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Lorentz transforming a momentum eigenstate

  1. Oct 20, 2015 #1
    Let's take a quantum state ##\Psi_p##, which is an eigenstate of momentum, i.e. ##\hat{P}^{\mu} \Psi_p = p^{\mu} \Psi_p##.
    Now, Weinberg states that if ##L(p')^{\mu}\,_{\nu}\, p^{\nu} = p'##, then ##\Psi_{p'} = N(p') U(L(p')) \Psi_{p}##, where ##N(p')## is a normalisation constant. How to construct an unitary operator like that, or how to even see that such an operator is to be found?

    Some thoughts.
    I define ## \Lambda^{\mu}\,_{\nu} \, k^{\mu} = p^{\mu}##. Component by component:
    \begin{align*}
    \Lambda_{00} &= \gamma \\
    \Lambda_{0i} = \Lambda_{i0} &= \gamma \beta_i \\
    \Lambda_{ij} = \Lambda_{ji} &= (\gamma - 1) \frac{\beta_i \beta_j }{ \beta^2 } + \delta_{ij}.
    \end{align*}
    Now let's try ##U(\Lambda) \Psi_{k,\sigma} := \sum_{\rho} D^{(j)}\,_{\rho,\sigma}\,(\Lambda) \Psi_{k, \rho}##, where ##D(\Lambda)## is Wigner's D-matrix, which gets different representations depending on the range of the state's quantum numbers ##\sigma##: ##\sigma = j, j-1, ..., -j+1, -j##, altogether ##2j+1## possible values.

    If we take ##\Psi_{k,\sigma}## to be a scalar particle for which ##\sigma = j = 0##, ##D^{(0)}\,_{0,0}\,(\Lambda) = 1## so that ##\Psi_{p,0} = N(p) \Psi_{k,0}##. In Dirac's notation, ##\Psi_{k} = | \boldsymbol{k} \rangle = \int d^3 \boldsymbol{x} \, | \boldsymbol{x} \rangle \langle \boldsymbol{x} | \boldsymbol{k} \rangle = \int d^3 \boldsymbol{x} \, | \boldsymbol{x} \rangle e^{-i \boldsymbol{k} \cdot \boldsymbol{x}}##, and ## \Psi_p = N(p) \Psi_k = \int d^3 \boldsymbol{x} \, | \boldsymbol{x} \rangle N(p) e^{-i \boldsymbol{k} \cdot \boldsymbol{x}}##. For these to be equal, I could choose ## N(p) = \int d^3 \boldsymbol{q} \, e^{-i \boldsymbol{p} \cdot \boldsymbol{q} + i \boldsymbol{k} \cdot \boldsymbol{q}}##, so that
    \begin{align*}
    N(p) \Psi_{k} &= \int d^3\boldsymbol{q} \, \int d^3\boldsymbol{x} \, e^{-i \boldsymbol{p} \cdot \boldsymbol{q} + i \boldsymbol{k} \cdot \boldsymbol{q} - i \boldsymbol{k} \cdot \boldsymbol{x}} |\boldsymbol{x} \rangle \\
    &= \int d^3\boldsymbol{x} \, \delta^{(3)}(\boldsymbol{p} - \boldsymbol{q}) e^{-i \boldsymbol{k} \cdot \boldsymbol{x}} |\boldsymbol{x} \rangle \\
    &= \int d^3\boldsymbol{x} \, e^{-i \boldsymbol{p} \cdot \boldsymbol{x}} |\boldsymbol{x} \rangle \\
    &= \Psi_{p,0}
    \end{align*}
    However, this choice for ##N(p)## also depends on ##\boldsymbol{k}##...
     
    Last edited: Oct 20, 2015
  2. jcsd
  3. Oct 21, 2015 #2
    ##U(\Lambda) \Psi_{k,\sigma} := \sum_{\rho} D^{(j)}\,_{\rho,\sigma}\,(\Lambda) \Psi_{k, \rho}##,
    this line is not correct, unless you assume that
    ## \Lambda^{\mu}\,_{\nu} \, k^{\mu} = k^{\mu}##,
    so the result is
    ##\Psi_{k,0} = N(p) \Psi_{k,0}##.
    and ##N(p)=1##
    actually N(p) is just a constant to fix the different normalization conventions between states, and it isn't so important here.

    the logic here is: if you assume the physical states have a lorentz symmetry, there must be a representation on the hilbert space.

    if you want to construct the explict form of ##U(\Lambda)##,you can refer to 2.4 of weinberg's book.
     
    Last edited: Oct 21, 2015
  4. Oct 30, 2015 #3
    That's right, sorry, I was sloppy adding that.
    Weinberg shows that
    $$ U(\Lambda) \Psi_{p,\sigma} = \sum_{\rho} D^{(j)}_{\rho \sigma}(W(\Lambda,p)) \Psi_{\Lambda p, \rho} $$
    where ##W(\Lambda,p)## is such that ##W^{\mu}_{\nu} k^{\nu} = k^{\mu}## so such transformations form a little group for every 'class' of momentum eigenstates. Division to 'classes' is obtained by noticing which quantities stay invariant under every proper orthocronous Lorentz transformation. Three physical 'classes' are obtained: the vacuum for which ##p^2 = 0, p^0=0##, a massless particle for which ##p^2 = 0, p^0 > 0## and a massive particle ##p^2 < 0, p^0 > 0##.
    Weinberg then goes on to explaining how to obtain a ##2j+1## dimensional irreducible representation from the generators: angular momentum operators. On the other hand he gives a possible decomposition of Wigner rotation ##W## into boosts and rotations that operate on four-vectors. But he doesn't seem to connect these two types of objects, ##W## and ##D(W)##. That is, how to obtain a representation for ##W = L^{-1}(\Lambda p) \Lambda L(p)## in ##2j+1## dimensions?

    Actually, as for massive particles ##W## is a spatial rotation, I should indeed be able to construct a representation for the rotations. But, I will not know how will it be related to boosts performed on four vectors. That is, I know ##D(\alpha, \beta, \gamma)## and a way to write ##W(\Lambda,p)## but not ##D(W(\Lambda, p))##.
     
    Last edited: Oct 30, 2015
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