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Duals in Dirac Notation

  1. Sep 10, 2008 #1
    Hi. I came across a problem in a book of mine that requires me to find the dual of a vector |x> = A |a> + B |b>. However, it's a bit sketchy about taking |x> to <x|. With a little algebra, I got

    |x>i = A |a>i + B |b>i

    So
    <x|i = |x>i*
    = (A |a>i + B |b>i)*
    = (A |a>i)* + (B |b>i)*
    = A* |a>i* + B* |b>i*
    = A* <a|i + B* <b|i

    So, finally
    <x| = A* <a| + B* <b|

    I just want to double check I'm not making any mistakes, since I'm still getting used to this wacky notation!
     
  2. jcsd
  3. Sep 10, 2008 #2

    CompuChip

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    Yep, that's right. However, what is a bit confusing about the notation, is the extensive use of the asterisk. For numbers like A, it really is conjugation: A* is the complex conjugate of A. But for ket's, like |x>, I would view an identity like: |x>* = <x| more like convenient notation, or the definition of |x>*, than as an actual conjugation.

    What you are actually doing is making a linear form (denoted by <x|) on the Hilbert space out of a vector (which is denoted by |x>). The nice thing about the notation is, that if you act with the linear form <x| on a vector |y>, you get the inner product between x and y, which is (often also outside of QM) written <x|y>. If you want to understand it better, you could start with this post of mine from an earlier thread (actually the question was about coordinate transformations, just skip over the parts that discuss those).

    PS Post 4 from this discussion is approximately the same, but more to the (this) point.
     
  4. Sep 10, 2008 #3
    What I meant by |x>i was the i-th component of the vector x. So the * would simply be Boring Old Complex Conjugation.

    I understand most of the ideas involved (a dual vector is a covector is a row vector is an operator from H -> C is isomorphic to a vector in H which is an operator from C -> H, blah balh blah). It's just that the notation just really throws me off. Especially when scalar product, inner product, outer product, operator application, and konecker product products can all written the same way! And I'm pretty sure that the bra-ket notation is inherently ambiguous from the standpoint of a formal grammar ;-0

    Thanks for setting me straight on this!
     
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