Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on an infinite-dimensional complex Hilbert $F$.
Let $T,S\in\mathcal{B}(F)$. The pair $(T,S)$ is said to be $\lambda$-commute if there exists $\lambda\in \mathbb{C}^*$ such that $TS=\lambda ST$.
If $\lambda=1$, the operators $T$ and $S$ are commuting.
There are of lot of pairs $T$ and $S$ which are $\lambda$-commute but they don't commute. It is possible to find a suitable conditions on $T$ and $S$ under which $\lambda$-commutativity implies commutativity?
Notice that in finite dimensional Hilbert spaces, since $tr(TS)=tr(ST)$, so if $tr(TS)$ is non-zero, then $\lambda=1$.
One idea would be to consider Hilbert-Schmidt operators
$$ \mathcal B_2(\mathcal H)=\lbrace B\in\mathcal B(\mathcal H)\,|\,\operatorname{tr}(B^\dagger B)<\infty\rbrace\subset\mathcal B(\mathcal H) $$
which form a Hilbert space under the scalar product $\langle A,B\rangle=\operatorname{tr}(A^\dagger B)$ (this can be seen as the "natural extension" of matrices to infinite-dimensions in some sense). Then we can use the trace argument from finite-dimensions as for $T,S\in\mathcal B_2(\mathcal H)$ now $TS=\lambda ST$ for some $\lambda\in\mathbb C$ together with $\operatorname{tr}(TS)\neq0$ implies
$$ \operatorname{tr}(TS)=\operatorname{tr}(\lambda ST)=\lambda \operatorname{tr}(ST)=\lambda \operatorname{tr}(TS)\ \Rightarrow\ \lambda=1 $$
so $T$ and $S$ commute.
As an introduction to the trace on infinite-dimensional Hilbert spaces, Hilbert-Schmidt operators etc. I recommend Chapter 3.4 of the book "Analysis Now" (1989) by Pedersen.