Tensor and Kronecker product I have a follow exercise. Let $\left|\Psi\right\rangle = \tfrac{1}{\sqrt{2}}\left(\left|0\right\rangle + \left|1\right\rangle\right)$. Write out $\left|\Psi\right\rangle^{\otimes 2}$ xplicitly, both
in terms of tensor products like $\left|0\right\rangle \otimes \left|1\right\rangle$ , and using the Kronecker product.
Using tensor products I get:
$$\frac{1}{2}\left(\left|00\right\rangle + \left|01\right\rangle + \left|10\right\rangle + \left|11\right\rangle\right).$$
How I will be able to express this in terms of the Kronecker product?
 A: Let $S : \mathbb{C}^2 \otimes \mathbb{C}^2 \to \mathbb{C}^4$ be the isomorphism that maps the tensor product $v \otimes w$ of $v$, $w \in \mathbb{C}^2$ to their Kronecker product. Then, in general, for $v = (v_1,v_2)^T$, $w = (w_1,w_2)^T \in \mathbb{C}^2$,
$$
 S\left(v \otimes w\right) = \begin{pmatrix} v_1 w \\ v_2 w \end{pmatrix} = \begin{pmatrix} v_1 w_1 \\ v_1 w_2 \\ v_2 w_1 \\ v_2 w_2 \end{pmatrix}.
$$
Thus, if $\left|0\right\rangle = (1,0)^T$ and $\left|1\right\rangle = (0,1)^T$ are spin up and spin down eigenvectors for the component of spin in the $z$ direction, then $\left|\Psi\right\rangle = \left(\tfrac{1}{\sqrt{2}},\tfrac{1}{\sqrt{2}}\right)^T$, and hence
$$
 S\left(\left|\Psi\right\rangle^{\otimes 2}\right) = \begin{pmatrix} \tfrac{1}{\sqrt{2}}\cdot\tfrac{1}{\sqrt{2}}\\ \tfrac{1}{\sqrt{2}}\cdot\tfrac{1}{\sqrt{2}}\\ \tfrac{1}{\sqrt{2}}\cdot\tfrac{1}{\sqrt{2}}\\ \tfrac{1}{\sqrt{2}}\cdot\tfrac{1}{\sqrt{2}}\end{pmatrix} = \frac{1}{2} \begin{pmatrix}1\\1\\1\\1\end{pmatrix}.
$$
This corresponds to your calculation above in $\mathbb{C}^2 \otimes \mathbb{C}^2$, since
$$
 S\left(\left|00\right\rangle\right) = S\left(\left|0\right\rangle \otimes \left|0\right\rangle\right) = \begin{pmatrix}1\\0\\0\\0\end{pmatrix}, \quad S\left(\left|01\right\rangle\right) = S\left(\left|0\right\rangle \otimes \left|1\right\rangle\right) = \begin{pmatrix}0\\0\\1\\0\end{pmatrix},\\
S\left(\left|10\right\rangle\right) = S\left(\left|1\right\rangle \otimes \left|0\right\rangle\right) = \begin{pmatrix}0\\1\\0\\0\end{pmatrix}, \quad
S\left(\left|11\right\rangle\right) = S\left(\left|1\right\rangle \otimes \left|1\right\rangle\right) = \begin{pmatrix}0\\0\\0\\1\end{pmatrix}.
$$
