Scalar Product of Momentum Eigenvectors in terms of Little Group Representation

[dmoney]

I'm trying to derive the equation for the scalar product of one particle momentum eigenvectors $\Psi_{p,\sigma}$ ($p$ is the momentum eigenvalue and $\sigma$ represents all other degrees of freedom), in terms of the little group of the Lorentz group with elements $W$ that take the standard four momentum $k$ into itself, and is given in Weinberg's The Quantum Theory of Fields on page 66 before Eq. (2.1.14) as:

$$(\Psi_{p',\sigma '},\Psi_{p,\sigma}) = N(p)N^*(p')D(W(L^{-1}(p),p'))^*_{\sigma \sigma '} \delta^3(\bold{k}'-\bold{k})$$

where

$$\Psi_{p,\sigma} \equiv N(p)U(L(p))\Psi_{k,\sigma},$$

$U(\Lambda)$ is an element of the unitary representation of the homogenous Lorentz group that acts on state vectors, $\Lambda$ an arbitrary homogenous Lorentz transformation, $L(p)$ is the Lorentz transformation defined by $p\equiv L(p)k\$ that takes $k$ into $p$, $k' \equiv L^{-1}(p)p'$, $$W(\Lambda,p)\equiv L^{-1}(\Lambda p)\Lambda L(p)$$, and the $D_{\sigma\sigma '}$ form a representation of the Little group with action on an eigenvector of the standard four momentum:

$$U(W)\Psi_{k,\sigma} = \sum_{\sigma '}D_{\sigma '\sigma}(W)\Psi_{k,\sigma '}$$.

which I use in the following derivation

$$(\Psi_{p',\sigma '},\Psi_{p,\sigma})=N(p)(\Psi_{p',\sigma '},U(L(p))\Psi_{k,\sigma})$$

$$=N(p)(U^\dagger (L(p)))\Psi_{p',\sigma '},\Psi_{k,\sigma})$$

$$=N(p)(U^{-1}(L(p))\Psi_{p',\sigma '},\Psi_{k,\sigma}))$$

$$=N(p)(U(L^{-1}(p))\Psi_{p',\sigma '},\Psi_{k,\sigma}))$$

$$=N(p)N^*(p')(U(L^{-1}(p))U(L(p'))\Psi_{k,\sigma '},\Psi_{k,\sigma})$$

$$=N(p)N^*(p')(U(L^{-1}(p)L(p'))\Psi_{k,\sigma '},\Psi_{k,\sigma})$$

$$=N(p)N^*(p')(\sum_{\sigma ''}D(L^{-1}(p)L(p'))\Psi_{k,\sigma ''},\Psi_{k,\sigma})$$

Now

$$W(L^{-1}(p),p')=L^{-1}(L^{-1}(p)p')L^{-1}(p)L(p')$$

which if it were equal to $L^{-1}(p)L(p')$ would allow me to continue the derivation as

$$=N(p)N^*(p')(\sum_{\sigma ''}D(W(L^{-1}(p),p'))\Psi_{k,\sigma ''},\Psi_{k,\sigma})$$

$$=N(p)N^*(p')D(W(L^{-1}(p),p'))^*_{\sigma \sigma '}\delta^3({\bold{k'}-\bold{k})$$

Where did I go wrong?

 

[eljose79]

There are a few issues with your derivation. Firstly, the definition of the Little group is that it is the subgroup of the Lorentz group that leaves the standard four-momentum invariant, not the subgroup of the Lorentz group that takes the standard four-momentum into itself. So the equation for W(L^-1(p),p') should be W(L^-1(p),k').

Secondly, the equation U(W)Ψk,σ = ∑σ′Dσ′σ(W)Ψk,σ′ is not the action of the Little group on an eigenvector of the standard four-momentum, but rather the action of the unitary representation of the homogeneous Lorentz group on a state vector. The Little group action on an eigenvector of the standard four-momentum is given by U(W)Ψp,σ = ∑σ′Dσ′σ(W)Ψp,σ′.

Finally, in the equation N(p)N*(p')(U(L^-1(p))U(L(p'))Ψk,σ′,Ψk,σ), you have not taken into account the fact that the state vectors Ψk,σ and Ψp,σ are related by a unitary transformation U(L(p)).

With these corrections, the derivation should proceed as follows:

(Ψp',σ',Ψp,σ) = N(p)(Ψp',σ',U(L(p))Ψk,σ)

= N(p)(U(L^-1(p))Ψp',σ',U(L(p))Ψk,σ)

= N(p)N*(p')(U(L^-1(p))U(L(p))Ψp',σ',Ψk,σ)

= N(p)N*(p')∑σ''Dσ''σ'(L^-1(p)L(p))Ψp',σ'',Ψk,σ

= N(p)N*(p')∑σ''Dσ''σ'(W(L^-1(p),k'))Ψp',σ'',Ψk,σ

= N(p)N*(p')Dσ'σ(W(L^-1(p),k'))Ψp',σ',Ψk,σ

= N(p)N*(p')Dσ'σ(W(L^-1(p),k'))Ψk,σ',Ψk