Can I Convert a Qubit in complex form to polar form?

Question

Let's say I have a qubit
$$\left| \psi \right> = (\alpha_1 + i\alpha_2 ) \left|0\right> + (\beta_1 + i\beta_2 )\left|1\right>$$

Can I able convert this to polar form $$|\psi \rangle = \cos\big(\frac{\theta}{2}\big) |0\rangle + e^{i\varphi} \sin\big(\frac{\theta}{2}\big) |1\rangle$$

and is it possible to convert back the polar form to complex form?

Answer 1

In polar form we have, $\alpha_1 + i\alpha_2 = r_1e^{i\phi_1}$, and $\beta_1 + i\beta_2 = r_2e^{i\phi_2}$

So,

$$|\psi\rangle = r_1e^{i\phi_1} |0\rangle + r_2e^{i\phi_2} |1\rangle = e^{i\phi_1}(r_1 |0\rangle + r_2e^{i(\phi_2 - \phi_1)} |1\rangle)$$

Lets, $\varphi = \phi_2 - \phi_1$ then,

$$|\psi\rangle = e^{i\phi_1}(r_1 |0\rangle + r_2e^{i\varphi} |1\rangle)$$

Assuming that $|\psi\rangle$ is a normalized quantum state (that is, $r_1^2 + r_2^2 = 1$) we can find an angle $\theta$ such that $\cos(\frac{\theta}{2}) = r_1$ and $\sin(\frac{\theta}{2}) = r_2$. That means,

$$|\psi\rangle = e^{i\phi_1}(\cos{\small(\theta/2)} |0\rangle + e^{i\varphi} \sin{\small(\theta/2)} |1\rangle)$$

And $\phi_1$ is a global phase.

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This code snippet calculates $\theta$ and $\varphi$ given two complex numbers $z_1=\alpha_1 + i\alpha_2$ and $z_2=\beta_1 + i\beta_2$

import math, cmath

r1, phi1 = cmath.polar(z1)
r2, phi2 = cmath.polar(z2)

theta = 2 * math.acos(r1)
phi = phi2 - phi1

print('θ =', theta)
print('φ =', phi)

Answer 2

Yes, if you can accept that states are equivalent up to a global phase. The second equation assumes that the $|0\rangle$ coefficient is real. We can fix the first equation by adding a global phase so that the $|0\rangle$ coefficient is also real.

If you can make the assumption that the coefficient $(\alpha_1 + i\alpha_2)$ is a real number, i.e. $\alpha_2 = 0$, then there will always be a solution for $\theta$ and $\varphi$