  For a commuting $d$- tuple of operators $\boldsymbol T$ defined on a complex
separable Hilbert space $\mathcal H$, let $\big [ \!\!\big [ \boldsymbol T^*,
\boldsymbol T \big ]\!\!\big ]$ be the $d\times d$ block operator $\big
(\!\!\big (\big [ T_j^* , T_i\big ]\big )\!\!\big )$ of the commutators $[T^*_j
, T_i] := T^*_j T_i - T_iT_j^*$. We define the determinant of $\big [ \!\!\big
[ \boldsymbol T^*, \boldsymbol T \big ]\!\!\big ]$ by symmetrizing the products
in the Laplace formula for the determinant of a scalar matrix. We prove that
the determinant of $\big [ \!\!\big [ \boldsymbol T^*, \boldsymbol T \big
]\!\!\big ]$ equals the generalized commutator of the $2d$ - tuple of
operators, $(T_1,T_1^*, \ldots, T_d,T_d^*)$ introduced earlier by Helton and
Howe. We then apply the Amitsur-Levitzki theorem to conclude that for any
commuting $d$ - tuple of $d$ - normal operators, the determinant of $\big [
\!\!\big [ \boldsymbol T^*, \boldsymbol T \big ]\!\!\big ]$ must be $0$. We
show that if the $d$- tuple $\boldsymbol T$ is cyclic, the determinant of $\big
[ \!\!\big [ \boldsymbol T^*, \boldsymbol T \big ]\!\!\big ]$ is non-negative
and the compression of a fixed set of words in $T_j^* $ and $T_i$ -- to a
nested sequence of finite dimensional subspaces increasing to $\mathcal H$ --
does not grow very rapidly, then the trace of the determinant of the operator
$\big [\!\! \big [ \boldsymbol T^* , \boldsymbol T\big ] \!\!\big ]$ is finite.
Moreover, an upper bound for this trace is given. This upper bound is shown to
be sharp for a class of commuting $d$ - tuples. We make a conjecture of what
might be a sharp bound in much greater generality and verify it in many
examples.
