I would like to start by saying that any comment or idea is highly appreciated.


Let us observe that for Hilbert-Schmidt operators $H\_1,H\_2$ on an infinite-dimensional separable complex Hilbert space $H$ and a bounded self-adjoint operator $T: H \rightarrow H$
$$\operatorname{Tr} \left(H\_1 ([T,H\_2])^\*\right) = \operatorname{Tr} \left([T,H\_1] H\_2^\* \right).$$ 


In other words, the operator $S\_T(H\_1):=[T,H\_1]$ is self-adjoint with respect to the Hilbert-Schmidt inner product.


Then, we can look at $0 \le S\_T^2(H\_1) = [T,[T,H]]$
and see that this operator is positive, as it is a square.


Now, I would like to ask the following questions: 


(i)Is there a self-adjoint bounded $T$ such that $S\_T^2$ has a spectral gap? (FALSE! See answer by Mateusz Wasilewski.)


(ii)Is there a finite collection of self-adjoint bounded $T\_1,...,T\_n$ such that $S\_{T\_1}^2 + ... + S\_{T\_n}^2$ has a spectral gap?


(iii) Is there an infinite collection of positive bounded $(T\_n)$ such that $\sum\_n T\_n$ converges pointwise such that $\sum\_n S\_{T\_n}^2$ has a spectral gap?


I say $S\_T^2$ has a spectral gap if
$$\operatorname{Tr}(S\_T^2(H\_1)H\_1^\*) \ge \lambda \operatorname{Tr}(H\_1H\_1^\*)$$ 


for some positive $\lambda>0$.


It looks to me that somebody with a operator algebra background may be more familiar with this problem. I assume the three questions are increasingly likely to be true?



Existence of spectral gap
The answer to question (i) is negative. First of all, I'd like to remark that the shouldn't be an $i$ in the definition of $S\_T$ therefore for me $S\_T^2(H)$ will be $[T,[T,H]]$.


Let us first consider the case in which $T=\textrm{diag}(c\_i)$ is a diagonal operator. Then $\textrm{Tr}(S\_T^2(H) H^{\ast}) = \sum\_{i,j} (c\_i-c\_j)^2 |h\_{ij}|^2$, where $H=(h\_{ij})$. Now $H=e\_{ii}$ is in the kernel of $T$, so there is no spectral gap.


To prove it in the general case, assume that $T$ is self-adjoint contraction such that $S\_T^2$ has a spectral gap $\lambda>0$, which means that for any $H$ with $\textrm{Tr}(H H^{\ast})=1$ we have $\textrm{Tr}(S\_T^2(H) H^{\ast}) \geqslant \lambda$. Now, by the Weyl-von Neumann theorem, we can write (up to unitary conjugacy which is harmless) $T= D + \varepsilon$, where $D=\textrm{diag}(d\_i)$ is diagonal and $\varepsilon$ is Hilbert-Schmidt with arbitrarily small Hilbert-Schmidt norm. We can compute $S\_T^2(H) = [D,[D,H]] + [\varepsilon,[D,H]]+[D,[\varepsilon,H]] + [\varepsilon,[\varepsilon,H]]$. By our assumption on $H$ and $T$, the norm of $D$ is at most slightly above, and $H$ has Hilbert-Schmidt norm equal to $1$, so the Hilbert-Schmidt norm of the last three terms can be made arbitrarily small. If we assume that the Hilbert-Schmidt norm of $\varepsilon$ is small enough, this would force $S\_D^2$ to have a spectral gap, which we excluded earlier. 


