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

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?

• The most succinct possible answer is "because $SU(2)$ is a double cover of $SO(3)$" but I imagine that isn't very helpful. Dec 30 '20 at 6:45
• I'd state "we can represent a single qubit on the Bloch sphere". As soon as you have more than one qubit (that is not entirely uncorrelated) you're going to need more than 3 dimensions. Dec 31 '20 at 20:13
• Qubits exist in projective 2D complex vector space. And the Bloch sphere only has two dimensions (it is the surface of the sphere). Jan 7 at 19:22

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.

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).

• I would add that Bloch vectors with norm smaller than one correspond to mixed states. Dec 30 '20 at 10:05
• Added clarification. Thanks! Dec 30 '20 at 15:44

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.

• Most concise answer! Jan 7 at 19:22
• Actually this one is more concise, and still would be even if a sentence about global phase was added (though that is not necessary). Apr 16 at 21:08

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.

• What about the global phase? Jan 7 at 19:19