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

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!

Mark Spinelli
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Max Rush
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1 Answers1

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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.

Richard Rodenbusch
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    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. – Richard Rodenbusch Sep 28 '22 at 00:53
  • 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 – Max Rush Sep 28 '22 at 12:39
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    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 $\sqrt{i} = \frac{\sqrt{2}}{2} (1 + i)$ – Richard Rodenbusch Sep 28 '22 at 15:59