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In quantum information contexts (where in most cases finite dimensions are considered) I have often seen the statement that the space of bounded linear operators $\mathcal{B}(\mathcal{H})$ on a Hilbert space is (isomorphic to) the tensor product of the Hilbert space $\mathcal{H}$ with its dual $H^*$ (space of linear functionals). Is this only true in finite dimensions (where all operators are bounded anyway) or does it also apply to infinite dimensions? May I ask for a not too-technical answer without a lot of math jargon (I am a physicist, my apologies!)

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Short answer: Within $\infty$ it does not hold.
When dealing with an infinite-dimensional Hilbert space, then a norm on the tensor product $\mathcal{H}\otimes\mathcal{H}^*$ is needed, followed by the corresponding norm completion. There is no unique choice for it, but you never get the full space of all bounded linear operators (which is a $C^*$-algebra).

The algebraic tensor product as it stands above is isomorphic to the algebra of finite-rank operators on $\mathcal H$, often denoted by $\mathscr F(\mathcal H)$. This is the complement of those operators whose image is infinite-dimensional, e.g., the identity.

Three important choices (but far from all) of a norm are the

  • projective tensor product $\,\mathcal{H}\hat{\otimes}_\pi\mathcal{H}^*$ which is isomorphic to the trace-class operators on $\mathcal H$,

  • Hilbert space tensor product $\,\mathcal{H}\hat{\otimes}\mathcal{H}^*$, isomorphic to the Hilbert-Schmidt operators on $\mathcal H$,

  • injective tensor product $\,\mathcal{H}\hat{\otimes}_\epsilon\mathcal{H}^*$ being isomorphic to $\mathscr K(\mathcal H)$, the compact operators.

In a sense to be made precise, the first and the last one are the smallest ($\otimes_\pi$), and the largest
($\otimes_\epsilon$) reasonable completions, respectively.
You can find more on this SE site, e.g., more technical info or here (regarding trace-class).

Summa summarum
$\mathscr B(\mathcal H)$ equals $\mathcal{H}\otimes\mathcal{H}^*\,$ if and only if $\;\mathcal{H}$ is finite-dimensional.

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  • $\begingroup$ Thank you very much! May I ask a (basic) question: with the “equality” $\mathcal{B}(\mathcal{H})=H\otimes H^*$ one actually means isomorphism, right? $\endgroup$ – quantumorsch Mar 29 '18 at 8:31
  • $\begingroup$ Yes, a "full isomorphism" is meant which also acknowledges the norm and all that. In the infinite-dim case it is crucial to take the norm into account, whereas in the finite-dim case you "may disregard them" coz there are all equivalent anyway. $\endgroup$ – Hanno Mar 29 '18 at 8:40
  • $\begingroup$ @Hanno when dealing with general Banach spaces, the projective and injective tensor norms are the simplest examples. However, when dealing with Hilbert spaces $𝑋$ and $𝑌$, there is a natural cross norm turning the tensor product $𝑋\otimes 𝑌$ into an inner product space (basically it's given by $\langle 𝑥_1 \otimes y_1,x_2 \otimes y_2 \rangle = \langle x_1,x_2\rangle\langle y_1,y_2\rangle$). The tensor product of Hilbert spaces is usually understood to be the completion of this inner product space. It is Hilbert space isomorphic to the space of Hilbert–Schmidt operators $X^* \to Y$. [...] $\endgroup$ – Josse van Dobben de Bruyn Feb 15 '19 at 22:38
  • $\begingroup$ [...] So when $X = \mathcal H^*$ and $Y = \mathcal H$, the Hilbert space tensor product is isomorphic (as a Hilbert space) to the space $L^2(\mathcal H)$ of all Hilbert–Schmidt operators on $\mathcal H$. This is a subspace of the space of all compact operators $K(\mathcal H)$, so in particular it is not all of $B(\mathcal H)$ whenever $\mathcal H$ is infinite-dimensional. $\endgroup$ – Josse van Dobben de Bruyn Feb 15 '19 at 22:43
  • $\begingroup$ Thank you all so much. I actually have a follow-up question: is the failure to get the full space of all bounded linear operators somehow related to the well-known fact in quantum information/physics that in infinite dimensions there are states (i.e. linear , positive and normalized functionals on $\mathcal{B(H)}$ ) that are are not representable as a density operator? $\endgroup$ – quantumorsch Jun 23 '19 at 16:55
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Is $ H$ a Hilbert space, then $H^* \otimes H \subset B ( H )$ is an isometric embedding in $B(H)$. It can be shown that with this identification the tensor product is the ideal $K(H)$ of all compact operators on $H$.

If $ \dim H < \infty$, then $K(H)=B(H)$. If $\dim H = \infty$, then $K(H)$ is a proper subset of $B(H)$.

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