new eq num method for Sec 2_11

2.11 Dual Spaces and Tensor Functions We denote dual-space vectors and tensors by Greek or script font letters. The dual space V* is by definition the space of linear functionals over V. If α ϵ V*, we can then write α : V → K α(v) = k ϵ K (2.11.1) where K is any field (but we always use the reals). Since α is a linear functional, α(v) is a linear function. In normal calculus, if f: V → R, one refers to f as a function, and f(v) as that function evaluated at some point in V, though loosely speaking f(v) is also called a function. To emphasize the distinction, we shall refer to f as a "functional" and f(v) as a "function". Comments: Much of the rest of this section will be repeated in later Chapters. We have found that the notations involved can be a major stumbling block, and feel it is important to exercise the notation in many ways to make the reader (and author) feel comfortable with it. As with most endeavors, it is a matter of practice. We also try to explain why certain notations are used. (a) The Dual Space V* in Matrix and Dirac Notation For every column vector v in V, there exists a row vector vT such that (vT)i = vi. For example, for N=2, v = = |v> vT = (a, b) = <v| (2.11.a1) Here we have snuck in the Dirac bra-ket notation where the ket |v> is a column vector and the bra <v| is the corresponding row vector. The notation vT means that the row vector is the Transpose of the column vector. We now have multiple ways to write the dot (inner, scalar) products of Section 2.9 : v v' = vTv' = (a, b) = aa'+bb' = <v | v' > . (2.11.a2) Because our vectors have real components, the above can also be written v' v = v'Tv = (a', b') = aa'+bb' = <v' | v > . (2.11.a3) We mention real components only because in the Dirac notation one has <a|b> = <b|a>* where * means complex conjugation, so if this scalar product is real, then <a|b> = <b|a> . We regard v or |v> as being a vector in the vector space V, while vT or <v| (the row vector) is a vector in the dual space V*. This is really a simple concept. Sometimes the dual-space vector vT = <v| is referred to as the covector of v = | v > . Suppose αT = <α | is a vector in the dual space V* . One can regard this dual-space vector <α | as being a functional which acts on vectors in the space V. Then, α = <α | = functional α(v) = <α | v> = α v = function = a scalar number (2.11.a4) α : V → K . // K = any real field, such as the real numbers Just as the space of column vectors V is a linear space (a vector space), so also the dual space of row vectors V* is a linear space, so we know that the functional <α | is a linear functional. That in turn implies that the function α(v) is a linear function, which we now show directly: α(sv) = α (sv) = s ( α v ) = s α(v) α(v+v') = α (v+v') = α v + α v' = α(v) + α(v') . (2.11.a5) (b) Functional notation We have now a slight notational conundrum. We like to write a scalar-valued function F(v) in non-bold font, whereas a vector-valued function would be F(v). Thus we have written α(v) above with a non-bold α, since α(v) is a scalar-valued function. On the other hand, α(v) is really a function of the vector α, so it seems misleading to refer to it as α(v), and we ought to call it α(v) so then α(v) = <α | v> has everything bolded on both sides. But then the functional would have to be called α = <α | . But this contradicts our notation earlier that α is a vector, αT is the transpose, and we should write αT = <α |. If we use α = <α | then we avoid that contradiction. This is what authors end up doing, writing a functional as a scalar entity which for us means an unbolded entity. A possible solution would be to say, fα = <α | = functional fα(v) = <α | v> = α v = function (2.11.b1) where f is non-bold, and the subscript label α is bold, but then we have introduced a new symbol f which seems superfluous. So the conclusion is this: α = <α| = functional, α(v) = <α | v> = function, and one must understand that α(v) is a function of the vector quantity α. Obviously there is a unique functional α(v) for