quantum tensor equation simplification I've started learning quantum computing of late and got interested in some secret sharing. In a recent post on the site https://quantumcomputing.stackexchange.com/questions/13195/grover-search-with-different-diffusion-operators i asked a question to which i got a very satisfactory answer. But i have some doubts relating to the tensor equation involved, which is
$$(H\otimes H)(\mathbb{I}\otimes \mathbb{I}-2|0\rangle\langle0|\otimes |1\rangle\langle 1|)(H\otimes H)(\mathbb{I}\otimes \mathbb{I}-2|1\rangle\langle 1|\otimes |0\rangle\langle 0|)(H\otimes H)|0\rangle|0\rangle=|00\rangle$$
Now, here $H$=hadamard gate, $\mathbb{I}$ is the identity operator, $\langle.||.\rangle$ is bra-ket notation in quantum mechanics. Can somebody explain how did the equation got reduced to $|00\rangle$? Can somebody suggest some refernces? Some hints?
 A: Late to the party and although I assume you figured this out by now, I'll give an answer for the sake of completeness. This is but a straightforward computation using
$$
H\otimes H=\Big(\frac1{\sqrt{2}}\begin{pmatrix}1&1\\1&-1\end{pmatrix}\Big)\otimes \Big(\frac1{\sqrt{2}}\begin{pmatrix}1&1\\1&-1\end{pmatrix}\Big)=\frac12\begin{pmatrix} 1&1&1&1\\1&-1&1&-1\\1&1&-1&-1\\1&-1&-1&1 \end{pmatrix}
$$
(by means of the Kronecker product) which gives
\begin{align}
&(H\otimes H)(\mathbb{I}\otimes \mathbb{I}-2|0\rangle\langle0|\otimes |1\rangle\langle 1|)(H\otimes H)(\mathbb{I}\otimes \mathbb{I}-2|1\rangle\langle 1|\otimes |0\rangle\langle 0|)(H\otimes H)=\\
&=\scriptsize\frac12\begin{pmatrix} 1&1&1&1\\1&-1&1&-1\\1&1&-1&-1\\1&-1&-1&1 \end{pmatrix}\begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}\frac12\begin{pmatrix} 1&1&1&1\\1&-1&1&-1\\1&1&-1&-1\\1&-1&-1&1 \end{pmatrix}\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&1\end{pmatrix}\frac12\begin{pmatrix} 1&1&1&1\\1&-1&1&-1\\1&1&-1&-1\\1&-1&-1&1 \end{pmatrix}\\
&=\begin{pmatrix}1&0&0&0\\0&0&1&0\\0&-1&0&0\\0&0&0&1\end{pmatrix}\,.
\end{align}
With this it's clear that $|0\rangle|0\rangle=|00\rangle=(1,0,0,0)^T$ is a fixed point of this operation, as claimed.
