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Dyade Dirac notation

  1. May 4, 2012 #1
    [tex]\{\vec{A},\vec{B}\}\cdot \vec{C}=\vec{A}(\vec{B}\cdot \vec{C})[/tex]
    [tex]\vec{C} \cdot \{\vec{A},\vec{B}\}=(\vec{C}\cdot \vec{A}) \vec{B}[/tex]

    I want to write dyade in Dirac notation.

    [tex](|\vec{A}\rangle\langle\vec{B}|)|\vec{C}\rangle= |\vec{A}\rangle\langle\vec{B}|\vec{C}\rangle[/tex]
    [tex]\langle\vec{C}|(|\vec{A}\rangle\langle\vec{B}|)=< \vec{C} |\vec{A}\rangle|\vec{B}\rangle[/tex]

    Why not

    [tex]\langle \vec{C} |\vec{A}\rangle\langle \vec{B}|[/tex]

    in last equation?
     
  2. jcsd
  3. May 4, 2012 #2

    dextercioby

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    It should be, it should have the same 'bra-keting' as the C in the LHS.
     
  4. May 4, 2012 #3
    I'm confused. In QM [tex]|\psi \rangle[/tex] and [tex]\langle \psi |[/tex] are vectors from some vector space and his dual respectively. But in some tensor analyses I don't see difference between [tex]|\vec{A}\rangle[/tex] and [tex]\langle \vec{A}|[/tex].
    In this case precisely. I have some number [tex]\langle|\rangle[/tex] which multiply vector.
     
  5. May 4, 2012 #4

    Fredrik

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    I don't follow what you're saying here. Yes, ##\langle C|A\rangle## is a number. Why is that a problem?

    There's no need to use the arrow notation for vectors here (unless what you have in mind is that ##\vec A## is a vector in some other vector space ##\mathbb R^3##), since kets are always vectors.
     
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