Symmetries in QM and QFT --- operator transformation laws

Question

In quantum mechanics, we implement transformations by operators $U$ that map the state $|\psi\rangle$ to the state $U|\psi\rangle$. Alternatively, we could transfer the action of $U$ onto our operators: $$ O\mapsto U^\dagger OU $$ These operators $U$ ought to constitute a representation of the transformation group of interest. The question we then ask is: which representation shall we choose? I have learnt that we pin down the matrices $U$ that we desire by asking that the position operator, or the position eigenstate, transforms in the appropriately geometric way: if $R$ is some rotation matrix in 3D, then we wish that $$ U(R)^\dagger \vec{x} U(R) = R\vec{x} $$

I have multiple questions regarding this procedure:

1) It is often argued that $U$ is unitary since 'probabilities must be conserved'. But it is clear from my studies that operators representing Lorentz boosts are not unitary. How are these two facts reconciled?

2) On a related note, we say that a transformation is a symmetry of the system if the Hamiltonian is preserved under the transformation. This makes sense to me in that it is the Hamiltonian that defines the system; if the Hamiltonian is unchanged, then the old solutions to Schrodinger's equation are still solutions. However, it is clear that whatever transformation operator implements a Galilean boost (in non-relativistic QM) or Lorentz boost (in relativistic QFT) is not going to leave the Hamiltonian unchanged. That doesn't mean that boosts aren't symmetries of our system, however. What's going on here? Why exactly is $U^\dagger H U = H$ the condition for $U$ to be a 'symmetry'?

3) Demanding that the operators $U$ satisfy the property (choosing translation for concreteness) $$ U^\dagger \vec{x} U = \vec{x} + \vec{a}$$ doesn't immediately tell us how the transformation will affect other operators, such as the momentum or angular momentum operators. For the case of a translation by $a$, we find that $$ U = \exp\left( -i\frac{\vec{a}\cdot\vec{p}}{\hbar}\right) $$ satisfies the above equation (work infinitesimally). This then tells us that $U^\dagger \vec{p} U = \vec{p}$. This is what we want --- translations in space do not affect the momentum. But we haven't told our operators $U$ that they ought to satisfy this property, so the fact that they do seems something of a coincidence. That is, requiring only that conjugating the position operator with $U(R)$ rotates it seems to force every vector operator to be rotated under conjugation with $U(R)$. Is there something fundamental underlying this?

4) In QFT, we wish to implement Lorentz boosts with operators $$ S = \exp \left( -i\frac{\omega_{\mu \nu} M^{\mu \nu}}{2\hbar}\right) $$ where $\delta^\mu_\nu + \omega^\mu{}_\nu$ is the infinitesimal Lorentz transformation we are representing. Now for $T$ representing a different, not necessarily infinitesimal, Lorentz transformation $\Lambda$, we have $$T^{-1} S T = T^{-1} \left(1 - i\frac{\omega_{\mu \nu} M^{\mu \nu}}{2\hbar} \right) T$$ But the condition that the matrices $S$ and $T$ must constitute a representation of the Lorentz group implies that the combination on the left hand side ought to equal $$\exp \left( -i\frac{\Omega_{\mu \nu} M^{\mu \nu}}{2\hbar}\right) $$ where $\Omega_{\mu \nu}$ defines the composite transformation: $$ \delta^\mu_\nu + \Omega^\mu{}_\nu \equiv (\Lambda^{-1})^\mu{}_\rho(\delta^\rho_\sigma + \omega^\rho{}_\sigma)\Lambda^\sigma{}_\nu $$ Comparing these equations then gives $$T^{-1} M^{\mu \nu} T = \Lambda^\mu{}_\rho \Lambda^\nu{}_\sigma M^{\nu \sigma}$$ using the raising and lowering properties of the Minkowski metric. My question is similar to that in 3) --- the object $M^{\mu \nu}$ is just a collection of 6 operators, and it is not at all obvious that those indices ought to be tensor indices. And yet somehow --- just by asking for the matrices $S$ to constitute a representation --- we've managed to reproduce the tensor transformation law (admittedly we have a $T^{-1}$ here rather than a $T^\dagger$ --- this relates to my first question). In other words, we've got the correct geometric action of the Lorentz transformation on the operators $M^{\mu \nu}$, even though we never asked for it!

As you can tell, I'm very confused by the whole situation. I apologise if the above discussion sounds a bit muddled, and I would appreciate any help that can be given!

Answer 1

Here is my answer to some of your questions - this is based purely on my understanding of these concepts and could be wrong.

(1) Whenever Lorentz transformations are a symmetry of any quantum system, they must necessarily be represented by unitary linear transformations on the quantum Hilbert space of the system. Operators representing Lorentz boosts on a relativistic quantum system are therefore unitary as pointed out in one of the links mentioned in the comments.

(2) The condition $ U^{\dagger}HU = H $ cannot, in general, be a necessary and sufficient condition for a $U$ to be a symmetry transformation in relativistic quantum mechanics. One way of seeing this is to consider a one particle eigen state of a free Klein-Gordon field $ |p\rangle $ and observe that under a Lorentz transformation, the expectation value $ \langle p|H|p\rangle $ (which gives the energy of this particle) should transform like the $0^{th}$ component of the energy-momentum four-vector. $$ \langle p|U^{\dagger}(\Lambda)HU(\Lambda)|p\rangle = {\Lambda^{0}}_{\mu}\langle p|P^{\mu}|p\rangle $$ and not be left invariant as $ U^{\dagger}HU = H $ would imply. At the same time, Lorentz transformation is a symmetry of this system precisely because similar transformation laws are valid for all the four-momentum components - which is equivalent to the scalar $ P_{\mu}P^{\mu} $ remaining invariant under a Lorentz transformation.

The point to note here is therefore that the requirement that a transformation be a symmetry does not, in general translate into commutation with the Hamiltonian. This happens in the special cases of the time-independent symmetry transformations (eg - spatial translations and rotations in non-relativistic QM ). The most general criterion is that the Action be invariant (upto a constant) under the concerned transformation.

(3) Refer to any good quantum mechanics text (eg. Sakurai, Modern Quantum mechanics chapters 2 and 4) that should answer your question.

(4) The result obtained here (that the generators of the Lorentz group should transform like tensors) is not very surprising. An elementary (though not very general) argument here is the following - for any observable which is classically a tensor (like the momentum 4-vector), there must exist a corresponding self-adjoint operator on a quantum mechanical Hilbert space. The expectation value of this operator wrt any state should necessary transform like the tensor observable itself. This is a consequence of the basic paradigm of Quantum mechanics (expectation values represent physically measurable quantities). Therefore one must have $$ T^{−1}S_{μν}T={Λ_{μ}}^ρ {Λ_{ν}}^σ S_{\rho σ} $$ quite generally for any tensor operator $ S_{\mu\nu}. $

PS : As mentioned at the beginning, this answer is based on my understanding of the concepts of 'symmetry' and 'lorentz covariance' - needless to say, I would appreciate constructive criticism.