edit section E.5

E.5 Matrix notation for dyadics We revert to developmental notation for a while. 1. Using matrix notation to replace dyadic notation for contravariant vectors For a vector V, certainly Vi = (VT)i, meaning the object in the ith row of V is the same as the object in the ith column of VT . Therefore one can express the dyadic product in this more down-to-earth manner, (ab)ij ≡ aibj = ai (bT)j = (abT)ij (E.5.1) or ab = abT . (E.5.2) The notation on the left is the dyadic notation, whereas matrix/vector notation is used on the right. Here one knows that ab is a "dyadic" because there is no other meaning for two bolded column vectors abutting each other with no intervening operator, so no special notation like [ab] is needed to indicate that ab is a dyadic. The object abT on the other hand has a well-defined meaning in matrix algebra, abT = (b1 b2) = = a matrix // same as matrix ab = ab (E.5.3) and one sees that in fact (ab)ij = (abT)ij = aibTj = aibj . (E.5.4) Meanwhile, the object aTb is just a number, aTb = (a1 a2) = a1b1 + a2b2 . . (E.5.5) This transpose notation can then be applied to the dyadic expansion of a 2x2 matrix A, A = Σij αij bibj = Σij αij bibjT = α11 b1 b1T + α12 b1 b2T ... . (E.5.6) In the special case that the bi are the unit vectors ui , and assuming N = 2 dimensions, one has A = Σnm Anm unum = Σnm Anm unumT = A11 u1 u1T + A12 u1 u2T + A21 u2 u1T + A22 u2 u2T = a matrix with A12 in the upper right corner (E.5.7) where un is a column unit vector and unT is the corresponding row unit vector (see comment starting Section 3.4 about "unit" vectors). For example, u1u2T= ( 0 1) = . (E.5.8) Obviously this matrix visualization is valid for any dimension N, not just N=2. For rank n > 2, however, this transpose-of-vector concept does not conveniently generalize. For n=3 the object uaubuc would be a cube of zeros with a single 1 located at coordinates a,b,c, and so on for n > 3. One cannot write this as uaubucT for example. The direct product or polyadic notation seems clearest for rank n > 2. 2. Application to both vector types. Suppose in (E.5.1) we want the first vector to be a covariant vector. Then we generalize the dyadic and matrix notation in this manner : (b)ij ≡ ibj = i (bT)j = (bT)ij (E.5.1)' or b = bT . (E.5.2)' and similarly for the two other cases a and . How would these last equations be translated into Standard Notation? (ab)ij ≡ aibj = ai (bT)j = (abT)ij (E.5.1)' or [ ab = abT]SN,ut (E.5.2)' [abT]SN,ut = (a1 a2) This is a pretty ugly notation. At least it gives the right covariant dot product a b = [abT]SN,ut = (a1 a2) = aibi So here is the ugliness we end up with in Standard Notation (ab)ij ≡ aibj = ai (bT)j = (abT)ij [ab = abT]SN,up . (ab)ij ≡ aibj = ai (bT)j = (abT)ij [ab = abT]SN,ut . (ab)ij ≡ aibj = ai (bT)j = (abT)ij [ab = abT]SN,dt (ab)ij ≡ aibj = ai (bT)j = (abT)ij [ab = abT]SN,dn a b = aibi = (ab)ii = (ab)ii = (abT)ii = (abT)ii I have no way to write a b in dyadic or matrix notation without using indices!! ************************************************* (ab)ij ≡ aibj = ai (bT)j = (abT)ij (E.5.1) or ab = abT . (E.5.2) From Section E.1 I know that, (ab)ij = aibj and in (E.1.5) I would write (ab) (a'b') = aibj a'i b'j = aa' bb' = a number This seems OK. Now what does this look like in dyadic notation? (ab) (a'b') = aibj a'i b'j = aa' bb' which still seems just fine. The problem I think arises when I try the matrix notation. (abT) (a'b'T) = dot product of two matrices?? In the matrix world, the above dot product has to mean this, (abT)ij (a'b'T)ij = aa' bb' and (abT)ii = ab Again, you MUST show indices to get a dot product in the matrix notation. Is that so bad? *********************** try again ******************************* E.5 Matrix notation for dyadics Consider the following matrix interpretation of the dyadic product, (ab)ij ≡ aibj = ai (bT)j = (abT)ij If we think of b as a column vector with components bj, then the corresponding row vector bT contains those same elements bj, so we set (bT)j = bj. The object (E.5.1) or ab = abT . (E.5.2) Here one knows that ab is a "dyadic" because there is no other meaning for two bolded column vectors abutting each other with no intervening operator, so no special notation like [ab] is needed to indicate that ab is a dyadic. The object abT on the other hand has a well-defined meaning in matrix algebra, abT = (b1 b2) = = a matrix // same as matrix ab = ab (E.5.3) Here we show the indices down on both the a and b components, but in fact they must match the index positions on *************************** What is wrong with using the bar in Standard Notation? (ab)ij ≡ aibj = ai (bT)j = (abT)ij [ab = abT]SN,up . (b)ij ≡ aibj = ai (bT)j = (bT)ij [ab = abT]SN,ut . (a)ij ≡ aibj = ai (bT)j = (aT)ij [ab = abT]SN,dt ()ij ≡ aibj = ai (bT)j = (T)ij [ab = abT]SN,dn a b = aibi = (ab)ii = (ab)ii = (abT)ii = (abT)ii We then have abT = (b1 b2) bT = (b1 b2) and so on. This is really the same bar used everywhere in dev notation. Then I can write ab = abT b = bT a = aT = T and then these are shorthands for the four cases. In all the above cases we have a matrix on the right side, and a "dyadic" on the left side, whatever that means. Well, really both sides of each equation are matrices. There is no other way it can be. We just write the matrix in "dyadic notation" or in "matrix notation". What is wrong with this idea? We can do the covariant dot product in this manner, a b = aibi = a T = aibi = bT and now I have an "index-free" way to write the dot product. Then I think this is OK, A = Σij A'ij eiej = Σij A'ij eiejT eiejT = ( (ej)1 (ej)2 ) Then Aab = Σij A'ij (eiej)ab = Σij A'ij (ei)a(ej)b I am beginning to think that the matrix approach is just plain wrong! The dyadic and cross product notations seem OK. But then is there some way for me to rescue my Hilbert Space idea? Aab = A'ij (ei)a(ej)b = (ei)a A'ij (ej)b Where is my "operator A" if I don't use any matrices? A = Σij A'ij eiej = Σij A'ij eiej = operator in direct product space A'ij = A eiej = Aab (ei)a (ej)b = coefficient in above expansion What is the difference between "operator" and "expansion" ? Go to the vector case A = Σi A'i ei = Σi A'i ei = operator in direct product space A'i = A ei = Aa (ei)a = coefficient in above expansion But really A is a Hilbert Space vector in the first line, it is not a Hilbert Space operator!!! In the general case, A is a vector in the direct product Hilbert Space, it is not an operator in that space. How Must Appendix E be changed? Again, I think everything is fine up to Section E.5. How would I continue the appendix ? Maybe I just eliminate E.5 completely! Or maybe I restrict sections E.5, E.6 and E.7 to Cartesian Space! Let's try a new E.5 *********************************8 E.5 Matrix notation for dyadics (Cartesian Space) In Cartesian space g = 1 so for any vector a, we have ai = ai. If we think of b as a column vector with components bj, then the corresponding row vector bT contains those same elements bj, so we set (bT)j = bj. Therefore one can express the dyadic product in this more down-to-earth manner, (ab)ij ≡ aibj = ai (bT)j = (abT)ij (E.5.1) or ab = abT . (E.5.2) Here one knows that ab is a "dyadic" because there is no other meaning for two bolded column vectors abutting each other with no intervening operator, so no special notation like [ab] is needed to indicate that ab is a dyadic. The object abT on the other hand has a well-defined meaning in matrix algebra, abT = (b1 b2) = = a matrix // same as matrix ab = ab (E.5.3) and one sees that in fact (ab)ij = (abT)ij = aibTj = aibj . (E.5.4) Meanwhile, the object aTb is just a number, aTb = (a1 a2) = a1b1 + a2b2 = a b . . (E.5.5) This transpose notation can then be applied to the dyadic expansion of a 2x2 matrix A, A = Σij αij bibj = Σij αij bibjT = α11 b1 b1T + α12 b1 b2T ... . (E.5.6) In the special case that the bi are the unit vectors ui , and assuming N = 2 dimensions, one has A = Σnm Anm unum = Σnm Anm unumT = A11 u1 u1T + A12 u1 u2T + A21 u2 u1T + A22 u2 u2T = a matrix with A12 in the upper right corner (E.5.7) where un is a column unit vector and unT is the corresponding row unit vector. For example, u1u2T= ( 0 1) = . (E.5.8) Obviously this matrix visualization is valid for any dimension N, not just N=2. For rank n > 2, however, this transpose-of-vector concept does not conveniently generalize. For n=3 the object uaubuc would be a cube of zeros with a single 1 located at coordinates a,b,c, and so on for n > 