# Describe |00> and |10> in terms of |0> and |1>

I came across following lines:

$$|00\rangle$$ : both quibts are in state of $$|0\rangle$$

since $$|00\rangle = [1 0 0 0]$$ in column vector and $$|0\rangle = [1 0]$$ in column vector, so if each single qubit is $$[1 0]$$ then how multi-qubit state $$|00\rangle = [1 0 0 0]$$ ?

Is it like each single qubit have 100% probability of being in first state basis so in multi-qubit system they have 100% probability of being in first state basis?

$$|10\rangle$$ : The qubit states are $$|1\rangle$$ (on the left) and $$|0\rangle$$ (on the right).

since $$|10\rangle = [0 0 1 0]$$

$$|0\rangle = [1 0]$$

$$|1\rangle = [0 1]$$

so if one qubit is in $$|0\rangle$$ and other in $$|1\rangle$$ then how in multi-qubit system $$|10\rangle = [0 0 1 0]$$ ?

## 1 Answer

The vector of a non-entangled multiple-qubits state is given by the tensor product of the one-qubits vectors : $$\begin{pmatrix}a\\b\end{pmatrix} \otimes \begin{pmatrix}c\\d\end{pmatrix} = \begin{pmatrix}a*c\\a*d\\b*c\\b*d\end{pmatrix}$$ thus : $$|00\rangle = |0\rangle \otimes|0\rangle = \begin{pmatrix}1\\0\end{pmatrix} \otimes \begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}1\\0\\0\\0\end{pmatrix}$$ and : $$|10\rangle = |1\rangle \otimes|0\rangle = \begin{pmatrix}0\\1\end{pmatrix} \otimes \begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}0\\0\\1\\0\end{pmatrix}$$ In a two qubits state $$\begin{pmatrix}\alpha\\ \beta\\ \gamma\\ \delta\end{pmatrix}$$:

• the probability of observing $$|00\rangle$$ is $$|\alpha|²$$
• the probability of observing $$|01\rangle$$ is $$|\beta|²$$
• the probability of observing $$|10\rangle$$ is $$|\gamma|²$$
• the probability of observing $$|11\rangle$$ is $$|\delta|²$$