Symmetry transformation in Heisenberg vs Schrödinger Picture

[hgandh]

Symmetry transformations are changes in our point of view that preserve the possible outcomes of experiment:
$$\Psi \rightarrow U(\Lambda) \Psi$$
In the Heisenberg picture, observables in a fixed reference frame evolve according to:
$$P(t) = U^\dagger (t)PU(t)$$ while in the Schrödinger picture, the state vector evolves as $$\Psi (t) = U(t) \Psi$$
Now at a time t, we change our reference frame. The expectation value of the observable in the two pictures is then
$$(U(\Lambda) \Psi, P(t)U(\Lambda) \Psi) $$ in the Heisenberg picture and $$(U(\Lambda) \Psi (t), PU(\Lambda) \Psi (t)) $$ in the Schrödinger picture. However, these do not give equivalent values. Can someone point out where my reasoning is wrong?

 

[strangerep]

Is your symmetry transformation time-independent? I'm guessing it probably is and, if so, then $U(\Lambda)$ commutes with $U(t)$, leading to equivalent expectation values, afaics.

HTH.

 

[hgandh]

Is your symmetry transformation time-independent? I'm guessing it probably is and, if so, then $U(\Lambda)$ commutes with $U(t)$, leading to equivalent expectation values, afaics.

HTH.

No it is a general Lorentz transformation. I believe my mistake was the the time evolution operator itself changes in the new reference frame since the Hamiltonian as seen in this new frame is in general not the same as in the original.
$$U(t) \to U(\Lambda)U(t)U^{-1} (\Lambda)$$