2
$\begingroup$

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:

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

$\endgroup$
2
$\begingroup$

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

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

| improve this answer | |
$\endgroup$
  • $\begingroup$ Why does $e^{i\varphi}|\beta| = \frac{\beta\alpha*}{|\alpha|}$? Where did the $|\beta|$ come from, or am I missing something? $\endgroup$ – Isaac Khor Mar 8 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 Mar 8 at 18:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.