is phi star symmetric v1 REVIEWED

Is it possible that φ* is symmetric? PhL 3.9.15 I think the answer is that φ* = R acting to the left (now in (10.7.17) ) and no, it is not symmetric. I could have played more with R acting to the right, but that seemed unnecessary. At this time I was thinking of φ* as a synonym for operator R but then it became the name of a function, not an operator. In general <a| Q |b> = <a |Qb> = <QTa|b> aTQb = aT (Qb) = (QTa)Tb Then if things are real we have <a |Qb> = <Qb |a> = <b|QT|a> So if you could show that <a| Q |b> = <b| Q |a> for an entire basis, then you would know that Q = QT. Comment: I think one generally assumes that the bras and the kets are in the same space. You can have basis change stuff like <p|x> in QM, so what exactly is that "space"? In QM it is the state of a particle. So consider: φ*T | xui> = Σj=1n Rij | tuj> Then think about <tus| φ*T | xui> = Σj=1n Rij <tus| tuj> = Ris = <xui| φ* | tus> So for the entire basis in both spaces we know that <tus| φ*T | xui> = <xui| φ* | tus> = Ris = a real number I don't think you could ever say that φ*T = φ* because these two operators act on different spaces acting to the right, for example. Specifically, you can see that φ* acts on x-space to the left, but acts on t-space to the right, and φ*T does just the opposite. What I know is valid is the following: (xui) = R(tui) i = 1..n // component transformation (xui)r = Σs=1nRrs(tui)s r = 1..m , but i = 1..n only This tells you how to compute the components of a vector in one space from the components of all the vectors in the other space. I think we should avoid trying to align the spaces. For example, you could try to write the following vector statement (xui) = Σj=1n (aij) (tuj) only for the subspace i = 1..n but I think this is illogical. You should not try to write an x-space vector as a lincomb of t-space vectors. Similarly this makes no sense: < tua| xui> = Σj=1n (aij) < tua| tuj> = aia because you are doing the scalar product between two different spaces. Now go back to this item from above φ*T | xui> = Σj=1n Rij | tuj> Both sides are vectors in t-space, so consistent with itself. Consider this graphical representation of the above equation just to have it in mind, Now here is the transpose. Again both sides are t-space vectors, <xui | φ* = Σj=1n Rij <tuj | i = 1,2..m Close on the right with | tur> to get <xui | φ*| tur> = Rir = (φ*)ir xuiT So in cross-space sense, we find that the matrix elements of the matrix φ* are simply (φ*)ir = Rir So in this mixed basis, it seems that we have simply φ* = R what could possibly be simpler. Write graphically as <xui | φ*| tur> = Rir = (φ*)ir : This tells me that φ* is not some abstract operator, it is a matrix with m rows and n columns. It is therefore a linear operator. It is non-square of course. What exactly are the elements of this matrix? [φ*T] (xui) ≡ Σj=1n Rij (tuj) i = 1..m (10.6.9) (xui)T [φ*T] (tur) = Rir These are the matrix elements in this mixed basis. We know that (xui) = Rut(tui) (xui) = R(tui) (xui) = Rij(tuj) = (Rut)ij (tuj) (xui) = Rdt(tui) (xui) = R(tui) (xui) = Rij(tuj) = (Rdt)ij (tuj) so (xui)T = (tui)TRutT where T means the matrix transpose Then I can write (tui)T RutT [φ*T] (tur) = Rir (tui)T [φ*Rut]T (tur) = Rir Is this of any use? Here is the Big Paradox: Now consider my definition of φ* φ*T | xui> = Σj=1n Rij | tuj> Thus must then be the same as RT | xui> = Σj=1n Rij | tuj> Can I prove this claim directly? I know that (xui) = R(tui) or |xui> = R| tui> So then I have to show that RTR| tui> = Σj=1n Rij | tuj> or <tus| RTR| tui> = Ris Maybe rewrite as <R tus | R tui> = Ris And then use | R tui> = | xui> < R tus| = < xus| Then the above is claiming that < xus| xui> = Ris This is the first appearance of this equation in this document. Nor does this appear in Section 10-6 v1. Maybe this is the missing link. But look at un um = gnm un um = δnm un um = gnm (7.18.3)' (2.4.2) This would argue that < xus| xui> = xus xui = δsi which then conflicts with < xus| xui> = Ris !!!