Why are half angles used in the Bloch sphere representation of qubits?

Question

Suppose we have a single qubit with state $| \psi \rangle = \alpha | 0 \rangle + \beta | 1 \rangle$. We know that $|\alpha|^2 + |\beta|^2 = 1$, so we can write $| \alpha | = \cos(\theta)$, $| \beta | = \sin(\theta)$ for some real number $\theta$. Then, since only the relative phase between $\alpha$ and $\beta$ is physical, we can take $\alpha$ to be real. So we can now write

$$| \psi \rangle = \cos(\theta) | 0 \rangle + e^{i \phi} \sin(\theta)| 1 \rangle$$

My Question: Why are points on the Bloch sphere usually associated to vectors written as $$| \psi \rangle = \cos(\theta/2) | 0 \rangle + e^{i \phi} \sin(\theta/2)| 1 \rangle$$ instead of as I have written? Why use $\theta /2$ instead of just $\theta$?

Answer 1

It is a convention, chosen so that $\theta$ is the azimuthal angle of the point representing the state in the Bloch sphere.

To see where this convention comes from, start from a state $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$. Remembering the normalisation constraint $|\alpha|^2+|\beta|^2=1$, and assuming without loss of generality $\alpha\in\mathbb R$, a natural way to parametrise the state is by defining an angle $\gamma$ such that $|\alpha|=\alpha=\cos\gamma$ and $|\beta|=\sin\gamma$. A generic state $|\psi\rangle$ thus reads $$|\psi\rangle=\cos\gamma|0\rangle + e^{i\varphi}\sin\gamma|1\rangle,$$ for some phase $\varphi\in\mathbb R$. Remember now that the Bloch sphere coordinates of a generic (pure) state $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$ have the explicit form \begin{align}\newcommand{\on}[1]{\operatorname{#1}}\newcommand{\bs}[1]{\boldsymbol{#1}} x\equiv\langle\psi|\sigma_x|\psi\rangle &= 2\on{Re}(\bar\alpha\beta),\\ y\equiv\langle\psi|\sigma_y|\psi\rangle &= 2\on{Im}(\bar\alpha\beta),\\ z\equiv\langle\psi|\sigma_z|\psi\rangle &= |\alpha|^2 - |\beta|^2. \end{align} Relating these with our previous parametrisation with $\gamma$ we find $$z=\cos^2\gamma-\sin^2\gamma=\cos(2\gamma).$$ But the canonical way to define spherical coordinates uses $z=\cos\theta$, so if we wish to interpret the coefficients of the state as angles in the Bloch sphere, we have to set $\gamma=\theta/2$.

See also the analogous question on physics.SE for more info.

Answer 2

If we use the convention $$| \psi \rangle = \cos(\theta) | 0 \rangle + e^{i \phi} \sin(\theta)| 1 \rangle$$ then the North ($\theta=0$) and the South ($\theta=\pi)$ are (physically) the same state $|0\rangle$;

If we use the convention$$| \psi \rangle = \cos(\theta/2) | 0 \rangle + e^{i \phi} \sin(\theta/2)| 1 \rangle$$

then North is $|0\rangle$ and South is $|1\rangle$ which is better.

Answer 3

Let $\hat{n}=(\cos\phi\sin\theta,\sin\phi \sin\theta,\cos\theta)$ i.e. the Cartesian coordinate vector for a point on the unit sphere with polar angle $\theta$ and azimuthal angle $\phi$. By sending a spin-1/2 particle through a Stern-Gerlach device with orientation $\hat{n}$, we can measure the observable

\begin{align} S_n:=\vec{S}\cdot \hat{n} &=S_x \cos\phi\sin\theta +S_y \sin\phi \sin\theta+S_z \cos\theta\\ &= \frac{\hbar}{2}\begin{pmatrix} \cos\theta & \cos\phi\sin\theta-i \sin\phi \sin\theta\\ \cos\phi\sin\theta+i \sin\phi \sin\theta & -\cos\theta\end{pmatrix} \\&= \begin{pmatrix} \cos \theta & e^{-i\phi}\sin\theta \\ e^{i\phi}\sin\theta & -\cos\theta\end{pmatrix} \end{align} in the $S_z$ basis. The obvious step is now to determine eigenvalues and eigenvectors. But if we denote the spin-up and spin-down eigenstates of $S_z$ as $|0\rangle$ and $|1\rangle$ respectively, then

$$| \psi \rangle = \cos(\theta/2) |0 \rangle + e^{i \phi} \sin(\theta/2)| 1 \rangle=\begin{pmatrix} \cos(\theta/2)\\ e^{i\phi}\sin(\theta/2)\end{pmatrix}$$ and therefore \begin{align} S_n |\psi\rangle &= \frac{\hbar}{2}\begin{pmatrix} \cos \theta & e^{-i\phi}\sin\theta \\ e^{i\phi}\sin\theta & -\cos\theta\end{pmatrix}\begin{pmatrix} \cos(\theta/2)\\ e^{i\phi}\sin(\theta/2)\end{pmatrix} \\ &= \frac{\hbar}{2}\begin{pmatrix} \cos(\theta)\cos(\theta/2)+\sin(\theta)\sin(\theta/2)\\ e^{i\phi}[\sin(\theta)\cos(\theta/2)-\cos(\theta)\sin(\theta/2)]\end{pmatrix}\\ &= \frac{\hbar}{2}\begin{pmatrix} \cos(\theta/2)\\ e^{i\phi}\sin(\theta/2)\end{pmatrix}=+\frac{\hbar}{2}|\psi\rangle \end{align} where in the second-to-last equality I've used the trigonometric product-to-sum formula. Hence $|\psi\rangle$ is the $S_n=+\hbar/2$ eigenstate. In other words: If a spin-1/2 particle passes through an SG device with orientation $\hat{n}$ and comes out deflected up, then $|\psi\rangle$ is the resulting spin state. (Correspondingly, one can show that $S_{-n}|\psi\rangle=-\hbar/2|\psi\rangle$ i.e. $|\psi\rangle$ will deflect down if the SG device is flipped.) The upshot is that $\theta,\phi$ are not angles in Hilbert space; rather, they're the angles in real space for the SG device for which $|\psi\rangle$ is the spin state of the upward-deflected beam.

Note that the above description is limited to points on the surface of the Bloch sphere i.e pure states. For points on the interior of the Bloch sphere, we need to go to the density matrix formalism as presented by gLs and I'll defer to that answer.

Answer 4

It is just a convention for $ 0 \le \theta \le \pi $. You can write it your way (indeed you can "include" a constant in a variable) but in that case $ 0 \le \theta \le \pi / 2 $.

But we take this convention for unique coordinates. If you refer to the Spherical coordinate system

you can see that if you want a unique set of spherical coordinates for each point of the sphere, you need to restrict their range.