Why must the Bogoliubov transform preserve anticommutation relations?

Question

$\mathbf{Background}$: In my research I am studying the Ising model, expressed in terms of Jordan-Wigner fermions:

$$ H = \sum_{j=1}^n(c_j - c_j^\dagger)(c_{j+1} + c_{j+1}^\dagger) + \lambda c_jc_j^\dagger - \lambda c_j^\dagger c_j $$ where the operators $\{c_j, c_j^\dagger\}_{j=1}^n$ obey the canonical anticommutation relations, \begin{align} \{c_j, c_k\} &= c_j c_k + c_k c_j = 0\\ \{c_j, c_k^\dagger\} &= c_j c_k^\dagger + c_k^\dagger c_j = \delta_{jk} \end{align}

This Hamiltonian is quadratic in the fermionic ladder operators, and hence has an associated quadratic form: $$ H = \vec c^\dagger \mathbf A \vec c $$ where $$ \vec c \equiv \begin{pmatrix} c_1 \\ \vdots \\ c_n \\ c_1^\dagger \\ \vdots \\ c_n^\dagger\end{pmatrix}, \ \ \vec c^\dagger \equiv \begin{pmatrix} c_1 & \dots & c_n & c_1^\dagger & \dots & c_n^\dagger\end{pmatrix} $$ and $\mathbf A$ is a symmetric (?), block matrix. The task of finding the ground state of such a theory is accomplished by computing the Bogoliubov transform: \begin{align} b_j &= \sum_{k=1}^n \left(U_{jk}c_k + V_{jk}c_k^\dagger\right)\\ b_j^\dagger &= \sum_{k=1}^n \left( U^*_{jk}c_k^\dagger + V_{jk}^*c_k\right) \end{align} The matrices $\mathbf U$ and $\mathbf V$ are chosen such that in the transformed coordinates, the Hamiltonian is diagonal: \begin{align} H &= \vec b^\dagger \mathbf D \vec b\\ &= \sum_{j} D_{jj} b_j^\dagger b_j + \text{const.} \end{align} Additionally, $\mathbf U$ and $\mathbf V$ have the constraint that $\{b_j,b_j^\dagger\}_{j=1}^n$ must satisfy the canonical anticommutation relations, so that they act as creation and annihilation operators for some kind of exotic quasiparticle.

Question: Why do we enforce that the anticommutation relations should be satisfied? Is this purely because we have knowledge and tools at our disposal for handling fermionic systems? I have read a few texts on the subject and none of them seem to give a good intuition as to why we wish to preserve this property. Perhaps this is a question related to general transforms and the study of quasiparticles as a whole, but this was the context in which I encountered the problem, so I figured I'd present the scenario I am familiar with.

Answer 1

That's essentially what we mean by a quasiparticle: a degree of freedom that has been otherwise decoupled from the rest of the system and which behaves exactly like a particle as far as quantum mechanics can tell. Since the (anti)commutation relations are the crucial part of the quantum mechanics of the relevant particles, those need to be carried as well.

Moreover, the canonical anticommutation relations are the best you could hope for anyways. This is because the transformation, \begin{align} b_j &= \sum_{k=1}^n \left(U_{jk}c_k + V_{jk}c_k^\dagger\right)\\ b_j^\dagger &= \sum_{k=1}^n \left( U^*_{jk}c_k^\dagger + V_{jk}^*c_k\right) , \end{align} requires the anticommutator to read \begin{align} \{b_i^\dagger,b_j\} &= \left\{ \sum_{k=1}^n \left( U_{ik}^*c_k^\dagger + V_{ik}^*c_k\right), \sum_{l=1}^n \left(U_{jl}c_l + V_{jl}c_l^\dagger\right) \right\} \\&= \sum_{k=1}^n \sum_{l=1}^n \left[ U_{ik}^*U_{jl} \left\{c_k^\dagger,c_l \right\} + V_{ik}^*V_{jl} \left\{c_k,c_l^\dagger\right\} \right] \\&= \sum_{k=1}^n \sum_{l=1}^n \left[ -U_{ik}^*U_{jl} + V_{ik}^*V_{jl} \right] \delta_{kl} \\&= \sum_{k=1}^n\left[ V_{ik}^*V_{jk} - U_{ik}^*U_{jk} \right] \end{align} in the most general case. Thus, in the worst-case scenario, you've bungled up the (anti)commutation relations, and you're going to need a nonzero (anti)commutator any time you need to slide a $b_j$ past a $b_i^\dagger$, even if $i\neq j$. Thus, you require at the very least that the anticommutator be diagonal.

Moreover, since $\{b,b^\dagger\}=bb^\dagger + b^\dagger b$ is the sum of two positive semidefinite hermitian operators, the diagonal anticommutator needs to be real and non-negative, so you've got two possibilities:

• it can be positive, in which case it can be rescaled to one, or
• it can be zero, in which case you're essentially losing information, probably via a degenerate matrix, and the algebra spanned by the $b_i,b_i^\dagger$ will be unable to fully span all of the $c_i,c_i^\dagger$.

More broadly speaking, you're obliged to "conserve the quantumness" of the problem (unless you explicitly want to make mean-field approximations or similar) and this requires you to maintain a full set of generators for the operator algebra and therefore to have nontrivial anticommutators between your new quasiparticle fermionic operators.

Answer 2

Maybe we can think from an "opposite" point of view, here we have a binary Hamiltonian: (a single particle operator in the language of second quantization) $$H=c^{\dagger}Ac$$ the Bogoliubov transformation is essentially a transformation trying to diagonalize the Hamiltonian, or in other words, find a new basis. In quantum mechanics, we usually seek for two normalised bases that are related by a unitary transformation: $$\begin{align} |i\rangle_b&=U_{i,j}|j\rangle_c \\ \rightarrow b^{\dagger}_i &=U_{i,j} c^{\dagger}_j \\ \rightarrow b_i &= c_jU^{\dagger}_{j,i} \\ \end{align}$$ with $U_{i,j}U^{\dagger}_{j,i'} = \delta_{i,i'}$, and this automatedly leads to the fact that: $$\begin{align} [b_i,b^{\dagger}_j]_+ &=[c_{\alpha},c^{\dagger}_{\beta}]_+ U^{\dagger}_{\alpha,i}U_{j,\beta} \\ &=U^{\dagger}_{\alpha,i}U_{j,\alpha} \\ &=\delta_{i,j} \end{align}$$ so we can see that the commutation trlation does not change after the transformation.