Second quantization

Remarks

The operator $d\Gamma(A)$ is self-adjoint, and $\Gamma(e^{itA}) = e^{it\, d\Gamma(A)}$ for all $t \in \R$, so $d\Gamma(A)$ is the generator given by Stone's theorem. Also $\Gamma(UV) = \Gamma(U)\Gamma(V)$.

Example: $d\Gamma(I) = N$ is the number operator, which acts on $\mathcal{H}^{(n)}_\pm$ as multiplication by $n$. If $A e_j = \lambda_j e_j$ for an orthonormal basis $(e_j)$ of $\mathfrak{h}$, then $d\Gamma(A) = \sum_j \lambda_j\, a^*(e_j)\, a(e_j)$ on finite-particle vectors in the span of the basis tensors, with the creation and annihilation operators.

Non-example: $d\Gamma(A)$ is additive over the particles, not multiplicative. For $A = \lambda I$ with $\lambda \in \R$, the operator $d\Gamma(A)$ acts on $\mathcal{H}^{(2)}_\pm$ as $2\lambda$, while $A \otimes A$ acts as $\lambda^2$. So $A \otimes A$ is not the second quantization of $A$ on two particles unless $\lambda \in \set{0, 2}$.

If $H_1 = -\frac{\hbar^2}{2m}\Delta$ on $\mathfrak{h} = L^2(\R^3)$, then $d\Gamma(H_1)$ is the total kinetic energy of a free gas of identical particles with any number of particles.