3. One cannot write this as uaubucT for example. The direct product or polyadic notation seems clearest for rank n > 2. In associating of M = abT with a rank-2 tensor, we must limit our interest to Cartesian space where g = 1. That is because abT is the single matrix shown above in (E.5.5). If g ≠ 1, then there are four different matrices Mij, Mij, Mij and Mij and the notation M = abT cannot support this fact. When g = 1 all four matrices are the same matrix. E.6 Large and small dots used with dyadics (Cartesian Space) Sometimes a small-size dot • is used to indicate the action of a dyadic (matrix) on a vector. If A is a dyadic (same symbol for matrix), and if c and d are vectors, then one defines: A • c ≡ Ac = a column vector => (A • c)i = (Ac)i = Aijcj => A • c = ΣijAijcj ui c • A ≡ cTA = a row vector => (c • A)i = (cTA)i = cjAji => c • A = ΣijcjAji ui d • A • c = dTAc = a number = diAijcj . (E.6.1) It then follows that, for the particular dyadic A = ab , (ab) • c ≡ (ab) c = (abT)c = a(bTc) = a (b c) = (b c) a = a column vector c • (ab) ≡ cT (ab) = cT(abT) = (cTa) bT = (c a) bT = a row vector d • (ab) • c = dT (ab) c = dTabT c = (dTa)( bT c) = (d a)(b c) = a number . (E.6.2) Here is more detail on the first line of the above group showing a skeletal matrix structure, (ab) c = (a bT)c = abTc = a(bTc) = a(bc) { } = {(b1 b2)} = (b1 b2) = { (b1 b2) } = bc (E.6.3) The same small dot is used to indicate the product of two dyadics, which is to say, matrix multiplication A•B ≡ AB . (E.6.4) Regarding this small size dot • : (1) from a matrix algebra point of view, it is completely superfluous except in the case c • A ≡ cTA ; (2) it is completely different from the dot used in bTc = bc. The next section provides an explanation of the small dot as part of an operator interpretation for dyadics. E.7 Operators and Matrices for Rank-2 tensors: the bra-ket notation (Cartesian Space) We continue with the assumption that x-space is Cartesian, g = 1. The discussion below applies only to rank-2 tensors. One of our goals is to make contact with the dyadic discussion of Morse and Feshbach, one of the few places dyadic notation appears in a general mathematical physics book. Operator concept. As discussed in Section 5.10, x-space and x'-space of Picture A are both N-dimensional real Hilbert Spaces, with a scalar product indicated by the large dot , and one can regard V as a vector in either space. Expressed as a "vector" in x-space one can expand V as in (7.18.7) with (Vb')i ≡ [V(b)]i , V = Σi [V(b)]i bi . // [V(b)]i are the coefficients of this expansion . (E.7.1) Moreover, one can regard a rank-2 tensor A as an "operator" in this Hilbert space, using notation (E.5.6), A = Σij [A(b)]ij bibjT . (E.7.2) In (E.2.10) and (E.5.6) the coefficients in these two expansions were called αi and αij, so here we are providing more descriptive names for these coefficients. Application of (bT)n on the left of (E.7.2) and bm on the right, and then a double use of (bT)nbi = bn bi = δni ( see 7.18.1) gives, [A(b)]nm = (bn)T A bm . (E.7.3) Here, one regards A as an operator in the x Hilbert space, whereas [A(b)]nm is a "matrix" which is associated with the operator A in the particular bn basis. If bn = un, then [A(u)]nm = Anm which is a particular matrix for this particular un basis. More generally, [A(b)]nm = (bT)n A bm = [(bT)n]i Aij [bm]j = [bT]i Aij [bm]j which is in general a completely different matrix formed by linearly combining elements Aij. Bra-ket Notation. For the author of this document, the bra-ket notation commonly used in quantum mechanics (Paul Dirac 1939) provides a useful way to look at a rank-2 tensor A as an operator. It is true that in quantum mechanics one usually deals with infinite-dimensional Hilbert spaces and complex numbers, but the formalism applies just as well to real Hilbert spaces with finite dimensions (used for example to study "spin"). Here is how the bra-ket notation works: bi → |bi> // vector biT → <bi| // transpose vector abT (E.5.3) → |a><b| // a matrix (outer product) aTb = a b (E.5.5) → <a | b> // a number (inner product) a b = b a (7.4.14) → <a | b> = <b | a> for any a,b // real Hilbert Space V = Σi [V(b)]i bi (7.18.7) → |V> = Σi [V(b)]i |bi> // vector expansion ... [V(b)]i = bi V (7.18.7) → [V(b)]i = <bi|V> // and coefficients bi bj = δij = δi,j (7.18.3) → <bi|bj> = δi,j // orthogonality = <bj|bi> = <bi|bj> = <bj|bi> Σi bi biT = 1 (7.18.6) → Σi |bi><bi| = 1 = Σi |bi><bi| // completeness A = Σij [A(b)]ij bibjT (E.7.2) → A = Σij [A(b)]ij | bi> <bj| // tensor expansion ... [A(b)]ij = (bi)T A bj → [A(b)]ij = <bi | A | bj > // and coefficients (E.7.4) In this notation, the N |bi> are a set of basis vectors which span an N-dimensional real Hilbert Space, while <bi| span the so-called adjoint (or transpose in our case) Hilbert Space. One then refers to [A(b)]ij = <bi | A | bj > as the "matrix element of the operator A in the bi basis ". In general, |bi> and |bi> are different vectors because bi and bi are different. For example en = g'nm em and our restriction to g = 1 certainly allows g' ≠ 1. In this notation, based on what was presented earlier, one can write, Anm = <un | A | um > = the x-space components of tensor A (basis un) raise/lower with g A'nm = <en | A | em > = the x'-space components of tensor A (basis en) raise/lower with g' [A(b)]nm = <bn | A | bm > = the matrix of A in the bn basis raise/lower with w (E.7.5) In the first of these three lines, one can raise and lower indices with gab and gab on both sides of the equation, but we are taking g = 1 so up and down here do not matter. On the second line this can be done with g'ab and g'ab. Eq (7.18.4) shows that bn = wnm bm and conversely bn = wnmbm where wnm is the metric tensor g'nm one would get for some underlying transformation Fb which causes bn to be its tangent base vectors en. So, on the third line above we can raise and lower indices on each side with wab and wab where wnm = bn bm as in (7.18.3). Notice in the last three equations that the operator A between the vertical bars is the exact same operator in each case. The matrices are different not because the operator has changed, but because the basis vectors are different. [A(b)]nm are the components of a rank-2 tensor for x' = F(x) in only two cases -- those shown in the first pair of equations above. In the first case Anm are components of a tensor in x-space, and in the second case the A'nm are components of a tensor in x'-space. In a more consistent notation one might write Anm = [A(u)]nm and A'nm = [A(e)]nm . Bases are related by a transformation. Consider again, [A(b)]nm = <bn | A | bm > = (bn)T A bm = [bn]i Aij [bm]j = <bn|ui><ui|A|uj><uj|bm> . (E.7.6) We lower index m on both sides (using wab as noted above) and reverse the j tilt to get [A(b)]nm = <bn | A | bm > = (bn)T A bm = [bn]i Aij [bm]j = <bn|ui><ui|A|uj><uj|bm> . (E.7.7) One could then define the following tensor-like object, Bni ≡ [bn]i . (E.7.8) The first index on B is raised and lowered by w, while the second is raised and lowered by g, so this object is a bit like R and S in its non-tensor nature. Lowering n and raising i then gives Bni = [bn]i = (BT)in , (E.7.9) where we use the notion of the transpose of a tilted matrix described in (7.9.3). One then has [A(b)]nm = Bni Aij(BT)jm . (E.7.10) Since all the matrices are tilted the same way and summed indices are contractions, this is one of the "legal" Standard Notation matrix multiplication forms like (7.8.6) and we then write, A(b) = BABT or more precisely [A(b) = BABT ]SN,dt (E.7.11) where SN,dt means Standard Notation, down-tilt, as described below (7.8.6). The matrix equation A(b) = BABT shows that the [A(b)]nm are related to the Aij by a "congruence transformation" with a matrix Bni = [bn]i whose rows are the basis vectors bn. When bm = um , matrix B is the identity matrix, and when bm = em one has Bni = [en]i = Rni, so that B = R in this case. In Standard Notation the general R matrix is real orthogonal, [RRT= 1]SN,dt and [RT= R-1]SN,dt (see (7.9.3) and following text) so in fact one has for the bm = em basis, A(e) = BABT = R A RT = R A R-1 = R A S . (E.7.12) Specifically in this case, [A(e)]nm = RniAijSjm = RniRmjAij = A'nm . (E.7.13) which is the expected result looking at the second line of (7.5.8). Warning: Note that for general R, one does not have RRT = 1 and RT = R-1 in developmental notation (DN) unless R happens to be a pure rotation. This fact was pointed out below (7.9.3). We