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Bra - ket?

  1. Oct 26, 2012 #1
    bra - ket??

    Hi, maybe a stupid question, but i would like to know if, if We have a real number, but we are i a vector space, and the operator is hermitian, is |a> is equal to < a |*?


    i assume this, because if a is the vector (1,0) (spin up), and only real entries.

    im trying to make ((Sy|a>) |b>) to ((<a|Sy)|b>) somehow..if both a and b are real..plz solve this mess

    I am talking about Sx, Sy, Sz is hermitian operators.. and these are operating..
    are these 3 hermitian conjugates to each other?
    i might add, that Sy can ONLY operate on the 'a' state ( <a| or |a>)

    and, finally, how would you write out <a|Sy?, is it a vector?
     
    Last edited: Oct 26, 2012
  2. jcsd
  3. Oct 26, 2012 #2
    Re: bra - ket??

    Can you give a little more information about what you mean when you talk about real numbers? Bras and kets are members of a Hilbert space, so the concept of "real number" doesn't mean anything when talking about them.
     
  4. Oct 26, 2012 #3
    Re: bra - ket??

    Do you mean that,

    Sy|a> = a |a> where a is real?

    <a| Sy = row vector times matrix = row vector
     
  5. Oct 27, 2012 #4
    Re: bra - ket??

    yes, i mean row vector times matrix, i would like to show that Sy|a> = <a|Sy, if Sy is hermitian..
     
  6. Oct 27, 2012 #5

    Fredrik

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    Re: bra - ket??

    If |a> is a member of a Hilbert space H, then so is Sy|a>. But <a| is a member of the dual space H*, defined as the set of continuous linear functions from H into ℂ. <a|Sy is another member of H*, so it can't be equal to any member of H, like e.g. Sy|a>.

    It would however make sense to ask if they have the same components, with respect to some basis {|i>} and its dual basis {<i|} (the latter being defined by <i|j>=δij). The ith component of <a|Sy is by definition <a|Sy|i>. (See this post for more about components of linear maps with respect to a basis). The ith component of Sy|a> is just its projection onto the ith basis vector, <i|Sy|a>.
     
  7. Oct 27, 2012 #6
    Re: bra - ket??

    another way to think of it,

    Sy|a> = <a|Sy

    Sy|a> = matrix times column vector = column vector


    <a|Sy = row vector times matrix = row vector

    it is not possible that Sy|a> = <a|Sy
     
  8. Oct 27, 2012 #7

    Fredrik

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    Re: bra - ket??

    This argument is fine when the vector space is finite.
     
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