# Reducing the number of parameters of a quantum state from 4 to 3

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.

• 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. Jun 14 at 0:34
• @DanielC ohhh, I see it now. Thanks for the clarification! Jun 14 at 0:46

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}
• $\vert\alpha\vert^2+\vert\beta\vert^2=1$ 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. Jun 14 at 4:14
• @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. Jun 14 at 4:16
• 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! Jun 14 at 4:18