A *state* is *something* that encodes our knowledge about the system.

And that's it.

There are many ways to encode a state in quantum mechanics. As a wavefunction ("Schrödinger representation"), as Fock momentum states ("Fock representation"), as a density matrix, as a ray in a Hilbert space, as a linear functional on the $C^*$-algebra of observables, as a point in a projective Hilbert space,...

Not all of these ways are always permissible. The "as a ray in a Hilbert space", "density matrix" and "functional on algebra of observables" are, to my knowledge, always possible, while e.g. the encoding as a wavefunction fails for quantum systems whose Hilbert space is finite-dimensional (e.g. qubits) because those don't have the usual position operator.

The representation as a density matrix generalizes to statistical quantum mechanics, that as a single ray in a Hilbert space or as a wavefunction does not. But whenever two of these descriptions are possible, they are, at the level of rigor of physicists at least, equivalent.

So the question "What really is a state in quantum mechanics?" doesn't have a single answer. Or any answer other than "it depends what you want to do with it".

But this should not surprise you, after all, it is the same in classical mechanics: You can have Newtonian or Lagrangian or Hamiltonian descriptions, and here a state is position and velocity, there a state is position and momentum, or even some generalized coordinates which have no direct physical meaning at all.

There is no truth to a state, be it classical or be it quantum, other than "it encodes all possible information about the physical system in a convenient way".

## Best Answer

There is no particularly interesting new physical significance to such a state vector. As you already stated, it represents

exactly the same physical state. The only difference is that, on taking the modulus squared, the new state gives an unnormalised probability distribution over possible measurement outcomes. You can easily extract the probability of obtaining a measurement outcome corresponding to the (possibly unnormalised) state $|\phi\rangle$ from an unnormalised state $|\psi\rangle$ by using the Born rule: $$ \mathrm{Pr}(\phi) = \frac{|\langle \phi | \psi \rangle |^2}{\langle \phi | \phi \rangle \langle \psi | \psi \rangle }. $$ Clearly, using normalised states is just a handy convention that avoids any worries about calculating the denominator above. There is nothing wrong with formulating quantum mechanics without normalising state vectors to unity. Indeed, most people avoid bothering with the overall normalisation factors until they are needed at the very end of the calculation, since they just add unsightly clutter to the mathematics and have no physical significance.