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scratch junk Maybe better to write this as Σn( n)a (bn)b = Σn( Bn )a (bn)b = Σn ac(Bn)c (bn)b = ac (Bn)c (bn)b So then completeness says ac (Bn)c (bn)b = δa,b and I avoid having floating around here. But this just does not look nice to me. There is no way to write this in matrix form. But suppose I apply gda to both sides. I get gda ac (Bn)c (bn)b = gdaδa,b or δdc (Bn)c (bn)b = gdb (Bn)d (bn)b = gdb or (Bn)a (bn)b = gab Maybe THIS is what I should be calling "completeness" ? The matrix form would be Σn Bn bnT = gdn and in Standard notation Σn bn bnT = gup Can I reverse the n tilt here? Feeding time. Back. How might I re-analyze the analogy to the bra-ket notation? I may have made an error again with the notation ion Appendix E . Lets review E.1 All OK, used covariant notation. E.2 OK down to (E.2.11), and I retain my red commend about trying to tilt the other way, put that off for the moment. Orthonormal Basis section: seems OK as is. I think all of E.2 is OK. E.3 Polyadic notation. Seems all OK, not dot products appear. E.4 OK, not dot products, keeping up and down indices in right places. E.5. Perhaps first signs of trouble. First problem is in (E.5.6) where I say aTb = a b = (a1 a2) = a1b1 + a2b2 = a scalar (if a and b are vectors) . (E.5.5) This is NOT the covariant dot product!!! Beware!! I can fix this section simply by replacing the above line with this aTb = (a1 a2) = a1b1 + a2b2 . (E.5.5) Yes, that rescues the entire Section E.5 so install that change right now. E.6 Now more big trouble with dot notation! I have highlighted in red chunks that are problematical, but most of it survives for now. E.7 Here is where I try to get bra-ket going. The big problem is this: Σi i biT = 1 (7.18.6) → Σi |bi><bi| = 1 = Σi |bi><bi| // completeness Let's try this rewrite of the above Σi i biT = 1 (7.18.6) → Σi |i><b| = 1 // completeness abT = (b1 b2) The thing on the left in abT is the column vector which is the |a> thing. I don't like the above idea because I would like to see the regular vector |bi> appear. Question 1: What happens if I transpose both sides of the left matrix equation? [ Σi i biT]T = 1T = 1 Σi ( i biT)]T == 1 Σi bi ( i)T == 1 This still is a matrix on the left in this sense abT = (b1 b2) = = a matrix // same as matrix ab = ab (E.5.3) Now lets try this connection Σi bi ( i)T == 1 → Σi |bi><i| = 1 The thing on the left in abT is the column vector which is the |a> thing. So at least in the above form I get the normal ket on the left which is |bi>. Question 2: Can I move the bar to the other term? Go way back to this Σn( n)a (bn)b = δa,b // completeness relation for the dual set {bn, Bn } (6.2.15) Go back to notes above: I am wondering now about my completeness thing Σn( n)a (bn)b = δa,b Σn n bnT = 1 Maybe better to write this as Σn( n)a (bn)b = Σn( Bn )a (bn)b = Σn ac(Bn)c (bn)b = ac (Bn)c (bn)b So then completeness says ac (Bn)c (bn)b = δa,b Now play with the last equation ac (Bn)c (bn)b = δa,b (Bn)c ac (bn)b = δa,b But this does NOT produce (Bn)c n = δa,b. So the answer is: you cannot just move the bar to the other term! Plan A. Let's try to construct the whole analogy from scratch assuming this connection from above, Σi bi ( i)T == 1 → Σi |bi><i| = 1 Here we go: bi → |bi> // vector biT → <bi| // transpose vector abT (E.5.3) → |a><b| // a matrix (outer product) aTb (E.5.5) → <a | b> // a number (inner product) a b = aibi → V = Σi [V(b)]i bi (7.18.7) → |V> = Σi [V(b)]i |bi> // vector expansion ... The next item is suspicious, since I have a covariant dot on the left side: [V(b)]i = bi V (7.18.7) → [V(b)]i = <bi|V> // and coefficients How can I fix this up? bi V = (i)a Va STOP and go back to dev notation Σn( n)a (bn)b = δa,b // completeness relation for the dual set {bn, Bn } (6.2.15) How do you translate this equation to standard notation?? Rule 1: n → bn Rule 2: contravariant indices go up We then get Σn( bn)a (bn)b = δab = δa,b = gab This looks good! Then in matrix notation this becomes Σn( bn)(bn)T = 1 and this is MUCH better. So what about (7.18.7) now? (7.13.1) Σn(n)i(en)j = δi,j Σn (bn)i(bn)j = δi,j // completeness (7.18.5) (7.13.1) Σn n enT = 1 Σn bn bnT = 1 // completeness (matrix form) (7.18.6) Remember from (7.13.1) that (n)i →(en)i , so the label goes up, the index stays down!!!