# Symmetries vs time evolution

My question is the following: when in quantum mechanics one introduces symmetry, says that a states and observables transform both, in order to mantain mean values intact (kind of like a change of coordinate system), i.e.:

$$|\psi>\rightarrow U|\psi>$$

and

$$O\rightarrow UOU^\dagger$$

while when one is concerned about time evolution, only states (Schrodinger picture) or observables (heisenberg picture) change?

The first thing to say is that if one doesnt' do that, mean values will never depend on time, which is rather strange..

tom.stoer
Look at the Heisenberg picture.

What you find is an operator equation where the time evolution of an operator O is related to the commutator [H,O] via

$$\frac{dO}{dt} = i[H,O]$$

which corresponds to the classical Poisson bracket. From this operator equation of motion one derives the time evolution operator

$$U(t) = e^{-iHt}$$

and the time evolution

$$O(t) = U^\dagger(t)\,O\,U(t)$$

In the Heisenberg picture the states are constant, i.e. not subject to time evolution generated via U(t), i.e.

$$|\psi\rangle \stackrel{\text{time}}{\to} |\psi\rangle$$

$$O \stackrel{\text{time}}{\to} O(t) = U^\dagger(t)\,O\,U(t)$$

Calculating expectation values means

$$\langle O(t) \rangle_\psi = \langle\psi|O(t)|\psi\rangle = \langle\psi|U^\dagger(t)\,O\,U(t)|\psi\rangle$$

What you are talking about are unitary transformations T(a) generated via hermitian operators Ω

$$T(a) = e^{-ia\Omega}$$

But w.r.t. these unitary transformations T(a) both operators and states do change

$$|\psi\rangle \stackrel{\text{trf}}{\to} |\psi\rangle_a = T^\dagger(a)|\psi\rangle$$

$$O(t) \stackrel{\text{trf}}{\to} O_a(t) = T^\dagger(a)\,O(t)\,T(a)$$

which means that T(a) drops out from expectation values trivially.

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Yes, I know these mathematics behind that, but my question is why do we do so? I mean, why one says the things you wrote in bold? Is it somehow related to active vs passive transformations?

tom.stoer