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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?

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    $\begingroup$ The most succinct possible answer is "because $SU(2)$ is a double cover of $SO(3)$" but I imagine that isn't very helpful. $\endgroup$ Dec 30 '20 at 6:45
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    $\begingroup$ 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. $\endgroup$
    – Carlo Wood
    Dec 31 '20 at 20:13
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    $\begingroup$ Qubits exist in projective 2D complex vector space. And the Bloch sphere only has two dimensions (it is the surface of the sphere). $\endgroup$ Jan 7 at 19:22
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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.

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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).

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

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

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  • $\begingroup$ What about the global phase? $\endgroup$ Jan 7 at 19:19

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