# ket zero $|0\rangle = [1,\, 0]$, why a qubit which is initialized with zero is 1 in first basis state?

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.

• 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. May 10 at 9:07

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

• thanks for this great explaination!! May 10 at 12:14
• 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 May 10 at 12:17
• Exactly! And in general, the probability of being in $0$ state is $|\alpha|^2$, and the probability of being in $1$ state is $|\beta|^2$, May 10 at 12:23