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We have a quantum state

$$ |\psi\rangle = \alpha|0\rangle + \beta|1\rangle, $$

where $\alpha$ and $\beta$ are complex numbers, i.e. $\alpha = a + bi$ and $\beta = c + di$. Therefore, our current parameter count is 4.

From this question, I understand how you can ignore one parameter because of global phase to go from 3 to 2 parameters. However, I don't understand how to go from 4 to 3 parameters with $\alpha^*\alpha + \beta^*\beta = 1$. Expanding this equation we get

$$ \begin{align} \alpha^*\alpha + \beta^*\beta &= (a-bi)(a+bi) + (c-di)(c+di) \\ &= a^2 + abi - abi -b^2i^2 + c^2 + cdi - cdi -d^2i^2 \\ &= a^2 + b^2 + c^2 + d^2 = 1 \end{align} $$

However, I don't understand how this expression helps us get rid of one parameter. I've been working with quantum computing for almost two years now, so I guess I'm missing to notice something elementary.

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  • $\begingroup$ Perhaps there is something really obvious here, but you're left with a constraint upon 4 real parameters, which means that 3 of them are independent. $\endgroup$
    – DanielC
    Jun 14 at 0:34
  • $\begingroup$ @DanielC ohhh, I see it now. Thanks for the clarification! $\endgroup$
    – epelaaez
    Jun 14 at 0:46
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Maybe a more conventional way to see this:

Write in polar form $\alpha=\vert\alpha\vert e^{i\phi_a}$, $\beta=\vert\beta\vert e^{i \phi_b}$. Then your state is $$ e^{i\phi_a}\left( \vert\alpha\vert \vert 0\rangle + e^{-i(\phi_b-\phi_a)}\vert \beta\vert\vert 1\rangle\right)\, . $$ You can eliminate the overall phase $e^{i\phi_a}$ as two states differing by an overall phase are equivalent. Your are then left with $3$ parameters: the magnitudes $\vert \alpha\vert$, $\vert \beta\vert$ and the phase difference $e^{i(\phi_b-\phi_a)}$.

To bring this down from $3$ to $2$, eliminate the overall phase $e^{i\phi_a}$ and write $\vert \alpha\vert=\cos\theta$, $\vert \beta\vert =\sin\theta$ so that $\vert\alpha\vert^2+\vert \beta\vert^2=1$, and $\varphi=\phi_b-\phi_a$. You then have \begin{align} \vert\psi\rangle\sim \cos\theta \vert 0\rangle + e^{i \varphi}\sin\theta \vert 1\rangle\, . \end{align}

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  • $\begingroup$ And then how do you insert the unitary constraint to get down to two parameters? $\endgroup$
    – epelaaez
    Jun 14 at 1:56
  • $\begingroup$ $\vert\alpha\vert^2+\vert\beta\vert^2=1$ so write $\vert\alpha\vert=\cos(\theta)$, $\vert\beta\vert=\sin(\theta)$ and $\varphi=\phi_b-\phi_a$. You now only have $\theta$ and $\varphi$ as parameters. $\endgroup$ Jun 14 at 4:14
  • $\begingroup$ @epelaaez To normalise the state, divide $\alpha$ and $\beta$ by $\sqrt{|\alpha|^2+|\beta|^2}$. Note that “unitary” is a property of operators, not of states. $\endgroup$
    – gandalf61
    Jun 14 at 4:16
  • $\begingroup$ Thank you @ZeroTheHero, that clears up my doubts! $\endgroup$
    – epelaaez
    Jun 14 at 4:16
  • $\begingroup$ Yes @gandalf61, I was just using the term “unitary” to refer to the fact that $|\alpha|^2 + |\beta|^2 = 1$. Probably should’ve referred to it differently. But thanks! $\endgroup$
    – epelaaez
    Jun 14 at 4:18

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