Tensor operators Applying the Baker-Hausdorff lemma We have used the vector notation for three component operators for example to express the scalar product, like eiGλAe−iGλ = i2λ2 p · x0 = p x0 + p y0 + p z0. A + iλ[G, A] + [G, [G, A]] + ··· x y z 2! Classically a vector is a quantity that under rotations inλn + [G, [G, [G, . [G, A]]] ...] + ··· transforms like V ∈ R3 (or ∈ C3), i.e. if R ∈ O(3), then n! 3 0 X we end up with the commutators Vi = RijVj. j=1 [Jj, [Jj, [··· [Jj,Vi] ···]]]. In quantum mechanics V is a vector operator provided that hV i ∈ C3 is a vector: These will be evaluated in turn into operators Vi and Vk (k 6= i, j). † Rhα|Vi|αiR = hα|D (R)ViD(R)|αi A vector operator (V ) is defined so that it satisfies the 3 commutation relation X = Rijhα|Vj|αi, j=1 [Vi,Jj] = i¯hijkVk. ∀|αi ∈ H,R ∈ O(3). We can easily see that for example p, x and J are vector Thus we must have operators. X In classical mechanics a quantity which under rotations D†(R)V D(R) = R V . i ij j transforms like j X X X Thus the infinitesimal rotations Tijk ··· −→ ··· Rii0 Rjj0 Rkk0 ··· Ti0j0k0···, 0 0 0 iJ · nˆ | {z } i j k D(nˆ) = 1 − n indeces ¯h is called a Cartesian tensor of the rank n. satisfy Example The dyad product of the vectors U and V iJ · nˆ iJ · nˆ 1 + V 1 + T = U V ¯h i ¯h ij i j i 2 is a tensor of rank 2. = Vi + (J · nˆVi − ViJ · nˆ) + O( ) ¯h Cartesian tensors are reducible, for example the dyad X = RijVj product can be written as j U · V (U V − U V ) or U V = δ + i j j i X i j 3 ij 2 Vi + [Vi, J · nˆ] = Rij(nˆ)Vj. i¯h UiVj + UjVi U · V j + − δ . 2 3 ij Substituting the explicit expressions for infinitesimal rotations, like We see that the terms transform under rotations 2 differently: 1 − 2 − 0 2 U · V R(zˆ) = , • δij is invariant. There is 1 term. 1 − 2 0 3 0 0 1 (U V − U V ) • i j j i retains its antisymmetry. There are 3 we get 2 terms. V + [V ,J ] = V − V + O(3). x i¯h x z x y UiVj + UjVi U · V Handling similarly the other components we end up with • 2 − 3 δij retains its symmetry and tracelessness. There are 5 terms. [Vi,Jj] = i¯hijkVk. We recognize that the number of terms checks and that the partition might have something to do with the Finite rotation angular momentum since the multiplicities correspond to A finite rotation specified by Euler angles is accomplished the multiplicities of the angular momenta l = 0, 1, 2. by rotating around coordinate axises, so we have to (k) consider such expressions as We define the spherical tensor Tq of rank k so that the argument nˆ of the spherical function iJjφ iJjφ exp Vi exp − . m ¯h ¯h Yl (nˆ) = hnˆ|lmi is replaced by the vector V : Under the infinitesimal rotations (k) m=q iJ · nˆ Tq = Yl=k (V ). D(nˆ) = 1 − ¯h Example The spherical function Y1: a spherical tensor behaves thus like r r r 0 3 3 z (1) 3 Y = cos θ = 7→ T = Vz 1 4π 4π r 0 4π iJ · nˆ (k) iJ · nˆ 1 + Tq 1 − r r ¯h ¯h ±1 3 x ± iy (1) 3 Vx ± iVy Y1 = ∓ √ 7→ T±1 = ∓ √ . k 4π 2r 4π 2 X (k) 0 iJ · nˆ = T 0 hkq | 1 + |kqi q ¯h Similarly we could construct for example a spherical q0=−k tensor of rank 2: k k X (k) 0 X (k) 0 = T 0 hkq |kqi + iT 0 hkq |J · nˆ|kqi, r 2 r q q ±2 15 (x ± iy) (2) 15 2 q0=−k q0=−k Y = 7→ T = (Vx ± iVy) . 2 32π r2 ±2 32π or (k) (k) X (k) 0 The tensors Tq are irreducible, i.e. there does not exist [J · nˆ,Tq ] = Tq0 hkq |J · nˆ|kqi. any proper subset q0 Choosing nˆ = zˆ and xˆ ± iyˆ we get {T (k),T (k),...