Does $\mathcal{B}(\mathcal{H})=\mathcal{H}\otimes\mathcal{H}^*$ in infinite dimensions?

Question

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!)

Answer 1

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.

Answer 2

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 $B(H)$.

Answer 3

Somewhat related: Using Hahn-Banach, for infinite dimensional $H$, you can make a state on $B(H)$ that vanishes on the compact operators but is 1 on the identity. That state does not come from a density operator.