Tensor product in dual-space

Question

I am a bid confused regarding the notation for tensor products when going into dual-space

If $\left| \Psi \rangle \right. = \left| A \rangle \right. \left| B \rangle \right.$ is $ \left. \langle \Psi \right| = \left. \langle A \right| \left. \langle B \right|$ or $\left. \langle B \right| \left. \langle A \right|$?

My guess is the last choice, since operators should act on

$\left| 0 \rangle \right. = \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) $

$\left| 1 \rangle \right. = \left( \begin{matrix} 0 \\ 1 \end{matrix} \right) $

so

$\left| 01 \rangle \right. = \left| 0 \rangle \right. \left| 1 \rangle \right. = \left( \begin{matrix} 0 \\ 1 \\ 0 \\ 0 \end{matrix} \right) $

But then, for me, it should be

$\left( \begin{matrix} 0 & 1 & 0 & 0 \end{matrix} \right) = \left \langle 10 \right|$ But this is not correct if i calculate the tensor product, since

$\left \langle 1 \right| \left \langle 0 \right| = \left( \begin{matrix} 0 & 0 & 1 & 0 \end{matrix} \right)$

For instance, shoulden't it be

$\left| 01 \rangle \right. \left \langle 10 \right| = \left( \begin{matrix} 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{matrix} \right) $ ?

I hope I have made my confusion clear

Answer 1

I believe the general case is more clear. If you have two space with inner products $(V,<,>_V),(W,<,>_W),$ then you can define inner product $<,>$ on $V\otimes W$ in such a way that for every $v_1,v_2\in V$ and $w_1,w_2\in W$ $$<v_1\otimes w_1,v_2\otimes w_2>=<v_1,v_2>_V\cdot<w_1,w_2>_W.$$ So given an orthonormal basis $\{v_i\}$ and $\{w_j\}$ respectively in $V$ nad $W$ you get that $\{v_i\otimes w_j\}$ is orthonormal in $V\otimes W.$

If in your question $|A>,|B>$ are such that $<A|A>=1$ and $<B|B>=1$ then $$<AB|AB>=<A|A>\cdot <B|B>=1,$$ which i believe is your notion of duality. If you are familiar with matrix notation, then check by yourself that this notion is just an abstract generalisation of matix one.