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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?

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  • $\begingroup$ 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. $\endgroup$
    – jacob1729
    Apr 24, 2021 at 19:39
  • $\begingroup$ I assume the a are bosons ... might make sense to say so? $\endgroup$ Apr 24, 2021 at 19:54
  • $\begingroup$ 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.) $\endgroup$ Apr 24, 2021 at 19:56

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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.

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    $\begingroup$ This doesn't explain what went wrong - it just gives another bit of evidence (after the question) that something did go wrong. $\endgroup$ Apr 24, 2021 at 19:30
  • $\begingroup$ @NorbertSchuch As far as I can tell, OP didn't make any mistakes in deriving the Hamiltonian matrix, but then applied a matrix diagonalization method that doesn't work for that matrix. That is precisely what went wrong, IMO. $\endgroup$
    – Anyon
    Apr 24, 2021 at 19:47
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    $\begingroup$ Then it seems that both the question and the answer can only be understood by people who know the method anyway ... :-/ $\endgroup$ Apr 24, 2021 at 19:53
  • $\begingroup$ I would suspect that the reason is one must not sum over all k, but rather only over k>0, since k and -k appear in tuples ... what you obverse is just a manifestation of the resulting ambiguity if you allow to split your sum to all k (which is highly ambiguous), and I would be surprised if all splittings except for the singular one would give the the same result. $\endgroup$ Apr 24, 2021 at 19:55
  • $\begingroup$ The diagonalisation method is explained in the linked question, physics.stackexchange.com/q/487767/128264. The trouble is the mixing of creation/annnihilation operators, which means that standard unitary diagonalisation is not the relevant operation, $\endgroup$ Apr 25, 2021 at 11:46

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