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I was reading qiskit's text book, there I found that for a Double quantum coin toss, we have negative probability amplitude for $|1\rangle$ state when we starts from $|1\rangle$ state.
Link : https://qiskit.org/textbook/what-is-quantum.html
My question is why it is so?, Why when starting from state 1 we get negative?, For $|0\rangle$ , it is positive and as $|0\rangle$ and $|1\rangle$ are almost similar. Why is it different?
Although it is not explained up to that point in the Qiskit textbook, the quantum toss is in reality applying the Hadamard gate, denoted $H$. In matrix form, this operator looks like:
$$ H = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} $$
Now, we express the basis states in column form as follows:
$$ \begin{gather} |0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \\ |1\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \end{gather} $$
Applying an operator to a basis states is equivalent to matrix-vector multiplication. Therefore, applying the Hadamard gate to $|0\rangle$ is $H|0\rangle = 1/\sqrt{2}(|0\rangle + |1\rangle)$. You can see that this is just taking the first column out of the $H$ matrix. Therefore, applying $H$ to $|1\rangle$ is the same as taking the second column of the matrix which gives $H|1\rangle = 1/\sqrt{2}(|0\rangle-|1\rangle)$.
Having this negative probability amplitude is one of the things that allow for interference on quantum computing. If, on the contrary, the quantum toss gave the same state for both basis states, then the quantum toss operator would not be reversible since given the output we would not be able to know the input state. And since we want quantum computations to be reversible, we cannot define the operator like that.
