From superoperator of unitary to the unitary itself

Question

I have a question about super operators. Let's say I have the super operator of some unitary matrix $u$ called $SU$ where $SU = u^\ast\otimes u$ (here $u^\ast$ is the complex conjugate of $u$). If I have $SU$, how can I get $u$ from this? Is there any tensor product relation I can use?

Answer 1

$\newcommand\SU{\text{SU}}$Vectorizing the matrix $u$ (by writing $a_i:=u_{k,l}$ for $i:=(k,l)$, so that the entries of the matrix $u^*\otimes u$ are $\bar u_{k,l}u_{r,s}=\bar a_i a_j$ for $i:=(k,l)$ and $j:=(r,s)$), we reduce the problem to the following:

For a nonzero complex vector $a=(a_i)$, we know the matrix $p=(p_{ij})$, where $p_{ij}=\bar a_i a_j$. Can we recover $a$ based on this information?

To answer this question, note that the condition that $a$ is nonzero implies that $a_i\ne0$ for some $i$. Fix any such $i$.

Then $p_{ii}=\bar a_i a_i=|a_i|^2\ne0$, and from the condition $p_{ij}=\bar a_i a_j$ we get $a_j=C_{ij}a_i$, where $C_{ij}:=p_{ij}/p_{ii}$ is a known complex number.

It also follows that $a_i=\sqrt{p_{ii}} e^{it}$ for some $t\in[0,2\pi)$ and hence
$$a_j=C_{ij}\sqrt{p_{ii}} e^{it}=\frac{p_{ij}}{\sqrt{p_{ii}}}\,e^{it}.$$ for all $j$.

Thus, we have determined all the $a_j$'s up to the constant factor $e^{it}$ of modulus $1$.

Thus, we have determined the matrix $u$ up to the constant factor $e^{it}$ of modulus $1$: $$u_{r,s}=\frac{p_{(k,l)(r,s)}}{\sqrt{p_{(k,l)(k,l)}}}\,e^{it}$$ for some $(k,l)$ such that $p_{(k,l)(k,l)}>0$ and for all $(r,s)$, where $p:=\SU$.

This is all that can be done: the matrix $u$ can be determined only up to a constant factor $e^{it}$ of modulus $1$. Indeed, if $u$ is a unitary matrix such that $u^*\otimes u$ is the given matrix $\SU$, then for any real $t$ we have that $v:=e^{it}u$ is a unitary matrix such that $v^*\otimes v=\SU$. $\quad\Box$

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At the OP's request, here is the link to a pdf image (two pages) of a Mathematica notebook providing an example showing how to get $u$ from $\SU$, up to a constant factor of modulus $1$.