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?

Answer 1

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:

$$ \sin^2\theta + \cos^2\theta = 1\tag{3} $$

so we can set $c_1=\sin\theta$ and $c_2=\cos\theta$.

However also remember that $|e^{\textrm{i}\phi}|=1$, so we can add that to one of the coefficients, so that: $c_1=\sin\theta$ and $c_2=e^{\textrm{i}\phi}\cos\theta$.

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.

Answer 2

Three real parameters are sufficient due to the constraint that

$$ |\alpha|^2 + |\beta|^2 = 1\tag1 $$

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 $\alpha = r e^{i\theta}$ and $\beta = s e^{i\zeta}$ 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. $\alpha\ge 0$). 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).

Answer 3

For vector representation of any qubit it is true that:

1. it has to be a unit vector
2. 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.

Answer 4

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.