Why is $\theta \over 2$ used for a Bloch sphere instead of $\theta$?

Question

I'm a beginner in studying quantum info, and I'm a little confused about the representation of a qubit with a Bloch Sphere. Wikipedia says that we can use $$\lvert\Psi\rangle=\cos\frac{\theta}{2} \lvert 0\rangle + e^{i\phi}\sin\frac{\theta}{2} \lvert 1\rangle$$ to represent a pure state, and map it to the polar coordinates of the sphere. What I'm not sure about is, where does the "$\frac{\theta}{2}$" come in?

I mean, in polar coordinate, the vector equals $\cos{\theta}\ \hat{z} + e^{i\phi}\sin{\theta}\ \hat{x}$, but even if we use $\hat{z}=\lvert 0\rangle$ and $\hat{x}=\lvert 0\rangle + \lvert 1\rangle$, it's still different from above. How could this be transformed into the formula above?

Or... does this mean that the sphere is simply a graphical representation of $\theta$ and $\phi$, while $\lvert 0\rangle$ and $\lvert 1\rangle$ do not geometrically correspond to any vector on the sphere? (but here it writes $\hat{z}=\lvert 0\rangle$ and $-\hat{z}=\lvert 1\rangle$...)

Answer 1

I) The main point is that the half-angle $\frac{\theta}{2}$ doubles when we go from the ket $$\tag{1} |\psi\rangle~=~\begin{bmatrix}\cos\frac{\theta}{2} \cr e^{i\phi}\sin\frac{\theta}{2}\end{bmatrix}, \qquad ||\psi||~=~1, $$ to the density matrix/operator $$\tag{2}\rho~=~| \psi\rangle \langle\psi | ~=~\frac{1}{2}\left({\bf 1}_{2\times 2}+ \vec{r}\cdot \vec{\sigma}\right),\qquad {\rm tr}\rho~=~1. $$ In eq. (2) $$\tag{3}\vec{r}~=~\begin{bmatrix}x\cr y\cr z\end{bmatrix}~=~\begin{bmatrix}r\cos\phi\sin\theta\cr r\sin\phi\sin\theta\cr r\cos\theta\end{bmatrix},\qquad r~=~1, $$ is the radius vector in spherical coordinates, and $\sigma_i$ are the Pauli matrices. (We mention for completeness that the Bloch sphere $S^2=\partial B^3$ of pure qubit states is the boundary of the Bloch ball $B^3$ of mixed qubit states.)

II) Alternatively, for a detailed group theoretical explanation of the presence of the half-angle, consult e.g. Ref. 1. In short, the 2-dimensional Hilbert space $H\cong\mathbb{C}^2$ of the qubit is a spinor/dublet representation of the $G=SU(2)$ Lie group, which is a double cover of the 3D rotation group $SO(3)$. The adjoint representation $$\tag{4}{\rm Ad}:~ G ~\longrightarrow~GL(su(2),\mathbb{R}), $$ given by $$\tag{5} {\rm Ad}(g)\sigma~=~g\sigma g^{-1}, \qquad g~\in~G, \qquad \sigma~\in~su(2)~\cong ~\mathbb{R}^3, $$ is a Lie group homomorphism, whose image $$\tag{6} {\rm Ad}(G)~\cong ~SO(3), \qquad {\rm Ad}(\pm {\bf 1}_{2\times 2})~=~{\bf 1}_{3\times 3},$$ is isomorphic to $SO(3)$. The doubling of angles implicitly takes place in formula (5). (This is similar to the fact that a half-spin particle requires a $4\pi$ rotation (rather than $2\pi$) to get back to the starting point.)

References:

1. G. 't Hooft, Introduction to Lie Groups in Physics, lecture notes, chapter 6. The pdf file is available here.

Answer 2

$\newcommand{\ket}[1]{\left|{#1}\right\rangle}$

Or... does this mean that the sphere is simply a graphical representation of $\theta$ and $\phi$, while $\lvert 0\rangle$ and $\lvert 1\rangle$ do not geometrically correspond to any vector on the sphere? (but here it writes $\hat{z}=\lvert 0\rangle$ and $-\hat{z}=\lvert 1\rangle$...)

This is not an artificial graphical representation. But this representation of $\Psi$ on the Bloch sphere is based on stereographic projections, it is not a "linear" representation. For example the Euclidean equality $\ket{1}=-\ket{0}$ that you have noted, occurs only for the representations of $\ket{1}$ and $\ket{0}$, not for the "true" $\ket{1}$ and $\ket{0}$.

The $\frac{\theta}{2}$ can be seen on the picture given in my answer below. The red vector $\xi$ is the key point.

Once one writes the pure qubits as $$\Psi=\cos\frac{\theta}{2} \lvert 0\rangle + e^{i\varphi}\sin\frac{\theta}{2} \lvert 1\rangle \qquad (\star),$$ it is obvious that the spherical coordinates provide a one-one correspondence (a homeomorphism) between a pure qubit and a point on the 2D-sphere (the Riemann sphere, or Bloch sphere in this context). But I want to show that this homeomorphism is not artificial.

