# Why must the Bogoliubov transform preserve anticommutation relations?

$\mathbf{Background}$: 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.

• According to the definition given in physics.stackexchange.com/q/268268/16689 the reason why the (anti-)commutation must be preserved is because otherwise, the transformation is not called a Bogoliubov (or canonical) transformation. End of it. Nothing forbids you to define a Zxv transformation allowing to change the physical nature of the quasiparticle thus generated. I'm curious how far you might go. Surely the Zxv transformation will destroy many properties of the system, as preservation of the phase-space volume, the structure of the equations of motion,their symplectic nature, ... – FraSchelle Jul 20 '17 at 5:46
• In general, considering more generic symmetries (especially the symmetry under the exchange of two identical particles) put strong restrictions on the (quasi-)particle statistics in 3D. Ultimately, it leads to the restriction toward the fermion/boson couple, i.e. the Fock space is a direct sum of the fermionic and bosonic spaces (the ones span by anticommuting and commuting creation/annihilation operators). But in 2D and 1D (D = space dimension), anyons are allowed. You can pursue your philosophical pilgrim about particle statistics following these keywords if you want to. – FraSchelle Jul 20 '17 at 5:53

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.

• I understand that in the case where the Bogoliubov transformation is a mapping from creation -> creation / annihilation -> annihilation operators, then the mapping is unitary. However, in the general case where BOTH operators are used, the mapping is not necessarily unitary: physics.stackexchange.com/questions/268268/…. I am more curious about this latter scenario, and what the reasons behind why we impose the anticommutation relations. – Zxv Jul 20 '17 at 0:25
• @Zxv Yes, I admit that here I only considered a specific case. My personal understanding about this question is that we don't want to change the property (fermionic or bosonic) of quasiparticles and this type of transformation might be the easiest thing we can do (I don't know mathematically if it is allowed that a linear transformation can give rise to a change of the (anti)commutation relation). – Chuan Chen Jul 20 '17 at 2:04
• Well, the Jordan-Wigner transformation performed to produce the fermionic Hamiltonian is a linear transformation chosen specifically so that the mixed commutation relations of the spin chain are re-expressed as purely anticommutation relations. I believe the transformation is unitary (or at the very least, invertible), and so it seems reasonable to conclude that more general linear transformations that do not preserve commutation relations exist. – Zxv Jul 20 '17 at 4:10
• Can you explain why the Jordan-Wigner transformation is a linear transformation? – Chuan Chen Jul 20 '17 at 8:13
• Oh, you're right. Of course it isn't. Perhaps it isn't possible, for some well justified reason. And that gives me the idea that the answer might be buried in the study of morphisms on Lie Algebras (or supposedly Jordan Algebras for the anticommutator), or in study of Clifford algebras. Thank you very much for the discussion; I think I have enough direction to know where to find the answer I am looking for. – Zxv Jul 21 '17 at 2:31

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.

• Thanks for the reply. Most of your answer is what I expected, so it's good to have confirmation on those fronts. I have two questions however: 1) In the expression for the anticommutators, for some reason I can't follow line 2 to 3. Isn't the anticommutator symmetric? Where did the minus sign come from? 2) "Thus, you require at the very least that the anticommutator be diagonal." - I think I understand the thought behind this statement, but could you elaborate a little further? I believe you are saying things quickly become difficult with a non-diagonal anticommutator? – Zxv Jul 20 '17 at 0:08