3

$\newcommand{\bra}[1]{\left<#1\right|}\newcommand{\ket}[1]{\left|#1\right>}\newcommand{\bk}[2]{\left<#1\middle|#2\right>}\newcommand{\bke}[3]{\left<#1\middle|#2\middle|#3\right>}$
In *Quantum Computation and Quantum Information* the authors make the following statement "[G]iven any basis states $\ket{a}$ and $\ket{b}$ for a qubit, it is possible to express an arbitrary state as a linear combination $\alpha\ket{a} + \beta\ket{b}$ of those states." They go on to say that if the states are orthonormal, then one can perform a measurement with respect to the $\ket{a}\!, \ket{b}$ basis.

My question is does this imply that any two distinct pure states $\ket{a}$ and $\ket{b}$ form a basis for the Bloch sphere? (I apologize if this might make more sense in reference to a Hilbert space or some other vector space but I haven't gotten to that point in the book yet and have no experience with it otherwise).

It appears that any two distinct pure states are linearly independent, so I would suspect yes, but I do not believe I know enough to prove it.

For example, I attempted to try to transform the state defined by $\ket{\psi} = \frac{\sqrt{2 + \sqrt{2}}}{2}\ket{0} + \frac{\sqrt{2 - \sqrt{2}}}{2}\ket{1}$ into the basis $\{\ket{0}\!, \ \ket{+}\}$ where $\ket{+} = \frac{\ket{0} + \ket{1}}{\sqrt{2}}$. I obtained the following: $$ \begin{align*} \ket{\psi} &= \alpha\ket{0} + \beta\ket{1} = \alpha\ket{0} + \beta(\sqrt{2}\ket{+}-\ket{0}) = (\alpha - \beta)\ket{0} + \sqrt{2}\beta\ket{+}\\ &= \frac{ \sqrt{2+\sqrt{2}} - i\sqrt{2-\sqrt{2}} }{2}\ket{0} + \frac{i\sqrt{2}\sqrt{2 - \sqrt{2}}}{2} \ket{+} \end{align*} $$

This basis is neither orthogonal nor normal, so we would not be able to make a measurement with respect to it, however I believe that this is still a valid way to represent a state.

As the requirement that the basis be orthonormal in order to perform a measurement with respect to it implies that such bases exist, I would be interested in an example of orthogonal but not normal basis. I believe that taking two orthogonal bases (antipodal with respect to the bloch sphere), then scaling them, would be sufficient to realize this type of example.

I would also be interested in an example of a normal but not orthogonal basis, however no simple examples come to mind (it may be that there are no simple examples).

Thanks for your answer! Am I correct in understanding that you are saying that two distinct states form a basis for a qubit only if we define distinct as not being equal up to the global phase? If so, this would seem to imply that not all two non-equal points on the Bloch sphere define a basis for a qubit. This is slightly counter intuitive when compared to the real case, but I imagine this is due to the fact that $\alpha$ and $\beta$ are complex, and multiplying by a complex number can change direction in more ways than multiplying by a real number. – Techmaster21 – 2020-04-24T01:15:48.927

I'm also not certain I got it correct when I said that $\left|0\right>!,\ \left|+\right>$ are not normal. They are both normal as I understand it, however in using them to represent a state that resulting state may not be normal with respect to the basis. Is that due to the fact that they are not orthogonal? – Techmaster21 – 2020-04-24T01:16:09.173

$|0\rangle$ and $|+\rangle$ are normal but not orthogonal. If you use a not-orthogonal basis, then the amplitudes do not have to have a sum-mod-square equal to 1. – DaftWullie – 2020-04-24T07:45:33.213