Mathematically, our qubits (pure quantum states) can be represented as a normalized vector (of length 1) in a complex Hilbert space.

is it because of the sum of squares of vertical and horizontal component of a circle with unit radius is always 1 and so is the sum of probabilities of qubit collapsing in either of basis state?

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1 Answer 1


If the coefficients $c_k$ of ket vectors are to be interpreted as giving outcome probabilities via $|c_k|^2$, then we clearly need $\sum_k |c_k|^2=1$, and thus we can interpret the points $(c_1,...,c_N)$ as elements of a "complex sphere". Note that a "complex sphere" is not really a sphere in the geometric sense, due to complex numbers being involved. Then again, there's different ways to get such (hyper)spheres from these coefficients:

  1. The set of absolute values $(|c_1|,..., |c_N|)$ of a normalised state $|\psi\rangle\in \mathbb{C}^N$ lie in an $(N-1)$-sphere (a sphere in $\mathbb{R}^N$ dimensions).
  2. The set of real and imaginary components of the coefficients $c_k$, that is, the points $$(\operatorname{Re}(c_1),..., \operatorname{Re}(c_n), \operatorname{Im}(c_1), ..., \operatorname{Im}(c_n)),$$ lie on a $(2n-1)$-sphere. But this representation forgoes the phase "gauge freedom" of states, so only specific subsets (submanifolds) of such spheres should be considered as directly representing states. This is essentially another way to state that $\mathbb{CP}^{n}\simeq S^{2n+1}/S^1$. As an aside, this process, in the $n=1$ case, gives rise to the Bloch sphere representation, as it just so happens that $S^3/S^1\simeq S^2$.

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