2P, 2 D states ΔM L = −ΔM S = 0, ±1 5.73 Lecture #25 NONLECTURE for HSO : COUPLED BASIS 2 S +1 LJ 2 SO H COUPLED S1/2 2 25 - 6 SO H COUPLED P1/2 & 3/2 (S (P (P (D (D 2 1/2 2 1/2 2 3/2 2 3/2 2 5/2 2 D3 2 and 5 2 ) ) ) ) ) L ! = ζ ns (0) = 0 2 ! = ζ np 2 J a 1 × 1 matrix with matrix element = 0 ⎛ −2 ⎜0 ⎜ ⎜0 ⎜ ⎜0 ⎜0 ⎜⎜ ⎝0 0 −2 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 0 0 0⎞ 0⎟⎟ 0⎟ ⎟ 0⎟ 0⎟ ⎟⎟ 1⎠ J = 1/2 (2 × 2) J = 3/2 (4 × 4) J( J +1) −L(L +1) −S(S +1) = 0 1/2 3/4 0 −3/4 0 1 1/2 3/4 −2 −3/4 −2 1 3/2 15/4 −2 −3/4 +1 2 3/2 15/4 −6 −3/4 −3 2 5/2 35/4 −6 −3/4 +2 J =3/2 H SO COUPLED ! = ζ nd 2 0 0 ⎛ −3 0 0 ⎜ 0 −3 0 0 0 ⎜ 0 −3 0 0 ⎜0 ⎜0 0 0 −3 0 ⎜ ⎜0 0 0 0 2 ⎜ 0 0 0 0 ⎜0 ⎜0 0 0 0 0 ⎜ 0 0 0 0 ⎜0 ⎜ 0 0 0 0 ⎜0 ⎜ ⎝0 0 0 0 0 center of gravity rule: trace of matrix = 0 (obeyed for all scalar terms in H) 6 0 0 0 0 0⎞ 0 0 0 0 0⎟ ⎟ 0 0 0 0 0⎟ 0 0 0 0 0⎟ ⎟ 0 0 0 0 0⎟ ⎟ 2 0 0 0 0⎟ 0 2 0 0 0⎟ ⎟ 0 0 2 0 0⎟ ⎟ 0 0 0 2 0⎟ ⎟ 0 0 0 0 2⎠ J = 5/ 2 (4 × 4) (6 × 6) 5.73 Lecture #25 2S+1 25 - 7 NONLECTURE for HSO : UNCOUPLED BASIS L 2 S H SO UNCOUPLED = !ζ ns (1 2 ⋅0) = (0) 2 P H SO UNCOUPLED = !ζ np × MJ ML MS 3/2 1 1/2 1/2 1/2 1 −1/2 0 1/2 0 1/2 −1/2 0 (1× 1) 0 0 0 −1/2 2−1/2 0 0 0 0 2−1/2 0 0 0 0 −1/2 0 0 0 0 2−1/2 0 1/2 0 0 0 0 0 0 −1/2 −1 −3/2 −1 −1/2 0 0 2−1/2 −1/2 0 0 0 1/2 Each box along main diagonal is for one value of MJ = ML + MS. 2 H SO UNCOUPLED D = !ζ nd × MJ ML MS 5/2 2 1/2 0 0 0 0 0 0 0 0 0 3/2 2 −1/2 0 −1 1 0 0 0 0 0 0 0 3/2 1 1/2 0 1 1/2 0 0 0 0 0 0 0 1/2 1 −1/2 0 0 0 −1/2 (3/2)1/2 0 0 0 0 0 1/2 0 1/2 0 0 0 (3/2)1/2 0 0 0 0 0 0 −1/2 0 −1/2 0 0 0 0 0 0 (3/2)1/2 0 0 0 1/2 0 0 0 0 0 (3/2)1/2 −1/2 0 0 0 −3/2 −1 −1/2 0 0 0 0 0 0 0 1/2 1 0 −3/2 −2 0 0 0 0 0 0 0 1 −1 0 −5/2 −2 −1/2 0 0 0 0 0 0 0 0 0 −1/2 −1 1/2 1 End of Non-Lecture 7 1 5.73 Lecture #25 25 - 8 4. Matrix Elements of H Zeeman = −γB z (L z + 2Sz ) A. Very easy in uncoupled representation Zeeman H uncoupled = −γ Bz L′M L′S′M S′ L z + 2Sz LM L SM S = −γ Bz ! ( M L + 2M S )δ L′Lδ S ′Sδ M L′ M L δ M S′ M S strictly diagonal B. Coupled representation Lz + 2Sz = J z + S z easy hard — no clue! can’t evaluate matrix elements in coupled representation without a new trick, discussed in item #5 5. If we wish to work in coupled representation, because it diagonalizes HSO, we need to find the transformation between coupled and uncoupled representations. JLSM J = Σ a M LM LSMS = M J − M L L ML lengthy procedure: use J ± = L ± + S± and orthogonality Always start with an extreme ML, MS basis state, where we are assured of a trivial 1 to 1 correspondence between basis sets: M L = L, M S = S, M J = M L + M S = L + S, J = L + S J = L + S LSM J = L + S = LM L = L SM S = S coupled uncoupled Now the fun begins … 8 5.73 Lecture #25 25 - 9 Apply J– to both sides of the equation: MJ J# ! " !" # J− | L + S LS L + S〉 = (L − + S− ) LML = L SM S = S ⎡( L + S )( L + S + 1) ⎤ ⎢ ⎥ ⎣− ( L + S)( L + S − 1)⎦ 1/ 2 L + S LS L + S − 1 = [ L( L + 1) − L (L − 1)] 1/2 + [S(S + 1) − S(S − 1)] 1/2 LL −1SS LLSS − 1 Thus we have derived an equality between one coupled basis state and a specific linear combination of two uncoupled basis states. There is only one other orthogonal linear combination that belongs to the same value of ML + MS = MJ: it must belong to the L + S − 1 LS L + S − 1 basis state. lowered J NONLECTURE Work this out for 2P using J– = L– + S– JLSM J = 3 / 2 1 1/ 2 JLSM J − 1 = 3 / 2 = LM LSM S = 1 1 1 / 2 1/ 2 \ 21/2 1 0 1 / 2 1/ 2 + 1 1 1 / 2 −1 / 2 1/2 3 now use orthogonality: − 1 0 1 / 2 1/ 2 + 21/2 1 1 1 / 2 −1 / 2 J − 1LSMJ − 1 = 1 / 2 1 1/ 2 1 / 2 = . 31/ 2 Continue laddering down to get all four J = 3/2 and all two J = 1/2 basis states. ⎛ 2⎞ 3 / 2 1 1 / 2 −1 / 2 = ⎜ ⎟ ⎝ 3⎠ 1/2 ⎛1⎞ 1 0 1 / 2 −1 / 2 + ⎜ ⎟ ⎝ 3⎠ 1/2 1 −1 1 / 2 1 / 2 3 / 2 1 1 / 2 −3 / 2 = 1 −1 1 / 2 −1 / 2 ⎛1⎞ 1 / 2 1 1 / 2 1 / 2 = −⎜ ⎟ ⎝3⎠ 1/2 ⎛ 2⎞ 1 0 1 / 2 −1 / 2 + ⎜ ⎟ ⎝ 3⎠ 1/2 You work out the transformation for 2D! Next step will be to evaluate HSO + HZeeman in both coupled and uncoupled basis sets and look for limiting behavior. 9 1 −1 1 / 2 1 / 2