Appendix D

written 11/22/15 Appendix D: A Unified View of Tensors and Tensor Functions 1 D.1 Basis 1 D.2 Basis change matrix 3 D.3 Transformations of tensors and tensor functions 5 D.4 When is a scalar not a scalar? 6 Appendix D: A Unified View of Tensors and Tensor Functions In this section multiindex notations are shown in red to the right. D.1 Basis In the bra-ket notation (Paul Dirac, 1939), a rank-k tensor functional T is represented by the bra <T| which is an element of the dual space V*k. Meanwhile, elements of the space Vk are written as kets which are a tensor product of smaller kets, | vi,vi, .... vi > = |vi> |vi> ..... |vi> . | vI> (D.1.1) Here the ir are labels, not components. Each vi is a vector having n components, where n ≥ k. The tensor function T(vi,vi, .... vi) is then represented by the application of the functional <T| to vectors in Vk so that, <T | vi,vi, .... vi > = T(vi,vi, .... vi) T(vI) = <T | vI > (D.1.2) Due to the tensor product (of vector spaces) construction of the "ket" shown in (D.1.1), the function shown in (D.1.2) is manifestly k-multilinear. The bra-ket notation represents an inner product (scalar product) so the spaces here are Hilbert spaces, not just vector spaces. Eq. (D.1.2) is similar to a basic quantum mechanics k-particle wavefunction in the coordinate representation, <ψ| r1, r2...rk> = ψ(r1, r2...rk) . (D.1.3) The object ψ is a functional in V*k which gets applied to |r1,r2....rk> = |r1>|r2>....|rk> and the resulting function ψ(r1,r2...rk) is called a "wavefunction" which describes the "probability amplitude" (density) that the k particles are at spatial locations r1, r2, ....rk. Digressive Comments: 1. It happens that in quantum mechanics literature it is the ket that is the functional in V*k and the bra which is the element of Vk. So in a physics text one sees equations like, ψ(r1, r2...rk) = <r1, r2...rk| ψ>. (D.1.4) This is a long-standing convention difference between the physics and math worlds. When talking about a functional f applied to a vector x, it seems natural to have f(x) = <f | x>, which is the math convention. The physics person writes <x|ψ> = ψ(x) and says that the state vector |ψ> is being projected onto the coordinate representation basis element <x|. Usually ψ is not called a "functional". A ket is thought of as a vector v, while the bra is a transpose vector vT and then <v1|v2> = v1Tv2 in a matrix notation sense, so here it seems logical to put "the vector", whether v, v2 or ψ, on the right. 2. Our functional T maps elements of Vk to the real numbers, and <a|b> = <b|a> = a b, so one can switch the role of which is the functional, and which is the ket acted upon by the functional. In quantum mechanics the functional maps to complex numbers, and <a|b> = a* b where * is complex conjugation. Then <b|a> = b* a = (a b*)* = a* b = <a|b>*. And <v1|v2> = v1T* v2 = v1†v2 . 3. If the k particles are electrons or other half-integral spin particles which are in a "symmetric spin state", then the wavefunction (D.1.4) must be replaced by [Alt(ψ)](r1, r2...rk) in order to make it be totally antisymmetric in the coordinates ri, as required by "Fermi statistics" for half-integral spin particles. We mention this just to show that the Alt operator and the permutation group in general have important applications in quantum mechanics. The covariant tensor Tii....i is, we claim, is this, Tii....i = <T | ei,ei, .... ei > . TI = <T | eI > (D.1.5) From this one would conclude from (D.1.2) that Tii....i = T(ei,ei, .... ei) TI = T(eI) (D.1.6) in agreement with our established fact (6.2.1a). The contravariant form is then Tii....i = T(ei,ei, .... ei) . TI = T(eI) (D.1.7) We now assume that the vectors vi in (D.1.1) form a basis for Vk. This allows for the existence of a dual basis {vi} where vi vj = δij as in (2.11.2) for the ei basis and its dual ej. This puts a small restriction on the function T(v1,v2...vk) which is that det[v1,v2...vk] ≠ 0 (matrix of column vectors vi). The vector arguments have to be linearly independent. Looking at our two equations from above, <T | vi,vi, .... vi > = T(vi,vi, .... vi) (D.1.2) <T | ei,ei, .... ei > = Tii....i (D.1.5) one can say that the tensor Tii....i and the tensor function T(vi,vi, .... vi) are both representations of the same abstract tensor T ~ T in two different bases, |vI> and |eI>. Recall T = ΣI TIλI ϵ V*k = a tensor functional T = ΣI TIeI ϵ Vk = a tensor T ~ T by the isomorphism V*k ~ Vk [see circa (2.11.12)] . (D.1.8) Notice that for the basis {vi}, <vI|vJ> = < vi|vj>< vi|vj> .... < vi|vj> = (vi vj)(vi vj) .... (vi vj) = δijδij...