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

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}##...

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# Lorentz transforming a momentum eigenstate

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