Solving a Bloch sphere where alpha is imaginary

Question

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

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!

Answer 1

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.