Let $H$ be an infinite-dimensional Hilbert space, and $a_1,b_1,\ldots,a_k,b_k$ be bounded selfadjoint operators on $H$. Can the sum $\sum_{i=1}^k[a_i,b_i]$ be a (pure imaginary) multiple of the identity ?
I know a result of Halmos which says that the identity is the sum of two commutators, but they are of the form $[P,P^\dagger]$ so cannot fit the bill.
Yes. Indeed, for self-adjoint $a,b$, the commutator $[a,b]$ is equal to $\frac{i}{2}[p,p^*]$ for $p=a+ib$. So the theorem by Halmos that you refer to (every hermitian operator is a sum $[p_1,p_1^*]+[p_2,p_2^*]$ for bounded operators $p_1,p_2$) implies that every anti-hermitian operator $X$ is a sum of $[a_1,,b_1]+[a_2,b_2]$ for self-adjoint operators.
I think that in general, it depends on the algebra of the operators.
If for example your operators span some Lie algebra, then no it cannot happen. Since then, the commutators will be primitive elements (with regards to the hopf structure of the universal enveloping algebra), and thus their sum will also be primitive, which a multiple of identity cannot be.
But for an arbitrary set of operators, i do not know.
