transformations in space and in dual space

Transformations in Space and in Dual Space PhL 1.16.16 1. In "regular space" we have a ket of this form | v1, v2.....vk> = |v1> |v2> ..... |vk> where the subscripts are labels, not components. This is an element of the tensor product space Vk. For the case k = 1 we have just |v1> 2. Under a transformation x' = F(x), with its v' = Rv linearization we have this transformation rule |v1'> = |Rv1> where (v1')i = Rij (v1)j ** must be true Digression: Apply this to a basis vector to get |ei'> = |Rei> = R | ei> We know that for a Type A transformation, (v1)'i = <ei | v'1> (v1)i = <ei | v1> Thus we can write (v1')i = Rij (v1)j as, <ei | v'1> = Rij <ej | v1> must be true Digression. Apply the above to a basis vector <ei | e'k> = Rij <ej | ek> = Rij δjk = Rik This is how a normal rank-1 tensor transforms as expressed in bra-ket notation. Now close ** with ei on the left to get <ei |v1'> = <ei |Rv1> must be true Digression. Apply the above to a basis vector <ei | e'k> = <ei | Rek> Therefore <ei |Rv1> = Rij <ej | v1> must be true Digression. Apply the above to a basis vector <ei | Rek> = Rij <ej | ek> = Rik or <RTei | ek> = Rij <ej | ek> or <RTei | = Rij <ej | or <ei | R = Rij <ej | We can supposedly move R from the ket vector to the bra vector by using RT so <ei |Rv1> = <RTei |v1> Then we get <RTei |v1> = Rij <ej | v1> It must then be true that <RTei| = Rij <ej| or <ei| R = Rij <ej| Digression: Close with |ek> from the right to get <ei| R |ek> = Rij <ej |ek> = Rik seems reasonable *** Transpose this equation from bra to ket to get (using result rule 2 below) | RTei > = Rij |ej > or RT| ei > = Rij |ej > Is this a correct or reasonable result?? Close with <ek| to get <ek| RT| ei > = Rik = < ei | R | ek > The right side here agrees with *** so I think all is OK. 3. Using matrix notation, we define <a | A | b> = aTAb // row vector * matrix * column vector = a scalar Since aTAb = aT(Ab) we can say <a | A | b> = <a |Ab> // aTAb = aT(Ab) Since this is true for any bra vector a, it must be that A | b> = |Ab> . // A b = (Ab) 1 Since aTATb = (Aa)T b we can say <a | AT |b> = <Aa|b> // aTATb = (Aa)T b Since this is true for any ket vector b, it must be that <a | AT = <Aa | // aTAT = (Aa)T 2 Transposing 1 we get <A b| = <b|AT . // (A b)T = bTAT Finally, we know that <a | A | b> = <a |Ab> = <Ab|a> = <b | AT |a> Here then is a summary of the bra-ket rules <a | A | b> = <a |Ab> = <ATa |b> rule 1 A | b> = |Ab> <Ab| = <b|AT rule 2 <a | A | b> = <b | AT |a> rule 3 asdf