each vector α in V (and thus for each αT in V*). The spaces V and V* have the same dimension n and are isomorphic to each other in the sense just noted. (c) Basis vectors for the dual space V* Now recall that our axis-aligned x-space basis vectors ui have dual basis vectors ui where ui uj = δij which is the idea of orthogonality in the covariant world (which might be non-Cartesian). In our notations above, ui uj = (ui)T uj = <ui | uj> = δij = ui uj = (ui)T uj = <ui | uj> = δij = δi,j . (2.11.c1) Since ui and ui are in general different column vectors in V, (ui)T and (ui)T are different row vectors in the dual space V*. Just as the column vectors |ui> and |ui> form two distinct bases for V, the row vectors <ui | and <ui | form two distinct bases for V*. Certainly dim(V) = dim(V*). Definition of λi Above we discussed α = <α| as a vector functional, and α(v) = <α|v> as the corresponding scalar function. Whereas <α| is some general vector in V*, we now consider in its place a basis vector <ui| in V*. With what notation shall we represent this functional? In analogy with α and α(v) we could use ui and ui(v) where the ui is unbolded to indicate a scalar function. Or we could use fu = <ui| and fu(v) = <ui| v> . The first notation is not uncommon (see wiki dual space where ui = ei), while the latter notation is unpleasant. Other common notations are v*i(v) or e*i(v) which for us would be u*i(v). We shall use the following notation, λi ≡ <ui| basis functional in V* // λi = (ui)T so λi(v) = <ui|v> basis function in V*f // λi(v) = (ui)Tv . (2.11.c2) The λ is unbolded, consistent with α(v). λ is a Greek letter consistent with our plan to use Greek or script letters for dual space objects. The index on λi is up, matching the index on ui in <ui|. Notice that λi(uj) = <ui|uj> = δij . (2.11.c3) We think of λi ≡ <ui| as being in the dual space V* while λi(v) = <ui|v> lies in a directly corresponding space of functions which we call V*f. Comment: For the dual space basis functionals Sjamaar used symbol λi in his 2006 notes (p 84), but changed to βi in his 2015 update (p 91). Spivak uses φi (p 76). Wiki (dual basis) uses basis vectors vi instead of ei so their λi is called vi. Wiki (dual space) uses ei while Lang Algebra uses fi (p 143). There seems to be no standard notation as in physics where F = ma is universally recognized as Newton's Second Law which would be hard to identify if written G = nb. Probably ui or ui (unbolded) is the most logical choice if the V basis vectors are ui, but it is so easy to confuse functional ui with the vector ui (especially when we drop our bolding of vectors starting in Chapter 3) that we shall stick with λi. Eq. (2.7.1) line 3 gives the expansion of a vector v onto the ui v = Σi vi ui where vi = ui v or |v> = Σi vi |ui> where vi = <ui| v > = ui v . (2.11.c4) Notice therefore that λi(v) = <ui|v> = ui v = vi . (2.11.c5) The function λi(v) is sometimes called "the ith coordinate function" since it projects out the ith component the vector v. As summarized in (2.7.17), since each dot product ui v is a scalar, the functions λi(v) i = 1..N transform as scalars despite the fact that the values of these scalars are the components of the vector vi. Transposing (2.11.c4) produces a vector functional expansion in V*, vT = Σi vi (ui)T or <v| = Σi vi <ui| = Σi vi <ui| . (2.11.c6) Using Greek letters for dual space objects we write this as α = <α| = Σn αi <ui| = Σn αiλi . (2.11.c7) Then, α = Σiαiλi functional (2.11.c8) α(v) = Σiαiλi(v) = Σiαivi = α v function (2.11.c9) or in bra-ket notation, <α| = Σiαi<ui| functional α(v) = Σiαi<ui|v> =Σiαivi α v = <α |v> function (2.11.c10) and we replicate the result (2.11.b1). Definition of λ'i We have defined λi ≡ <ui| as a notation for a certain basis functional in dual x-space. We would like to somehow define an object λ'i which is a basis functional in dual x'-space. How should this be done? One might intuitively feel that one should set λ'i ≡ <u'i| . Or one might think that once λi is defined as above, then the meaning of λ'i is forced upon