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Unitary Operator: Exponential Form

  1. Aug 19, 2014 #1
    1. The problem statement, all variables and given/known data
    I'm working on this problem:

    Let [itex]\hat{U}[/itex] an unitary operator defined by:
    [itex]\hat{U}=\frac{I+i\hat{G}}{I-i\hat{G}}[/itex] with [itex]\hat{G}[/itex] hermitian. Show that [itex]\hat{U}[/itex] can be written as: [itex]\hat{U}=Exp[i\hat{K}][/itex] where [itex]\hat{K}[/itex] is hermitian.



    2. Relevant equations
    [itex]\hat{U}=\frac{I+i\hat{G}}{I-i\hat{G}}[/itex] , [itex]\hat{U}=Exp[i\hat{K}][/itex]




    3. The attempt at a solution
    My attempt at a solution: I have to show who is [itex]\hat{K}=\hat{K}(\hat{G})[/itex] (as a function) so after several algebra manipulation,equating the two relevant equations I arrive to:

    [itex]\hat{G}=tan(\frac{\hat{K}}{2})[/itex]

    I would like to simply apply the inverse of tan, in that way:

    [itex]\hat{K}=2 arctan(\hat{G})[/itex]. I do not know if it is arctan defined for an operator, if it is, I think that is via taylor Series.

    Some help please
     
  2. jcsd
  3. Aug 20, 2014 #2

    ShayanJ

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    Gold Member

    Try taylor-expanding [itex] \hat U [/itex] w.r.t. [itex] \hat G [/itex]!
     
  4. Aug 20, 2014 #3
    I've tried, but i think this is not the way because when you expand both sides there is no clear relation between G and K.
     
  5. Aug 28, 2014 #4

    DEvens

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    Education Advisor
    Gold Member

    Don't forget those important words: Hermition and unitary. You probably need to use those.

    You can get there in a couple ways. Expansion by a Taylor series is one way. Be careful that you don't depend things commuting if they don't.
     
  6. Aug 28, 2014 #5
    I agree. And I used those properties. Thanks
     
  7. Sep 2, 2014 #6
    You may be over-thinking this problem.

    Simply equate Exp[i K] = (I +iG)/(I-iG) and solve by taking the log of both sides. (Note issues of phase in your equality after taking the log etc.)

    Now you have a form of K in terms of G, which you must show is Hermitian. This is where a series expansion comes into play. Expand your K as a series expansion, and show that by daggering (taking the Hermitian conjugate of) each term in the expansion you leave the expansion unchanged, thus showing that K is Hermitian, and showing the solution.
     
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