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ket zero $|0\rangle = [1,\, 0]$

The $[1,\, 0]$ is telling us that the probability amplitude for being in the first basis state is $1$, and the probability amplitude for being in the second basis state is $0$.

why a qubit which is initialized with zero is $1$ in first basis state? I mean since it is zero then it should be zero in all basis state, IMO... but of course I am wrong, please help to understand this.

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    $\begingroup$ The basis states being $|0\rangle$ and $|1\rangle$ is just a formalism and has nothing to do with the numbers $0$ and $1$. You might as well call them $| a \rangle$ and $| b \rangle$, but using $|0\rangle$ and $|1\rangle$ is standard notation (and has some convenient properties for calculations). Compare that to a standard bit, which is "off" (0) or "on" (1). Even if it is "off", that does not mean it has vanished. $\endgroup$
    – cheetah
    May 10 at 9:07

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A qubit is a two-level quantum system. These two levels (two states) are usually denoted as $|0\rangle$ and $|1\rangle$. That means we can write a general qubit state as $$\alpha|0\rangle + \beta |1\rangle$$ where $\alpha$ and $\beta$ are complex number such that

$$\alpha^2 + \beta^2 = 1$$

So, we can write the qubit state as a vector in two dimensional vector space spanned by the two basis states $|0\rangle$ and $|1\rangle$ as $\begin{bmatrix}\alpha \\ \beta\end{bmatrix}$

You can easily now see that $|0\rangle$ is equivalent to the case when $\alpha = 1$ and $\beta = 0$. That is $|0\rangle = \begin{bmatrix}1 \\ 0\end{bmatrix}$.

On the other hand, the vector $\begin{bmatrix}0 \\ 0\end{bmatrix}$ is not a valid qubit state because it does not satisfy the condition $\alpha^2 + \beta^2 = 1$

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  • $\begingroup$ thanks for this great explaination!! $\endgroup$ May 10 at 12:14
  • $\begingroup$ here as α = 1 does it mean that there is 100% probability of finding this qubit in 1st basis state and 0% probability in 2nd basis state? @Egretta $\endgroup$ May 10 at 12:17
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    $\begingroup$ Exactly! And in general, the probability of being in $0$ state is $|\alpha|^2$, and the probability of being in $1$ state is $|\beta|^2$, $\endgroup$ May 10 at 12:23

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