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This construction is found in Methods of modern mathematical physics vol. 1. See page 49

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However it seems like this construction is flawed, if $(\cdot,\cdot)$ indeed is an inner product on $\mathcal{E}$ (for which he constructs $H_1\otimes H_2$ as its completion). For any $a\in\mathbb{C}$ and $\phi\in H_1,\psi\in H_2$ $$ a\phi\otimes \psi(x,y)= (x, \overline{a}\phi_1)(y,\phi_2) = (\bar{a}\phi_1)\otimes \phi_2(x,y). $$ for any $(x,y)\in H_1\times H_2$. That is, $a(\phi\otimes \psi) = (\bar{a}\phi)\otimes \psi$.

But if $(\cdot,\cdot)$ is a inner product it is conugate linear in the second argument. Thus $$ a\langle \eta\otimes \mu , \phi \otimes \psi\rangle =\langle \eta\otimes \mu , \overline{a}(\phi \otimes \psi)\rangle=\langle \eta\otimes \mu , (a\phi) \otimes \psi)\rangle = ( \eta, a\phi)(\mu,\psi) = \overline{a}( \eta, \phi)(\mu,\psi)=\overline{a}\langle \eta\otimes \mu , \phi \otimes \psi\rangle $$ for any $\eta,\phi\in H_1$ and $\mu,\psi\in H_2$.

What am I not understanding?

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    It is common in physics to define inner products to be conjugate linear in the *first* argument rather than the second. You should double check which convention is used in this book. – 2017-02-02
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    Note that, in the notation of this book, $(\cdot,\cdot)$ denotes an inner product, but $\langle \cdot, \cdot \rangle$ does **not**. – 2017-02-02
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    @Omnomnomnom Thanks mate, I totally missed that. I guess it all works out alright then if i define $\phi_1\otimes\phi_2(\psi_1,\psi_2)=(\phi_1,\psi_1)_{H_1}(\phi_2,\psi_2)_{H_2}$ in terms of regular inner products. But does this also entail that i should define the inner product on $\mathcal{E}$ by $(\phi\otimes \psi,\eta\otimes\mu)=\overline{(\phi,\eta)(\psi,\mu)}= (\eta,\phi)(\mu,\psi)$, if i want to use normal inner products? Or do the rest of the construction carry out exactly as he states when using normal inner products? – 2017-02-02
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    I don't see a mistake in your calculation. So, if you use "normal" inner products, something needs to get switched. – 2017-02-02
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    Well upon further inspection, it seem like I only need to define the simple tensor as $\phi\otimes \psi(x,y)= (\phi,x)_{H_1}(\psi,y)_{H_2}$ such that is also becomes a conjugate bilinear map (if I assume that $(\cdot,\cdot)_{H_1}$ is a normal inner product). Furthermore I don't think i should change anything in the definition of the inner product on $\mathcal{E}$ because if i define $(\phi\otimes \psi,\eta\otimes\mu):= (\eta,\phi)_{H_1}(\mu,\psi)_{H_2}$ in terms of normal products on the RHS, then we get conjugate linearity in the first argument again, so we must still define – 2017-02-02
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    $(\phi\otimes \psi,\eta\otimes\mu):= (\phi,\eta)_{H_1}(\psi,\mu)_{H_2}$ even with normal inner products on the RHS. Do you agree? Btw please post your comment as an answer, and i will award you the points. – 2017-02-02
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    Wasn't planning on posting an answer. You could post one on your own question. Otherwise, I'll probably make an answer summing up what I said in the comments tomorrow, when I get the chance – 2017-02-02

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