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I asked something like this on Math StackExchange, but now that I think about it, this probably belongs better over here.

I want to find all linear operators (non necessarily hermitian) $\{\hat{A}\}$ other than the identity operator for a particular wavefunction such that

$$\hat{A}\psi = \psi$$

Basically, I'm trying to extract symmetries from a wavefunction in order to reduce the information necessary to represent it.

For example, consider a spherically symmetric function $f$. One way to represent it is in the Cartesian basis as $$f(x,y,z) = e^{-\left(x^2 + y^2 + z^2\right)}$$

However, if you recognize that some kind of rotation operator applied to this function leaves it unchanged, you can rewrite it in the basis of spherical coordinates as:

$$f(r) = e^{-r^2}$$

Thus a function of 3 dimensions has been reduced to a function of 1 dimension by extracting a symmetry from the system. I'm trying to identify all such operators that can be applied to a wavefunction, so I can change its basis into one that gives a minimal representation of that function.

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  • $\begingroup$ In general such symmetries do not exist any more than in classical mechanics. Where they do exist in classical mechanics, these symmetries are linked to the notion of integrability in Hamiltonian systems, I would suggest to look at e.g. physics.stackexchange.com/q/76783 for clues. $\endgroup$
    – CuriousOne
    Commented Dec 30, 2014 at 20:55

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In mechanics (meaning classical and quantum), the notion of symmetry is not just of states, but it is linked to the relevant Hamiltonian. Since your question is mainly about quantum mechanics, I will analyse this scenario. An operator that implements a symmetry must commute with the Hamiltonian operator. It might then be possible to find an operator with the property you have requested, but it might fail to commute with the Hamiltonian of the system. Hence this is not defining a symmetry.

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