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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)?

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  • $\begingroup$ 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. $\endgroup$ – knzhou Jan 26 '18 at 8:27
  • $\begingroup$ 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. $\endgroup$ – knzhou Jan 26 '18 at 8:28
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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.

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  • $\begingroup$ I think I understand the first paragraph now. However, I still don't get your second paragraph completely. Have a look at this PDF on Quantum Computing: web.mit.edu/zoya/www/quantComp.pdf. They say: "Quantum states are unit vectors in a Hilbert space (a complex, norm-preserving, vector space). The bases used for this space depend on a choice of observables - they are the eigenvectors of the observables. When we say that a measurement collapses the state of a vector, we mean that the vector is projected onto one of the basis vectors with probability equal to the projected norm." $\endgroup$ – user182786 Jan 26 '18 at 7:37
  • $\begingroup$ 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 $1$. $\endgroup$ – user182786 Jan 26 '18 at 7:37
  • $\begingroup$ @user182786 one can make a convention of always choosing a unit vector to represent the state since it simplifies a few formulae (which is very often adopted) but what I wrote there is nevertheless true. For instance, there is a version of the Born rule for non-normalized states, see e.g. physics.stackexchange.com/q/156367/50583 $\endgroup$ – ACuriousMind Jan 26 '18 at 8:01

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