Unitary representation of dilatations in Minkowski space Consider a field theory in Minkowski space (Lorentzian signature). I'm looking at the Dilatation operator $D$ which presumably satisfies:
$$[D,\phi(0)]=\tilde{\Delta}\phi(0)$$
where $\phi(0)$ is a Hermitian operator located at the origin of space-time. If $D$ is Hermitian then $\tilde{\Delta}$ must be real: if $|d\rangle$ is an eigenvector of $D$ with eigenvalue $d$ then
$$D \phi(0)|d\rangle = (\tilde{\Delta}+d) \phi(0) |d \rangle$$
so $\tilde{\Delta}+d$ is an eigenvalue of $D$ and $\tilde{\Delta}$ must be real. But then I get that exponentiating the dilatation operator gives me
$$e^{i D \lambda} \phi(0) e^{-i D\lambda} = e^{i \tilde{\Delta}\lambda} \phi(0)\,.$$
I was under the impression that if $U(\lambda)$ generates dilatations we want it to satisfy
$$U(\lambda) \phi(0) U^{-1}(\lambda) = e^{\Delta \lambda} \phi(0)\,.$$
My conclusion would then be that the latter definition of dilatations is a non unitary transformation. Is this correct? I have never seen this discussed in the literature.
 A: In Minkowski signature $D$ is anti-Hermitian. Indeed, consider a real scalar $\phi$, then we have
$$
[D,\phi(x)]=(x\cdot \partial_x+\Delta_\phi)\,\phi(x)
$$
where the right-hand side is a Hermitian operator. Therefore, we should have a Hermitian operator on the left, which is only possible when $D$ is anti-Hermitian. I.e.
$$
[D,\phi]^\dagger=[\phi^\dagger,D^\dagger]=-[D^\dagger,\phi].
$$
This solves your problem because the finite unitary rescaling is then $e^{\lambda D}$.
However, this leads to an apparent contradiction: the state
$$
|\psi\rangle = \phi(0)|0\rangle
$$
is an eigenstate with of $D$ with real eigenvalue $\Delta_\phi$, even though the operator is anti-Hermitian, which would suggest pure imaginary eigenvalues. The resolution is that the way one proves the latter is by considering the expectation value of $D$ in $|\psi\rangle $. This doesn't work in this situation because $|\psi\rangle $ is not normalizable,
$$
\langle\psi|\psi\rangle =\langle 0|\phi(0)\phi(0)|0\rangle=undefined.
$$
In fact, the Hermitian operator which has the discrete spectrum of eigenvalues equal to operator dimensions in Minkowski signature is not $D$ but the so-called conformal Hamiltonian $H=P^0+K^0$. To see this, note that this operator corresponds to the vector field
$$
v_\mu=-\delta^0_\mu+2x_\mu x^0-x^2\delta^0_\mu.
$$
The points $x^\mu=(\pm i,0,0,\ldots)$ are fixed by this vector field. Thus this operator satisfies
$$
[H,\phi(-i,0,0,\ldots)]\propto \Delta_\phi\phi(-i,0,0,\ldots),
$$
where the constant of proportionality is real. Note that the state
$$
|\phi\rangle\equiv \phi(-i,0,0,\ldots)|0\rangle=e^{-P^0}\phi(0)|0\rangle
$$
is well-defined, because $P^0$ is energy and thus bounded from below, and thus has finite norm
$$
\langle\phi|\phi\rangle=\langle 0|\phi(i,0,0,\ldots)\phi(-i,0,0,\ldots)|0\rangle < \infty.
$$
Therefore, it gives a well-defined eigenstate of $P^0+K^0$.
