Which set of basis states can a quantum system of qubits actually collapse to?

Question

I was watching a video on "How Does a Quantum Computer Work?".

I'm confused about what they mean by: "Although the qubits can exist in any combination of states, when they are measured they must fall into one of the basis states."

From what I know about linear algebra, if we represent the state of a qubit by $|\psi\rangle$ it can be written like $\alpha|x\rangle + \beta |y\rangle$ (where $|x\rangle$ and $|y\rangle$ form a basis) or $\gamma (|x\rangle+|y\rangle) + \delta (|x\rangle-|y\rangle)$ or even $A(|x\rangle+|100y\rangle) + B |y\rangle$! What I mean is that no set of basis states is unique.

So, in reality which set of basis states can a qubit (or more generally a quantum system of qubits) actually collapse to? Can an actual measurement land us with a basis state like $(|0\rangle + |1\rangle)$ or $(|0\rangle - |1\rangle)$ ? Or is only $|0\rangle$ and $|1\rangle$ possible? Also does the basis vector which a qubit can land up in have to have norm $1$ (i.e. must it be an element of an orthonormal basis)?

Answer 1

A standard projective quantum measurement of an observable - a self-adjoint operator on the Hilbert space - results in the measured object being in an eigenstate of said operator, with the corresponding eigenvalue being the value measured. (Whether you want to call this "collapse" is interpretation-dependent.) So there is no generic answer as to what states are possible as the outcomes of measurements, it entirely depends on what the observables of the system you are considering are.

Asking whether the resulting state is orthonormal or not is actually a meaningless question, since states are not vectors but rays in Hilbert space - every scalar multiple of a vector represents the same physical state as the original vector.