# Explicit form of a qubit state and relative phase

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.

• To elaborate more on ZeroTheHero's great answer, I asked a similar question about the meaning of the relative phase in the wave function a couple of years ago, and the answer there shows how the relative phase can change measurement results: physics.stackexchange.com/questions/177588/… Apr 13, 2020 at 9:30

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.