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Finite and infinitesimal Unitary transformations

  1. Dec 15, 2009 #1
    Hi

    I have a question regarding unitary operators:

    If an infinitesimal operation (such as a rotation) is unitary does this guarantee that a finite transformation will also be unitary?

    thanks

    M
     
  2. jcsd
  3. Dec 16, 2009 #2

    haushofer

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    Well, how do you go from an infinitesimal transformation to a finite transformation? You use the exponential map. An infinitesimal transformation on a field is generated by the generator of the corresponding symmetry group, which is an element of the Lie algebra of that group.

    So to answer your question, you should write down in your convention an infinitesimal transformation, say that it's unitary, and then use the exponential map to make it a finite transformation. Then you should check if this expression is als unitary.

    Write an infinitesimal transformation on a field phi as

    [tex]
    \delta\phi = \Theta \phi
    [/tex]

    where Theta is an element of the Lie algebra. Your assumption is that

    [tex]
    \Theta^{\dagger} = \Theta^{-1}
    [/tex]

    Now I use the exponential mapping to generate a finite transformation (formally you should take a sum over all generators, but let's keep this easy) with the convention of using an "i":

    [tex]
    \phi' = e^{i\Theta}\phi
    [/tex]

    Now calculate:

    [tex]
    [e^{i\Theta}]^{\dagger} = e^{-i\Theta^{\dagger}}
    [/tex]

    This finite transformation is unitary if

    [tex]
    e^{-i\Theta^{\dagger}} = e^{-i\Theta}
    [/tex]

    So in this case a unitary generator means a unitary group element.

    Hope this helps :)
     
  4. Dec 16, 2009 #3

    haushofer

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    By the way, making statements about the generators when you have a group element at your disposal can be done by considering the expression

    [tex]
    \frac{d}{dt}e^{it\Theta}|_{t=0}
    [/tex]

    Because the Lie algebra is a vector space you're garanteed that [itex]e^{it\Theta}[/itex] is a group element whenever [itex]e^{i\Theta}[/itex] is.
     
  5. Dec 16, 2009 #4

    George Jones

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    [tex]
    e^{-i\Theta^{\dagger}} \left( e^{-i\Theta} \right)^{-1}= e^{-i\Theta^{\dagger} + i \Theta},
    [/tex]

    so the condition on [itex]\Theta[/itex] needed for unitarity of [itex]e^{i \Theta}[/itex] is [itex]\Theta = \Theta^\dagger[/itex] (hermiticity), not [itex]\Theta^{\dagger} = \Theta^{-1}[/itex] (unitarity).
     
  6. Dec 16, 2009 #5
    Hi

    thanks for the replies

    I guess my question is:
    I have a result for a finite transformation that is in fact a CP map
    to understand the problem better I calculated the generators of the group
    these appear to be hermitian which I thought would imply a unitary map

    does this mean that my result is wrong? or is there a theorem like
    a finite map is unitary iff the corresponding infinitesimal map is unitary?

    thanks
     
  7. Dec 16, 2009 #6

    George Jones

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    I am confused.
    An infinitesimal rotation operator (for physicists) is Hermitian, not unitary.
    It does.
     
  8. Dec 16, 2009 #7

    haushofer

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    Yes, you're right. I'm mixing up things. But I hope my point was clear though ;)
     
  9. Dec 16, 2009 #8
    The answer is no.
    As an example:
    [tex]
    \begin{pmatrix}
    \cos \theta & \sin \theta + 1-\exp(\theta^2)\\
    -\sin \theta+1-\exp(\theta^2) & \cos \theta
    \end{pmatrix}
    [/tex]
    You can obtain the infintesimal operation by expanding to first order:
    [tex]
    \begin{pmatrix}
    1 & \delta\theta\\
    -\delta\theta & 1
    \end{pmatrix} = I - i\delta\theta\begin{pmatrix}
    0 & i\\
    -i & 0
    \end{pmatrix}
    [/tex]
    The infintesimal operation is unitary - it looks just like it was generated from a hermitian generator.

    It is easy to verify that the the finite operation is not unitary.

    What is take home message? Many transformations can share the same tangent space at the identity. Only for lie groups does the exponential map generate the rest of the set of transformations.

    It is as simple as pointing out that exp(x) is not the only function with a gradient of 1 at when x = 0.
     
  10. Dec 16, 2009 #9

    Fredrik

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    When we're dealing with Lie groups whose members are matrices, there's a very simple way to express the relationship between members of the Lie group and members of the Lie algebra. A matrix X belongs to the Lie algebra if and only if exp(itX) belongs to the Lie group for all real numbers t. In this context, the exponential is defined as a power series.

    So what's the Lie algebra of U(n)? To answer this, we must find all X such that exp(itX) is a member of U(n). Such an X must obviously be an n×n matrix, and it must satisfy

    [tex]I=(e^{itX})^\dagger e^{itX}=e^{-itX^\dagger}e^{itX}[/tex]

    [tex]e^{-itX}=e^{-itX^\dagger}[/tex]

    Now take the derivative of both sides with respect to t and then set t=0 (or just expand in a series and match term by term). We get

    [tex]-iX=-iX^\dagger[/tex]

    [tex]X=X^\dagger[/tex].

    Check out this book for more about this.
     
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