# Does $\{\left|\varphi\right>\left<\psi\right|+\left|\psi\right>\left<\varphi\right||\varphi\in\{\psi\}^{\perp}\}$ split $\mathfrak{S}_1$?

Let $$\mathfrak{S}_1$$ be the space of trace-class self-adjoint operators on $$L^2(\mathbb{R}^n)$$, and $$\psi\in L^2(\mathbb{R}^n)$$ such that $$\int |\psi|^2 = 1$$. Is there a projection from $$\mathfrak{S}_1$$ onto $$\left\{ \left|\varphi\right>\left<\psi\right|+\left|\psi\right>\left<\varphi\right|| \varphi\in\{\psi\}^{\perp}_{L^2}\right\} ?$$

If needed, one can change $$\mathfrak{S}_1$$ for $$\mathfrak{S}_1 \cap \{ tr \cdot = 0 \}$$ its subspace of vanishing-trace operators. A first question would be whether $$\left\{ \left|\varphi\right>\left<\psi\right|+\left|\psi\right>\left<\varphi\right|| \varphi\in\{\psi\}^{\perp}_{L^2}\right\}$$ is closed in $$\mathfrak{S}_1$$.

• This is self-adjoint rank $2$ operators (which is certainly a closed space) with an extra condition, which doesn't ruin anything. If $T$ is such an operator, then $\varphi = T\psi$, so if $T_n\to T$ (and in fact strong convergence is already sufficient, you don't need the trace norm), then $\varphi_n\to\varphi$. – Christian Remling Jun 3 '19 at 21:48

Yes, there is an $${\mathbb R}$$-linear projection of norm one. Let $$P^\perp$$ denote the orthogonal projection onto $$\{\psi\}^\perp$$. The projection is given by $$\Theta\colon S \mapsto |P^\perp \frac{S+S^*}{2}\psi \rangle \langle \psi |\, +\, | \psi\rangle \langle P^\perp \frac{S+S^*}{2}\psi |$$ for $$S\in{\mathfrak S}_1$$. To prove $$\Theta$$ has norm one, we may assume that $$S=S^*$$. By considering the decomposition $$L^2({\mathbb R})=\{\psi\}^{\perp\perp}\oplus\{\psi\}^\perp$$, the map $$\Theta$$ looks like $$S=\left[ \begin{matrix} \alpha & \langle\phi| \\ |\phi\rangle & B \end{matrix}\right] \mapsto \left[ \begin{matrix} 0 & \langle\phi| \\ |\phi\rangle & 0 \end{matrix}\right] =|\phi \rangle \langle \psi |\, +\, | \psi\rangle \langle \phi|.$$ One has $$2\|\phi\|^2=|\langle\phi|S\psi\rangle|+|\langle\psi|S\phi\rangle|\le\|S\|_1\|\psi\|\|\phi\|$$ because $$\phi\perp\psi$$. One the other hand, $$\|\,|\phi \rangle \langle \psi |\, +\, | \psi\rangle \langle \phi|\, \|_1 = 2\|\phi\|$$ by, e.g., considering the two dimensional subspace spanned by $$\phi$$ and $$\psi$$. By combining these, one sees $$\Theta$$ has norm one.