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Operators - semantics

  1. Apr 20, 2005 #1
    I'm just looking at another quantum computation question. It is stated like so:

    The operators Y and Z on [tex]C^2[/tex] are defined by:

    [tex]Y|0\rangle =i|1\rangle ; Y|1\rangle = -i|0\rangle[/tex]
    [tex]Z|0\rangle = |0\rangle ; Z|1\rangle = -|1\rangle[/tex]

    Write Z in diagonal form
    Write Y in Dirac form with respect to the basis [tex]\{ 0\rangle , |1\rangle\}[/tex]

    Now, I'm confusing myself something silly. I'm under the impression that the diagonal form of an operator is given by:

    [tex]A=\sum \lambda_{n}|n\rangle\langle n|[/tex]

    where [tex]|n\rangle[/tex] are the eigenvectors and [tex]\lambda_n[/tex] are the eigenvalues of A.

    But I would also take this to be the Dirac form, so I'm clearly missing something.

    The eigenvalues of Z are clearly [tex]\{1,-1\}[/tex] with eigenvectors [tex]\{ |0\rangle ,|1\rangle\}[/tex], so the diagonal form is [tex]Z=|0\rangle\langle 0|-|1\rangle\langle 1|[/tex].

    I suppose my question breaks down to 'What is meant by the Dirac form of an operator?'

    Any hints?

    Edited to remove me being stupid and working out eigenvectors incorrectly.

    Edit: Or, by Dirac form of an operator, do they mean the matrix representation which, for Y, is given by:

    [tex]Y=\left(\begin{array}{cc}0&-i\\i&0\end{array}\right)[/tex]
     
    Last edited: Apr 20, 2005
  2. jcsd
  3. Apr 20, 2005 #2
    My guess is that by "diagonal form" they mean write Z as a diagonal matrix. And by "Dirac form" they mean write Y as a sum of projection operators as you did for Z. But I agree with you, in that I would say the "Dirac Form" version of Z written in terms of eigenvectors is also in "diagonal form".
     
  4. May 26, 2005 #3
    To anyone who's remotely interested, it transpires that given an operator, A, in matrix respresentation, then the Dirac form of the operator is:

    [tex]\hat A=\sum_{ij} A_{ij} |i\rangle\langle j|[/tex]

    and given the eigenvalues [itex]\lambda_i[/itex] and eigenvectors [itex]|\lambda_i\rangle[/itex] of an operator, the diagonal representation is (of course):

    [tex]\hat A = \sum_i \lambda_i |\lambda_i\rangle\langle\lambda_i |[/tex]

    As a 'learning point', note that the Dirac form of an operator is identical to the diagonal form for a diagonal matrix. An obvious result, but perhaps it will be useful for others.
     
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