Irreducible Tensor Operators April 14, 2010 1 Cartesian Tensors We know that vectors may be rotated by applying a rotation matrix. Beginning with vectors, we can build other objects that transform simply under rotations by taking outer products: T = ~v ⊗~w Ti j = viw j 0 1 v1w1 v1w2 v1w3 = @ v2w1 v2w2 v2w3 A v3w1 v3w2 v3w3 By adding objects of this sort, we can build arbitrary matrices. By choosing enough different vectors, v1;v2;:::;vn, we can build any matrix, 1 2 3 4 5 6 n−1 n Mi j = vi v j + vi v j + vi v j + ::: + vi v j This tells us how matrices transform under rotations. Since each of the vectors rotates according to v˜i = ∑Ri jv j j The matrix Mi j rotates to ˜ 1 2 n−1 n Mi j = v˜i v˜j + ::: + v˜i v˜j ! ! 1 2 n−1 n = ∑Rimvm ∑Rikvk + ::: + ∑Rimvm ∑Rikvk m k m k 1 2 n−1 n = ∑∑RimRik vmvk + ::: + vm vk k m = ∑∑RimRikMmk k m Therefore, Mi j transforms with two copies of the rotation matrix Ri j, one on each index. We can build up more general objects in the same way. Taking outer products of n vectors, ~u;~v;:::;~w, we build rank-n tensors: T = ~u ⊗~v ⊗ ::: ⊗~w Ti j:::k = uiv j :::wk and these will also transform with one factor of Ri j on each index. A more systematic way to build these tensors is to start with a basis of unit vectors, nˆ(1) = (1;0;0) nˆ(2) = (0;1;0) nˆ(3) = (0;0;1) 1 These are vectors. The labels in parenthesis just tell us which vector we’re using. Notice that, for example,n ˆ(1) has components n = ˆ(1) i d1i We can build a general vector by taking a linear combination of the three basis vectors, v = ∑vinˆ(i) i The three numbers vi = v1;v2;v3 are the Cartesian components of v. This works for higher rank tensors as well. Take the nine possible outer products of the basis vectors, nˆ(i) ⊗ nˆ( j) This is a matrix with a 1 in row i, column j, and zeros everywhere else. Then an arbitrary matrix may be written as M = ∑Mi jnˆ(i) ⊗ nˆ( j) i; j where Mi j is just the number in row i and column j. Think of this object as rotationally invariant. It is the components Mi j which change under rotation. For a general rank-n tensor, we have T = ∑Ti j:::knˆ(i) ⊗ nˆ( j) ⊗ ::: ⊗ nˆ(k) i; j n and the 3 numbers Ti j:::k are the Cartesian components. 2 Irreducible tensors The Cartesian components of tensors are mixed by the rotation transformations, Ri j, but not all components of a given tensor mix with all the others. For example, we have seen how a matrix may be broken into rotationally independent pieces as 1 1 1 1 M = tr (M)d + (M − M ) + M + M − d tr (M) i j 3 i j 2 i j ji 2 i j ji 3 i j where tr (M) = ∑k Mkk is the trace of M. Each of these three parts, 1 Mtr = tr (M)d i j 3 i j 1 MA = (M − M ) i j 2 i j ji 1 1 MS = M + M − d tr (M) i j 2 i j ji 3 i j is preserved by rotations, in the sense that 0 tr Mi j = ∑ RikRilMkl i;k;l t = ∑ RkiRilMkl i;k;l −1 = ∑ Rki RilMkl i;k;l = ∑ dklMkl i;k;l = ∑Mkk 2 = (tr (M))0 0 1 MA = R R (M − M ) i j ∑ ik jl 2 kl lk 1 = R R M − R R M 2 ∑ ik jl kl ik jl lk 1 0 = M − R R M 2 i j ∑ jl ik lk 1 0 = M − M0 2 i j ji 0 A = Mi j 0 1 1 MS = R R M + M − d tr (M) i j ∑ ik jl 2 kl lk 3 kl 1 1 = R R M + R R M − R R d tr (M) ∑ 2 ik jl kl ik jl lk 3 ik jl kl 1 1 = M0 + R R M − R R−1tr (M) 2 i j ∑ jl ik lk 3 ∑ ik k j 1 1 = M0 + M0 − d tr (M) 2 i j ji 3 i j 0 S = Mi j For example, the components of the antisymmetric part mix only with the components of the antisymmetric part. These are the irreducible parts of the tensor under rotations. 