Let $H$ be a Hilbert space, $\xi, \eta \in H$. Understanding the bounded linear operator $ \xi \otimes \eta^* $?

Question

Let $H$ be a Hilbert space with inner product $\langle \cdot \mid \cdot \rangle$ and let $\xi, \eta$ be vectors in $H$.

We define the map $\xi \otimes \eta^*: H \to H$ by $\xi \otimes \eta^*(\zeta) = \langle \zeta \mid \eta \rangle \xi$ for each $\zeta \in H$.

I am trying to show that $\xi \otimes \eta^*$ is a bounded linear operator on $H$ and that $|| \xi \otimes \eta^* || = || \xi|| \cdot || \eta ||$.

I understand the idea, I think. It's clear that $\xi \otimes \eta^*(\zeta) = \langle \zeta \mid \eta \rangle \xi = c \xi$, for some $c \in \mathbb{F}$ that depends on $\zeta, \eta$. So it makes sense that it's a bounded linear map.

But I'm being conceptually thrown off by the notation - I don't really understand what $\eta^*$ is referring to, if anything. The notation makes me think of the adjoint of an operator on Hilbert space, but I don't understand how/why a vector has an adjoint. I also don't really understand how this relates to tensor products.

Does this operator have a well-known name that I can search to read more about it? Or can anyone explain the intuition behind why the notation might be this way?

Answer 1

If we would write $\xi \otimes \eta$ instead, this would suggest that $(\xi, \eta) \mapsto \xi \otimes \eta$ is bilinear. However, this is not the case since we have antilinearity in the second factor. To emphasise this and because the adjoint has the property $(\lambda x)^* =\overline{\lambda}x^*$, we use the notation $\xi \otimes \eta^*$ instead. Perhaps it is safer to use $\xi \otimes \overline{\eta}$ instead.

I'm not aware of any special name, since these operators are just the rank-one operators.

Note that $$\|\xi \otimes \eta^*\| = \sup_{\|\zeta\| \le 1} \|\langle \zeta,\eta\rangle\xi\| = \|\xi\| \sup_{\|\zeta\| \le 1} |\langle \zeta, \eta\rangle| \le \|\xi\|\|\eta\|$$ by the Cauchy-Schwarz inequality. To see the other inequality, note that $$\left\|(\xi \otimes \eta^*)\left(\frac{\eta}{\|\eta\|}\right)\right\|= \|\xi\|\|\eta\|$$ when $\eta \ne 0$.

Answer 2

You can identify $\eta$ with the linear map $ \mathbb C \to H$ which sends $1$ to $\eta$. The adjoint of this map is a linear functional on $H$. Then the conposition $\xi \eta^*$ is an operator on $H$.

Answer 3

There are some isomorphisms hidden in this notation. The Hilbert space $H$ is canonically isomorphic to $B(\mathbb{C},H)$ und the map $T$ that maps $\xi$ to $z\mapsto z\xi$. The adjoint $T(\xi)^\ast$ is then given by $T(\eta)^\ast\xi=\langle \xi\mid\eta\rangle$. Thus $T(\xi)T(\eta)^\ast\zeta=\langle \zeta\mid \eta\rangle\xi$. That was already explained by Blazej.

But why the tensor product? The Hilbert space $H$ is also canonically isomorphic to $\mathbb{C}\otimes H$ and $H\otimes\mathbb{C}$ under the maps $\Phi\colon \xi\to 1\otimes \xi$ and $\Psi\colon \xi\mapsto \xi\otimes 1$. Thus $$ (\Psi^{-1}\circ (T(\xi)\otimes T(\eta)^\ast)\circ\Phi)(\zeta)=\Psi^{-1}(\xi\otimes\langle\zeta\mid\eta\rangle)=\langle\zeta\mid \eta\rangle\xi. $$ So instead of $\xi\otimes\eta^\ast$ we should really write $\Psi^{-1}\circ (T(\xi)\otimes T(\eta)^\ast)\circ\Phi$. But this notation is quite cumbersome, so we forget about all the isomorphisms.