Spherical Tensor Operators in NMR Muhammad Sabieh Anwar March 5, 2004 1. Quick Review of Vector Operators: Considering an infinitesimal rotation through ± ˆ ˆ ˆ ˆ 0 about the axis Ju = J:u, the components of a vector operator V, transform Vi ! Vi according to the following transformation rules: ˆ D(R) ˆ 0 ˆ y Vi ¡¡¡! Vi = D(R)ViD (R) (1a) ± = Vˆ ¡ i [Vˆ ; Jˆ ] (1b) i ¯h i u Moreover X y D (R)VˆiD(R) = Rij Vˆj (1c) j also ± D(R) = 1 ¡ exp (¡i Jˆ ) (1d) ¯h u The equality in (1c) holds because of the following concepts: (a) The expectation value of a component h®jVˆij®i remains unchanged in the rotated frame 0 ˆ 0 0 ˆ (with respect to the transformed operator), i.e., h® jVi j® i = h®jVij®i where j®i ¡! 0 ˆ 0 D(R)j®i = j® i and Vi is the operator transform. The expectation value in the two bases must be the same, because we are assuming that space is isotropic, and all physical observables and physical laws must remain invariant under the rotation of the entire ˆ 0 0 ˆ system, including the measuring apparatus. In such a scenario, Vi j® i = Vij®i. (b) The expectation value h®jVˆij®i is unchanged with respect to transformed operator and kets, but with respect to the kets and operators in the un-rotated frame, it trans- ˆ P ˆ forms like the components of a cartesian vector, i.e., h®jVij®i ¡! j Rijh®jVjj®i = 0 0 y h® jVˆij® i = h®jD (R)VˆiD(R)j®i. This last equation is not to be confused with the 0 0 0 equality h® jVˆ j® i = h®jVˆij®i, in which both the operator and the kets are transformed. i P (c) Since the matrix with elements Rij is orthogonal, i.e., k RikRjk = ±ij, (1c) can also be written as: X y D(R)VˆiD (R) = VˆjRji (1e) j The above equation will be used as our starting point for defining tensor operators. 1 The defining transformation rule for vector operators can also be written in a more compact form: [Vˆi; Jˆj] = i¯h²ijkVˆk (1f) We have observed that scalar and vector operators are defined through the behaviour under rotations. Likewise we can also define tensor operators by characterizing their behaviour under transformations — scalar and vector operators being particular cases of these tensor operators (ranks 0 and 1 respectively). 2. Elaboration of point (b) Rotations in Quantum Mechanics present some confusion be- cause they involve a rotation of both the variables as well as the states. Another confusion arises because of the possibility of active and passive rotations. We shall be concerned with passive rotations, which involve a rotation of the co-ordinate axes – affording what is called a basis transformation. The active conceptualism is totally equivalent, however it pays to focus on just either of the two! Consider a state represented by a ket jÃi in the Hilbert space. This state will be depend on a dynamic variable, like the position co-ordinate ¡!r .A complete description of the state is thus jÃ(¡!r )i. Now we rotate the co-ordinate axes such that the position vector transforms as ¡!r 7! ¡!r 0. We assume that this rotation in the physical space is represented by a rotation matrix D, such that D¡!r = ¡!r 0. In the new basis, the state will also have been modified. We define the transformation of the state jÃi 7! jÃ0i, such that: jÃ0(¡!r 0)i = jÃ(¡!r )i (2a) (2a) would strictly be true upto some global phase e iÁ. It expresses our intuition that all observables and spectra of eigenvalues remain unchanged when both the variable and the 0 ˆ 0 0 ˆ state are rotated: the same idea as has been expressed in [1b] as h® jVi j® i=h®jVij®i. From (2a), we continue: jÃ0(¡!r 0)i = jÃ(D ¡1¡!r 0)i (2b) As ¡!r 0 is just an arbitrary variable, we can replace it by ¡!r : jÃ0(¡!r )i = jÃ(D ¡1¡!r )i = DjÃ(¡!r )i (2c) D is unitary because the norms of Djá!r i and já!r i are the same. Moreover from the last equality in (2c), note that we are using the same rotation matrix D for rotations of both the co-ordinates and the states. If we are considering rotations in the real, physical space, all elements of the matrix representing D are real and hence its representation is a symmetric matrix. In general Hilbert space, the matrix representation of an arbitrary rotation R is denoted as D(R). 