I am trying to prove that the class of Hilbert-Schmidt operators $\mathcal{J}_2(\mathcal{H}_1,\mathcal{H}_2))$ where $\mathcal{H}_1,\mathcal{H}_2$ are complex, separable Hilbert spaces ( scalar products linear in the first, conjugate linear in the second factor) is a realisation of the (non_categorical) tensor product $\mathcal{H}_1\otimes\mathcal{H}_2$. For this purpose I defined two mappings by: $$\Phi:\mathcal{H}_1\otimes\mathcal{H}_2\rightarrow\mathcal{J}_2(\mathcal{H}_1,\mathcal{H}_2),\varphi\otimes\psi\mapsto\langle\varphi,.\rangle_{\mathcal{H}_1}\psi$$ i.e. $\Phi(\varphi\otimes\psi)\eta=\langle\varphi,\eta\rangle_{\mathcal{H}_1}\psi$ for all $\eta\in\mathcal{H}_1$ and $$\tilde{\Phi}:\mathcal{H}_2\otimes\mathcal{H}_1\rightarrow\mathcal{J}_2(\mathcal{H}_2,\mathcal{H}_1),\psi\otimes\varphi\mapsto\langle.,\psi\rangle_{\mathcal{H}_2}\varphi.$$ i.e. $\tilde{\Phi}(\psi\otimes\varphi)\xi=\langle\xi,\psi\rangle_{\mathcal{H}_2}\varphi$.
Since $\eta\mapsto\langle\varphi,\eta\rangle_{\mathcal{H}_1}$ is a bounded functional there exists $\varphi^*\in\mathcal{H}_1$ so that $\langle\varphi,\eta\rangle_{\mathcal{H}_1}=\langle\eta,\varphi^*\rangle_{\mathcal{H}_1}$ and then we have for the adjoint of $\Phi(\varphi\otimes\psi)$: $$\left(\Phi(\varphi\otimes\psi)\right)^*=\tilde{\Phi}(\psi\otimes\varphi^*)$$as the following calculation shows: $$\langle\Phi(\varphi\otimes\psi)\eta,\xi\rangle_{\mathcal{H}_2}=\langle\langle\varphi,\eta\rangle_{\mathcal{H}_1}\psi,\xi\rangle_{\mathcal{H}_2}=\langle\varphi,\eta\rangle_{\mathcal{H}_1}\langle\psi,\xi\rangle_{\mathcal{H}_2}$$ on the one hand and $$\langle\eta,\tilde{\Phi}(\psi\otimes\varphi^*)\xi\rangle_{\mathcal{H}_1}=\langle\eta,\langle\xi,\psi\rangle_{\mathcal{H}_2}\varphi^*\rangle_{\mathcal{H}_1}=\langle\eta,\varphi^*\rangle_{\mathcal{H}_1}\overline{\langle\xi,\psi\rangle_{\mathcal{H}_2}}=\langle\varphi,\eta\rangle_{\mathcal{H}_1}\langle\psi,\xi\rangle_{\mathcal{H}_2}$$ on the other hand. Now the Hilbert-Schmidt product of two such rank-one operators is given by: $$\langle\Phi(\varphi_1\otimes\psi_1),\Phi(\varphi_2\otimes\psi_2)\rangle_{HS}=\operatorname{tr}(\Phi(\varphi_1\otimes\psi_1)(\Phi(\varphi_2\otimes\psi_2))^*)=\operatorname{tr}(\Phi(\varphi_1\otimes\psi_1)\tilde{\Phi}(\psi_2\otimes\varphi_2^*))$$ and I wanted to compute this trace. I took an orthonormal basis $\{\xi_j\}_{j=1}^{\infty}$ of $\mathcal{H}_2$ and computed first $$\Phi(\varphi_1\otimes\psi_1)\tilde{\Phi}(\psi_2\otimes\varphi_2^*)\xi_j=\Phi(\varphi_1\otimes\psi_1)\langle\xi_j,\psi_2\rangle_{\mathcal{H}_2}\varphi_2^*=\langle\varphi_1,\langle\xi_j,\psi_2\rangle_{\mathcal{H}_2}\varphi_2^*\rangle_{\mathcal{H}_1}\psi_1$$ and then $$\operatorname{tr}\Phi(\varphi_1\otimes\psi_1)\tilde{\Phi}(\psi_2\otimes\varphi_2^*)=\sum_{j=1}^{\infty}\langle\varphi_1,\langle\xi_j,\psi_2\rangle_{\mathcal{H}_1},\varphi_2^*\rangle_{\mathcal{H}_1}\langle\psi_1,\xi_j\rangle_{\mathcal{H}_2}=\langle\varphi_1,\varphi_2^*\rangle_{\mathcal{H}_1}\sum_{j=1}^{\infty}\overline{\langle\xi_j,\psi_2\rangle_{\mathcal{H}_2}}\langle\psi_1,\xi_j\rangle_{\mathcal{H}_2}$$ $$=\langle\phi_2,\phi_1\rangle_{\mathcal{H}_1}\sum_{j=1}^{\infty}\langle\psi_2,\xi_j\rangle_{\mathcal{H}_2}\langle\xi_j,\psi_1^*\rangle_{\mathcal{H}_1}=\langle\varphi_2,\varphi_1\rangle_{\mathcal{H}_1}\langle\psi_2,\psi_1^*\rangle_{\mathcal{H}_2}=\langle\varphi_2,\varphi_1\rangle_{\mathcal{H_1}}\langle\psi_1,\psi_2\rangle_{\mathcal{H}_2}$$ where $\psi_1^*$ is the element of $\mathcal{H_2}$ satisfying $\langle \xi,\psi_1^*\rangle_{\mathcal{H_2}}=\langle\psi_1,\xi\rangle_{\mathcal{H}_2}$ for all $\xi\in\mathcal{H}_2$. Now this is not what I wanted. Of course I wanted to achieve: $$...=\langle\varphi_1,\varphi_2\rangle_{\mathcal{H}_1}\langle\psi_1,\psi_2\rangle_{\mathcal{H}_2}=\langle \varphi_1\otimes\psi_1,\varphi_2\otimes\psi_2\rangle_{\mathcal{H}_1\otimes\mathcal{H}_2}$$ and I can't find a mistake and seem to have lost the sight of the wood for the trees.If someone could tell me where I went wrong? Many thanks for any help in advance!