Relation between Hilbert space norm and trace operator $\DeclareMathOperator\tr{tr}$I am currently reading a paper where I encountered an equation that doesn't make sense to me.
Let $H$ be a Hilbert space and $x\in H$. The equation that is used is:
$$\lVert x\rVert_H^2 = \tr(xx^*)$$
where $\tr$ denotes the trace operator and $x^*$ the adjoint of x.
I am unsure how to interpret this, as I don't know any definition for an adjoint of a Hilbert space element, only of operators on $H$.
This is clearly supposed to be a generalization of the equation for $\mathbb{R}^n$:
$$\lVert x\rVert_{\mathbb{R}^n}^2 = \tr(xx^t)$$
Does this make sense to anyone?
 A: Elements of a Hilbert space are routinely seen as linear operators via the Riesz Representation Theorem. That is, if $x\in H$ then $x^* :H\to\mathbb C$ is the linear functional $x^*y=\langle y,x\rangle$. This is coherent with the finite-dimensional picture where this would be literally the matrix product $x^*y$.
Then $zx^*$ (often denoted $z\otimes x$) is the rank-one operator $zx^*(y)=\langle y,x\rangle z$. The $zx^*$ notation has the advantage that it correctly captures the fact that the $x$ conjugates scalars, in the sense that $(\alpha a)\otimes(\beta x)=\alpha\overline\beta\,(z\otimes x)$. The notation is also coherent with $x^*y=\langle y,x\rangle$. So one can correctly write $$(zx^*)y=z (x^*y)=\langle y,x\rangle\,a.$$ Incidentally, this exactly what physicists do with the bra-ket notation: $zx^*$ is precisely $|z\rangle\langle x|$.
It's an easy exercise that the operator norm of $zx^*$ is $\|zx^*\|=\|z\|\,\|x\|$. In particular, $\|xx^*\|=\|x\|^2$. If you calculate the trace using an orthonormal basis that contains $x/\|x\|$ it is clear that $$\operatorname{tr}(xx^*)=\Big\langle (xx^*)\frac x{\|x\|},\frac x{\|x\|}\Big\rangle=\|x\|^2.$$ Alternatively one can note that $\big(\frac x{\|x\|})(\frac x{\|x\|})^*$ is a rank-one projection and so it's trace is $1$.