3. Cartesian Tensor Operator Motivated by the definition of a vector operator presented in (1e) , we can define a tensor operator T with components fTsg which transforms according to the equation: X 0 ˆ y Ti ¡! Ti = D(R)TiD (R) = TsDs;i(R) (3) s 2 Given two vector operators Vˆ and Wˆ , we can form a cartesian tensor with 9 components VˆiWˆ j where i; j = 1; 2; 3. This cartesian tensor, however is not irreducible and it is not convenient to work with it. The component VˆiWˆ j can be reduced into a scalar, a tensor product (or antisymmetric tensor of rank 1) and a traceless, symmetric tensor of rank 2 as expressed below: Vˆ :Wˆ (Vˆ Wˆ ¡ Wˆ Vˆ ) ¡Vˆ Wˆ + Wˆ Vˆ Vˆ :Wˆ ¢ Vˆ Wˆ = ± + i j i j + i j i j ¡ ± (4) i j 3 ij 2 2 3 ij The number of components corresponding to these tensors of ranks 0; 1 and 2 are 1; 3 and 5 respectively, which add up to 9. These numbers correspond well to 2l + 1 components of the m spherical harmonic functions Yl of rank l where l = 0; 1 and 2. In fact the irreducible tensor operators transform just like the spherical harmonic functions and hence they are most often called the spherical tensor operators. 4. Transformation of Spherical Harmonics under Rotations The spherical harmonic function parametrically depends on the angles θ and Á in the spatial co-ordinate basis and is given by the wavefunction representation in space as: m Yl (θ; Á) = hnˆjl; mi (5) where nˆ is a unit vector that points in the θ,Á direction. Now consider the rotation jnˆi 7¡! jnˆ0i = D(R)jnˆi. Taking the inner product with the bra hl; mj, we arrive at the relationship: hl; mjnˆ0i = hl; mjD(R)jnˆi hnˆ0jl; mi = hnˆjD(R)yjl; mi =) Y m(nˆ0) = hnˆjD(R)yjl; mi l X = hnˆjl; m0ihl; m0jD(R)jl; mi (6a) m0 where in (6a), we have suppressed the sum over all possible values of l, because of the invariance of the E (l) sub-space. Finally (6a) can written in the more amenable form as: X m 0 m0 l Yl (nˆ ) = Yl (nˆ)Dm0;m(R) (6b) m0 5. Definition of Spherical Tensor Operators Being motivated by the transformation rela- tions of tensors (3) and spherical harmonics (5), we can define a tensor operator of rank k as an operator whose 2k + 1 components transform according to the relations given below: Xk y k k k D (R)Tq D(R) = Dq;q0 (R)Tq0 (7a) q0=¡k If we write the transpose of (7a), we obtain the following definition of a tensor operator of rank k: Xk k y k k D(R)Tq D (R) = Tq0 Dq0;q(R) (7b) q0=¡k 3 k Note the similarity of (7b) with (1e). The tensor operators Tq transform in a similar fashion m to the spherical harmonic wavefunctions Yl with k = l and q = m. Some authors prefer to q use the notation Tk , whereas I shall use the former. A more convenient operational definition of a spherical tensor operator, analogous to the relation given in (1f) is given by the following commutation rleations that must be satisfied by a tensor operator: ˆ k k [Jz;Tq ] = ¯hq Tq (8a) and p ˆ k k [J§;Tq ] = ¯h k(k + 1) ¡ m(m § 1) Tq§1 (8b) 6. Spherical Tensor Operator Reduction of a Rank 1 Tensor (Vector) A vector Aˆ has components fAˆx; Aˆy; Aˆzg. These components do not form an irreducible representation. However, we can find the following so-called standard components of Aˆ that transform like 0;§1 the components of a rank 1 tensor, T1 : 1 ˆ T0 = Az (9a) 1 1 T = ¡p (Aˆx + iAˆy) (9b) 1 2 1 1 T = p (Aˆx ¡ iAˆy) (9c) ¡1 2 It can be verified that this spherical tensor operator of rank 1 satisfies the commutation relations given in (8). A simple recipe for the formation of tensor operators is by using the m m the general forms of the spherical tensor operators, Yl (θ; Á) = Yl (nˆ) that are generally presented in standard tables. The unit vector nˆ can be replaced by an arbitrary vector Vˆ , with components fVˆx; Vˆy; Vˆzg. The individual components of the unit vector are replaced according to the relations (nˆ)z = z=r 7! Vˆz,(nˆ)x = x=r 7! Vˆx and (nˆ)y = y=r 7! Vˆz. We consider the example of the formation of the components of a rank 1 tensor operator from 0;§1 the spherical harmonics Y1 : r r 3 3 z Y 0 = cos θ = =) T 1 = aVˆ (10a) 1 4¼ 4¼ r 0 z r r 3 3 x § iy Y §1 = ¨ e§iÁsin θ = ¨ =) T §1 = ¨b (Vˆ § Vˆ ) (10b) 1 8¼ 8¼ r 0 x y In deriving (10), we have also made use of the substitutions commonplace in the spherical coordinate system, i.e., x = r cos Á sin θ, y = r sin Á sin θ and z = r cos θ.