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Mar 4, 2019 at 9:26 comment added Robert Rauch Well, yes: given $v:\mathfrak{h}\to\mathfrak{h}^*$ (note the order of $\mathfrak{h}$ and $\mathfrak{h}^*$), then there is a Bogoliubov transformation $U$ on $\mathcal{F}$ with $u=0$ and the given $v$. The main difficulty here is to construct the image $U\Omega$ of the vacuum $\Omega\in\mathcal{F}$. A proof can be found e.g. in Solovejs lecture Notes on Many Particle Quantum Mechanics (Theorem 9.5). Note that in the infinite-dimensional case $V^*V$ is required to be trace class ("Shale Stinespring condition").
Mar 1, 2019 at 14:55 comment added InfiniteLooper If i give you a $v : h^* \to h$ could you give me a preimage for the matrice with coeeficients $u=0$ and $v$ ?
Mar 1, 2019 at 13:44 history edited Robert Rauch CC BY-SA 4.0
Fix typo in definition of Bogoliubov transformation
Feb 26, 2019 at 22:47 history edited YCor
edited tags
Feb 26, 2019 at 22:11 history asked Robert Rauch CC BY-SA 4.0