Retirement of old Section A.11 on Dirac Alt REVIEWED

Dirac Alt idea PhL 1.11.15 5/16/16. I actually had this idea installed as an Appendix but then decided it was not very useful. I have other docs on this same subject somewhere. Trying to do something with this: < Alt f | 1,2...k> = < f | Alt | 1,2...k> = <f | Alt 1,2...k> . A.11 The Alt Operator in Dirac bra-ket notation Consider this example of (A.2.1) and its Dirac notation equivalent, g(i1,i2) = [ Alt(f)](i1,,i2) = (1/2) [ f(i1,i2) - f(i2,i1)] (A.11.1) <g | i1,i2> = <Alt f | i1,i2> = (1/2) [ <f | i1,i2> - <f | i2,i1> ] (A.11.2) where, as in Section A.10, the integers i1 and i2 are stand-ins for vectors having i1 and i2 as labels. Using the notation of (2.11.g.10) and (2.11.g.10) we write this as <Alt f | i1,i2> = <f | AltT | i1,i2> where AltT is the transpose of some Dirac space operator Alt. Comparing with (A.11.2) we then have AltT | i1,i2> = (1/2) [ | i1,i2> - | i2,i1> ] Continuing to follow Section (2.11) (g) we would then write AltT | i1,i2> = | (AltT) i1,i2> = (1/2) [ | i1,i2> - | i2,i1> ] In vector notation the last equation appears as (AltT) = (1/2) [ - ] = (1/2) // meaning e.g. (1/2) so (AltT) = (1/2) = (Alt) and now we have a matrix representation for Alt where AltT = Alt. STOP. I don't like this at all. I am replacing VV with VV and it is just illogical. I was trying to get a matrix for Alt, but this is stretching things too far. But if I just say Alt is an operator, then AltT has no meaning if there is no corresponding matrix. I would have to say instead. <Alt f | i1,i2> = <f | Alt | i1,i2> Alt | i1,i2> = (1/2) [ | i1,i2> - | i2,i1> ] = definition of the Dirac space Alt operator Then I would have to claim <f | Alt | i1,i2> = < Alt f | i1,i2> = < g | i1,i2> But that does not seem much of a claim since we already have written g = [Alt(f)] . So I think this whole section is pointless. For archiving, I will save the old section below ****************************** A.10 The Alt Operator in Dirac bra-ket notation We won't make use of what follows, but it seems a reasonable way to incorporate operators like Alt and Sym into the Dirac bra-ket notation. In the above discussion, when we say g = Alt(f), what we mean is that, g(1,2...k) = Alt(f)(1,2...k) = (1/k!) ΣP(-1)S(P) f(P(1),P(2)....P(k) ) . (A.10.1) The stripped down statement g = Alt(f) is somewhat vague because there are no visible labels for Alt to act upon. We can give g = Alt f a more concrete meaning using the Dirac bra-ket notation. Write, g(1,2...k) = <g | 1,2...k> <g | = functional (A.10.2) Alt(f)(1,2...k) = < Alt f |1,2...k > < Alt f | = functional (A.10.3) Here |1,2...k > is a ket in some generic permutation space Gk . Next we define Alt |1,2...k> ≡ (1/k!) ΣP(-1)S(P) |P(1),P(2)...P(k)> (A.10.4) < Alt f | 1,2...k> ≡ <f | Alt 1,2...k> . (A.10.5) In (A.10.5) we are pretending that Alt is a symmetric (self-adjoint) operator AltT = Alt which can be swung for free from the bra space to the ket space. It is just notation. Then we can interpret the stripped statement g = Alt(f) to mean < g | = < Alt f | . (A.10.6) When both sides are closed with the ket |1,2,...k> we get the intended result, < g | 1,2,...k> = < Alt f |1,2...k> = <f | Alt 1,2...k> = <f | [(1/k!) ΣP(-1)S(P) |P(1),P(2)...P(k)>] = (1/k!) ΣP(-1)S(P) <f | P(1),P(2)...P(k)> (A.10.7) which then says g(1,2..k) = (1/k!) ΣP(-1)S(P) f(P(1),P(2)....P(k) ) . (A.10.1) (A.10.8) In our two applications we have tensor: | 1,2...k> = | ei, ei ...ei> <T| 1,2...k> = <T|ei, ei ...ei> = Tii...i (A.10.9) tensor function: | 1,2...k> = | vi, vi ...vi> <T| 1,2...k> = <T|vi, vi ...vi> = T(vi, vi ...vi) . (A.10.10) Example: Prove that Alt(Alt(f)) = Alt(f) in Dirac notation. That is, show < Alt Alt f | = <Alt f |. Proof: Let | i1,i2...ik> be an arbitrary permutation of | 1,2...k> ( an arbitrary ket in Gk ) < Alt Alt f | i1,i2...ik> = < Alt f | Alt i1,i2...ik> = < Alt f | (1/k!) ΣP(-1)S(P) iP(1),iP(2)...iP(k)> = <f | Alt (1/k!) ΣP(-1)S(P) iP(1),iP(2)...iP(k)> = (1/k!) ΣP(-1)S(P) <f | Alt iP(1),iP(2)...iP(k)> = (1/k!) ΣP(-1)S(P) <f | (1/k!) ΣQ (-1)Q iQP(1),iQP(2)...iQP(k)> = (1/k!)2 ΣP <f | ΣQ (-1)S(QP) iQP(1),iQP(2)...iQP(k)> = (1/k!)2 ΣP <f | ΣQ (-1)S(Q) iQ(1),iQ(2)...iQ(k)> // rearrangement theorem (A.1.3) = (1/k!)2 (k!) <f | ΣQ (-1)S(Q) iQ(1),iQ(2)...iQ(k)> // ΣP(1) = k! = <f | Alt i1,i2...ik> = <Alt f | i1,i2...ik> (A.10.11) Since this is true for all kets in Gk we conclude that < Alt Alt f | = <Alt f |. QED