Let $(e\_j)$ be a orthonormal basis (ONB) of a separable Hilbert space $(\mathcal{H}, \langle\cdot, \cdot\rangle\_{\mathcal{H}})$ and $(\mathcal{S\_H}, \langle\cdot, \cdot\rangle\_{\mathcal{S\_H}})$ be the separable Hilbert space of Hilbert Schmidt operators. Then, e.g. $(\phi\_{ij})$ is a ONB of $\mathcal{S\_H}$ with $\phi\_{ij}(e\_k) = e\_i\delta\_{jk},\,\forall k,$ where $\delta\_{jk}$ denotes the Kronecker-Delta.


Moreover, for $p \in \mathbb{N},$ let's consider the $p$-dimensional case. $(\mathcal{S}^p\_{\mathcal{H}},\langle\cdot, \cdot\rangle\_{\mathcal{S}^p\_{\mathcal{H}}})$ can be considered as the space of vectors where each component can be considered as a Hilbert Schmidt operater from $\mathcal{H} \rightarrow \mathcal{H}$ (in other words every component is an element of $\mathcal{S\_H})$ and for $\boldsymbol{s} \colon\!= (s\_1, ..., s\_p)^T,$ $\boldsymbol{t} \colon\!=(t\_1, ..., t\_p)^T \in \!\mathcal{S}^p\_{\mathcal{H}}\colon \langle\boldsymbol{s}, \boldsymbol{t}\rangle\_{\mathcal{S}^p\_{\mathcal{H}}} \colon\!= \sum\_{i=1}^p\langle s\_i, t\_i\rangle\_{\mathcal{S\_H}}.$ Also, $(E\_j)\_j = ((E^{(1)}\_j\!\!\!, ..., E^{(p)}\_j)^T)$ is a ONB of $(\mathcal{H}^p, \langle\cdot, \cdot\rangle\_{\mathcal{H}^p})$ where $\langle\boldsymbol{x}, \boldsymbol{y}\rangle\_{\mathcal{H}^p} \colon\!= \sum\_{i=1}^p\langle x\_i, y\_i\rangle\_{\mathcal{H}}, \boldsymbol{x} \colon\!= (x\_1, ..., x\_p)^T \in\mathcal{H}^p$ and $\boldsymbol{y} \colon\!=(y\_1, ..., y\_p)^T \in \!\mathcal{H}^p.$


**Question:** Am I able (and how?) to construct a ONB of $\mathcal{S}^p\_{\mathcal{H}}$ given a ONB $(E\_j)$ of $\mathcal{H}^p$ ?


Thank you in advance!



Construction of orthonormal basis of the Hilbert space $\mathcal{S}^p_{\mathcal{H}}$ of vectors of $p \in \mathbb{N}$ Hilbert Schmidt operators
Thanks to "Jan-Christoph Schlage-Puchta", who edited my question before, for giving me a hint to write it down correctly, hopefully ;).


If $(E\_j)$ is a ONB of $\mathcal{H}^p,$ then I can choose a subsequence $(j\_k)\_k \subseteq \mathbb{N}$ such that $(E^{(1)}\_{j\_k})\_k$ is maximal linear independent. Then the orthonormalised version of $(E^{(1)}\_{j\_k})\_k,$ denoted as $(e\_k),$ is a ONB of $\mathcal{H}$ which leads to a ONB $(\phi\_{ij})$ of $\mathcal{S\_H}$ defined by $\phi\_{ij}(e\_k) = e\_i\delta\_{jk},\,\forall i,j,k.$ 


**Answer** A ONB of $\mathcal{S}^p\_{\mathcal{H}}$ given $(E\_j)$ could be: $(\Phi\_{ijl}) \colon\!= \big((0, \dots, \phi\_{ij}, 0, ..., 0)^T\big),$ where $\phi\_{ij}$ is located at the $l$-th position, since $\forall i, i', j, j' \in \mathbb{N}$ and $\forall l, l' \in \{1, ..., p\}\colon$ $$\left\langle \Phi\_{ijl}, \Phi\_{i'j'l'}\right\rangle\_{\mathcal{S}^p\_{\mathcal{H}}} = \sum\_{k=1}^p \left\langle \Phi^{(k)}\_{ijl}, \Phi^{(k)}\_{i'j'l'}\right\rangle\_{\mathcal{S\_H}} \!\!\!= \delta\_{ll'} \left\langle\Phi^{(l)}\_{ijl}, \Phi^{(l)}\_{i'j'l}\right\rangle\_{\mathcal{S\_H}} \!\!\!= \delta\_{ll'}\left\langle \phi\_{ij}, \phi\_{i'j'}\right\rangle\_{\mathcal{S\_H}} = \delta\_{ll'}\delta\_{ii'}\delta\_{jj'},$$ where the last step follows from the fact that $(\phi\_{ij})$ is a ONB.