might then write [RRT≠ 1]DN except for a rotation. This is why we tend to avoid the RT notation in Standard Notation, but it was useful above to show that, for the down-tilt components of a mixed rank-2 tensor A, a basis change from basis un to bn can be thought of as a congruence transformation by a matrix B whose rows are the vectors bn. This is a standard concept in linear algebra where one operates in Cartesian x-space. More on bra-ket notation and its relation to the small dyadic dot. Consider the following facts, <d | A | c > = dT A c = dT [A c ] = <d |Ac > <d | A | c > = dT A c = [ dT A] c = [AT d]T c = <ATd | c> (E.7.14) where |(Ac) > = a new Hilbert space vector which results when operator A is applied to |c> = A|c> <(ATd) | = a new transpose Hilbert space vector which results when A is applied to <d| = <d|A . (E.7.15) So one has this general idea that <d | A | c > = <d |Ac > = <ATd | c> A | c > = |(Ac)> <d | A = <(ATd) | . (E.7.16) In this last line, the isolated A's are the same operator A sitting in the Hilbert space. This operator can "act" either to the right or to the left as shown. The object |(Ac)> ≡ |e> is some different vector in the Hilbert space (different from |c>), call it |e>, and the grouping (Ac) labels this vector. Similarly, <(ATd) | is some vector <f| in the transpose Hilbert space. The distinction between A as an abstract operator in the Hilbert space, and the A in (Ac) and (ATd) = (dTA)T as vectors in the Hilbert space is a subtle one. It is just this distinction that is implied by the small dot in the dyadic notation discussed in the previous section, and here is the correspondence between the dyadic notation and the bra-ket notation: A • c = Ac d • A = (ATd)T = dTA d • A • c = dTAc A • B c A | c > = |Ac> <d | A = <(ATd)| <d | A | c > = <d | Ac > AB| c > . (E.7.17) In the rightmost column operator B is applied first to |c> to get vector |(Bc)>, and then operator A is applied to |(Bc)> to give yet another vector | (ABc)>. In bra-ket notation the product of two abstract operators is given just as AB, but in dyadic notation it is written A • B. Dyadics as operators. According to the above discussion, one can regard a dyadic (AB), being a rank-2 tensor, as an operator and not as a matrix. The matrix Tnm = (AB)nm = AnBm is specific to the un basis in x-space (again, one might have g ≠1) Tnm = (AB)nm = <un |(AB)| um > = (un)T A BT um = [(un)T]a Aa (BT)b [um]b = δna Aa Bb δmb = AnBm . (E.7.18) In the generic bn basis one has [(AB)(b)]nm = <bn |(AB)| bm > . (E.7.19) It is to emphasize this operator view of a dyadic that Morse and Feshbach use fancy letters like U to represent dyadics. Then their small-dot notation U • B emphasizes the idea of an operator acting on a vector, equivalent to U| B>. Here are a few samples from their Section 1.6 on dyadics. Under each clip we have tried to relate the stated equation(s) to our notation above. Like us, Morse and Feshbach are working in a Cartesian space, so up and down indices are the same for x-space objects. We will therefore lower our indices to make the comparisons below as close as possible. Note that un = un when g = 1. The M&F symbols U and an are our A and un : p 55 A • b = Σmn um Amn bn b • A = Σmn bm Amn un (E.6.1) or A |b> = Σmn |um><um|A|un><un|b> <b|A = Σmn<b|um><um|A|un><un| (E.7.20) p 55 A = Σnm unAnm um (E.5.7) (E.7.21) A-1 • A = A • A-1 = 1 // A-1 defined if matrix [A(u)]mn = Amn is invertible (E.7.22) p 57 0 • f = 0 1 • f = f ; 1 = Σn ununT (E.7.4) with bn = un or 0 | f> = |0> ; 1 | f> = | f> ; 1 = Σn | un><un| (E.7.4) (E.7.23) p 59 Us = Σnm [Us]nm unum = [Us]11u1u1 + [Us]12u1u2 + ... = [Us]11 i i + [Us]12i j + .. where i = u1, j = u2 , k = u3 and where [Us]ij = = a symmetric matrix (E.7.24) Notice the impressive name "idemfactor" for the identity operator 1 = Σi | ai><ai| = Σi aiaiT = Σiaiai where Σiaiai uses the dyadic notation (E.5.2) . Comment: A favorite bra-ket notation trick is to obtain useful results by inserting 1 = Σn | un><un | in opportune places, as in (E.7.20) above : A |b> = 1 A 1 |b> = [Σm | um><um | ] A [Σn | un><un | ] |b> = ΣnΣm | um><um |A| un><un |b> = ΣnΣm | um> Amn bn = Σnm Amn bn |um> (E.7.25) The vector A |b> = |Ab> has thus been expanded on the basis | um>.