us by some equation like λ'i = Rijλj . Both these notions are not what we want to do. We are not forced to say λ'i ≡ <u'i| just because λi ≡ <ui| since we are making two separate definitions. And λ'i = Rijλj is complete nonsense for the following reason. The N functionals λi for i = 1..N are each vectors in V*, so {λi} is a set of vectors, not a set of numbers, whereas when one tries to write λ'i = Rijλj one is implying that λj is a set of numbers which form a vector. Recall that the ui are the "axis-aligned" basis vectors in x-space since (ui)j = (ui)j = δi,j. Recall that the e'i are the "axis-aligned" basis vectors in x'-space since (e'i)j = (e'i)j = δi,j. This suggests that the proper definition of λ'i is the following: λ'i ≡ <e'i| (2.11.c11) One then finds that, for v' a vector in x'-space, λ'i(v') ≡ <e'i| v'> = v'i (2.11.c12) which is then analogous to λi(v) ≡ <ui| v'> = vi (2.11.c5) In both cases then λi and λ'i are the "ith coordinate functions", projecting out the ith coordinate from a vector. Since v'i = Rijvj we can certainly write λ'i(v') = Rij λj(v) (2.11.c13) as a statement relating two vectors of scalars. Notice this does not say λ'i = Rijλj which we already noted above does not even make sense. If we display the fact that Rij in general is Rij(x) then λ'i(v') = Rij(x) λj(v) (2.11.c14) Since this does not fit into any of the molds shown in (2.1.16), one cannot quite claim that λj(v) transforms as a vector field, but the transformation is similar. We can study (2.11.c13) in Dirac notation as follows (see below for Dirac notation details). λ'i(v') = <e'i|v'> = <ei|v> = <ei| 1 | v> = <ei|uj >< uj| v> = ei uj λj(v) = Rijλj(v) (2.11.c15) where the last step comes from (2.4.3). Once we have a functional λ'i, we can define a general rank-1 functional α' in dual x'-space as follows: α' = Σiα'iλ'i functional in V'* (2.11.c16) α'(v') = Σiα'iλ'i(v') = Σiα'iv'i = α' v' function in V'*f (2.11.c17) It then follows that α'(v') = α' v' = α v = α(v) (2.11.c18) and in some sense one could say that α(v) transforms as a scalar field, where v plays the role normally occupied by the position vector x. On the other hand, the vector |v> and the dual vector (functional) α = <α| transform as vectors and so α is a vector functional. We now define α(v) to be a "rank-1 tensor function". Spivak would call it a "1-tensor". We have this seeming contradiction that α(v) is a rank-1 tensor function, yet that function transforms as a rank-0 scalar. The rank-1 description really applies to the functional α = <α| which is in fact a vector and transforms as a vector. When this is closed with the ket |v> one obtains the scalar object α(v) = <α | v>. (2.11.c19) Vector space names: V, V*, V*f and V', V'*, V'*f (2.11.c20) Here we have associated vector space names V and V* with x-space in Picture A (2.1.1), while V' and V'* are associated with x'-space. All these spaces have the same dimension N and all are isomorphic. There is a 1-to-1 relationship between V and dual space V* as noted above, and there is a 1-to-1 relationship between V and V' since for every vector v in x-space there is a unique corresponding vector v' = Rv in x'-space. We refer to V* as dual x-space and V'* as dual x'-space. Associated with the dual space V* of functionals is the space V*f of corresponding functions, and similarly for V'* and V'*f. (d) Rank-2 functionals and tensor functions A rank-2 tensor may be represented as T = Σab Tab ua ub (2.11.d1) V = Σa Va ua where on the second line for comparison we show a general rank-1 tensor (vector). In Dirac notation, we write | ua, ub> ≡ |ua> |ub> ↔ ua ub (2.11.d2) which represents any of the n2 basis vectors of the tensor product space V2 = VV. We could write this as | ua, ub>2 ≡ |ua>1 |ub>1 to distinguish the fact that some kets are in V1 and others in V2, but the contents of the ket usually make it obvious to which vector space a ket belongs. In Dirac notation, the tensor T is written |T> = ΣabTab |ua> |ub> = ΣabTab | ua, ub> (2.11.d3) and this is a general element of the space V2. The