Question: How do symmetries and time evolution interact in quantum mechanics?

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Discussion Overview

The discussion revolves around the interaction between symmetries and time evolution in quantum mechanics, exploring how states and observables transform under these concepts. It includes theoretical considerations and mathematical formulations related to different pictures of quantum mechanics, such as the Schrödinger and Heisenberg pictures.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions the treatment of symmetries and time evolution, noting that maintaining mean values requires transformations of both states and observables.
  • Another participant explains the Heisenberg picture, detailing how the time evolution of an operator is governed by the commutator with the Hamiltonian, leading to a specific form of time evolution for observables.
  • A third participant seeks clarification on the rationale behind the mathematical treatment of symmetries and time evolution, suggesting a connection to active versus passive transformations.
  • One participant distinguishes between the effects of time evolution and symmetry transformations, arguing that time evolution alters observables while leaving states invariant, whereas symmetry transformations change both states and observables in a way that cancels out in expectation values.

Areas of Agreement / Disagreement

Participants express differing views on the implications and interpretations of symmetry transformations versus time evolution, indicating that the discussion remains unresolved with competing perspectives on the significance of these transformations.

Contextual Notes

Some participants reference mathematical formulations and relationships, but there is no consensus on the underlying reasons for the treatment of symmetries and time evolution, nor on the implications of these transformations.

giova7_89
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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 (Schrödinger 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..
 
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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?
 
The time evolution U is derived via the Heisenberg equstion of motion, the symmetry transformation T(a) are nothing else but a change of the basis vectors in Hilbert space. The only thing which relates them is the fact that both are unitary transformations.

But U(t) really does something, it transforms O and leaves the states invariant whereas T(a) essentially does nothing, it transforms O but this compensated by the change of the states.
 

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