Pure qubits are rays

One usually defines a qubit as a vector in the complex plane $\mathbb{C}^2$ $$ \Psi = v_0 \ket{0} + v_1 \ket{1} = v_0 \begin{pmatrix} 1 \\ 0 \end{pmatrix}+ v_1 \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} v_0 \\ v_1 \end{pmatrix} $$ where $v_0$ and $v_1$ are complex numbers satisfying ${|v_0|}^2+{|v_1|}^2=1$ (then $\Psi$ is said to be a normalized vector). The space of qubits has dimension $3$.

But, when $\Psi$ and $\Psi'$ are two qubits differing by a complex proportionnality factor $z$ (necessarily having modulus $1$, hence $z=e^{i\alpha}$ and called a phase factor): $$ \Psi' = z \Psi = \begin{pmatrix} zv_0 \\ zv_1 \end{pmatrix} $$ they define the same "logic" through the Born rule (that also means that $\langle \Psi, A\Psi\rangle = \langle \Psi', A\Psi'\rangle$ for self-adjoint operators $A$), and considering only qubits having form $$\Psi=\cos\frac{\theta}{2} \lvert 0\rangle + e^{i\varphi}\sin\frac{\theta}{2} \lvert 1\rangle \qquad (\star),$$ called the pure qubits, is enough when we look at qubits up to a complex proportionnality factor.

The space of pure qubits defined in this way (qubit up to a proportionnality factor) is also known as the space of rays or the complexe projective space $\mathbb{C}\mathbb{P}^1$. This it the mathematical formalism behind pure qubits, and I will come back to this point.

Homeomorphism with the Riemann sphere

It is obvious that the expression $(\star)$ provides an homeomorphism between the space of pure qubits and the Riemann sphere with the help of spherical polar coordinates. Obviously this homeomorphism is not linear; for example it is clear that $\ket{1} \neq -\ket{0}$ while this relation can be seen on the representations of $\ket{1}$ and $\ket{0}$ on the Riemann sphere. And it is clear that the linear combination $$ \Psi=\cos\frac{\theta}{2} \lvert 0\rangle + e^{i\varphi}\sin\frac{\theta}{2} \lvert 1\rangle $$ does not occur on the 3D Euclidean representation.

Nevertheless, this homeomorphism is not an artifical one. In the sequel, let us carefully distinguish between $\Psi$ and its representation (the $\Psi$ shown on the sphere).

It is well known that the Riemann sphere $\mathbb{S}^2$ is a representation (is homeomorphic to) the space $\bar{\mathbb{C}}$ of complex numbers "plus a point at infinity" through the stereographic projection. The stereographic projection of the representation of $\Psi$ in the $(xy)$-plane is the vector $$ \xi = \tan\frac{\theta}{2} e^{i\varphi}, $$ shown in red on the figure below.

Interpreting the $(xy)$-plane as the space of complex numbers, note that $\xi$ actually lies in $\bar{\mathbb{C}}$ because $|1\rangle$ at the Southern pole is sent to the point at infinity (whereas $\ket{0}$ is sent to the origine of the plane). Denote by $\textit{Stereo1}$ this usual stereographic projection: $$ \textit{Stereo1}\colon \mathbb{S}^2 \to \bar{\mathbb{C}}, $$ which sends the representation of $\Psi$ to the red vector $\xi$.

The point is the following one. As said before, the space of qubits is the complex projective space $\mathbb{C}\mathbb{P}^1$. And this one is known to be homeomorphic to the Riemann sphere too. This homeomorphism is called the stereographic projection too: $$ \textit{Stereo2}\colon \mathbb{C}\mathbb{P}^1 \text{(the space of pure qubits)} \to \bar{\mathbb{C}}, $$ and it is given by $$ \textit{Stereo2}(\Psi) = \frac{v_1}{v_0} = \frac{e^{i\varphi}\sin \frac{\theta}{2}}{\cos\frac{\theta}{2}}= \tan\frac{\theta}{2} e^{i\varphi} = \xi. $$

This is why I said the homeomorphism provided by spherical polar coordinates is not an artifical one: it is a natural homeomorphism because of the relation $$ \textit{Stereo1}(\text{representation of $\Psi$}) = \textit{Stereo2}(\Psi), $$ that is to say $$ \text{representation of $\Psi$} = {\textit{Stereo1}}^{-1} \bigl(\textit{Stereo2}(\Psi)\bigr). $$

Summary card

Answer 3

$|0\rangle$ and $|1\rangle$ are perpendicular states. Geometrically, that means they should be 90 degrees apart; at right angles to each other. To get them to be 180 degrees apart, like they are on the Bloch sphere, we had to double all the angles.

