Is there a way to make a $4\times4$ matrix composed of Pauli/Identity matrices that has determinant equal to $1$? Something of a general formula? It's interesting to me.
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1What exactly do you mean by "composed of" pauli/identity matrices? – Ben Grossmann Sep 18 '19 at 11:22
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You might find this post interesting – Ben Grossmann Sep 18 '19 at 11:24
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1Suppose $\mathbf{P}$ is our $4\times4$ matrix. We can write $\mathbf{P}$ as a $2\times2$ matrix composed of four elements, each of which is a matrix. – rami_salazar Sep 18 '19 at 11:25
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Okay, so in other words you want to have Pauli/identity matrices as submatrices of $P$ – Ben Grossmann Sep 18 '19 at 11:25
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Yes, that's exactly right. – rami_salazar Sep 18 '19 at 11:26
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To go one step further, the matrix has to have the form $\boldsymbol{P} = \begin{bmatrix}\boldsymbol{A}\quad \overline{\boldsymbol{B}} \ \boldsymbol{B} \quad \overline{\boldsymbol{A}}\end{bmatrix}$ – rami_salazar Sep 18 '19 at 11:27
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Have you seen the Kronecker product of two matrices before? – Ben Grossmann Sep 18 '19 at 11:28
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Let us continue this discussion in chat. – rami_salazar Sep 18 '19 at 11:29
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If you would like $P$ to be Hermitian, perhaps it makes more sense to have the upper-right entry be $\mathbf B^\dagger (z)$ – Ben Grossmann Sep 18 '19 at 11:29
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No, out of the $256$ possible $4\times 4$ block matrices that you can obtain from matrices of the form $$ \begin{pmatrix} 1&0\\ 0&1 \end{pmatrix},\,\,\, \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix},\,\,\, \begin{pmatrix} 0&-i\\ i&0 \end{pmatrix},\,\,\, \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix} $$ there are only $6$ possible determinant values, namely $0,-2,2,-2i,2i$ and $4$. Informally speaking, it is not hard to write a small code to check this.
sam wolfe
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