corresponding rank-2 linear functional in the dual space V*2 is given by <T| = ΣabTab <ua | <ub| = ΣabTab < ua, ub| . (2.11.d4) This is done in analogy with the vector case |V> = Σa Va |ua> <V| = Σa Va<ua | (2.11.d5) where we are careful to have the index "tilt" have the form of a contraction, even though we are not really contracting indices on a tensor. The rank-2 functional <T| is linear in both V* spaces of V*V*, so it is called a bilinear functional. If we let (subscript 1 and 2 are labels of two vectors, not components of v ) | v1, v2> ≡ |v1> |v2> (2.11.d6) represent an arbitrary (but pure) element of V2 = VV, then we may construct T = <T| rank-2 tensor functional T(v1,v2) = <T | v1, v2> rank-2 tensor function (a Spivak "2-tensor") . (2.11.d7) It follows that T(v1,v2) = <T | v1, v2> = ΣabTab < ua, ub| v1, v2> = ΣabTab <ua| v1> <ub| v2> = ΣabTab (v1)a (v2)b (2.11.d8) where we have used the fact that the scalar product for elements of V*2 with elements of V2 is the product of two V*-with-V scalar products, as seen for example in (2.9.13). In the last line above we see that the tensor function T(v1,v2) is the contraction of a rank-2 tensor with two rank-1 tensors, and so is a scalar. Thus, T'(v'1,v'2) = T(v1,v2) (2.11.d9) and a rank-2 tensor function transforms as a "scalar field of two arguments". The "rank-2" description applies to the tensor functional <T| , and when this is closed with an element of V2 the result is a scalar. Note from (2.11.d8) and (2.4.1) that (ui)a = δia , T(ui,uj) = ΣabTab (ui)a (uj)b = ΣabTab δia δjb = Tij (2.11.d10) so the tensor function evaluated at the basis vectors gives a corresponding element of the tensor. Using λi = < ui| as defined above in (2.11.c2), we can rewrite (2.11.d4) <T| = ΣabTab <ua | <ub| as rank-2 tensor functional T = ΣabTab λa λb (2.11.d11) which is in analogy to <α | = Σa αa <ua| rank-1 tensor functional α = Σa αa λa . (2.11.d12) We continue to use script or Greek fonts to represent functionals, such as α and T. Taking the special case of a rank-2 functional which is just λa λb we construct the following rank-2 tensor function, (λa λb)(v1,v2) = < ua, ub| v1, v2> = <ua| v1> <ub| v2> = (v1)a (v2)b = λa(v1) λb(v2) . (2.11.d13) This function is manifestly linear in both arguments, since λa(v1) is linear, so it is a bilinear function. For example, (λa λb)(v1+v1',v2) = (v1+v1')a (v2)b = (v1)a (v2)b + (v'1)a (v2)b = (λa λb)(v1,v2) + (λa λb)(v'1,v2) . (2.11.d14) Whereas <T| shown above is a general rank-2 tensor functional, we can consider the special case of a pure rank-2 functional formed from two vector functionals α = <α| and β = <β| . In that case one finds, (α β) = <α| <β| = <α, β | rank-2 functional (α β)(v1,v2) = <α, β | v1, v2> = <α | v1> <β | v2> = α(v1)β(v2) rank-2 tensor function (α β)(ui,uj) = α(ui)β(uj) = αiβj = (α β)ij rank-2 tensor (2.11.d15) where the very last item is αiβj expressed in the tensor product notation of (2.8.9). Once again, evaluation of a tensor function at two basis vectors creates an element of the tensor. Comment on vertical bars in the Dirac Notation Let |a> be a vector in V, and <b| a vector in the dual space V*. Notice that <b| |a> = <b||a> = <b|a> . (2.11.d16) The official notation for the scalar product is <b | a> not <b || a> so one replaces the || with | . The same replacement is made for example doing a scalar product between elements of V*2 and V2 <a| <b| |c> |d> = <a,b||c,d> = <a,b | c,d> or <a| <b| |c> |d> = ( <a| |c> ) ( <b| |d> ) = <a|c><b|d> . (2.11.d17) (e) Rank-k functionals and tensor functions It is a simple matter to generalize from k = 2 to k = k, so the vector space is Vk and the dual space is V*k, Vk ≡ VxVx....xV k factors // Cartesian product of k spaces Vk ≡ VV....V k factors // tensor product of k vector spaces V*k ≡ V*V*....V* k factors // tensor product of k dual spaces . (2.11.e1) We then have as a most general element of Vk (a rank-k tensor), T = Σii....i Tii....i (ui ui ..... ui ) . T = ΣITI uI (2.11.e2) with (ui ui ..... ui) = |ui> |ui >..... |ui> = | ui, ui ....., ui >k (2.11.e3) = | ui, ui ....., ui > . uI ≡ ui ui ..... ui On the right in red we show our equations expressed in the multi-index notation introduced in (2.10.17-22). The letter Z which appears below is used to represent the set of integers 1,2...k. Then the rank-k tensor T in Vk is represented in Dirac notation as |T> = Σii....i Tii....i | ui, ui ....., ui > . |T> = ΣITI |uI> (2.11.e4) The rank-k tensor functional <T| of V*k is then <T | = Σii....i Tii....i < ui, ui ....., ui | <T| = ΣITI <uI| or (2.11.e5) T = Σii....i Tii....i λi λi ..... λi . T = ΣITI λI A general pure element of Vk is specified by |v1, v2, ...vk> = |v1> |v2> .... |vk> . |vZ> = |v1> |v2> .... |vk> (2.11.e6) The corresponding rank-k tensor function is given by T(v1, v2, ...vk) = <T | v1, v2, ...vk> = Σii....i Tii....i < ui, ui, ..., ui | v1, v2, ...vk> = Σii....i Tii....i < ui|v1>< ui|v2> .... < ui|vk> (2.11.e7) = Σii....i Tii....i (v1)i (v2)i....(vk)i T(vZ) = ΣITI (vZ)I This shows that the rank-k tensor function is a linear combination of the products of the argument components weighted by the components of the corresponding rank-k tensor. Since this is the contraction of a rank-k tensor with k rank-1 tensors, the result transforms as a scalar, so then T'(v'1, v'2, ...v'k) = T(v1, v2, ...vk) . T'(v'Z) = T(vZ) (2.11.e8) That is to say, the rank-k tensor function transforms as a scalar field, where the term "rank-k" is associated with the functional T = <T| which is an element of the dual space V*k . Finally we see that T(uj,uj, .... uj) = <T | uj,uj, .... uj > = Σii....i Tii....i (uj)i (uj)i....(uj)i = Tjj....j . T(uJ) = TJ (2.11.e9) From (2.11.e7) one sees that the tensor function T(v1, v2, ...vk) is manifestly k-multilinear, which is the generalization of linear for k = 1 and bilinear for k = 2. Once can construct a rank-k tensor functional purely from the dual basis vectors, (λiλi ... λi) = <ui| <ui| ... <ui| rank-k tensor functional λI = <uI| (2.11.e10) (λiλi ... λi)(v1,v2....vk) = λi(v1)λi(v2)....λi(vk) = (v1)i(v2)i ... (vk)i rank-k tensor function λI(vZ) = (vZ)I (2.11.e11) (λiλi ... λi)(uj,uj, .... uj) = λi(uj)λi(uj)....λi(uj) = (uj)i(uj)i ... (uj)i = δjiδji ... δji evaluated at basis vectors . λI(uJ) = δIJ (2.11.e12) As an alternative to the most general rank-k tensor functional T and the all-basis-vector rank-k tensor functional (λiλi ... λi), one can consider a "pure" rank-k tensor functional constructed from k dual vectors which we shall call <αi| . In this case we find, <α1, α2....αk| = <α1| <α2| .... <αk| pure rank-k tensor functional = α1 α2... αk = (α1α2...αk) (2.11.e13) (α1α2...αk)(v1, v2, ...vk) = α1(v1)α2(v2) ....αk(vk) = (α1v1)(α2v2) ....(αkvk) pure rank-k tensor function (2.11.e14) (α1α2...αk)(uj,uj, .... uj) = α1(uj)α2(uj) ....αk(uj) = (α1 uj) (α2 uj) ... (αk uj) evaluated at ur = (α1)j(α2)j...(αk)j = (α1α2...αk)jj... j outer product notation (2.11.e15) Hopefully after this long slog, the following paragraph makes some sense to the reader: A rank-k tensor function is the bra-ket closure (inner product) of a rank-k dual tensor functional <T| of V*k with a pure rank-k non-dual tensor |v1,v2...vk> of Vk such that T(v1,v2,...vk) = <T|v1,v2...vk>. The tensor function is k-multilinear in its arguments, and transforms as a scalar field with k vector arguments. When the rank-k tensor function is evaluated at the basis vectors ur, it replicates the non-dual rank-k tensor with which is it associated, which is to say, T(uj,uj, .... uj) = Tjj....j . Spivak on page 75 refers to a rank-k tensor function as a "k-tensor". (2.11.e16) As we shall see later, the motivation for using tensor functions is their crashingly simple description of the tensor product of an arbitrary rank-k tensor with an arbitrary rank-k' tensor to produce a rank-(k+k') tensor : k<T | v1,v2...vk>k k'< S| vk+1,vk+2...vk+k'>k' = [ k<T k'<S| ] [|v1,v2...vk>k |vk+1,vk+2...vk+k'>k'] = k+k'<TS | v1,v2...vk+k'>k+k' (2.11.e17) or T(v1,v2,...vk) S(vk+1,vk+2,...vk+k') = (TS)(v1,v2 .... vk+k') . (2.11.e18) This equation appears below as (6.6.13) and also appears in Spivak