Suppose $Q \in L(U)$ is non negative operator with $Tr Q < \infty$ where $(U, \langle,\rangle)$ is Hilbert space. Suppose that $e_k, k=0...$ is an orthonormal base of $U$ consisting of eigenvectors of $Q$ with $\lambda_k$ eigenvalues. Suppose that $U_0=Q^{1/2}(U)$ is the Hilbert space with scalar product $$\langle u,v\rangle_0=\langle Q^{-1/2}u,Q^{-1/2}v\rangle$$ Call $(L_2^0=L_2(U_0,H),||\cdot||_{L_2^0})$ the Hilbert Schmidt operators from $U_0$ to $H$ where $H$ is another Hilbert space.
Then by the Parseval identity $$||\Psi||_{L_2^0}^2=\sum_{k,h}^\infty |\langle\Psi g_h,f_k\rangle|^2=\sum_{k,h}^\infty \lambda_h |\langle\Psi e_h,f_k\rangle|^2=\sum_{k,h}^\infty |\langle\Psi Q^{1/2} e_h,f_k\rangle|^2=||\Psi Q^{1/2} ||_{L_2(U,H)}^2$$
Now why does the following hold [Da prato - Stochastic equations in infinite dimensions 1999] ?
$$||\Psi||_{L_2^0}^2=Tr (\Psi Q \Psi^*)$$
where $g_k=\sqrt{\lambda_k}e_k, k=0...$ is an orthonormal base of $U_0$ and $f_k,k=0...$ is an orthonormal base of $H$
Using the definition: \begin{align} \|\Psi\|^2_{L_2^0}&=\|\Psi Q^{1/2}\|^2_{L_2(U,H)}=\operatorname{Tr}((\Psi Q^{1/2})^*\Psi Q^{1/2})\\[0.3cm]&=\operatorname{Tr}(Q^{1/2}\Psi^*\Psi Q^{1/2})=\operatorname{Tr}(\Psi Q\Psi^*). \end{align} In case the problem is the apperance of the trace, $$ \sum_{h,k}|\langle \Phi g_h,f_k\rangle|^2=\sum_h\|\Phi g_h\|^2=\sum_h\langle \Phi^*\Phi g_h,g_h\rangle=\operatorname{Tr}(\Phi^*\Phi). $$