Consider the most general possible many-body quantum Hamiltonian in second quantized form (one species of spinless particle in $d\geq 1$ spatial dimensions):

$$H_{\rm int} = \sum_{n,m\geq 0} \int _{\mathbb{R}^{(n+m)d}} h_{nm}(\vec k’_1,\ldots,\vec k’_n,\vec k_1,\ldots,\vec k_m) \, a^\dagger(\vec k’_1)\cdots a^\dagger(\vec k’_n) \, a(\vec k_1)\cdots a(\vec k_m) $$

where I have suppressed the integration measure ${\rm d}^d\vec k’_1\cdots {\rm d}^d\vec k’_n \, {\rm d}^d\vec k_1\cdots{\rm d}^d\vec k_m$ and the coefficients $h_{nm}$ are distributions, constrained only by requirement the the total Hamiltonian is Hermitian, $H^\dagger = H$.

Define unitary operators $U(\vec p)$ and $U(R,\vec a)$ implementing Galilean boost and Euclidean transformations, respectively, \begin{align} U(R,\vec a) \, a(\vec k) \, U^\dagger(R,\vec a) & = e^{-i \vec k \cdot a} a(R\vec k) \\ U(\vec p) \, a(\vec k) \, U^\dagger(\vec p) & = a(\vec k + \vec p) \end{align} A sufficient (and I believe necessary) condition for $H_{\rm int}$ to be Galilean invariant is that $h_{nm}$ satisfy \begin{align} & h_{nm}(\vec k’_1 + \vec p,\ldots,\vec k’_n + \vec p,\vec k_1+\vec p,\ldots,\vec k_m + \vec p) = h_{nm}(\vec k’_1,\ldots,\vec k’_n,\vec k_1,\ldots,\vec k_m) \\ & h_{nm}(R\vec k’_1,\ldots,R\vec k’_n,R\vec k_1,\ldots,R\vec k_m) = h_{nm}(\vec k’_1,\ldots,\vec k’_n,\vec k_1,\ldots,\vec k_m) \\ &\sum_{i=1}^n \vec k’_i \neq \sum_{j=1}^m \vec k_j \implies h_{nm}(\vec k’_1,\ldots,\vec k’_n,\vec k_1,\ldots,\vec k_m) = 0 \end{align} The physically interesting case, however, is Galilean covariance for some mass parameter $m>0$, $$H \longmapsto H - \vec P \cdot \frac{\vec p}{m} + \frac{|\vec p|^2}{2m} N$$ where $\vec{P}$ and $N$ are the total momentum and number operators, respectively.

A candidate Galilean covariant Hamiltonian is clearly $$H = H_0 + H_{\rm int}$$ where $H_{\rm int}$ is Galilean invariant and $$ H_0 = \int_{\mathbb{R}^d} \frac{|\vec k|^2}{2m} a^\dagger(\vec k)a(\vec k) $$

Is it true that all Galilean covariant Hamiltonians are of the above form? If so, is there a well-known proof of this statement?

  • 2
    $\begingroup$ There should be a theorem with a similar statement in a book by J.M. Jauch, if I remember well.... $\endgroup$ Sep 25, 2022 at 16:00


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