Which operator can implement a fermionic Bogoliubov Transformation?

Question

In bosonic systems (for example in Quantum Optics), two-mode Bogoliubov transformations are implemented via squeezing operators as $$\hat{S}_2(\xi)=\exp(\xi^{*}\hat{a}\hat{b}-\xi\hat{a}^\dagger\hat{b}^\dagger),$$ with $\xi=re^{i\theta}$, such that $$\hat{S}_2^\dagger(\xi)\hat{a}\hat{S}_2(\xi)=\hat{a}\cosh(r)-e^{i\theta}\hat{b}^\dagger\sinh(r),$$ $$ \hat{S}_2^\dagger(\xi)\hat{b}\hat{S}_2(\xi)=\hat{b}\cosh(r)-e^{i\theta}\hat{a}^\dagger\sinh(r).$$

I'd like to know:

• Is there an equivalent operator $\hat{S}_2^{\text{fermion}}(\xi)$ that can implement Bogoliubov transformations in fermionic creation/annihilation operators? What is its expresion?
• If it exists, is its interpretation equivalent to that of the squeezing operator in bosonic systems?

Answer 1

When $\hat a$, $\hat a^\dagger$, $\hat b$, $\hat b^\dagger$ obey the fermion algebra $$ \{ \hat a, \hat a^\dagger\}= \{ \hat b, \hat b^\dagger\}=1, \quad \{ \hat a, \hat a\}=\{ \hat b, \hat b\}=\{ \hat a, \hat b\}= \{ \hat a, \hat b^\dagger\}=0, $$ we have $\hat a^2 =(\hat a^\dagger)^2=0$, so there is no fermion analogue of the boson single-mode squeezing operator. We can, however, still construct a two-mode operator $$ U[z]=\exp\{z \hat a^\dagger \hat b^\dagger - z^* \hat b\hat a\}\nonumber\\ = \exp\{(e^{i\theta} \tan |z|) \hat a^\dagger\hat b^\dagger\} \exp\{(\ln\cos |z|)[ (\hat a^\dagger \hat a+{\textstyle \frac 12})+( \hat b^\dagger \hat b+{\textstyle \frac 12 })]\}\exp\{(-e^{-i\theta} \tan |z|) \hat b \hat a\} $$ which implements a Bogoliubov-Valatin transformation $$ U[z]\hat a U^\dagger[z] = (\cos |z|) \hat a - (e^{i\theta}\sin |z|) \hat b^\dagger, \nonumber\\ U[z]\hat b U^\dagger[z] = (e^{i\theta}\sin |z|) \hat a^\dagger +(\cos |z|) \hat b. $$ The right-hand-side is now a compact ${\rm SU}(2)$ rotation rather than a non-compact ${\rm SU}(1,1)$ transformation.