is phi star symmetric REVIEWED

Is it possible that φ* is symmetric? PhL 3.9.15 5.17.16: Most of the problem below is confusion about the covariant transpose which I think I now have under control, both in wedge doc and tensor doc! Also, I am still doing battle with φ* as an operator and not as a function. This was a long battle, I remember it now. Most of what a researcher does is wrong! 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. Perhaps if we restrict xui to i = 1,2..n then we might thing of both sides being the same space just spanned by different set of n basis vectors. We do know for example that (xui) = R(tui) i = 1..n // component transformation (xui)r = Σs=1nRrs(tui)s r = 1..m , but i = 1..n only Is there some corresponding vector transformation? I keep coming back to this notion. (xui) = Σj=1n (aij) (tuj) only for the subspace i = 1..n | xui> = Σj=1n (aij) | tuj> STOP. The above states that an x-space vector is a linear combination of t-space vectors. How can you claim that for vectors in completely different space? < tua| xui> = Σj=1n (aij) < tua| tuj> = aia But < tua| xui> = ??? = Σr=1n ( tua)r ( xui)r does not conform! I need a new notation to represent just the first part of xui = ( xu'i, xu"i) n m-n Then go back and rewrite: (xu'i) = R(tui) i = 1..n // component transformation (xu'i)r = Σs=1nRrs(tui)s r = 1..n and i = 1..n only Plan A. Let's break things up graphically. We start with the full situation and we break that up into two situations where R' is a square n x n matrix. In this new language, maybe write out some of our old equations. How about this one: φ*T | xui> = Σj=1n Rij | tuj> Consider this graphical representation, Now if I take the contravariant component of both sides, I know that (tuj)s = δjs. So if I think of the above graphics as being a contravariant vector equation, then it says Plan B. Go back to <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 this mixed basis, 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 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 Plan C. What we have here is an isolated "linear algebra problem" <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 this mixed basis, we find that the matrix elements of the operator φ* are simply (φ*)ir = Rir So in this mixed basis, it seems that we have simply φ* = R 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 and that seems impossible just based on scaling. Comment: I have some basic misunderstanding of linear algebra here. This problem has a long history with me going deep back into quantum mechanics and operators, I have encountered this confusion many times. I don't think it matters whether you use Dirac notation or vector/matrix notation. The confusion is present in any notation, but of course I don't know what that confusion really is. This problem really has nothing to do with wedge doc or differential forms. It is a modularized confusion within linear algebra itself. I have brought Chapter 10.6 to a stable form, showing all the things I wanted to show. However, I feel that some major piece of knowledge is missing which would make the section about 10x simpler if I knew what it was. For example I end up with what looks like φ*| vn,vn...vn> = | Rvn,Rvn...Rvn> . This is completely mysterious right now, but I know it will be very simple when I have it figured out. Today is March 9 and I have run out of time because I have to do the MRL taxes very soon and get it mailed to her and all that stuff. I have to buy the tax package, etc etc. So I am now shutting down wedge operations probably for 2-3 weeks. Then I can try to detangle this mystery. A related problem is to get the transpose T operator more under control, with a covariant and matrix version looking the same and various ambiguities floating around in various docs.