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

Question

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 $$\begin{equation} \tag{1} \begin{pmatrix} 2(ac + bd) \\ 2(-bc + ad)\\ a^2+b^2-c^2-d^2 \end{pmatrix} \end{equation} $$ 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).

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Cross-posted on physics.SE

Answer 1

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.