Explicit form of a qubit state and relative phase

Question

In quantum computing, we can write our wave function as: $$ |\psi\rangle=\alpha|0\rangle+\beta|1\rangle $$ which can be rewritten as $$ |\psi\rangle=\cos(\theta/2)|0\rangle+e^{i\phi}\sin(\theta/2)|1\rangle $$ ignoring the global phase. This can be represented in a Bloch sphere where theta and phi are angles on the sphere. That being said, my question is, how is this rewritten equation found and why is there a $e^{i\phi}$ term multiplied by the $\sin(\theta/2)$? And what does this additional phase represent? And finally, why is it on the sine rather than the cosine? Thank you! To preface, I am currently an undergraduate physics junior who is doing research on QC/QI.

Answer 1

This is covered in very many textbooks. One example is the book of John Townsend. Anyways the point is if you write $$ \vert\psi\rangle = \alpha \vert 0\rangle +\beta \vert 1\rangle $$ you need $\vert\alpha\vert^2+\vert\beta\vert^2=1$ for normalization. Thus $$ \alpha=e^{i a}\cos\frac{1}{2}\theta\, ,\qquad \beta=e^{i b}\sin\frac{1}{2}\theta $$ will do the trick so one writes \begin{align} \vert\psi\rangle =e^{i a}\left(\cos(\textstyle\frac{1}{2}\theta)\vert 0\rangle + e^{i\phi} \sin(\textstyle\frac{1}{2}\theta)\vert 1\rangle\right) \tag{1} \end{align} with $\phi=b-a$ and choose $a=0$ since the factor $e^{i a}$ is an overall phase. It is conventional to choose the coefficient of $\vert 0\rangle$ to be real.

The half angle is related to the rotation properties of spin states. Alternatively, opposite spin states in 3D space are orthogonal in spin-1/2 space so set $\theta=0$ (North pole state) to get $\vert 0\rangle$ and $\theta=\pi$ (South pole state) to get $\vert 1\rangle$, which are indeed orthogonal.

This is the most general form. The relative phase $e^{i\phi}$ controls the interference between the two states. You can see its effects by computing and comparing the average values of $\hat \sigma_x$ and $\hat \sigma_y$ on (1): these average values are explicitly $\phi$-dependent.

Moreover the eigenstates of $\hat \sigma_x$ and $\hat \sigma_y$ only differ by a relative phase factor.