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

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)?

• It collapses to whatever basis you measure it in. If the computational basis is, say, the spin of a spin 1/2 particle along the z axis, your proposed other states would result from measuring the spin along the x axis. – knzhou Jan 26 '18 at 8:27
• By convention in quantum computing we usually talk about computational basis measurements because (1) it makes it look more like regular computing, with its zeros and ones and (2) you don’t lose anything, i.e. with my example an x spin measurement is the same as applying a field to rotate x to z and then measuring z spin. – knzhou Jan 26 '18 at 8:28

• That clearly contradicts your statement that "states are not vectors but rays in Hilbert space". Also they mention that quantum states are unit vectors i.e. they must have norm $1$. – user182786 Jan 26 '18 at 7:37