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Lorentz-Transforming a Newton-Wigner State

  1. Jan 27, 2010 #1
    Hello

    I recently stumbled upon an article about Newton Wigner states (cf 10.1103/RevModPhys.21.400). It is repeatedly mentioned in the literature that acting with a Lorentz transformation on a Newton Wigner state completely delocalizes the state. However I was not able to verify this, albeit I found an article by J. Mourad (arXiv:gr-qc/9310018v1), which deals with exactly this question.

    The Newton Wigner eigenstates are given by
    [tex]\psi (x) = \sqrt{\omega (\vec{p})} e^{-i\vec{p}\cdot\vec{q_0}[/tex]
    (a state localized at position [tex]q_0[/tex] at a time t).

    Then the transformed state can be written as (in a basis [tex]{|\vec{q}\rangle}[/tex] of Newton Wigner eigenstates)
    [tex]\langle \vec{q} | \psi '\rangle = \int \frac{d^3p}{(2\pi)^{3/2}\omega} \langle \vec{q}|\vec{p}\rangle \langle{\vec{p}|\psi '\rangle[/tex]
    where [tex]\psi'[/tex] denotes the Lorentz transformed state. The integral measure has a factor of [tex]\omega[/tex] in the denominator in order to be Lorentz invariant and [tex]\langle \vec{q}|\vec{p}\rangle = (2\pi)^{-3/2} e^{i\vec{p}\cdot\vec{q}}[/tex].
    Unfortunately I do not succeed in evaluating the last inner product.
    The whole integral in Mourad reads
    [tex]\langle \vec{q} | \psi '\rangle = \sqrt{\gamma} \int \frac{d^3p}{(2\pi)^{3}} \sqrt{\left(1-\frac{\vec{\beta}\cdot\vec{p}}{\omega}\right)}e^{i\vec{p}\cdot (\vec{q}-\vec{q_0'})}[/tex]
    where [tex]\gamma^2 = 1-\beta^2[/tex].

    Any help would be highly appreciated :). Thanks
     
  2. jcsd
  3. Jan 27, 2010 #2
    Hi l-ame,

    welcome to the Forum!

    I haven't seen these kinds of integrals evaluated explicitly. The usual argument for the de-localization in the moving frame is the qualitative statement given also in the Mourad's paper:

    "The function (3.4) cannot vanish outside a bounded domain because it is the Fourier transform of a non-analytic function. This is due to the presence of square roots in the integrand."

    A similar argument is used to justify the "superlumial spreading of wave packets", i.e., de-localization due to time translation. See works by Hegerfeldt and others.

    Eugene.
     
  4. Jan 27, 2010 #3
    Hi Eugene

    Thanks for your reply.
    Unfortunately I already struggle with the derivation of formula 3.4 in Mourad's paper. I can't see where the root in the integral is coming from?
     
  5. Jan 27, 2010 #4
    In order to get the Lorentz transform of a wave function in the position space you need to perform three steps:

    1. Change to the momentum representation (Fourier transform)
    2. Lorentz transform of the momentum-representation wave function.
    3. Change back to the position representation (inverse Fourier transform).

    The step 2. can be found in eq. (2.5.23) of S. Weinberg, "The quantum theory of fields", vol. 1. This eq. shows the appearance of the square root. Note also that Mourad and Weinberg use different normalization (scalar product) conventions. In Weinberg's book momentum eigenfunctions are normalized to delta function (2.5.19). In Mourad's paper the scalar product is given by (3.2). I am in favor of the Weinberg's approach. You can find the application of this approach to the de-localization in moving frames in subsection 11.1.2 of http://www.arxiv.org/abs/physics/0504062v12

    Eugene.
     
  6. Jan 27, 2010 #5
    Great! Thank you so much for your help.

    l-ame
     
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