Bogoliubov Transformation

Question

Let $\mathcal{A}_{CAR}(\mathcal{H})$ be a CAR algebra over a Hilbert space $\mathcal{H}$.

Consider a linear $S$ and an antilinear $T$ both bounded operators acting on $\mathcal{H}$ satisfying: $$S^*S+T^*T=1=SS^*+TT^*\\ S^*T+T^*S=0=ST^*+TS^*$$

Define a Bogoliubov transformation by: $$\gamma[a(f)]:=a(Sf)+a^*(Tf)\text{ and }\gamma[1]:=1$$ extended on the span of $a(f)$, i.e.: $$\gamma[\lambda a(f)]:=\lambda\gamma[a(f)]\\ \gamma[a(f)+a(g)]:=\gamma[a(f)]+\gamma[a(g)]\\ \gamma[a(f)a(g)]:=\gamma[a(f)]\gamma[a(g)]\\ \gamma[a(f)^*]:=\gamma[a(f)]^*$$ Similarly define its backtransformation by: $$\delta[a(f)]:=a(U^*f)+a(V^*f)^*\text{ and }\delta[1]:=1$$

Question 1: Why are the Bogoliubov transformation and its backtransformation well defined?

My problem is that the $a(f)$ are not independent to each other: $$a(\lambda f)=\overline{\lambda}a(f)\text{ and }a(f+g)=a(f)+a(g)\\ \{a(f),a(g)\}=0\text{ and }\{a(f),a(g)^*\}=\langle f,g\rangle 1$$

Question 2: Why are the Bogoliubov transformation and its backtransformation inverse to each other?

Question 3: Why are the Bogoliubov transformation and its backtransformation continuous?

Answer 1

The Bogoliubov transformation is well defined since it respects the antilinearity: $$a(U(\lambda f))+a(V(\lambda f))^*=\gamma[a(\lambda f)]=\gamma[\overline{\lambda}a(f)]=\overline{\lambda}\gamma[a(f)] \\=\overline{\lambda}a(Uf)+\overline{\lambda}a(Vf)^*=a(U(\lambda f))+a(V(\lambda f))^*\\ a(U(f+g))+a(V(f+g))^*=\gamma[a(f+g)]=\gamma[a(f)+a(g)]=\gamma[a(f)]+\gamma[a(g)] \\=a(Uf)+a(Ug)+a(Vf)^*+a(Vg)^*=a(U(f+g))+a(V(f+g))^* $$

and the anticommutation relations: \begin{align} 0&=\gamma[0]=\gamma[\{a(f),a(g)\}]=\{\gamma[a(f)],\gamma[a(g)]\}\\ &=\{a(Uf),a(Ug)\}+\{a(Uf),a(Vg)^*\}+\{a(Vf)^*,a(Ug)\}+\{a(Vf)^*,a(Vg)^*\}\\ &=0+\langle Uf,Vg\rangle 1+\langle Ug,Vf\rangle 1+0=\langle f,(U^*V+V^*U)g\rangle 1=0\\ \langle f,g\rangle 1&=\gamma[\langle f,g\rangle 1]=\gamma[\{a(f),a(g)^*\}]=\{\gamma[a(f)],\gamma[a(g)]^*\}\\ &=\{a(Uf),a(Ug)^*\}+\{a(Uf),a(Vg)\}+\{a(Vf)^*,a(Ug)^*\}+\{a(Vf)^*,a(Vg)\}\\ &=\langle Uf,Ug\rangle 1+0+0+\langle Vg,Vf\rangle 1=\langle f,(U^*U+V^*V)g\rangle 1=\langle f,g\rangle 1 \end{align}

The Bogoliubov backtransformation is well defined too since it also respects the antilinearity: $$a(U^*(\lambda f))+a(V(\lambda f))^*=\delta[a(\lambda f)]=\delta[\overline{\lambda}a(f)]=\overline{\lambda}\delta[a(f)] \\=\overline{\lambda}a(U^*f)+\overline{\lambda}a(V^*f)^*=a(U^*(\lambda f))+a(V^*(\lambda f))^*\\ a(U^*(f+g))+a(V^*(f+g))^*=\delta[a(f+g)]=\delta[a(f)+a(g)]=\delta[a(f)]+\delta[a(g)] \\=a(U^*f)+a(U^*g)+a(V^*f)^*+a(V^*g)^*=a(U^*(f+g))+a(V^*(f+g))^* $$

