# Complex conjugate of the Hilbert space

Consider a Hilbert space $H=L^2(\mathbb{R}_+)$, take its conjugate $\overline{H} := \left\{f^{+}, f \in H \right\}$, where $+$ stands for the conjugation. Space $\overline{H}$ is a Hilbert space with an inner product $\left<f^{+},g^{+} \right>= \int_{\mathbb{R}_+} \bar{f}(t)g(t) dt$. The space $H$ is (antilinear) isometric isomorphic to $\overline{H}$. How about the space $H \otimes \overline{H}$, $\overline{H} \otimes \overline{H}$, $H \otimes H$ ? Do they isometric isomorphic only anti-linearly? When we can replace $\overline{H}$ with just $H$? We can identify conjugate of $H$ with its dual, but in the literature we can always see that instead of the conjugate space, again, the space $H$ is considered. Why can we do it? The isomorphism is anti-linear, not linear? Thank you for the explanation.

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What does $f^{+}$ mean? – Christopher A. Wong Nov 30 '12 at 21:09

If $H$ is a Hilbert space then $H \otimes \bar{H}$ can be identified with Hilbert Schmidt operators on $H$ (sometimes denoted by $B_2(H)$), $\left| A \right> \left< B \right| \in B_2(H)$ corresponds to $A \otimes B^+ \in H \otimes \bar{H}$. I hope it answers most of your questions.