3 Arbitrary angular momentum 1 Consider the 2-dimensional space of spin- 2 vectors, 1 ;m 2 1 for m = ± 2 . Rotations act on these linearly and homogeneously, ij nˆ· 1 e 2 s ;m 2 To simplify the notation, we just write the two states as 1 1 jmi = j±i = ;± 2 2 1 when we know that j = 2 . Just as in the real case, we may take outer products of them, jm1i ⊗ jm2i ⊗ ::: ⊗ jmki By taking linear combinations of these, we may write general SU (2) tensors, T = ∑ Tm1m2:::mk jm1i ⊗ jm2i ⊗ ::: ⊗ jmki m1:::mk ij nˆ· These objects transform multilinearly and homogeneously, with one transformation matrix, e 2 s , on each index. 3 However, we know that this Cartesian basis is not irreducible. Not only that – we also know how to find the irreducible parts in general. We can write a product of two vectors as a sum of a triplet and a singlet, j1;1i = j+ij+i 1 j1;0i = p (j+ij−i + j−ij+i) 2 j1;−1i = j−ij−i and 1 j0;0i = p (j+ij−i − j−ij+i) 2 We may invert these relationships, j+ij+i = j1;1i 1 j+ij−i = p (j1;0i + j0;0i) 2 1 j−ij+i = p (j1;0i − j0;0i) 2 j−ij−i = j1;−1i Notice that the coefficients are the same. This corresponds to the reality of the Clebsch-Gordon coefficients, h j;mj j1; j2;m1;m2i = h j1; j2;m1;m2j j;mi Formally, the change of basis may be written as j j;mi = ∑ j j1; j2;m1;m2ih j1; j2;m1;m2j j;mi m1;m2 j j1; j2;m1;m2i = ∑ h j;mj j1; j2;m1;m2ij j;mi j;m for any j in the range j j1 − j2j ≤ j ≤ j1 + j2. Notice that we have produced a vector basis for the product space. Instead of thinking of the product j j1;m1i ⊗ j j2;m2i as a matrix, we treat it as a single, higher-dimensional vector, j j1; j2;m1;m2i = j j1;m1i ⊗ j j2;m2i and change the basis to a combination of the j j;mi kets by inserting an identity: j j1;m1i ⊗ j j2;m2i = j j1; j2;m1;m2i ! = ∑ j j;mih j;mj j j1; j2;m1;m2i j;m = ∑ h j;mj j1; j2;m1;m2ij j;mi j;m We trade off higher numbers of indices for higher-dimensional vectors. This means that we may write tensors in a more condensed form. Suppose we have a rank-2 tensor, T = ∑ Tm1m2 jm1i ⊗ jm2i m1:::mk 4 Using the change of basis, this becomes T = Tm1m2 jm1i ⊗ jm2i = ∑ ∑ Tm1m2 h j;mj j1; j2;m1;m2ij j;mi j;m m1:::mk = ∑ Tm1m2 h0;0j j1; j2;m1;m2ij0;0i + ∑ ∑ Tm1m2 h1;mj j1; j2;m1;m2ij1;mi m1=−m2 m m1:::mk We still know how these transform under rotations; moreover, the first and second terms in the last line transform independently – they are the irreducible tensors that make up the original T. It makes sense to treat the irreducible parts separately. The first tensor, 0 T0 = ∑ Tm1m2 h0;0j j1; j2;m1;m2ij0;0i m1=−m2 is a scalar, invariant under SU (2) rotations. The second is a triplet, 1 Tm = ∑ Tm1m2 h1;mj j1; j2;m1;m2ij1;mi m1;m2 ij n·J This transforms with one factor of e 2 , but now we use the 3 × 3 representation of J. These two tensors are irreducible spherical tensors. Notice that the components Tm1m2 transform as h ij n·s i h ij n·s i T˜m m = e 2 e 2 Tm m 1 2 ∑ m m m m 3 4 m3;m4 1 3 2 4 0 1 with two copies of the 2-dimensional generators, Ji = si. By contrast, the irreducible tensors T0 ;Tm transform as h ij (1) i ˜ 0 2 n·J 0 0 T0 = e Tm0 = 1T0 ∑ mm0 m3;m4 h ij (3) i ˜ 1 2 n·J 1 Tm = e Tm0 ∑ mm0 m3;m4 where J(1) = 1 and J(3) are the 1 × 1 and 3 × 3 representations of the rotation generators, respectively. We can continue this process for higher rank tensors. In general, an irreducible spherical tensor of rank k transforms linearly under the (2k + 1)-dimensional representation of SU (2), k h ij i ˜ k 2 n·J k Tq = e Tq0 ∑ qq0 q0=−k and this expression is suitable for defining them: any object that transforms in this way is an irreducible spherical tensor. This means that, rather than having more and more indices on our (Cartesian) tensors, we represent higher and higher rank tensors as larger and larger vectors. In a basis, the general rank-k spherical tensor may be written as k k k T = ∑ Tq jk;qi q=−k which clearly has 2k + 1 independent components.