Explicit form of this unitary transformation

Question

Disclaimer: This question has its motivation from physics. It is probably not entirely rigorous at the moment. I just want to clarify some steps and try to make the arguments rigorous afterwards, if possible.

The following was extracted from page 66 of Statistical Field Theory. Suppose we have families of fermionic creation and annihilation operators $\{a_{x}^{*}\}_{x\in \mathbb{Z}}$ and $\{a_{x}\}_{x\in \mathbb{Z}}$, respectively. These satisfy anticommutation relations $\{a_{x}^{*},a_{y}^{*}\} = \{a_{x},a_{y}\} = 0$ and $\{a_{x},a_{y}^{*}\} = \delta_{xy}$, where the latter is just the Kronecker delta. In the mentioned book, the authors introduce a unitary operator $U$ which is defined by its action, as follows: $$U^{-1}a_{x}^{*}U = \frac{1}{2}i(a_{x+1}^{*}-a_{x}^{*}+a_{x+1}+a_{x}) \tag{1}\label{1}$$ $$U^{-1}a_{x}U = \frac{1}{2}i(-a_{x+1}^{*}-a_{x}^{*}-a_{x+1}+a_{x}) \tag{2}\label{2}.$$ This operator $U$ is never made explicit, and it is difficult for me to understand what is going on. My question is: can someone please help me understand what is being done in the above formulas? It seems that these are some sort of Bogoliubov transformations, but I am not quite sure. Most importantly: can we write these operators explicitly, as exponential maps? For example, is there a realization $U = e^{Q}$, where $Q$ is a self-adjoint quadratic form on the creation and annihilation operators, from which $e^{-Q}a_{x}^{\#}e^{Q}$ reproduces (\ref{1}) and (\ref{2})?

Add: To make it more clear, I am not very interested about existence of the operator $U$ itself; I strongly believe such an operator exists. The point is to write it explicitly, in such a way that (\ref{1}) and (\ref{2}) can be obtained.

Answer 1

Let me define (for $x\in\mathbb{Z}$) the Hermitian operators $$\gamma_{2x-1}=a_x+a_x^\dagger,\;\;\gamma_{2x}=i(a_x-a_x^\dagger).$$ The $\gamma_n$'s are known in physics as Majorana operators. Each fermion operator $a_x$ is mapped onto a pair of Majorana operators $\gamma_{2x-1}$ and $\gamma_{2x}$. Notice that $\gamma_n\gamma_m=-\gamma_m\gamma_n$ for $n\neq m$, while $\gamma_n^2=1$. This implies that $2^{-1/2}(\gamma_n+\gamma_m)$ is unitary for $n\neq m$.

I seek a unitary operator $U$ which translates $\gamma_n$ to $\gamma_{n+1}$: $$U\gamma_{2x-1}U^{-1}=\gamma_{2x},\;\;U\gamma_{2x}U^{-1}=\gamma_{2x+1}.$$ This is equivalent to the transformation in the OP, $$Ua_xU^{-1}=\tfrac{1}{2}U\bigl(\gamma_{2x-1}-i\gamma_{2x}\bigr)U^{-1}=\tfrac{1}{2}\bigl(\gamma_{2x}-i\gamma_{2x+1}\bigr)=\tfrac{1}{2}i(a_x-a_x^\dagger-a_{x+1}-a_{x+1}^\dagger).$$

I place the Majorana operators $\gamma_n$ on a ring, taking $n\in\{1,2,\ldots 2N\}$ and identifying $\gamma_{2N+1}\equiv \gamma_1$. Then the desired translation operator is given by the unitary operator product $$U=2^{1/2-N}(\gamma_1+\gamma_2)(\gamma_2+\gamma_3)\cdots(\gamma_{2N-1}+\gamma_{2N}).$$

Answer 2

Denote by $b_x^*$ and $b_x$ the right-hand sides of (1) and (2). They satisfy the anticommutation relations, and therefore there is an isomorphism from the $C^*$-algebra generated by the $a_x$'s onto the $C^*$-algebra generated by the $b_x$'s sending $a_x$ to $b_x$.

However, this isomorphism is not implemented by a unitary. That is, there is no unitary transformation of the antisymmetric Fock space that satisfies your relations.

Indeed, if such a $U$ existed, then evaluating (2) at the vacuum state would give that $$ \| a_x (U\Omega)\| = \frac{1}{2}\| U (a_{x+1}^* + a_x^* + a_{x+1} -a_x)\Omega\| = \frac{1}{2}\| a_{x+1}^* \Omega + a_x^* \Omega\| = \frac{1}{\sqrt 2}.$$ The contradiction comes from the fact that for every vector $\xi$ in the antisymmetric Fock space, $\lim_{x \to \infty} \|a_x \xi\| = 0$ ($\|a_x \xi\| = 0$ if $x$ does not appear anywhere in $\xi$).

Two remarks:

• I am not even sure that the $C^*$-algebra generated by the $b_x$'s is not a strict subalgebra of the $C^*$-algebra generated by the $a_x$'s. This would give another, even stronger, reason why $U$ does not exist.

• As mentionned in the comments, what you claim becomes true if you replace $\mathbf{Z}$ by $\mathbf{Z}/N\mathbf{Z}$. Indeed, in that case, comparing the dimensions we see that $C^*( (b_x)_x) = C^*((a_x)_x) \simeq M_{2^N}$, and since automorphisms of $M_{2^N}$ are inner you get such a unitary $U$.