I’ve read Ballentine where he derives the conserved observable operators (momentum, energy, ...) from symmetries of space-time.
Can I read up such a derivation in more detail somewhere else or even one for the Poincaré group?
And more importantly, the derivation uses gauge invariance of the wavefunction. Is there a derivation with the density matrix formalism where you wouldn’t need complex exponentials for gauge invariance?
Yes, there is:
A unitary 1-parameter group of transformation $U(s)$ satisfies $U(0)=1$, $U(-s)=U(s)^*$, and $U(s+s')=U(s)U(s')$ for all real $s,s'$. In the Heisenberg picture, $U(s)$ transforms an observable $A$ into the conjugate observable $A(s)=U(s)AU(-s)$, thereby preserving the spectrum of $A$.
The transformation is called a continuous symmetry of a quantum system if it preserves the Hamiltonian $H(t)$, i.e., $U(s)H(t)U(-s)=H(t)$ for all $s$, $t$. Differentiation with respect to $s$ gives $U'(s)H(t)U(-s)-U(s)H(t)U'(-s)=0$. Taking $s=0$ and introducing the generator $K=\dot U(0)/i\hbar$ of the symmetry, we find $K H(t)-H(t)K=0$. Thus $K$ commutes with the Hamiltonian at all times $t$. Conversely, if $K$ commutes with the Hamiltonian at all times, then $U(s)= e^{isK/\hbar}$ also commutes with the Hamiltonian at all times, so that $U(s)H(t)U(-s)=U(s)U(-s)H(t)=H(t)$. Thus a 1-parameter group is a group of symmetries iff its generator commutes with the Hamiltonian at all times.
A quantum system is called time invariant if the Hamiltonian $H$ is independent of time. In this case, $K=H$ trivially commutes with $H$, defining the symmetry group of time translations $U(t)=e^{itH/\hbar}$. Using this group, one can define for an arbitrary observable $A$ the timeshifted observable $A(t)=U(t)AU(-t)$, describing the time-dependence in the Heisenberg picture.
In the Schroedinger picture, one refers everything to time zero by
writing time dependent expectations $\langle A\rangle_t=Tr\ \rho A(t)$
with respect to a fixed Heisenbeg state $\rho$ as expectations
$\langle A\rangle_t=Tr\ \rho(t) A$. Using the definition of $A(t)$ and
the properties of the trace, the resulting condition
$Tr\ \rho A(t)=Tr\ \rho(t) A$ gives (for time independent systems)
the formula $\rho(t)=U(-t)\rho U(t)$, with sign of $t$ opposite as in
the Heisenberg picture! Differentiation gives
$\dot\rho(t)=-U'(-t)\rho U(t)+U(-t)\rho U'(t)$, and since
$U'(t)=iHU(t)/h_bar=iU(t)H/h_bar$, we find the quantum Liouville
equation $\dot \rho(t) = -i/\hbar Tr\ [H,\rho]$. (In the time-dependent
case, this equation takes the form
$\dot \rho(t) = -i/\hbar Tr\ [H(t),\rho]$.)
A quantum observable $A$ is called conserved if, for all states $\rho$, its expectation $\langle A\rangle_t$ is independent of time. Differentiation gives the condition $0=Tr\ \dot \rho(t) A = -i/\hbar Tr\ [H(t),\rho] A$, which is equivalent with $0=Tr\ [H(t),\rho] A = Tr\ H\rho A-Tr\ \rho H A = Tr\ \rho (HA-AH) =Tr\ \rho [H,A]$. Since this must hold for all $\rho$, the observable $A$ is conserved iff it commutes with $H$.
This proves Noether's theorem that the generator of a 1-parameter group of symmetries is conserved.
If you specialize this to the case where $A$ is the energy operator (the generator of time translations), a component of the momentum operator (the generators of space translations), a component of the angular momentum operator (the generators of rotations), you get from invariance under the Poincare group the conservation of energy, momentum, angular momentum.
