Hello(adsbygoogle = window.adsbygoogle || []).push({});

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

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Lorentz-Transforming a Newton-Wigner State

**Physics Forums | Science Articles, Homework Help, Discussion**