Is the Process of Projection of a Generic State Onto a Subspace Impossible? I can define (in a standard way) the process of projection of a generic state onto the subspace $\mathcal{G}$ as a process that takes a generic state $|\psi\rangle$ of the Hilbert space $\mathcal{H}$ and returns $\sum_g|g\rangle\langle g|\psi\rangle\in\mathcal{G}$ where $\{|g\rangle\}$ forms a complete orthonormal basis for $\mathcal{G}$.
It appears self-evident that a measurement process cannot exist such that for every state $|\psi\rangle$ in the Hilbert space $\mathcal{H}$, the resulting state, arising out of performing the said measurement process on $|\psi\rangle$, lies in a subspace $\mathcal{G}$ of $\mathcal{H}$. As for the existence of such a measurement, a Hermitian operator would need to exist whose eigenstates do not span the full Hilbert space $\mathcal{H}$ but span only its subspace $\mathcal{G}$. This is clearly impossible. And thus, it seems clear that no measurement process can be a projection process. 
Furthermore, a unitary evolution operator certainly cannot perform such a shady job. This seems to be demonstrated in the following simple argument: There would always exist an eigenstate $|E_m\rangle$ of the Hamiltonian which doesn't belong to $\mathcal{G}$ (due to the Hermiticity of the Hamiltonian). Since $\langle E_m|\hat{U}_t|\psi\rangle=\sum_n\langle E_m |e^{-iE_nt}|E_n\rangle\langle E_n|\psi\rangle=e^{-iE_m t}\langle E_m|\psi\rangle$, it is explicitly clear that if $\langle E_m|\psi\rangle \neq 0$ and thus $|\psi\rangle\notin\mathcal{G}$ then $\hat{U}_t|\psi\rangle\notin\mathcal{G}$ as well.
Thus, it seems that neither a measurement process nor a unitary evolution can act as the process of projection in the sense of the phrase defined here. (And no other kind of process exists in the world than a measurement process and a unitary evolution.)  This seems pretty counter-intuitive based on daily facts like the existence of polaroids, etc.! Do such instruments that seem to perform the process of projection do so only in some approximate manner? 
 A: In any model that is meant to include the microscopic dynamics of the measurement process itself, so that the details of the measurement equipment are part of the initial state, the occurrence (or not) of a measurement depends on the initial state $|\psi\rangle$, just like it does in the real world. The occurrence of a measurement is very much state-dependent. 
(By "model," I mean a model that respects the usual general principles of quantum theory. Non-relativistic quantum electrodyamics is one "model." Quantum chromodynamics is another "model." And so on. Just standard quantum physics, with no speculative modifications.)
A unitary operator can't achieve a projection because that directly contradicts the definition of "unitary": Unitary operators preserve inner products. Projections don't. 
What can happen in quantum theory is that an appropriately chosen initial state $|\psi\rangle$ in a sufficiently rich model can evolve, through ordinary unitary time evolution, into a state of the form $\sum_n |n\rangle$, where the $|n\rangle$ are almost exactly orthogonal to each other, and remain almost exactly orthogonal after applying any projection corresponding to any practically feasible future measurement (see the Example, below). This last part is quantum theory's way of telling us that we might as well apply a projection onto one of those terms $|n\rangle$. Quantum theory can't tell us which one (it can only tell us their relative frequencies via Born's rule), but empirically we experience only one, and that's the one we should project onto. 

Thus, it seems that neither a measurement process nor a unitary evolution can act as the process of projection in the sense of the phrase defined here.

That's right. Unitary  evolution certainly can't achieve a projection, and neither can a measurement process if the latter is described as a physical process whose microscopic dynamics are governed by quantum theory. In quantum theory, projection is only something we do on paper to account for what we learned from actually doing the measurement in the real world; it is a bookkeeping device, not a physical process. Whatever this situation might say about the limitations of quantum theory is a separate subject, part of the infamous measurement problem, which is beyond the scope of Physics SE.

 Example 
Exactly solvable examples that include the dynamics of the measurement process are rare and usually contrived; but with the help of a little mathematical intuition, more-realistic examples can be deduced. To illustrate the idea, here is one example.
In a measurement, the object-of-interest must influence its surroundings in some way, because otherwise no information about the object would be gained. So for QM to describe the measurement process, we must use some kind of multi-particle model so that the surroundings can be influenced by the object of interest. 
As an example of such a model, consider a polaroid filter that passes photons of one polarization (say $H$) and absorbs photons of the orthogonal polarization (say $V$). We can think of this as a way of measuring the photon's polarization. To describe the measurement process, we need to use a model that includes the microscopic constituents of the filter (and air, etc), so that the photon's polarization has an opportunity to influence the microstate of the filter (etc). 
We can distinguish between the microstate of the filter after an $H$ photon passes through (call that microstate $h$) and the microstate of the filter after a $V$ photon is absorbed (call that $v$). The absorption of the photon affects the filter's microstate, and even if that effect is initially localized, it cascades and ends up affecting the microstate in a way that is too complicated for mortals like us to track — kind of like displacing a single molecule in the atmosphere has a cascading effect that quickly makes the microstate very different than it would have been otherwise. So the microstates $h$ and $v$ are very different from each other, even if the difference is macroscopically imperceptible.
Now, suppose for simplicity that the filter is unaffected when the photon passes through $H$. Before the photon passes reaches the filter, we can take the initial state to be $c_H|Hh\rangle+c_V|Vh\rangle$, where the two terms differ only in the polarization of the photon. These terms are orthogonal, but they differ in a very simple way, because the filter's microstate is the same in both terms. After the photon reaches the filter, the state is $c_H|Hh\rangle+c_V|Vv\rangle$. Now the filter's microstate is different in the two terms, and the difference is so complicated that those terms will remain orthogonal no matter what else we try to do. That's essentially the definition of a measurement, and it means that we might as well project onto just one of those two terms, even though quantum theory can't tell us which one.
Again, this is all ordinary quantum theory, but using a model that includes the microscopic constituents of the larger system, not just the photon-of-interest. This is necessary if we want to describe the dynamics of the measurement process itself, because measurement — by definition — is all about how the object-of-interest (the photon's polarization in this example) affects everything else.
