What are the constraints for the coefficents of the basis states in quantum computing?

Question

$\newcommand{\ket}[1]{\left|#1\right>}$ It's known that the Kolmogorov axioms characterise a probability distribution:

1. Probability of an event is a non-negative real number.
2. The sum of all probabilities is 1.
3. The probability of disjoint events is the sum of their probabilities.

But when dealing with a quantum system, as far as I know:
the first axiom is relaxed to allow negative & complex numbers;
The second axiom is changed to "The sum of the squares of the coefficients is 1".

But if that was all, wouldn't this allow for the following 1-qubit system: $$\ket{\phi}=i\ket{0}+\sqrt{2}\ket{1}$$ The probabilities do sum to 1: \begin{align*} i^2+\sqrt{2}^2 &= -1 + 2\\ &= 1 \end{align*} But interpreting the 'probabilities' as probabilities separately seems to be nonsensical: \begin{align*} P(\ket{0}) &= -100\%\quad(?)\\ P(\ket{1}) &= \phantom{-}200\%\quad(?) \end{align*}

So can this system exist in actuality, or is merely a mathematical curiosity?

Assuming the latter, and assuming we disallow such a system by positing "The magnitude of the squares of the coefficients must lie in the $[0,1]$ interval."

But also consider the following 2-qubit system: $$\ket{\psi}=\frac{1+\sqrt{3}i}{2}\ket{00}+\frac{1-\sqrt{3}i}{2}\ket{01}+\ket{10}+\ket{11}$$ Again the probabilities do sum to 1: \begin{align*} \Big(\frac{1+\sqrt{3}i}{2}\Big)^2 + \Big(\frac{1-\sqrt{3}i}{2}\Big)^2 + 1^2 + 1^2 &= \frac{1+2\sqrt{3}i-3}{4} + \frac{1-2\sqrt{3}i-3}{4} + 1 + 1\\ &= \frac{1-3+1-3}{4} + 2\\ &= \frac{-4}{4} + 2\\ &= 1 \end{align*} And again, we have two states with probability of $100\%$.

Did I miss something?

Answer 1

$\newcommand{\ket}[1]{\left|#1\right>}$ I did miss something, assuming the following system: $$\ket{\chi} = \sum_i{\alpha_i\ket{\chi_i}}\quad; \alpha_i\in\mathbb{C}$$ The probability of the system being in state $\ket{\chi_j}$ is $|\alpha_i|^2$, not $\alpha_i^2$ (where $|x|$ is the magnitude/amplitude of $x$).

So the system $\ket{\phi}$ cannot exist, because the probabilities of the square of the amplitudes is equal to 3 not 1:\begin{align*} |i|^2+|\sqrt{2}|^2 &= 1^2 + \sqrt{2}^2\\ &= 1 + 2\\ &= 3 \end{align*}

Likewise for the system $\ket{\psi}$:\begin{align*} \Big|\frac{1+\sqrt{3}i}{2}\Big|^2 + \Big|\frac{1-\sqrt{3}i}{2}\Big|^2 + |1|^2 + |1|^2 &= 1^2 + 1^2 + 1^2 + 1^2\\ &= 1 + 1 + 1 + 1\\ &= 4 \end{align*}

Answer 2

Kolmogorovian Probability Theory [Paragraph 3.3]

Let S be a set, understood for the moment as the state-space of a physical system, and let Σ be a σ-algebra of subsets of S. We can regard each partition E of S into countably many pair-wise disjoint Σ-measurable subsets as representing a “coarse-grained” approximation to an imagined perfect experiment that would reveal the state of the system. Let AΣ be the test space consisting of all such partitions. Note that the outcome set for AΣ is the set X=Σ∖{∅} of non-empty Σ-measurable subsets of S. Evidently, the probability weights on AΣ correspond exactly to the countably additive probability measures on Σ.

Thus the last sentences explains that there is only 100% Probability of the overall system. And cannot be count together 1 + 1 = 2 in terms the possible maximum is 1.

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@Sofviic: I saw you found a solution while writing my answer, I will post it in terms it can help also from another perspective. Also like your answer - upvote! Well done!