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0

For a non-relativistic system with Hamiltonian $\hat H$, the time-evolution of an arbitrary state $|\Psi\left(\textbf{r},t\right)\rangle$ is governed by $$i\hbar\frac{\partial}{\partial t}|\Psi\left(\textbf{r},t\right)\rangle = \hat H |\Psi\left(\textbf{r},t\right)\rangle \tag{1}\label{SE}.$$ We can take the adjoint of Eq.\eqref{SE} and it reads $$-i\hbar\...


0

Here's a nitpick (maybe the crucial one?): $U(t)$ is not a $C_0$ semigroup because it fails the second condition: $\forall t,r\ U(t+r)=U(t)\,U(r)$ since the choice of $t_0$ from which we evolve the system affects the second product differently than the first: The first operator evolves the system to time $t_0+t+r$ whereas the second (in the best case ...


3

The conditions needed for Stone's theorem require that $$U(t,t') = U(t-t',0)$$ for any $t,t'$. This is certainly not satisfied by a general time-dependent Hamiltonian $H(t)$. (Consider e.g. turning on a potential at time $t = 0$, then $U(s+\delta s,s)$ will depend on whether $s < 0$ or $s>0$.)


-1

The operational definition of Entropy is roughly proportional to the logarithm of the number of accessible microstates on any given equilibrium macrostate. For positive energy systems, the number of allowed microstates grow quickly with increasing system energy. If one expects that a negative-energy regime to be a mirrored version of the positive energy ...


4

Under a unitary $U$, for which $U^\dagger U = UU^\dagger = \mathbb I$, any arbitrary operator $A$ transforms as $$ A \mapsto A' = U^\dagger AU. $$ You're being asked to show that if $C=[A,B]$, then $C'=[A',B']$, or in other words, that $$ U^\dagger[A,B]U = [U^\dagger AU, U^\dagger BU]. $$ The proof is simple but it's for you to work out.


7

It's not #1, because it's easy to write down QFTs which are unstable against pair production. One simple example is $$\mathcal{L} = \frac12 (\partial_\mu \phi)^2 + \frac12 m^2 \phi^2$$ which corresponds to particles with negative $m^2$. Since $E^2 = p^2 + m^2$, it is energetically favorable to produce infinitely many particles, since the rest energy is ...


3

Option 3 is the closest match, but it's a bit like saying "Nothing guarantees that spacetime has a Lorentzian signature." We normally only consider spacetimes that do, because so much else depends on it. It's a requirement, not a theorem. Similarly, for relativistic QFT in flat spacetime, we normally only consider QFTs whose total energy has a finite lower ...


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