Let $A$ and $B$ be $C^{\ast}$-algebras. Let $A \otimes^h B$ denotes the Haagerup tensor product Of $A$ and $B$. In this paper it is given that for $x \in A \otimes B$,
$\|x\|_h=\text{inf} \{\|\sum_{i=1}^na_ia_i^*\|^{1/2 }\|\sum_{i=1}^n b_i ^*b_i\| ^{1/2}, x= \sum_{i=1}^n a_i \otimes b_i \in A \otimes B \}$
I know the definition of Haagerup tensor product for operator spaces but I don't see how does one obtains this beautiful expression of norm in case of $C^{\ast}$-algebras.
Let $E,F$ be operator spaces. As indicated in the comments, the Haagerup tensor product norm satisfies $$ \|x\|_h=\inf\left\{\|a\|\|b\|: x=\sum_{j=1}^n a_{1j}\otimes b_{j1}\right\}. $$ Here $a\in M_{1,n}(E)$ and $b\in M_{n,1}(F)$. If $E\subset B(H)$ and $F\subset B(K)$, then $M_{1,n}(E)\subset B(H^n,H)$ and $M_{n,1}(F)\subset B(K,K^n)$ and we can use the $C^\ast$-identity for bounded operators to get \begin{align*} \|a\|^2&=\|a a^\ast\|=\|a_{11}a_{11}^\ast+\dots+a_{1n}a_{1n}^\ast\|,\\ \|b\|^2&=\|b^\ast b\|=\|b_{11}^\ast b_{11}+\dots+b_{n1}^\ast b_{n1}\|. \end{align*} Thus $$ \|x\|_h=\inf\left\{\left\lVert\sum_{j=1}^n a_{1j} a_{1j}^\ast\right\rVert^{1/2}\left\lVert\sum_{j=1}^n b_{j1}^\ast b_{j1}\right\rVert^{1/2}:x=\sum_j a_{1j}\otimes b_{j1}\right\} $$ as desired.