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

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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 $|\delta|²$
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