page 75. As noted by Benn and Tucker page 2, the relationship between the vector space Vk and the dual vector space V*k is a reciprocal one. One could, as they say, perversely regard V*k as the starting vector space and then Vk would be the dual space of V*k. This amounts to swapping bra ↔ ket in the Dirac notation outlined above. Instead of having a functional α(v) = <α|v>, one would have a functional v(α) = <v|α>. We find that things are hard enough to understand without doing this "perverse" swapping of things right off the bat as they do. They refer to a rank-k tensor as a tensor of degree k, while other authors refer to rank as the order of a tensor. We us the term rank and promise not to confuse it with the different notion of the rank of a matrix which is the number of linearly independent rows or columns, or with various other meanings of the word "rank" in mathematics. (f) The Covariant Transpose Whereas the matrix transpose of a matrix Mab would be (MT)ab = Mba (swap the rows and columns), it is the covariant transpose (MT)ab = Mba that is significant in covariant notation. We quote from Tensor where M is a general rank-2 tensor while R and S are the "differentials" of (2.1.2), (MT)ab = Mba (RT)ab = Rba (ST)ab = Sba (MT)ab = Mba (RT)ab = Rba (ST)ab = Sba (MT)ab = Mba (RT)ab = Rba (ST)ab = Sba (MT)ab = Mba (RT)ab = Rba (ST)ab = Sba . (7.9.3)' (2.11.f1) Equations in any column can be obtained by lowering one or both indices in the top equation, so that the covariant transform MT is a rank-2 tensor if M is a rank-2 tensor. For all-up or all-down indices, the two kinds of transposes are the same: (MT)ab = (MT)ab = Mba. The transpose always has the indices reflected in a vertical line between the indices. This subject is discussed in Tensor Section 7.9 where all claims are proved. We quote some of the conclusions: det(M) = det(MT) = det(MT) (7.9.7)' (2.11.f2) RRT = RTR = 1 SST = STS = 1 RS = SR = 1 RT = R-1 = S ST = S-1 = R . (7.9.8)' (2.11.f3) (g) Linear Dirac Space Operators Consider these three ways of writing the same real number, where M is a matrix sandwiched between vector b on the right and transpose vector a on the left, aT (Mb) M acts to the right ( * * *) [ ] (aTM)b M acts to the left, and note that (aTM) = (MTa)T [ (* * *) ] aTM b can think of M acting either to the right or to the left. (2.11.g1) In writing these equations, one normally thinks of M as being a matrix Mij = (M[u])ij or M = M[u] . By default, the matrix elements are taken in the axis-aligned ui basis on both left and right (and this applies to all indices as discussed at the end of Section 2.4) so that (ui)TM (uj) = Σa,b (ui)a Mab (uj)b = Σa,b δia Mab δjb = Mij // = (M[u])ij. (2.11.g2) One could, however, do this in some other basis, for example, (ei) TM (ej) = Σa,b (ei)a Mab (ej)b = Σa,b Ria Mab Rjb = (RMRT)ij // = (M[e])ij (2.11.g3) and the result is a completely different matrix. In this case the matrices are related by a covariant similarity transformation by R M[e] = R M[u]RT . (2.11.g4) It is useful to think of the object M as being a basis-independent abstract linear operator which, when sandwiched between certain basis vectors, has certain matrix elements. Different types of basis vectors yield different matrices. We could even have mixed basis elements, (ui) TM (ej) = Σa,b (ui)aMab(ej)b = Σa,b δiaMabRjb = (MRT)ij // = (M[u,e])ij (2.11.g5) so in this case we get M[u,e] = MRT . (2.11.g6) The abstract operator M only becomes a matrix when it is properly sandwiched. This notion of thinking of the object M as a basis-independent linear operator becomes more pronounced in the Dirac notation. We restate the above equations as follows, all of which evaluate to the same real number, <a| ( M |b>) M acts to the right = <a | Mb > (<a|M ) |b> M acts to the left, and note that <a|M = <MTa | = <MTa | b > <a| M |b> can think of M acting either to the right or to the left. (2.11.g7) The space between the vertical bars is inhabited by abstract linear operators like M. The matrix elements shown above are then <ui | M | uj> = (M[u])ij = Mij <ei | M | ej> = (M[e])ij = (RMRT)ij <ui | M | ej> = (M[u,e])ij = (MRT)ij (2.11.g8) To emphasize this notion of abstract operator, we shall write the operator in a different font, so M is a matrix and M is a Dirac-space operator, and then <a| M |b> = <a| ( M |b>) = <a |M b> = a scalar product of two vectors <a| M |b> = (<a|M ) |b> = <MTa | b > = a scalar product of two vectors <ui | M | uj> = (M[u])ij = Mij etc . (2.11.g9) Here then is a review of the matrix and Dirac notations, a' = (Ma) = (M)a (b')T = (Mb)T = bT MT matrix notation |a'> = |Ma> = M|a> <b'| = <Mb| = <b|MT . Dirac notation (2.11.g10) Then consider the following claim Fact: <a | M | b> = <b| MT | a> (2.11.g11) where both M and MT are the names of abstract linear operators. Proof: <a | M | b> ≡ <a | M b> = a (Mb) = ai(Mb)i = ai[ Mijbj] = ai Mij bj = bj Mij ai = bj (MT)ji ai = bj [MTa]j = b (MTa) = <b| MTa> = <b| MT |a> . Operator M is defined by its action on an arbitrary ket vector M | b> = | M b> Operator MT is defined by its action on an arbitrary ket vector MT | b> = | MT b> Notice in the proof that the covariant transpose MT is the correct transpose to use since Mij = (MT)ji. Exercise: Show that wv is a scalar under any transformation x' = F(x) : w'v' = <w' | v'> = <Rw|Rv> = <w|RTR|v> = <w| 1 |v> = <w|v> = wv . (2.11.g12) In this example R is a matrix, whereas R is the corresponding Dirac space operator. The statement RTR = 1 (2.11.g13) is the operator version of our (2.11.f3) matrix statement RTR = 1 (2.11.g14) which we verify as follows, (RTR)ac = (RT)abRbc = Rba Rbc = δac // (2.1.9) #1 (2.11.g15) and which is valid for any transformation differential matrix Rij. One may represent a Dirac operator M in various ways M = Σij | ui> Mij <uj| = Σij | ei> [M[e]] ij <ej| = Σij | ui> [M[u,e]] ij <ej| (2.11.g16) as can be verified by closing with the appropriate basis vectors. For example, for the last line above, <ua | M | eb> = <ua | { Σij | ui> [M[u,e]] ij <ej|} | eb> = Σij <ua | ui> [M[u,e]] ij <ej| eb> = Σij δai [M[u,e]] ij δjb = [M[u,e]] ab . (2.11.g17) When M = 1 we find 1 = Σij | ui> δij <uj| = Σi | ui><ui| (2.11.g18) which is just a statement that the | ui> basis is complete. Finally, we make a comparison between the abstract Dirac operator M and the abstract rank-2 "vector" M, M = Σij | ui> Mij <uj| // Dirac operator M = Σij Mij ui uj // (2.8.10), "vector" in vector space V2 or |M> = Σij Mij | ui> | uj> . (2.11.g19) The first object M is an operator in the Dirac Hilbert Space V. The second object M or |M> is a vector in the tensor product space V V. M and M are completely different objects, though they both involve the same matrix elements Mij. In each case, we can project out those matrix elements in an appropriate fashion: <ua | M | ub> = <ua | { Σij | ui> Mij <uj|} | ub> = Mab [< ua| < ub| ] | M > = [< ua| < ub| ] Σij Mij | ui> | uj> = Mab . (2.11.g20) The above discussion is presented implicitly for a square matrix M, but only small adjustments are needed for it to apply to a non-square matrix. In this case, in aTM b one thinks of vectors a and b as having different dimensions. Perhaps b lies in x-space which is Rn while a lies in x'-space which is Rm with m > n, and then Mij is an m x n matrix. The x'-space V' has n basis vectors |u'i> while the x-space V has m basis vectors |ui>. Then one would have, for example, <ui | M | u'j> = Mij M = Σi=1m Σj=1n | ui> Mij <u'j| |M> = Σi=1m Σj=1n Mij | ui> | u'j> 1' = Σi | u'i><u'i| completeness in V' 1 = Σi | ui><ui| completeness in V (2.11.g21) This is exactly the situation we shall encounter in Chapter 9 where the matrix R is an m x n matrix. We shall not use a script font to represent Dirac space operators, but one should keep in mind that any non-scalar object which is seen to be operating to the left on a bra, or to the right on a ket, or which is found between the two vertical bars in < | | > is a Dirac space operator, regardless of the font used to represent it. Such operator objects are not matrices, but their matrix elements form matrices.