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Why are qubits represented as $$\left|{q}\right\rangle = \alpha\left|{0}\right\rangle+\beta\left|{1}\right\rangle\equiv\alpha\left[{1 \ 0}\right]^T+\beta\left[{0 \ 1}\right]^T; \alpha,\beta\in\mathbb{C}$$ rather than $$\left|{q}\right\rangle = a\left[{1 \ 0 \ 0}\right]^T+b\left[{0 \ 1 \ 0}\right]^T+c\left[{0 \ 0 \ 1}\right]^T; a,b,c\in\mathbb{R}\text{ ?}$$ Wouldn't these to representations be equivalent (assuming $\alpha^2+\beta^2=1$ and $a^2+b^2+c^2=1$)? If so, what is the mathematical (not physical) reason for choosing the former representation?

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To all that are interested in Quantum Information: The Quantum Information and Foundations proposal is currently in commitment phase. –  draks ... May 8 '12 at 17:51
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up vote 3 down vote accepted

Since $|q\rangle$ is an element of $\mathbb{C}^2$, $\alpha$ and $\beta$ are complex you have 4 degrees of freedom in total. The group acting on qubits is the unitary group $\text{U}(2)$, with a 4 dimensional Lie algebra, given you the possibility to alter all coefficients $\alpha=\alpha_1+i\alpha_2$ and $\beta=\beta_1+i\beta_2$.

Your real representation has only 3 dimensional and is therefore not enough. I also found it strange, since 1 of these degrees, the complex phase $e^{-i\phi}$ is physically not measurable, but if you work with more than 1 qubit, it becomes important, for example $\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$ and $\frac{1}{\sqrt{2}}(|00\rangle-|11\rangle)$ are orthogonal, but differ only in the phase of one qubit.

If you work with density operators $|q\rangle\langle q|$ a phase factor $e^{-i\phi}$ (1 degree of freedom) cancels and gives a 3 dimensional representation for 1 qubit.

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