I'm trying to get a more concrete idea how symmetry is understood in quantum theories, as broad as possible.

Consider a infinitesimal transformation of states in quantum physics of the form:

$$ |\phi'\rangle=(1+i\epsilon G)|\phi\rangle, \tag{1}$$

where $G$ is a (not necessarily Hermitian) operator, which may be associated with the transformation of any Hermitian operator $M$:

$$M'=M+i\epsilon G^\dagger M-i\epsilon MG, \tag{2}$$

so that the expectation value

$$\langle\phi'|M'|\phi'\rangle=\langle\phi|M|\phi\rangle \tag{3}$$

is invariant.

  • If we consider the transformation $G$ to be symmetry, then any operator will create a symmetry transformation of the quantum physics. To what extend is this statement true?

  • Generally symmetry is considered by group of symmetries. There is a group of finite transformation of $\epsilon G$ if and only if there is a closed algebra containing $G$. Then if we can construct a closed algebra of operators, can we associate it with a particular kind of symmetry?

  • Finally, the whole above discussion is considered under quantum framework. How can we relate to the classical one?

  • 2
    $\begingroup$ Have you heard of Lie groups and their algebras? That's the framework symmetries are usually discussed in, and since you mention them with nary a word, I thought I'd ask. Also, I don't understand your first bullet point - not every operator will be a symmetry, even though every operator transforms under the symetry. $\endgroup$ – ACuriousMind Aug 8 '14 at 15:45
  • $\begingroup$ As I see $G$ can be used to defined a transformation that reserve the scalar-the amplitude of the physics. Isn't that good enough to call it a symmetry> $\endgroup$ – Leaning Aug 8 '14 at 16:30
  • $\begingroup$ Careful with the terminology! Scalar already means that the thing called scalar is invariant under transformation, i.e. transforms as an element of the trivial representation of the transformation group. "the amplitude of the physics" - if you mean leaving all expectation values of observables the same then yes, that's a symmetry. But if it has an "infinitesimal version", that is equivalent to saying the symmetry group is a Lie group. In classical physics, you must leave the Lagrangian/action/equation of motion invariant. (things doing the latter, not the former, are called anomalous) $\endgroup$ – ACuriousMind Aug 8 '14 at 16:49
  • 1
    $\begingroup$ (The above is a comment since I do not think that the question, as it stands, has good answers. It is not a bad question, but it seems to use a very peculiar understanding of symmetry/operators, and I believe most of the question itself goes away when translating it into "regular" physics terminology) $\endgroup$ – ACuriousMind Aug 8 '14 at 16:51
  • $\begingroup$ Comment to the question (v3): What are OP's definitions of symmetry inside the quantum framework in the first bullet point in order to have a non-trivial statement? $\endgroup$ – Qmechanic Aug 8 '14 at 17:06

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