Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on an infinite-dimensional complex Hilbert $F$.



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> Let $T,S\in\mathcal{B}(F)$. The pair $(T,S)$ is said to $\lambda$-commute if there exists $\lambda\in \mathbb{C}^\*$ such that $TS=\lambda ST$. I am looking for necessary and sufficient conditions on the operators $S$ and $T$ such that $(T,S)$ $\lambda$-commute, then they already commute.
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I know already that $S,T \geq 0$ or $S,T \leq 0$ is a sufficient condition for $S$ and $T$ to commute if they $\lambda$-commute, but this is far from necessary.



When $\lambda$-commutativity implies commutativity?
I don't see which kind of condition you are looking for, as there are a lot of pairs $T,S$ such that $TS=\lambda ST$ and $\lambda\ne1$, even in finite dimension. Such pairs are said to $\lambda$-commute.


An interesting case happens when $\lambda$ is root of unity, say of order $r$. Then (Potter's Theorem) $(T+S)^r=T^r+S^r$.


Still when $\lambda^r=1$, here is a construction. Choose $T={\rm diag}(I\_{n\_1},\lambda I\_{n\_2},\ldots,\lambda^{r-1}I\_{n\_r})$ and $S$ a blockwise cyclic matrix
$$S=\begin{pmatrix} 0\_{n\_1} & M\_1 & 0 & \cdots & 0 \\
\vdots & \ddots & \ddots & \ddots & \vdots \\
\vdots & & \ddots & \ddots & 0 \\
0 & & & \ddots & M\_{r-1} \\
M\_r & 0 & \cdots & \cdots & 0\_{n\_r} \end{pmatrix}.$$
This pair $\lambda$-commutes.


