- #1
hgandh
- 27
- 2
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 Schrodinger 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 Schrodinger picture. However, these do not give equivalent values. Can someone point out where my reasoning is wrong?
$$\Psi \rightarrow U(\Lambda) \Psi$$
In the Heisenberg picture, observables in a fixed reference frame evolve according to:
$$P(t) = U^\dagger (t)PU(t)$$ while in the Schrodinger 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 Schrodinger picture. However, these do not give equivalent values. Can someone point out where my reasoning is wrong?