Results 1 to 4 of 4

Thread: Commutation Relation

  1. #1
    Member Mauritzvdworm's Avatar
    Joined
    Aug 2009
    From
    Pretoria
    Posts
    122

    Commutation Relation

    This has to be something stupid which I'm just not seeing. Let R_{\theta}(g)(z)=g(e^{2\pi i \theta}z) with g\in C(\mathbb{T}) and z\in\mathbb{T}
    Further, let M_f be the multiplication operator on B(L^2(\mathbb{T})) with f\in C(\mathbb{T})

    R_{\theta} can be extended to a unitary operator on B(L^2(\mathbb{T}))

    show that the following commutation relation holds
    R_{\theta}M_{id}=e^{2\pi i \theta}M_{id}R_{\theta}

    Where id denotes the identity function

    This is what I have:
    (R_{\theta}M_{id})(g)(z)=R_{\theta}(M_{id}(g)(z))=  R_{\theta}(id(z)g(z))=id(e^{2\pi i \theta}z)g(e^{2\pi i \theta}z)

    M_{id}R_{\theta}(g)(z)=M_{id}(R_{\theta}(g)(z))=M_  {id}(g(e^{2\pi i \theta}z))=id(e^{2\pi i \theta}z)g(e^{2\pi i \theta}z)

    but here the commutation relation does not hold. Am I using the definitions wrongly?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    Quote Originally Posted by Mauritzvdworm View Post
    This has to be something stupid which I'm just not seeing. Let R_{\theta}(g)(z)=g(e^{2\pi i \theta}z) with g\in C(\mathbb{T}) and z\in\mathbb{T}
    Further, let M_f be the multiplication operator on B(L^2(\mathbb{T})) with f\in C(\mathbb{T})

    R_{\theta} can be extended to a unitary operator on B(L^2(\mathbb{T}))

    show that the following commutation relation holds
    R_{\theta}M_{id}=e^{2\pi i \theta}M_{id}R_{\theta}

    Where id denotes the identity function

    This is what I have:
    (R_{\theta}M_{id})(g)(z)=R_{\theta}(M_{id}(g)(z))=  R_{\theta}(id(z)g(z))=id(e^{2\pi i \theta}z)g(e^{2\pi i \theta}z)

    M_{id}R_{\theta}(g)(z)=M_{id}(R_{\theta}(g)(z))=M_  {id}(g(e^{2\pi i \theta}z))=id(e^{2\pi i \theta}z)g(e^{2\pi i \theta}z)

    but here the commutation relation does not hold. Am I using the definitions wrongly?
    The calculation for (R_{\theta}M_{\text{id}})(g)(z) is correct. The other one is not. The operator M_{\text{id}} takes a function and multiplies its value at z by the independent variable z. So M_{\text{id}}R_{\theta}(g)(z)=M_{\text{id}}(R_{\th  eta}(g)(z))=M_{\text{id}}(g(e^{2\pi i \theta}z))=zg(e^{2\pi i \theta}z).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member Mauritzvdworm's Avatar
    Joined
    Aug 2009
    From
    Pretoria
    Posts
    122
    So then (R_{\theta}M_{id})(g)(z)=e^{2\pi i \theta}zg(e^{2\pi i \theta})
    and
    (M_{id}R_{\theta})(g)(z)=zg(e^{2\pi i \theta}z)
    in which case the commutation relation holds

    The only part that is bothering me is the multiplication operator M_{id} or more generally M_{f},f\in C(\mathbb{T})

    The way I have it is that it is defined in the following way
    M_{f}(g)(z)=f(z)g(z)

    now in the second case we had
    M_{f}(g)(e^{2\pi i \theta}z)
    so why do we not use the rotated coordinate but just z?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    Quote Originally Posted by Mauritzvdworm View Post
    So then (R_{\theta}M_{id})(g)(z)=e^{2\pi i \theta}zg(e^{2\pi i \theta})
    and
    (M_{id}R_{\theta})(g)(z)=zg(e^{2\pi i \theta}z)
    in which case the commutation relation holds

    The only part that is bothering me is the multiplication operator M_{id} or more generally M_{f},f\in C(\mathbb{T})

    The way I have it is that it is defined in the following way
    M_{f}(g)(z)=f(z)g(z)

    now in the second case we had
    M_{f}(g)(e^{2\pi i \theta}z)
    so why do we not use the rotated coordinate but just z?
    The definition of M_f is M_f g(z) = f(z)g(z). Now suppose you have a function h(z) defined by h(z) = g(e^{2\pi i \theta}z). It is still true that M_f h(z) = f(z)h(z) = f(z)g(e^{2\pi i \theta}z). In this case h = R_\theta g so that h(z) = R_\theta g(z).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. gl(3) commutation relations question
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Sep 8th 2011, 05:59 PM
  2. Replies: 1
    Last Post: Apr 6th 2011, 11:46 PM
  3. Commutation relation of operators
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Aug 2nd 2010, 01:48 AM
  4. fourier transform/commutation relation
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Apr 12th 2010, 02:18 PM
  5. Commutation relation - angular momentum operator
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: Apr 26th 2009, 11:57 PM

Search Tags


/mathhelpforum @mathhelpforum