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From Weinberg's QFT Vol.1, eq(2.5.11):

$$U(\Lambda)\Psi_{p,\sigma}=({N(p)\over N(\Lambda p)})\sum_{\sigma'}D_{\sigma'\sigma}(W(\Lambda,p))\Psi_{\Lambda p ,\sigma '}.\tag{2.5.11}$$

However, this is not a unitary representation. If we take the normalization factor $N(p)$ to be 1, then according to eq(2.5.17):

$$(\Psi_{p',\sigma '},\Psi_{p,\sigma})=\delta_{\sigma ' \sigma}({p^0\over k^0})\delta^3({\bf p}'-{\bf p}).\tag{2.5.17}$$

The basis vector isn't normalized uniformly, so $U(\Lambda)$ isn't unitary even if $D_{\sigma ' \sigma} $ is.

Conversely, if we take the normalization to be $N(p)=\sqrt{k^0/p^0}$, we have:

$(\Psi_{p',\sigma '},\Psi_{p,\sigma})=\delta_{\sigma ' \sigma}\delta^3({\bf p}'-{\bf p})$

which is uniformly normalized.

In this case, the factor $({N(p)\over N(\Lambda p)})$ remains, which again renders the transformation non-unitary.

There was a post which raised exactly the same question but left unanswered.

Why should the infinite-dimensional representation of Poincare group induced by the unitary representation of little group be unitary?

So I posted my own version.

I think it might be related to the fact that $p$ lies in a hyperbolic space, so the fourier analysis becomes different. But i can't clarify the issue.

Qmechanic
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Bababeluma
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1 Answers1

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The prominent issue can be picked out by considering a scalar representation. For the relativistically normalized kets: $\langle k\vert p \rangle=(2\pi)^32\omega_p\delta^3(k-p)$, the representation is:

$$U(\Lambda) \vert p \rangle = \vert \Omega p \rangle$$

The identity operator is:

$$\mathbb{1}=\int{d^3p\over (2\pi)^3 2\omega_p}\vert p\rangle\langle p\vert$$

The unitarity of $U(\Lambda)$ is formally proved as follow:

$$U(\Lambda)U(\Lambda)^{\dagger}=U(\Lambda)\int{d^3p\over (2\pi)^3 2\omega_p}\vert p\rangle\langle p\vert U(\Lambda)^{\dagger}=\mathbb{1}$$

The technical problem seems to arise when we consider representation in the positional space:

$$\phi (x)=\int{d^3p\over (2\pi)^3 2\omega_p}\alpha(p)e^{-ipx}$$

which means we are integrating these momentum space bases against arbitrary fourier components $\alpha(p)$. The transformation is induced by: $\phi(x)\rightarrow \phi(\Lambda x)$

$$\phi(\Lambda x)=\int{d^3p'\over (2\pi)^3 2\omega_{p'}}\alpha(p')e^{-ip'\Lambda x} \xrightarrow{p'=\Lambda p}\int{d^3p\over (2\pi)^3 2\omega_p}\alpha(\Lambda p) e^{-ipx}$$

So we have the transformation of the fourier components:

$$\alpha '(p)=\alpha(\Lambda p)=\int d^3p' \alpha(p')\delta^3(p'-\Lambda p)$$

Although the coefficients $\alpha(p)$ transform unitarily, the basis vectors aren't properly normalized. Thus, this isn't a unitary transformation.

However, a unitary representation isn't equivalent to a unitary transformation. The real basis of representation here is $\alpha(p)$, which indeed furnish a unitary representation.

The same can be said about the $\Psi_{p,\sigma}$ in Weinberg's book, they are bases of representation rather than basis vectors of a unitary transformation which need to be uniformly normalized. These are distinct concepts and the origin of confusion.

So, we have constructed a mathematical entity $\phi(x)$ which furnishes an infinite-dimensional unitary scalar representation of the Lorentz group without any resort to quantum mechanics.

This shows why and how we can combine the creation and annihilation operators, which by definition furnish a unitary representation of the Lorentz group, into the field operator.

The extension to non-scalar representation is trivial.

Bababeluma
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