3
$\begingroup$

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).


Cross-posted on physics.SE

$\endgroup$
2
  • 2
    $\begingroup$ This is a duplicate of physics.stackexchange.com/q/606545. $\endgroup$
    – benrg
    Jan 10 at 7:42
  • $\begingroup$ I think it's two communities, so I post this question in two places. $\endgroup$
    – narip
    Jan 10 at 7:57
3
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

There's a few different ways to interpret the question. For starters, it is worth noting that the titular question cannot be interpreted at face value: $\mathbf{SU}(2)$ and $\mathbf{SO}(3)$ are not vector spaces, and thus their elements are not vectors. Rather, they are (among other things) groups, and more specifically Lie groups.

Nonetheless, these groups can indeed be made to act on vector spaces. In particular, they admit "natural" representations of the form $$\phi_{\mathbf{SU}(2)}:\mathbf{SU}(2)\to\operatorname{GL}(2,\mathbb C), \\ \phi_{\mathbf{SO}(3)}:\mathbf{SO}(3)\to\operatorname{GL}(3,\mathbb R).$$ Here $\operatorname{GL}(n,\mathbb F)$ denotes the group of invertible $n\times n$ matrices over a field $\mathbb F$. One also often defines these Lie groups as matrix groups, that is, presenting their elements directly as $2\times2$ special unitary and $3\times 3$ special orthogonal matrices. In which case, this paragraph can be safely ignored for our purposes.

These two groups are also tightly related: there is a (surjective) homomorphism $\phi:\mathbf{SU}(2)\to\mathbf{SO}(3)$, whose kernel is $\{\pm I\}\subset\mathbf{SU}(2)$. A nice way to define it is via the Lie algebras, as $$\phi\left(\exp\Big(\sum_{i=1}^3 c_i t_i\Big)\right) \equiv \exp\left(\sum_{i=1}^3 c_i J_i\right),$$ where $c_i\in\mathbb R$, and $t_i\in\mathfrak{su}(2)$ and $J_i\in\mathfrak{so}(3)$ denote basis elements for the Lie algebras. This homomorphism is thus not injective, but rather is "two-to-one". An easy way to verify this directly is computing $\exp[\alpha(-\frac{i}{2}\sigma_x)]$ and the corresponding image through $\phi$, which is $\exp(\alpha J_x)$. Note that here $-\frac{i}{2}\sigma_x\equiv t_x$ is a basis element of $\mathfrak{su}(2)$, and $J_x\equiv |3\rangle\!\langle2|-|2\rangle\!\langle3|$ is the corresponding basis element of $\mathfrak{so}(3)$.

So with this in mind, how should the question be interpreted? I'd say the point is whether $\phi$ can be leveraged to find some $f:\mathbb C^2\to\mathbb R^3$ which connects the spaces on which the two groups are represented, that is, one such that $$f(U\psi)=\phi(U)f(\psi),$$ for all $U\in\mathbf{SU}(2)$ and $\psi\in\mathbb C^2$. Note that, because $\phi$ is not injective, $f$ will probably also not be: the RHS won't change upon changing $U\to -U$, but the LHS might change, unless $f$ is chosen appropriately.

To make this more concrete, consider unitaries of the form $U=e^{\theta t_z}$, where $t_z\equiv -i\sigma_z/2$. Then, $$\psi\equiv\begin{pmatrix}\psi_0 \\\psi_1\end{pmatrix} \mapsto U\psi = \begin{pmatrix}e^{-i\theta/2}\psi_0 \\e^{i\theta/2}\psi_1\end{pmatrix} = e^{-i\theta/2}\begin{pmatrix}\psi_0 \\e^{i\theta}\psi_1\end{pmatrix}.$$ On the other hand, $\phi(U)=e^{\theta J_z}$, and thus $$ \phi(U)f(\psi) =\begin{pmatrix}\cos(\theta) & -\sin(\theta) & 0\\\sin(\theta)&\cos(\theta)&0\\0&0&1\end{pmatrix} \begin{pmatrix}f_1(\psi)\\ f_2(\psi) \\ f_3(\psi)\end{pmatrix} = \begin{pmatrix}c_\theta f_1-s_\theta f_2\\ s_\theta f_1+c_\theta f_2 \\ f_3\end{pmatrix}, $$ with some shorthand notation in the last equation.

I'm not completely sure what's an a priori way to get to the explicit form of $f$, but one can readily observe that the Hopf map does the job. This is the map defined as $$f(\psi) = \begin{pmatrix}2\Re(\bar\psi_0 \psi_1) \\ 2\Im(\bar\psi_0 \psi_1) \\ |\psi_0|^2 - |\psi_1|^2\end{pmatrix}. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.