# Bogoliubov transforms for a Heisenberg antiferromagnet: inconsistencies with the excitation spectrum

Consider the standard 1D Heisenberg antiferromagnet (J>0) $$\mathcal{H} = J \sum_i S_i \cdot S_{i+i}$$ We apply the standard Holstein-Primakoff expansion about the classical ground state,

$$S_i = \begin{cases} (0,0,S) & i \text{ even} \\ (0,0,-S)& i \text{ odd} \end{cases}$$

After a linearised Holstein-Primakoff transformation, it can be shown that

$$\mathcal{H} = -JS^2 + JS \sum_k e^{ika} a_k a_{-k} + e^{-ika} a^\dagger_k a^\dagger_{-k} + a^\dagger_k a_{k} + a^\dagger_{-k} a_{-k}$$

where $$a_k$$ are momentum-space bosons satisfying canonical commutators, $$[a_k, a_k^\dagger] = 1$$

Here lies the rub. Naively applying a Bogoliubov transform to the summand in a similar manner to to this question, one re-expreses the summand as $$\begin{pmatrix}a^\dagger_k & a_{-k} \end{pmatrix} \begin{pmatrix} 1 & e^{ika} \\ e^{-ika} & 1\end{pmatrix}\begin{pmatrix}a_k \\ a^\dagger_{-k} \end{pmatrix} -[a_{-k}, a_{-k}^\dagger]$$

Note that this matrix mixes creation and annihilation operators, so cannot simply be unitarily rotated into a new basis - such a relation will not preserve the commutation relations. Instead, we define $$c_k = u_k a_k + v_k a_{-k}^\dagger$$, for complex numbers $$u, v$$ and require $$[c_k, c_k^\dagger] = 1 \Rightarrow |u|^2 - |v|^2 = 1$$, or in other words $$\begin{pmatrix}c_k \\ c_{-k}^\dagger \end{pmatrix} = \begin{pmatrix} u_k & v_k \\ v_{-k}^* & u_{-k}^*\end{pmatrix} \begin{pmatrix}a_k \\ a_{-k}^\dagger \end{pmatrix}$$

However, if one recognises that the sum over k-space is symmetric,

$$\mathcal{H} = - JS^2 + JS \sum_{k>0} (e^{ika} + e^{-ika}) a_k a_{-k} + (e^{-ika} + e^{ika}) a^\dagger_k a^\dagger_{-k} + 2a^\dagger_k a_{k} + 2a^\dagger_{-k} a_{-k}$$ the BdG matrix reads $$\begin{pmatrix} 1 & \cos(ka)\\ \cos(ka)& 1 \end{pmatrix}$$

For this matrix, the dispersion relation obtained $$E(k) = \sqrt{1-\cos^2(ka)} = |\sin(ka)|$$, agrees with the textbook answer.

Since the dispersion relation is a physically observable thing (e.g. by inelastic neutron scattering), there is a paradox here. What made the first approach fail?

• Haven't looked closely so this is just a guess. What normally happens when you don't fix $k>0$ that can go wrong is that you have two copies of the term $a_{-k}a_{+k}$, so you have linearly dependent terms in your sum. Apr 24, 2021 at 19:39
• I assume the a are bosons ... might make sense to say so? Apr 24, 2021 at 19:54
• I'd suspect that the whole transformation only makes sense if you restrict to k>0, since k and -k appear jointly in the sum. So you must group your Hamiltonian to a k>0 sum. (Once you split them you get an ambiguity, and your matrix is only one of many ways to make this ambiguous choice, and likely all will have different outcomes.) Apr 24, 2021 at 19:56

The issue is that your matrix has vanishing determinant, $$\det \begin{pmatrix} 1 & e^{ika} \\ e^{-ika} & 1\end{pmatrix} = 0,$$ and thus diag(1, -1) H only has one linearly dependent eigenvectors, whereas a proper diagonalization of a bosonic Hamiltonian requires linearly independent solutions, c.f. van Hemmen (1980) especially Sec. V. In general, for this type of diagonalization to be meaningful, the matrix should be positive definite - at least somewhere in k-space.