# Why can the most general state of a qubit be written as $|\Psi\rangle=\cos(\frac\theta2)|0\rangle+e^{i\phi}\sin(\frac\theta2)|1\rangle$?

Why we can express a most general qubit as $$|\Psi\rangle = \cos{\left(\frac{\theta}{2}\right)}|0\rangle + e^{i \phi} \sin{\left(\frac{\theta}{2}\right)} |1\rangle$$? Is there any formal proof for this?

• Generally the state of s qubit can be written as $\alpha |0\rangle + \beta |1\rangle$ (a bit of $|0\rangle$ with some probability and maybe some $|1\rangle$ with some probability ). Since we're talking probabilities here, so the sum should be 1, that means $\alpha + \beta = 1$. Maybe If you want to work it out yourself, you can try to check trigonometry rules and show that $\cos{\left(\frac{\theta}{2}\right)} + e^{i \phi} \sin{\left(\frac{\theta}{2}\right)} =1$ (knowing that we can have complex numbers in the coefficient) Nov 7 at 17:56

The most general pure state of a qubit can be written as $$|\Psi\rangle=a|0\rangle+b|1\rangle$$ where $$a,b\in\mathbb{C}$$. The amplitudes $$a$$ and $$b$$ can be written in polar form as $$a=re^{i\alpha}$$ and $$b=se^{i\beta}$$ where $$r,s\in[0,\infty)$$ and $$\alpha,\beta\in[0,2\pi)$$. Thus, $$|\Psi\rangle$$ is described by four real parameters $$r,s,\alpha,\beta$$ as

$$|\Psi\rangle=re^{i\alpha}|0\rangle+se^{i\beta}|1\rangle.\tag1$$

However, there are two constraints. First, the squares of the absolute values of the amplitudes are probabilities and therefore sum to one

$$|a|^2+|b|^2=r^2+s^2=1.\tag2$$

Consequently, $$r,s\in[0,1]$$. Now, for any real number $$x\in[0,1]$$ there exists a unique $$\varphi\in[0,\pi)$$ such that $$x=\cos\frac{\varphi}{2}$$. Let $$\theta\in[0,\pi)$$ be such that $$r=\cos\frac{\theta}{2}$$. Then from $$(2)$$ we have that $$s=\sin\frac{\theta}{2}$$ and substituting into $$(1)$$, we obtain

$$|\Psi\rangle=e^{i\alpha}\cos\frac{\theta}{2}|0\rangle+e^{i\beta}\sin\frac{\theta}{2}|1\rangle.\tag3$$

The second constraint arises from the fact that the global phase is unobservable. This allows us to fix the phase on $$|0\rangle$$ to be a positive real number. We can force this by dividing the amplitudes in $$(3)$$ by $$e^{i\alpha}$$

$$|\Psi\rangle\equiv\cos\frac{\theta}{2}|0\rangle+e^{i(\beta-\alpha)}\sin\frac{\theta}{2}|1\rangle\tag4$$

where $$\equiv$$ denotes equality up to global phase. Defining $$\phi=\beta-\alpha$$ we have

$$|\Psi\rangle\equiv\cos\frac{\theta}{2}|0\rangle+e^{i\phi}\sin\frac{\theta}{2}|1\rangle\tag5$$

where we can take $$\phi\in[0,2\pi)$$.