} ⊂ {T (k)|q = −k, . , +k}, p1 p2 q (k) (k) which would remain invariant under rotations. [Jz,Tq ] = ¯hqTq Transformation of spherical tensors and Under the rotation R an eigenstate of the direction (k) p (k) [J±,Tq ] = ¯h (k ∓ q)(k ± q + 1)T . transforms like q±1 0 Example Decomposition of the dyad product. |nˆi −→ |nˆ i = D(R)|nˆi. We form spherical tensors of rank 1 from the vector operators U and V : The state vectors |lmi, on the other hand, transform under the rotation R−1 like U0 = Uz,V0 = Vz, −1 X 0 (l) −1 Ux ± iUy Vx ± iVy D(R )|l, mi = |l, m iD 0 (R ). m m U±1 = ∓ √ ,V±1 = ∓ √ . m0 2 2 So we get Now m 0 0 † (0) U · V U+1V−1 + U−1V+1 − U0V0 Yl (nˆ ) = hnˆ |lmi = hnˆ|D (R)|lmi T0 = − = , −1 X 0 (l) −1 3 3 = hnˆ|D(R )|lmi = hnˆ||lm iDm0m(R ) (U × V )q 0 (1) m Tq = √ , X m0 (l) −1 i 2 = Yl (nˆ)Dm0m(R ) (2) m0 T±2 = U±1V±1, ∗ X m0 (l) (2) U±1V0 + U0V±1 = Yl (nˆ)Dmm0 (R). T±1 = √ , m0 2 (2) U+1V−1 + 2U0V0 + U−1V+1 We define a tensor operator Y m(V ) so that T0 = √ . l 6 † m X m0 (l)∗ D (R)Yl (V )D(R) = Yl (V )Dmm0 (R). In general we have m0 (k1) (k2) Theorem 1 Let Xq1 and Zq2 be irreducible spherical (k) tensors of rank k1 and k2. Then Generalizing we define: Tq is a (2k + 1)-component X X spherical tensor of rank k if and only if T (k) = hk k ; q q |k k ; kqiX(k1)Z(k2) q 1 2 1 2 1 2 q1 qq q q k 1 2 † (k) X (k)∗ (k) D (R)T D(R) = D 0 (R)T 0 q qq q is a (irreducible) spherical tensor of rank k. 0 q =−k (k) Proof: We show that Tq transforms like or equivalently k † (k) X (k)∗ (k) (k) † Pk (k) (k) D (R)Tq D(R) = Dqq0 (R)Tq0 . D(R)Tq D (R) = 0 D 0 (R)T 0 . q =−k q q q q0=−k Now Proof: Due to the property † (k) [J ,T (k)] = ¯hqT (k) D (R)Tq D(R) z q q X X = hk1k2; q1q2|k1k2; kqi we have q1 q2 0 0 0 (k) (k) hα , j m |[Jz,Tq ] − ¯hqTq |α, jmi ×D†(R)X(k1)D(R)D†(R)Z(k2)D(R) q1 q2 0 0 0 0 (k) X X X X = [(m − m)¯h − q¯h] × hα , j m |Tq |α, jmi = 0, = hk1k2; q1q2|k1k2; kqi q q 0 0 so 1 2 q1 q2 hα0, j0m0|T (k)|α, jmi = 0, (k1) (k1) −1 (k2) (k2) −1 q ×Xq0 Dq0 q (R )Zq0 Dq0 q (R ) 1 1 1 2 2 2 if m0 6= q + m X X X X X X X = hk1k2; q1q2|k1k2; kqi Theorem 3 (Wigner-Eckardt’s theorem) The matrix 00 q q 0 0 00 0 k 1 2 q1 q2 q q elements of a tensor operator between eigenstates of the 0 0 00 0 angular momentum satisfy the relation ×hk1k2; q1q2|k1k2; k q i 00 00 00 (k ) −1 (k1) (k2) 0 0 (k) ×hk1k2; q1q2|k1k2; k q iDq0q00 (R )Xq0 Zq0 , 0 0 0 (k) 0 0 hα j kT kαji 1 2 hα , j m |Tq |α, jmi = hjk; mq|jk; j m i , p2j + 1 where we have substituted the Clebsch-Gordan series expansion where the reduced matrix element hα0j0kT (k)kαji depends neither on the quantum numbers m, m0 nor on q. (j1) (j2) (k) D 0 (R)D 0 (R) = m1m1 m2m2 Proof: Since Tq is a tensor operator it satisfies the X X X condition hj1j2; m1m2|j1j2; jmi j m m0 (k) p (k) [J±,Tq ] = ¯h (k ∓ q)(k ± q + 1)Tq±1, 0 0 0 (j) ×hj j ; m m |j j ; jm iD 0 (R) 1 2 1 2 1 2 mm so 0 0 0 (k) hα , j m |[J±,Tq ]|α, jmi Taking into account the orthogonality of the (k) =h ¯p(k ∓ q)(k ± q + 1)hα0, j0m0|T |α, jmi. Clebsch-Gordan coefficients q±1 Substituting the matrix elements of the ladder operators X 0 0 hj1j2; m1m2|j1j2; jmihj1j2; m1m2|j1j2; j m i we get m1m2 p 0 0 0 0 0 0 0 (k) = δjj0 δmm0 (j ± m )(j ∓ m + 1)hα , j , m ∓ 1|Tq |α, jmi = p(j ∓ m)(j ± m + 1)hα0, j0, m0|T (k)|α, j, m ± 1i we get q + p(k ∓ q)(k ± q + 1)hα0, j0, m0|T (k) |α, jmi. † (k) q±1 D (R)Tq D(R) If we now substituted j0 → j, m0 → m, j → j , m → m , X X X X X 0 0 00 0 1 1 = δ 00 δ 00 hk k ; q q |k k ; k q i kk qq 1 2 1 2 1 2 k → j2 and q → m2, we would note that the recursion k00 q0 q0 q00 q0 1 2 formula above is exactly like the recursion formula for the 00 (k ) −1 (k1) (k2) Clebsch-Gordan coefficients, ×Dq0q00 (R )Xq0 Zq0 , 1 2 p (j ∓ m)(j ± m + 1)hj1j2; m1m2|j1j2; j, m ± 1i which can be rewritten as p = (j1 ∓ m1 + 1)(j1 ± m1) † (k) D (R)Tq D(R) ×hj1j2; m1 ∓ 1, m2|j1j2; jmi p + (j2 ∓ m2 + 1)(j2 ± m2) X X 0 0 0 (k1) (k2) = hk1k2; q1q2|k1k2; kq iXq0 Zq0 1 2 ×hj1j2; m1, m2 ∓ 1|j1j2; jmi.