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

Question

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 $\mathfrak{S}_1$.

Answer 1

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.