# Vector from SU(2) to SO(3)?

I know how to change the special unitary matrix in $$SU(2)$$ to the matrix in $$SO(3)$$, and I found one way to change the state(vector) from $$2\times 1$$ to $$3\times 1$$, but I don't know why.

The method is shown below:

For any vector$$|\psi\rangle \equiv (a+bi,c+di)^T$$, the corresponding one is $$$$\tag{1} \begin{pmatrix} 2(ac + bd) \\ 2(-bc + ad)\\ a^2+b^2-c^2-d^2 \end{pmatrix}$$$$ The process is :$$\langle\psi|\vec{\sigma} |\psi\rangle$$, and the $$x$$ component of it is the first line in $$(1)$$, the $$y$$ component of it is the second line in $$(1)$$ and so on, where $$\sigma$$ stands for Pauli operator

$$\sigma _x = \begin{pmatrix}0&1\\1&0\end{pmatrix},\sigma _y = \begin{pmatrix}0&-i\\i&0\end{pmatrix}, \sigma _z=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$$

For example, if the $$2\times 1$$ vector is $$\begin{pmatrix}\cos(\theta/2)\\ \sin(\theta /2)e^{i\phi}\end{pmatrix}$$ then using transformation $$(1)$$ the corresponding $$3\times 1$$ vector is

$$\begin{pmatrix}\sin(\theta)\cos(\phi)\\ \sin(\theta)\sin(\phi)\\ \cos(\theta)\end{pmatrix}.$$

However, I don't understand how to prove (1).

Cross-posted on physics.SE

## TL;DR

You have rediscovered the Bloch sphere! :-)

## Interesting special cases

Before we prove that the map defined by equation $$(1)$$ is the representation of single-qubit quantum states as points on the Bloch sphere, it is instructive to consider a few special cases. Let us denote the map with $$\phi: \mathcal{H} \to \mathbb{R}^3$$ where $$\mathcal{H}$$ is the Hilbert space of a qubit. We calculate that

$$\phi(|0\rangle) = \phi\left(\begin{pmatrix}1 \\ 0\end{pmatrix}\right) = \begin{pmatrix} 2 \cdot (1 \cdot 0 + 0 \cdot 0) \\ 2 \cdot (-0 \cdot 0 + 1 \cdot 0) \\ 1^2 + 0^2 - 0^2 - 0^2 \end{pmatrix} = \begin{pmatrix} 0\\0\\1 \end{pmatrix} \\ \phi(|1\rangle) = \phi\left(\begin{pmatrix}0 \\ 1\end{pmatrix}\right) = \begin{pmatrix} 2 \cdot (0 \cdot 1 + 0 \cdot 0) \\ 2 \cdot (-0 \cdot 1 + 0 \cdot 0) \\ 0^2 + 0^2 - 1^2 - 0^2 \end{pmatrix} = \begin{pmatrix} 0\\0\\-1 \end{pmatrix} \\ \phi(|+\rangle) = \phi\left(\begin{pmatrix}\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}\end{pmatrix}\right) = \begin{pmatrix} 2 \cdot \left(\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + 0 \cdot 0\right) \\ 2 \cdot \left(-0 \cdot \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \cdot 0\right) \\ \left(\frac{1}{\sqrt{2}}\right)^2 + 0^2 - \left(\frac{1}{\sqrt{2}}\right)^2 - 0^2 \end{pmatrix} = \begin{pmatrix} 1\\0\\0 \end{pmatrix} \\ \phi(|-\rangle) = \phi\left(\begin{pmatrix}\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}}\end{pmatrix}\right) = \begin{pmatrix} 2 \cdot \left(\frac{1}{\sqrt{2}} \cdot \left(-\frac{1}{\sqrt{2}}\right) + 0 \cdot 0\right) \\ 2 \cdot \left(-0 \cdot \left(-\frac{1}{\sqrt{2}}\right) + \frac{1}{\sqrt{2}} \cdot 0\right) \\ \left(\frac{1}{\sqrt{2}}\right)^2 + 0^2 - \left(-\frac{1}{\sqrt{2}}\right)^2 - 0^2 \end{pmatrix} = \begin{pmatrix} -1\\0\\0 \end{pmatrix} \\ \phi(|{+i}\rangle) = \phi\left(\begin{pmatrix}\frac{1}{\sqrt{2}} \\ \frac{i}{\sqrt{2}}\end{pmatrix}\right) = \begin{pmatrix} 2 \cdot \left(\frac{1}{\sqrt{2}} \cdot 0 + 0 \cdot \frac{1}{\sqrt{2}}\right) \\ 2 \cdot \left(-0 \cdot 0 + \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}\right) \\ \left(\frac{1}{\sqrt{2}}\right)^2 + 0^2 - 0^2 - \left(\frac{1}{\sqrt{2}}\right)^2 \end{pmatrix} = \begin{pmatrix}0\\1\\0 \end{pmatrix} \\ \phi(|{-i}\rangle) = \phi\left(\begin{pmatrix}\frac{1}{\sqrt{2}} \\ -\frac{i}{\sqrt{2}}\end{pmatrix}\right) = \begin{pmatrix} 2 \cdot \left(\frac{1}{\sqrt{2}} \cdot 0 + 0 \cdot \left(-\frac{1}{\sqrt{2}}\right)\right) \\ 2 \cdot \left(-0 \cdot 0 + \frac{1}{\sqrt{2}} \cdot \left(-\frac{1}{\sqrt{2}}\right)\right) \\ \left(\frac{1}{\sqrt{2}}\right)^2 + 0^2 - 0^2 - \left(-\frac{1}{\sqrt{2}}\right)^2 \end{pmatrix} = \begin{pmatrix}0\\-1\\0 \end{pmatrix}.$$

