How do I figure out the column-vector representation of a many-qubit state such as $|1101\rangle$? I know that we can express, for example the single-qubit state $\left|0\right>$ as $\begin{pmatrix} 1\\ 0 \end{pmatrix}$ and the state two-qubit state $\left|01\right>$ as $\begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}$. But how do I figure out the column-vector representation of a state such as $\left|1101\right>$ or $\left|100111\right>$? How do I know where to put the 1?
Any help is appreciated!
 A: You need for that the idea of Kronecker product $A\otimes B$.  If
\begin{align}
A=\left(\begin{array}{c} a_1 \\ a_2\\a_3 \end{array}\right)\, ,\qquad 
B=\left(\begin{array}{c} b_1 \\ b_2\end{array}\right)
\end{align}
then
\begin{align}
A\otimes B = \left(\begin{array}{c} a_1 \\ a_2\\a_3\end{array}\right)\otimes 
\left(\begin{array}{c} b_1 \\ b_2\end{array}\right)
= \left(\begin{array}{c} a_1b_1 \\ a_1b_2 \\a_2 b_1 \\ a_2 b_2 \\ a_3b_1 \\a_3b_2\end{array}\right)
\end{align}
so that, for instance, $\vert{01}\rangle$ yields
\begin{align}
\vert{01}\rangle= \vert 0\rangle \otimes \vert 1\rangle = 
\left(\begin{array}{c} 1\\ 0\end{array}\right)\otimes 
\left(\begin{array}{c} 0 \\ 1\end{array}\right)
= \left(\begin{array}{c} 0 \\ 1\\ 0 \\ 0\end{array}\right)\, .
\end{align}
This is associative so
\begin{align}
A\otimes B\otimes C = (A\otimes B)\otimes C
\end{align}
so you'd build up $\vert 1101\rangle=\vert 1\rangle \otimes \vert 1\rangle \otimes \vert 0\rangle \otimes \vert 1\rangle$ by repeated application of the above rule.
