Suppose that $\mathbb H$ is a separable Hilbert space. I am interested in the tensor $x\otimes y$, where $x,y\in\mathbb H$. From what I understand, the tensor $x\otimes y$ can be viewed as a bounded linear operator from $\mathbb H$ to $\mathbb H$. However, I see different definitions of this operator. It is defined as $$ (x\otimes y)(z)=\langle z,y\rangle x $$ or $$ (x\otimes y)(z)=\langle x,z\rangle y. $$ There are two things that need to be considered: (1) the element that is going to be outside of the inner product and (2) the order of the elements in the inner product if $\mathbb H$ is complex. In principle, we could have $4$ different definitions of $x\otimes y$.
Is one of the possible definitions of $x\otimes y$ more natural or more preferable than the other definitions? Are there any arguments to use some specific definition?
Any help is much appreciated.
The 4 possible definitions you mention are reduced to 2 when you choose whether the inner product is linear in its first or second argument. Let's first assume it's linear in its first argument.
There are good reasons to go for the convention: $$(x\otimes y)(z):=\langle z,y\rangle x~.$$ Reason one: When $H$ is finite dimensional, we can then see $x\otimes y$ as the matrix $$ \begin{pmatrix} x_1 \\\ \vdots \\\ x_n \end{pmatrix} (\overline{y_1},\ldots,\overline{y_n})~, $$ acting on column vectors $z$. This is much nicer than having the order of $x$ and $y$ reversed!
Reason two: In physics, the inner product is linear in its second argument. One denotes the operator $x\otimes y$ by $|x\rangle\langle y|$, with the very natural definition $$|x\rangle\langle y|(z):=x\langle y , z\rangle =\langle y,z\rangle x~.$$ This is another reason to go for the convention above, because the only difference is in the inner product.