δij // see (2.3.2) for basis {vr} with dual basis {vr} = δIJ . // orthonormal basis in the multiindex notation (D.1.9) This result applies as well to the basis |eI>, so <eI|eJ> = <vI|vJ> = δIJ . (D.1.10) D.2 Basis change matrix The basis-change transformation matrix between the |vI> and |eI> bases is given by, MIJ ≡ < ei,ei, .... ei | vj,vj, .... vj > MIJ = <eI|vJ> (D.2.1) = < ei|vj>< ei|vj> .... < ei|vj> // see (2.9.17) = (ei vj)(ei vj) .... (ei vj) = λi(vj)λi(vj) .... λi(vj) // see (2.11.3) = (vj)i (vj)i ...(vj)i // see (2.11.7) = (vJ)I . // using a multiindex notation shown below (7.8.2) (D.2.2) Entirely in multiindex notation, MIJ = <eI|vJ> = (vJ)I // mixed, see (2.1.6) line 2 or (D.2.3) MIJ = <eI|vJ> = (vJ)I . // pure covariant, see (2.1.6) line 4 The transpose is then, (MT)JI = MIJ = <eI|vJ> = <vJ| eI> = (vJ)I // Hilbert Space is real (MT)JI = MIJ = <eI|vJ> = <vJ| eI> = (vJ)I (D.2.4) In the bra-ket notation completeness of an orthonormal basis is expressed this way: 1 = ΣJ |eJ><eJ| = ΣJ |eJ><eJ| = ΣJ |vJ><vJ| = ΣJ |vJ><vJ| (D.2.5) Proof: (example) Consider a general Vk tensor T : (1) |T> = 1|T> = ΣJ |eJ><eJ| T> = ΣJ TJ |eJ> // so basis |eJ> must be complete (2) |eI> = 1|eI> = ΣJ |eJ><eJ|eI> = ΣJ |eJ>δJI = |eI> // why orthonormal is needed Therefore the up-tilt basis-change matrix M is real orthogonal, meaning MMT = 1 or MT = M-1 : (MMT)IK = ΣJ MIJ(MT)JK = ΣJ <eI|vJ><vJ| eK> = <eI| (ΣJ|vJ><vJ| )eK> = <eI | 1 | eK> = <eI | eK> = eI eK = δIK // see (2.11.2) or (D.2.6) MMT = 1 . // real orthogonal in the multi-index sense In quantum mechanics with complex Vk, one gets instead MM† = 1 (unitary) . The connection then between the tensors and tensor functions is given by, TI = <eI| T> = <eI| 1 | T> = <eI| ΣJ |vJ><vJ| T> = ΣJ <eI|vJ><vJ| T> = = ΣJ MIJ T(vJ) . // MIJ = (vJ)I (D.2.7) Going the other direction, T(vI) = <vI | T > = <vI | 1 | T > = <vI | ΣJ |eJ><eJ| | T > = ΣJ <vI|eJ> <eJ|T > = ΣJ (MT)IJ TJ . // (MT)IJ = (vI)J (D.2.8) Example of (D.2.7): Tii = Σjj (vj)i(vj)i T(vj,vj) or Tij = Σab (va)i(vb)j T(va,vb) . (D.2.7a) Example of (D.2.8): T(vi,vi) = Σjj (vi)j (vi)jTjj or T(vi,vj) = Σab (vi)a (vj)bTab . (D.2.8a) Comment: These examples can be compared to a simple quantum mechanics case. Let |x> be a basis vector describing a 1D particle at location x (coordinate representation), and let |p> be a basis vector describing a plane-wave particle having momentum p (momentum representation). Then it turns out that the basis change matrix is <x|p> = ψp(x) = C eipx where C is a normalization constant. So the basis change "matrix" (continuous matrix subscripts p and x) is a function of p, just as the basis change matrix in (D.2.8a) is a function of vi and vj. D.3 Transformations of tensors and tensor functions Since the ei form a complete basis for V, we can express vectors vi ϵ V as linear combinations of the ei where the coefficients form a matrix R (i and j are labels, not component indices), vi = Σj Rijej or |vi> = Σj Rij |ej> . j = 1,2...k (D.3.1) Then, since we showed below (D.1.2) that T is k-multilinear, one finds T(vi,vi, .... vi) = T( ΣjRijej, ΣjRijej, .... ΣjRijej) T(vI) = T(RIJeJ) = Σjj...j RijRij .... Rij T(ej,ej, .... ej) . = RIJ T(eJ) (D.3.2) But T(ej,ej, .... ej) = Tjj....j from (D.1.6) so, now with implied sums on the jr, T(vi,vi, .... vi) = RijRij .... Rij Tjj....j . T(vI) = RIJTJ (D.3.3) We can regard the coefficient matrix R as being the differential of the linear transformation x' = F(x) = Rx which maps x-space to x'-space as shown in Fig (2.1.1). For a rank-k tensor we know then that the transformation rule is, similar to the fourth line of (2.1.6), T'ii....i = RijRij .... Rij Tjj....j . T'I= RIJTJ (D.3.4) Comparison of (D.3.4) with (D.3.3) provides this interpretation of T'ii....i, T'ii....i = T(vi,vi, .... vi) . T'I = T(vI) (D.3.5) Since we earlier assumed that the vectors vi form a basis for V, we can define arbitrary vectors v'i as linear combinations v'i = ΣjRijvj where the coefficients now define a new matrix R. We take this to be the differential of x' = F(x) = Rx. Equation (D.3.2) then becomes (or just use k-multilinearity), T(v'i,v'i, .... v'i) = RijRij .... Rij T(vj,vj, .... vj) . T(v'I) = RIJ T(vJ) (D.3.6) Therefore (implied sums 1 to k for all ir ), T '(vi,vi, .... vi) ≡ T(Rijvj,Rijvj, .... Rijvj) va' = Rabvb x' = Rx = T(v'i,v'i, .... v'i) = RijRij .... Rij T(vj,vj, .... vj) . (D.3.7) We can interpret this as "the transformation rule for a rank-k tensor function" which is akin to "the transformation rule for a rank-k tensor", T'ii....i = RijRij .... Rij Tjj....j va = Rabeb x' = Rx . (D.3.8) In multiindex notation we now rewrite and compare transformation equations (D.3.7), (D.3.2) and (D.3.8). We have reordered the terms a bit, and all three underlying transformations are x' = F(x) = Rx . The first two equations are transformations for tensor functions, the third is for a tensor T: T '(vI) = RIJ T (vJ) = T(RIJvJ) = T(v'I) va' = Rabvb (D.3.7) T '(eI) = RIJ T (eJ) = T(RIJeJ) = T(vI) va = Rabeb (D.3.2) T'I = RIJ TJ va = Rabeb (D.3.8) (D.3.9) Comments: 1. A rank-k tensor field of multiple arguments transforms as T'I(x',y', ...) = RIJ TJ(x,y,...) x' = Fx, y' = Fy, ... see (2.1.15) (d) (D.3.10) 2. The components of vectors ei, vi and v'i have not been mentioned (so far) in this appendix. D.4 When is a scalar not a scalar? For fun, we present this as a sort of mystery, but the mystery is soon cleared up. We start with a k = 1 example. Recall equation (2.11.10) concerning a rank-1 functional α, α(v) = Σiαiλi(v) = Σiαivi = α v = Σiαivi (2.11.10) (D.4.1) If we move from x-space to x'-space as described in Fig. (2.1.1), we expect to find that α(v) = α v = α' v' = α'(v') (D.4.2) so we would describe α(v) as a "scalar field" (rank-0), similar to (2.1.15) (e). On the other hand, if we set T = α in the discussion of Section D.3 just above, we find that α(v'i) = Rij α(vj) (D.4.3) which says that α(v) transforms as a tensor function of rank-1 (vector). So is α a scalar or is it a vector? The same issue of course arises at the k = 2 level. Recall now this equation, T(v1,v2) = Σab Tab (v1)a(v2)b (2.11.18) (D.4.4) We claim that T(v1,v2) is scalar field of two arguments, as in the example (2.1.15) item (f). This is so because it is a contraction of a rank-2 tensor Tab against two vectors with upper a and b indices. For the doubtful, here is a direct derivation of this fact: (v1')a = Rar(v1)r (v2')b = Rbs(v2)s T'ab = RadRbeTde (D.4.5) T'(v'1,v'2) = Σab T'ab (v1')a(v2')b = [RadRbeTde] [ Rar(v1)r ][ Rbs(v2)s] // insert the above, all implied sums = RadRbeRarRbs Tde(v1)r(v2)s = (RadRar)(RbeRbs) Tde(v1)r(v2)s = δdrδesTde(v1)r(v2)s // (2.1.8) item 1, twice = Tde(v1)d(v2)e = T(v1,v2) (D.4.6) On the other hand, from the discussion of Section D.3 we find that T(v'i,v'i) = RijRijT(vj,vj) or T(v'a,v'b) = RaiRbjT(vi,vj) . (D.4.7) This says that T(vi,vj) transforms as a rank-2 tensor function. So is this T a scalar, or is it a rank-2 tensor? Doubtless the reader has spotted the explanation for these paradoxes. Recall the strong-sense of the word "tensor" as discussed below (4.1.8): a true tensor is a tensor with respect to some underlying transformation. In our k = 1 example above, in (D.4.1) we use this transformation, (v'i)a = Rab (vi)b . (D.4.8) With respect to this component transformation (and a similar one for αi), α(v) does indeed transform as a scalar field. But in (D.4.3) the transformation is v'i = Rij vj (D.4.9) which is a completely different vector transformation which does not involve components. With respect to this transformation, α(v) does indeed transform as a vector function. For k = 2 one has the same situation. The component transformations are shown in (D.4.5) which lead to the function T(v1,v2) transforming as a scalar field. But it is the vector transformation v'i = Rij vj which is used to get (D.4.7), and with respect to this different transformation, T(v1,v2) transforms as a rank-2 tensor function.