Though not a proof, the above equalities are very suggestive.

Moreover, these equalities explain why the three components of $$\langle\psi|\vec{\sigma}|\psi\rangle$$ correspond to the three components of $$\phi(|\psi\rangle)$$. For example, $$|+\rangle$$ and $$|-\rangle$$ are eigenvectors of $$\sigma_x$$, so $$\phi$$ maps them to vectors that are fixed by the corresponding 3D rotation of $$\mathbb{R}^3$$.

## Proof in the general case

You have almost obtained the proof in the general case when you switched from representing an arbitrary quantum state as $$|\psi\rangle=(a + bi, c + di)^T$$ to representing it as $$|\psi\rangle=(\cos\frac{\theta}{2}, e^{i\phi}\sin\frac{\theta}{2})$$ and calculated that

\begin{align} \phi(|\psi\rangle) &= \phi\left(\begin{pmatrix}\cos\frac{\theta}{2} \\ e^{i\phi}\sin\frac{\theta}{2}\end{pmatrix}\right) \\ & =\begin{pmatrix} 2 \cos\frac{\theta}{2} \sin\frac{\theta}{2} \cos\phi \\ 2 \cos\frac{\theta}{2} \sin\frac{\theta}{2} \sin\phi\\ \cos^2\frac{\theta}{2} - \sin^2\frac{\theta}{2} \cos^2\phi - \sin^2\frac{\theta}{2} \sin^2\phi \end{pmatrix} \\ &= \begin{pmatrix} \sin\theta\cos\phi \\ \sin\theta\sin\phi \\ \cos\theta \end{pmatrix}. \end{align}

Now the only missing step is to notice that the last result is the point on the unit sphere in $$\mathbb{R}^3$$ with azimutal angle $$\phi$$ and polar angle $$\frac{\theta}{2}$$, exactly as on the Bloch sphere.

## Relationship to the Lie group homomorphism $$SU(2)\to SO(3)$$

A key property of the Bloch sphere is that for any unitary $$U \in SU(2)$$ and any state $$|\psi\rangle \in \mathcal{H}$$, the 3-vector $$\phi(U|\psi\rangle)$$ can be obtained from the 3-vector $$\phi(|\psi\rangle)$$ by rotating the latter using $$\Phi(U)\in SO(3)$$ where $$\Phi: SU(2) \to SO(3)$$ denotes the Lie group homomorphism that maps single-qubit unitaries in $$SU(2)$$ to 3D rotations in $$SO(3)$$.

This fact can be expressed as

$$\phi(U|\psi\rangle) = \Phi(U)\phi(|\psi\rangle)\tag2$$

or by saying that the following diagram

$$\begin{array}{ccccc} &\mathcal{H} & \xrightarrow{\phi} & S^2 & \\ U&\Bigg\downarrow & & \Bigg\downarrow & \Phi(U) \\ &\mathcal{H} & \xrightarrow[\phi]{} & S^2 & \\ \end{array}$$

commutes for every $$U\in SU(2)$$. The proof of $$(2)$$ is a straightforward, but somewhat lengthy calculation.