How do I figure out the column-vector representation of a many-qubit state such as $|1101rangle$?

Question

I know that we can express, for example the single-qubit state $left|0right>$ as $begin{pmatrix} 1 0 end{pmatrix}$ and the state two-qubit state $left|01right>$ as $begin{pmatrix} 0  1  0  0 end{pmatrix}$. But how do I figure out the column-vector representation of a state such as $left|1101right>$ or $left|100111right>$? How do I know where to put the 1?
Any help is appreciated!

ZeroTheHero · Accepted Answer

You need for that the idea of Kronecker product $Aotimes B$.  If
begin{align}
A=left(begin{array}{c} a_1  a_2a_3 end{array}right), ,qquad 
B=left(begin{array}{c} b_1  b_2end{array}right)
end{align}
then
begin{align}
Aotimes B = left(begin{array}{c} a_1  a_2a_3end{array}right)otimes 
left(begin{array}{c} b_1  b_2end{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_2end{array}right)
end{align}
so that, for instance, $vert{01}rangle$ yields
begin{align}
vert{01}rangle= vert 0rangle otimes vert 1rangle = 
left(begin{array}{c} 1 0end{array}right)otimes 
left(begin{array}{c} 0  1end{array}right)
= left(begin{array}{c} 0  1 0  0end{array}right), .
end{align}
This is associative so
begin{align}
Aotimes Botimes C = (Aotimes B)otimes C
end{align}
so you'd build up $vert 1101rangle=vert 1rangle otimes vert 1rangle otimes vert 0rangle otimes vert 1rangle$ by repeated application of the above rule.

How do I figure out the column-vector representation of a many-qubit state such as $|1101rangle$?

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