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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?

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  • $\begingroup$ 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) $\endgroup$
    – user206904
    Nov 7 at 17:56
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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 $\phi\in[0,2\pi)$.

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