Let $A$ and $B$ two $C^*$-algebras, $H_1$ and $H_2$ complex Hilbert spaces and $\pi_1:A\to B(H_1)$, $\pi_2:B\to B(H_2)$ two $*$-representations. Then there is a $*$-representation $\pi_1\otimes \pi_2:A\odot B\to B(H_1\otimes H_2)$, defined as follows:
1.We have $\pi_1\odot \pi_2:A\odot B\to B(H_1)\odot B(H_2)$, $a\otimes b\mapsto \pi_1(a)\otimes \pi_2(b).$ ($\odot$ denotes the tensor product as $*$-algebras)
2.We have the canonical embedding $$\iota: B(H_1) \odot B(H_2)\to B(H_1 \otimes H_2),$$ where $T\in B(H_1)$ and $S\in B(H_2)$ will be mapped to $T\otimes S\in B(H_1\otimes H_2)$ with $T\otimes S(v\otimes w)=T(s)\otimes S(w)$ ($H_1 \otimes H_2$ is the tensor product as a Hilbert space).
Finally, we set $\pi_1\otimes \pi_2:=\iota \circ \pi_1\odot \pi_2$.
Conversely, I'm searching for an example which demonstrates that if you have a $*$-representation $\pi:A\odot B\to B(H_1\otimes H_2)$, then $\pi$ is does not have to be in the form of $\pi_1\otimes \pi_2$, where $\pi_1:A\to B(H_1)$, $\pi_2:B\to B(H_2)$ are two $*$-representations.
I.e. the claim is: there are $*$-representations $\pi:A\odot B\to B(H_1\otimes H_2)$ on the $*$-algebraic tensor product of $A$ and $B$, which are not induced by $*$-representations of $A$ and $B$.
One suggestion is to choose the flip: $A=B=B(\ell^2)$, $H_1=H_2=\ell^2$, and $$\pi:B(\ell^2)\odot B(\ell^2)\to B(\ell^2\otimes \ell^2),$$ $$T\otimes S\mapsto S\otimes T.$$ But I'm stuck to give a formal proof that it is such an example (although I think it can't be too difficult...). Do you have an idea how to do it?