# Definition of points on Bloch sphere

In the definition of the Bloch sphere, one demands that $\theta \in [0, \pi]$ ans $\phi \in [0, 2\pi)$ so that any state on the Bloch sphere can be represented by

$$|\phi \rangle= \cos(\theta/2)|0 \rangle+ e^{i \phi} \sin(\theta/2)|1 \rangle.$$

But I was wondering why the representation is chosen to be like this since in my opinion the natural way to choose this representation would be

$$|\phi \rangle=\cos(\theta)|0 \rangle+ e^{i \phi} \sin(\theta)|1 \rangle,$$ with $\theta \in [0, \pi], \phi \in [0, 2\pi)$.

If one chooses this representation, you would get in trouble since for example the states $|\phi_1\rangle$ with $\theta_1=\pi/4$ and $\phi_1=0$ and $|\phi_2\rangle$ with $\theta_2=3\pi/4$ and $\phi_2=\pi$ would both lead (when neglecting an irrelevant phase) to the representation

$$|\phi_1 \rangle=|\phi_2\rangle=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle).$$

But imagine that the Axiom of Quantum mechanics, that irrelevant phases can be neglected, didn't exist, would it then be possible to choose mapping as I proposed it? Is there any mathematical constraint that I forgot about?

Are there other possibilities to choose a representation on the Bloch sphere?

• I don't understand the question. You already state that your representation is useless because it has the same state for different values of $\theta,\phi$. Why do you want to imagine that quantum mechanics doesn't work like it does? Are you just asking about how to parametrize a sphere, forgetting about the physical situation? Sep 4, 2016 at 13:41
• If phases were relevant, the state space wouldn't be the 2-sphere, it would be the 3-sphere.
– user5174
Sep 4, 2016 at 20:22

The unit ball has a very nice property for representing the states of a qubit. If you perform a measurement around the $z$ axis that reports $1$ if the state is spin up and $-1$ if the state is spin down, then the expected value of this measurement is precisely its $z$ coordinate.

This means that the pure state $\alpha |0 \rangle + \beta| 1 \rangle$ must correspond to a point on the sphere with $z$ coordinate $|\alpha|^2 - |\beta|^2$.

For a point on the sphere the $z$ coordinate is precisely $\cos(\theta)$, and so we must have $|\alpha|^2 - |\beta|^2 = \cos(\theta)$ and $|\alpha|^2 + |\beta|^2 = 1$, and thus

$$|\alpha|^2 = \frac{1 + \cos(\theta)}{2}$$ $$|\beta|^2 = \frac{1 - \cos(\theta)}{2}$$

which we recognize as $|\alpha| = |\cos(\theta / 2)|$ and $|\beta| = |\sin(\theta / 2)|$.

(the same is true for the $x$ and $y$ coordinate as well, and a similar statement can be made about any axis)

The reason for the $\theta/2$ is because, when you hit a state by the rotation matrix $R(\theta) = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$, the state's position on the block sphere moves by $2 \theta$ radians instead of just $\theta$ radians. So if you want a rotation that moves the position by just $\theta$ radians, you should rotate by $\theta/2$ radians.

This ultimately comes down to the fact that $|0\rangle$ and $|1\rangle$ are perpendicular states. Geometrically speaking, that means they should be 90 degrees apart; at right angles to each other. To get them to be 180 degrees apart, to be up-vs-down instead of X-vs-Y, we had to double all the angles.

(The reason we want them to be 180 degrees apart, instead of 90, is that it frees up an axis and then makes such a nice analogy with rotations in 3d space. Every single-qubit quantum operation corresponds to a rotation around the Bloch sphere times a global phase factor. If we had stuck with the 90-degrees-apart thing, we'd have needed a fourth dimension to make the rotation analogy work.)

The reason why the irrelevant phase is neglected, is because it has no physical significance. One cannot measure it and it does not have any effect on the way things work. So if we neglect this phase the alternative representation that you give has this problem that the points one the Bloch sphere are not unique.

There are also other reasons why the representations of the Bloch sphere is the way it is. One reason off the top of my head is that the spinors can be combined into vectors. This is a property of irreducible representations of Lie group [SU(2) in this particular case].

Consider the spinor $$\eta = [\exp(i\phi/2)\cos(\theta/2), \exp(-i\phi/2)\sin(\theta/2)] .$$ One can show that the vector formed by contracting these spinors with the aid of the Pauli matrices gives a correctly parametrized vector $$\mathbf{v} = [\eta\sigma_x\eta^{\dagger}, \eta\sigma_y\eta^{\dagger}, \eta\sigma_z\eta^{\dagger}] = [\cos(\phi)\sin(\theta),\sin(\phi)\sin(\theta),\cos(\theta)] .$$ If one uses $\theta$ instead of $\theta/2$ in the arguments in the spinor, then this would would not work out right.