What makes representing qubits in a 3D real vector space possible?

Question

Qubits exist in a 2D complex vector space, but we can represent qubits on the Bloch sphere as a 3D real vector space. Mathematically, what makes this possible – why don't we need 4 real dimensions?

user1271772 · Accepted Answer

Mathematically a qubit's coefficients $c_1$, $c_2$ must have the following properties:
begin{align}
|c_1|^2 + |c_2|^2 =1 tag{1}
|c_1|, |c_2| in [0,1], tag{2}
end{align}
because Born's rule tells us that the modulus squared is a (classical) probability, and classical probabilities must add up to exactly 1 and must be 0, 1 or in between 0 and 1.
However we know:
$$
sin^2theta + cos^2theta = 1tag{3}
$$
so we can set $c_1=sintheta$ and $c_2=costheta$.
However also remember that $|e^{textrm{i}phi}|=1$, so we can add that to one of the coefficients, so that: $c_1=sintheta$ and $c_2=e^{textrm{i}phi}costheta$.
Any more factors like $e^{textrm{i}x}$ for different real-valued angles $x$ won't make a noticeable difference to any measurements, as the angles can be combined with each other, and remember that global phases do not make a difference to any measurements.
Therefore there's only two real-valued angles necessary: $theta$ and $phi$. We can add more, but they can always be factored out into what is known as a "global phase" which is something that doesn't make any difference in the outcomes of measurements.

jayheme · Answer

That is because we have a condition on the two complex amplitudes. The normalization condition. So it eliminates one real number and hence we can use the Bloch sphere picture.
And its not exactly a 3D real space. Its similar but not exactly same.

Adam Zalcman · Answer

Three real parameters are sufficient due to the constraint that
$$
|alpha|^2 + |beta|^2 = 1tag1
$$
where $alpha$ and $beta$ are the two components of a 2D complex vector describing the qubit state. This constraint ultimately derives from the fact that $|alpha|^2$ and $|beta|^2$ are probabilities of the two possible outcomes of the computational basis measurement.
In order to see how the constraint $(1)$ implies that three real parameters are sufficient write $alpha = r e^{itheta}$ and $beta = s e^{izeta}$ and substitute into $(1)$ to get
$$
r^2 + s^2 = 1.tag{2}
$$
This means that $r, theta, zeta in mathbb{R}$ are sufficient to specify $alpha, beta in mathbb{C}$ satisfying $(1)$.
Note that the global phase is unobservable and can be ignored, so we can in fact choose $theta = 0$ (i.e. $alphage 0$). This means that two real parameters are in fact sufficient. This is why a pure state of a qubit can be represented as a point on the 2D Bloch sphere (mixed states occupy the interior).

Martin Vesely · Answer

For vector representation of any qubit it is true that:

it has to be a unit vector
global phase does not matter and can be fixed at any value

As a result two degress of freedom are eliminated and as a result you are left with only two free parameters.
Note that although you represent a qubit on Bloch sphere, the sphere has unit radius. So, actually only 2D space is necessary to describe a qubit.

What makes representing qubits in a 3D real vector space possible?

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