# Solving a Bloch sphere where alpha is imaginary

$$\begin{array}{l} |\psi\rangle=\frac{\sqrt{3 i}}{2}|0\rangle-\frac{1}{2}|1\rangle \\ |\psi\rangle=0.924|0\rangle-0.382 i |1\rangle \end{array}$$

Basically I’m trying to convert these to a standard form/Bloch sphere. I thought the Bloch sphere was supposed to use real numbers so that $$\cos(\theta/2)=\alpha$$ and same for $$\beta$$. But I’m not sure how you would do it with complex numbers and it’s really hard to find resources. Thanks!

• can i just multiple the whole thing by i because psi equals 1 so it wouldnt change anything?? Sep 27, 2022 at 22:41

The bloch sphere can be confusing and intimidating for sure. There are a couple of key things I always try to remember when I want to work with it.

1. Euler's formula is always our friend: $$e^{ix}=\cos{x}+i\sin{x}$$
2. The convention of using half-angles makes the math simpler when you start to manipulate quantum states around the bloch sphere. But the bloch sphere is just a spherical coordinate system with radius of 1. These spherical coordinates are where $$\theta$$ and $$\psi$$ come from. Wikipedia is helpful to keep $$\theta$$ and $$\frac{\theta}{2}$$ straight.
3. For a single qubit, quantum state, the global-phase is 'immeasureable'. This means we can really only measure the relative phase difference between the two eigenstates $$\lvert 0 \rangle \text{ and } \lvert 1 \rangle$$. So that means that $$\lvert \psi \rangle$$ and $$e^{i\gamma}\lvert\psi\rangle$$ will 'measure' the same $$\forall \gamma \in [0,2\pi)$$. We say that $$\psi$$ and $$e^{i\gamma}\lvert\psi\rangle$$ are equvialent "up to a global phase". This lets us 'arbitrarily' change the global phase without changing any outcomes.

It is nice if our quantum state is already in the form $$\psi\text{ = }\alpha\lvert 0\rangle + \beta e^{\phi i}\lvert 1 \rangle \text{ } \alpha,\beta,\phi \in \mathbb{R}$$ We can then get $$\theta\text{ or }\frac{\theta}{2}$$ from $$\alpha$$. If our state is not in that form, such as your examples, we are going to exploit a global phase to map $$\psi$$ to an equivalent state $$\psi^{'}$$ which will be in the form we want.

So we start with: $$\psi\text{ = }\alpha \lvert 0 \rangle + \beta\lvert 1\rangle \text{ = }(\alpha_0 + \alpha_1 i)\lvert 0 \rangle + (\beta_0 + \beta_1 i)\lvert 1 \rangle$$ and we just rewrite $$\psi$$, using Euler's formula, to: $$\psi = r_\alpha e^{\gamma i}\lvert 0 \rangle + r_\beta e^{\rho i}\lvert 1 \rangle$$ then map $$\psi$$ to the equivalent $$\psi^{'}$$ using the global phase $$e^{-\gamma i}$$ which gives us: $$\psi^{'}=r_\alpha\lvert 0 \rangle + r_\beta e^{(\rho-\gamma)i}\lvert 1 \rangle$$

Now we have an equivalent state in the form that we like. Some trigonometry will get us to $$\theta$$,$$\frac{\theta}{2}$$ and $$\phi$$ that you can map onto your bloch sphere.

I hope that helps some.

• I just saw your follow-up question. You can't multiply by i to get what you want because that will change the relative phase between $\lvert 0\rangle$ and $\lvert 1 \rangle$. But you can multiply by $e^{\gamma i}$ to get to where you want. Sep 28, 2022 at 0:53
• Thanks a lot buddy! It’s hard to find information on this stuff lol, my quantum computing class is supposed to be the easy part right now Sep 28, 2022 at 12:39
• Yea it can all be overwhelming. One thing I want to point out is that in your first state, you seem to have a $\sqrt{i}$ . If i is truly under the radical, remember to expand it out with $\sqrt{i} = \frac{\sqrt{2}}{2} (1 + i)$ Sep 28, 2022 at 15:59