I understand that given a pure state $ |\psi\rangle$, we can express it in terms of two angles $\theta$ and $\varphi$ such that $|\psi\rangle = \cos(\theta/2)|0\rangle + \mathrm{e}^{i\varphi}\sin(\theta/2)|1\rangle $, and this is derived by converting from $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$ into their representations in terms of $r\mathrm{e}^{i\theta}$, and then factoring and rearranging that.

But how do I convert between the two representations given arbitrary states? I know that $|0\rangle = (\theta,\varphi) = (0,0), |1\rangle = (\pi,0), |+\rangle = (\pi/2, 0)$ etc, but how do I get it for an arbitrary state $|\psi\rangle$?

So far, I have:

  1. If $\alpha$ is complex, shift the entire state by phase $\bar{\alpha}$, where $\bar\alpha$ is the complex conjugate of $\alpha$, to end up with $\alpha\bar\alpha |0\rangle + \bar\alpha\beta|1\rangle$
  2. Use the formulas:

$$ \theta = 2 \arccos(\alpha\bar\alpha) \\ \varphi = ??? $$


1 Answer 1


As in if given $ |\psi\rangle=\alpha |0\rangle+\beta |1\rangle$, and you want it in the form $\cos(\theta/2) |0\rangle+e^{i\phi}\sin(\theta/2) |1\rangle$?

Assume the state is normalised $|\alpha|^2+|\beta|^2=1$.

I'd first start by multiplying by $\frac{\alpha^*}{|\alpha|}$ which is a phase (complex number unit length), and use $\alpha \alpha^*=|\alpha|^2$.

$$\frac{\alpha^*}{|\alpha|}|\psi\rangle=|\alpha| |0\rangle+\frac{\beta\alpha^*}{|\alpha|}|1\rangle =\cos(\theta/2) |0\rangle+e^{i\phi}\sin(\theta/2) |1\rangle$$

So $\theta=2\arccos(|\alpha|)$ or $\theta=2\arcsin(|\beta|)$.

Then $\frac{\beta\alpha^*}{|\alpha|}=e^{i\phi}|\beta|$ or $\frac{\beta\alpha^*}{|\alpha||\beta|}=e^{i\phi}$,

so that $\phi=\arg\left(\frac{\beta\alpha^*}{|\alpha||\beta|}\right)$ and it depends on how you want to calculate that, which branch of the loagarithm to take/where to measure angles from.

You could do something like $\phi=-i\ln\left(\frac{\beta\alpha^*}{|\alpha||\beta|}\right)$ etc.

In summary try

$$ \theta = 2 \arccos(|\alpha|) \\ \varphi = \arg\left(\frac{\beta\alpha^*}{|\alpha||\beta|}\right) $$

  • $\begingroup$ Why does $e^{i\varphi}|\beta| = \frac{\beta\alpha*}{|\alpha|}$? Where did the $|\beta|$ come from, or am I missing something? $\endgroup$
    – Isaac Khor
    Commented Mar 8, 2020 at 18:00
  • 1
    $\begingroup$ @IsaacKhor I could've written it a bit clearer, if $\theta=2\arcsin(|\beta|)$ which you get from equating the $|1\rangle$ coefficients, then $|\beta|=\sin(\theta/2)$. So again comparing the $|1\rangle$ coefficients you get that $\frac{\beta \alpha^*}{|\alpha|}=e^{i\phi}\sin(\theta/2)=e^{i\phi}|\beta|$ $\endgroup$
    – snulty
    Commented Mar 8, 2020 at 18:43

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