Since all the angles are doubled on the Bloch sphere, if you're thinking in terms of those angles you have to remember to halve them when returning to just the raw states. That's where the $\theta/2$ comes from. The $\theta$ is a Bloch sphere angle, so it has to be halved to get the actual proper angle.

Why do we put up with a representation where perpendicular things are 180 degrees apart? Because it makes such a nice analogy between single-qubit quantum operations and rotations in 3d space. The Bloch sphere also happens to save not one but two degrees of freedom when plotting. If we had stuck with the 90-degrees-apart thing then we'd be plotting in four-dimensional space, and wouldn't be able to use distance from the center to represent purity.

Yes there are deeper mathematical reasons for all of this, covered by the other answers, but pragmatically speaking it comes down to having to double the angles to get the nice analogy with rotations.

Answer 4

From the way it is defined $\left| \Psi \right\rangle$ is not a vector on the sphere, but rather a vector along the z-axis between $-\hat{z}$ and $\hat{z}$, because it is a linear combination of $\left|0\right\rangle$ and $\left|1\right\rangle$ which are both vectors along the z-axis.

Now we want $\left|\Psi(\theta = 0 , \phi =0)\right\rangle = \left|0\right\rangle$, and $\left|\Psi(\theta = \pi , \phi =0)\right\rangle = \left|1\right\rangle$ which gives $\left|\Psi(\theta , \phi)\right\rangle = \text{cos}\frac{\theta}{2}\left|0\right\rangle + \text{sin}\frac{\theta}{2}e^{i\phi}\left|0\right\rangle$, because cos$\frac{\pi}{2}=0$ and sin$\frac{\pi}{2}=1$.

Answer 5

For example here you can find a detailed explanatiion of why it's $\theta/2$ and not $\theta$, where $\theta\in [0,\pi]$.

Briefly, if you had $\theta$, due to the global phase invariance the antipodal points on the sphere would correspond to the same state (they would differ only by a sign). So only the upper hemisphere ($\theta\in [0,\pi/2]$) is needed to represent all the possible qubit states. But having a full sphere is more convinient (it has the full rotational symmetry, and the rotation operators can represent unitary operations on the qubit states), so we map the upper hemisphere to the full sphere (Bloch sphere) by $\theta'\to 2\theta$, on which the antipodal points are now corresponding to two mutually orthogonal qubit states.

Answer 6

For a long time (let’s say the 20th century), the prototypical 2-level quantum object was the angular momentum of a spin-$\frac12$ particle (electron, nucleon, nucleus). The transition to qubits has only happens during the late 1990’s, and mainly in quantum information. The Bloch sphere was introduced in 1946 (paper,$$) by Felix Bloch to study the quantum state spins $\frac12$, and its $(θ,φ)$ coordinates reflect the physical reality of such systems, for the mathematical reasons defined in the other answers.

The spin is a kind of intrinsic angular momentum, the classical version of it being represented by a vector, with a direction corresponding to the axis of rotation, an orientation corresponding to the rotation direction (clockwise vs counter-clockwise) and a magnitude corresponding to the total angular momentum.

For quantum spin-$\frac12$ particles, an angular momentum measurement has a binary output $±\frac{ℏ}{2}$, fixing the magnitude of the vector. The spin state of such a particle can therefore be represented by a vector of constant length, pointing to the direction $(θ,φ)$ of (the north pole of) its axis of rotation. These $(θ,φ)$ define here a point a sphere, the Bloch’s sphere, and correspond to an actual direction in 3D space.

When one measures the spin according to a given direction (a quantification axis), the two possible answer correspond to antipodal points on the sphere (opposite directions). The orthogonality of quantum states therefore does not correspond to right angles in this representation, hence the factor of 2 in $θ$. You can easily see this in the canonical basis, where the state $\lvert↑\rangle=|0\rangle$ corressponds to the north pole ($θ=0$), and the orthogonal state $\lvert↓\rangle=|1\rangle$ corresponds to the south pole ($θ=π$).

Complementary (and optional) note: This historical explanation in term of spin-$\frac12$ is only part of the story. This representation has been widely used because it proved useful (see other example). A proof of this is the (quasi-)total absence of the Poincaré sphere in quantum information, widely used in classical optics to study polarization. For single photons, a widely used kind of qubits, it is essentially a Bloch sphere with a factor of 2 in $φ$, in order to match with the physical polarization orientation. However, even people working with polarized photons as qubit prefer to use the Bloch sphere despite this factor of 2 discrepancy, because it corresponds better with the “true qubit-geometry”, aka $SU(2)$.

Answer 7

A much simpler answer why $\theta/2$.

Ignore the phase for a moment, to have a normalized state, you need a vector $(\cos\alpha, \sin\alpha)$, where $\alpha$ is between 0 and $\pi/2$. This range of $\alpha$ covers all the possible cases, concerning magnitudes.

Here it is convenient to define $\alpha=\theta/2$, so that the range of $\theta$ is between 0 and $\pi$. This way one can represent the state on a sphere.