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?


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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.