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Action of an operator on a state vector collapses the wave function to any of the eigenstate of that operator , So we get resulting state of the system as some base ket. But mathematically action of operator on a state vector change state vector to another state vector which is the superposition of all basis eigenkets, So how can we incorporate collapse of wave function and action of operator?

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Not all operators transform a ket into a superposition of kets: for instance $$ \hat \Pi=\vert\psi\rangle\langle\psi\vert $$ acting on a state $\vert \Psi\rangle$ will produce $\vert\psi\rangle\langle\psi\vert\Psi\rangle$, which is not a normalized ket and not a linear combination of kets either.

An operator like $\hat \Pi$ is a projection operator and is idempotent, i.e. $\hat\Pi\hat \Pi=\hat \Pi$; projection operators serve as basic prototypes for measurement operators.

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You have a misunderstanding here. The action of an observable operator like $x$ or $p= -i\frac{\partial}{\partial x}$ does not change the wave function into a collapsed state. The operator which does that is a projector, like:

$$|x=1\rangle \langle x=1|$$

or
$$|x=-12.531\rangle \langle x=-12.531|$$

or

$$|p=0\rangle \langle p=0|$$

Which projector you apply depends on the measurement result and what you are measuring. The ones above apply to the results $x=1$, $x=-12.531$, and $p=0$, respectively (I left out units since they're not important to the point).

If you want to calculate these projectors: they are outer products of the eigenvectors of the corresponding operators. For example, $$\hat{x}|x=-12.531\rangle = -12.531 |x=-12.531\rangle$$

The projectors above will not leave you with a superposition (at least not in the x, x, and p basis, respectively). In general, whether a vector is in a superposition depends on the basis you express it in.

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