and the anticommutation relations: \begin{align} 0&=\delta[0]=\delta[\{a(f),a(g)\}]=\{\delta[a(f)],\delta[a(g)]\}\\ &=\{a(U^*f),a(U^*g)\}+\{a(U^*f),a(V^*g)^*\}+\{a(V^*f)^*,a(U^*g)\}+\{a(V^*f)^*,a(V^*g)^*\}\\ &=0+\langle U^*f,V^*g\rangle 1+\langle U^*g,V^*f\rangle 1+0=\langle f,(UV^*+VU^*)g\rangle 1=0\\ \langle f,g\rangle 1&=\delta[\langle f,g\rangle 1]=\delta[\{a(f),a(g)^*\}]=\{\delta[a(f)],\delta[a(g)]^*\}\\ &=\{a(U^*f),a(U^*g)^*\}+\{a(U^*f),a(V^*g)\}+\{a(V^*f)^*,a(U^*g)^*\}+\{a(V^*f)^*,a(V^*g)\}\\ &=\langle U^*f,U^*g\rangle 1+0+0+\langle V^*g,V^*f\rangle 1=\langle f,(UU^*+VV^*)g\rangle 1=\langle f,g\rangle 1 \end{align}

They're inverse to each other: \begin{align} \delta[\gamma[a(f)]]&=\delta[a(Uf)+a(Vf)^*]=\delta[a(Uf)]+\delta[a(Vf)]^*\\ &=a(U^*Uf)+a(V^*Uf)^*+a(U^*Vf)^*+a(V^*Vf)\\ &=a((U^*U+V^*V)f)+a((U^*V+V^*U)f)^*=a(f)\\ \gamma[\delta[a(f)]]&=\gamma[a(U^*f)+a(V^*f)^*]=\gamma[a(U^*f)]+\gamma[a(V^*f)]^*\\ &=a(UU^*f)+a(VU^*f)^*+a(UV^*f)^*+a(VV^*f)\\ &=a((UU^*+VV^*)f)+a((UV^*+VU^*)f)^*=a(f) \end{align}

Moreover both are continuous: $$\|\gamma[A]\|^2=\|\gamma[A^*A]\|=\sup\sigma(\gamma[A^*A])\leq\sup\sigma(A^*A)=\|A^*A\|=\|A\|^2\\ \|\delta[B]\|^2=\|\delta[B^*B]\|=\sup\sigma(\delta[B^*B])\leq\sup\sigma(B^*B)=\|B^*B\|=\|B\|^2$$ (Besides, they are therefore even isometries: $\|\gamma[A]\|=\|A\|\text{ and }\|\delta[B]\|=\|B\|$)

Finally, by the uniform extension principle it follows: $$\bar{\delta}[\bar{\gamma}[A]]=\bar{\delta}[\bar{\gamma}[\lim_n A_n]]=\bar{\delta}[\lim_n\gamma[A_n]]=\lim_n\delta[\gamma[A_n]]=\lim_n A_n=A\\ \bar{\gamma}[\bar{\delta}[A]]=\bar{\gamma}[\bar{\delta}[\lim_n A_n]]=\bar{\gamma}[\lim_n\delta[A_n]]=\lim_n\gamma[\delta[A_n]]=\lim_n A_n=A$$

Answer 2

Although this is an old question, I would still like to add that the Bogoliubov transform is actually induced by a unitary transform on the fermion Fock space $\mathscr{F}$, i.e., for every $S,T$ which satisfy $\operatorname{tr}{(TT^*)}<\infty$, there exists a unitary operator $W:\mathscr{F} \rightarrow \mathscr{F}$ such that \begin{equation} W a^*(f)W^* =a^* (Sf)+a(Tf) \end{equation}

Your questions then follow from properties of the unitary transform.

Just a comment. I personally think that this is very important to understanding the Bogoliubov transform, but unfortunately, I have almost never seen it mentioned in textbooks or papers on the subject. The only reference I know that proves this rigorously is "Bach, V., Lieb, E.H. & Solovej, J.P. Generalized Hartree-Fock theory and the Hubbard model. J Stat Phys 76, 3–89 (1994)." (see Theorem 2.2) (though I admit that there are a few steps in the proof that I have not checked carefully)