Energy as charge with respect to time translations in QM Consider a non relativistic quantum mechanical system with Hamiltonian $\mathcal{H}$, and denote the states by $\psi \equiv \psi(t) \equiv | \psi(t) \rangle$.


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*From the Schrödinger equation we know how $\psi$ is modified by an infinitesimal time-translation:
$$ \tag{A} \delta \psi = - i \delta t \mathcal{H} \psi$$
Moreover, if we denote with $\mathcal{O}(t)$ some operator acting on $\psi$ we can write its time evolution in the Heisenberg picture as:
$$ \tag{B} \delta \mathcal{O} = i \delta t \left[\mathcal{H},\mathcal{O} \right]$$

*In a scalar field theory
$$ \tag{C} \mathscr{L} = \partial_\mu \phi^* \partial^\mu \phi - m^2 \phi^* \phi$$
the infinitesimal form of a global phase trasnformation has the form (in appropriate units)
$$ \tag{D} \delta \phi = -i \delta \alpha e \phi $$
The transformation law (D) is very similar to (A). I understand that this probably arises from the fact that $\mathcal{H}$ is the generator of time-translations but how far does the analogy go?
In particular:


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*Can I talk of an eigenstate $\psi_E$ obeying
$$ \delta \psi_E = - i \delta t E \psi_E $$
as being charged with respect to time translations, with charge $E$? Can I still talk of $\psi_E$ as being in a representation of an $U(1)$ Lie group? Also the fact that $\psi$ must be complex seems to find a natural justification here: it must be so for exactly the same reason that a real scalar field $\varphi$ is not charged while a complex one $\phi$ is.

*Can I talk of $\psi$ as being in a "fundamental representation" of the symmetry group of time-translations?

*Also the Heisenberg evolution equation (B) seems to be stating, at least formally, that $\mathcal{O}$ belongs to the adjoint representation of the group generated by $\mathcal{H}$:
$$ \delta \mathcal{O} = i \delta t \mathcal{H}_{Adj} \mathcal{O} $$
with an inverse evolution with respect to (A) which is readily explained by the necessity of (A) and (B) to describe the same time-evolution of average values. Does this make any sense? Is it just a coincidence or is there some deeper reason behind it?

 A: None of this is coincidental. But there is also not much mysterious going on:
The energy of any state is the expectation value of the Hamiltonian in that state. The Hamiltonian generates indeed a one-parameter group, the time translations, but it is not $\mathrm{U}(1)$, but rather $(\mathbb{R},+)$ - it is abelian, but not compact (since otherwise we could go forward in time and come back to where we started). Both have $(\mathbb{R},+)$ as their Lie algebra.


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*The space of states is, since the Hamiltonian acts on it, a representation space for this group. Since the abelian Lie groups have only one-dimensional irreducible representations, the space of states splits nicely into one-dimensional subspaces, each of which carries a (possibly different) representation of $\mathbb{R}$. These are, of course, the spaces generated by the individual energy eigenstates - since, by definition, $H\psi_E = E\psi_E$ is a representation of the Lie algebra, lifting directly to a representation of the Lie group by the exponential as usual: $U(t)\psi_E = \mathrm{e}^{\mathrm{i}Ht}\psi_E = \mathrm{e}^{\mathrm{i}Et}\psi_E$, where you can say that $E$ is the "charge" labelling the irreducible representation of $\mathbb{R}$. But we have learned...precisely nothing. $\mathbb{R}$ as a Lie group is just boring.

*The term fundamental representation has slightly different meanings in physics and mathematics. The mathematical description is highly abstract, while the physicists simply mean that, given a matrix group with multiplication as group structure, the representation is just applying the matrix itself. But...translations are not naturally matrix groups, they have no fixed points, so no map into the linear maps on a space (which is what a representation is) can ever be the identity! So our naive notion of fundamental representation fails to apply. Naturally, translations act by addition, not multiplication. Furthermore, since all irreducible representation of this group are one-dimensional, it doesn't really make sense to use our notion, because usually it selects the smallest-dimensional faithful (i.e. injective) representation. It is best to abandon the idea of "fundamental" representation for this group. (I suspect that the mathematical definition does not apply either, since it is explicitly about (semi-)simple Lie groups, which are defined to be non-Abelian. It just doesn't make much sense to throw group theory onto $\mathbb{R}$)

*Yes, the operators transform in the adjoint representation. But that is not suprising, since $H$ is an operator itself, and so embeds into the Lie algebra of operators, just as the symmetry group of time translations embeds into the Lie group that is the Lie integration of that operator algebra. Since a Lie group naturally acts on its algebra by the adjoint, it follows as a special case that the time translations act by the adjoint on operators.
