About energy

[XiankangTang]

Dear Andrey,

I've encountered a new problem. After checking all the energies obtained from the BdG transformation using the colpa method, I found that the four resulting eigenvalues are all positive, but only the two largest values are returned. However, theoretically, the magnon spectrum should contain both a particle band (energy > 0) and a hole band (energy < 0).

Looking forward to your reply.
Xiankang

[adrybakov]

Dear Xiankang,

In the case of bosons the eigenvalues of the quadratic LSWT Hamiltonian are all positive. See Colpa; 1978 and Ohashi, Kobayashi, Kawaguchi; 2020 for details.

In short, the quadratic boson Hamiltonian is written as

$H = A^{\dagger} D A,$

where $D$ is a $2M \times 2M$ grand-dynamical matrix, $M$ is the amount of magnetic sites in the unit cell (i. e. the amount of expected magnon modes). The diagonalization of this matrix shall be performed by preserving bosonic commutation relations, which introduces extra complications compared to the direct diagonalization of the matrix appearing in the fermionic case.

Now, the matrix $gD$ has M positive eigenvalues and M negative eigenvalues, which are the diagonal elements of the matrix $L$ (See equation (2.11) of Colpa or equation (12) of OKK). Here matrix $g = \sigma_3\otimes I$ and $\sigma_3$ is the third Pauli matrix, $I$ is $M \times M$ identity matrix (in other words $g$ is a diagonal matrix with first $M$ elements equal to 1 and second $M$ elements equal to $-1$),

Then, to obtain the eigenvalues of the actual grand-dynamical matrix $D$ one multiplies $L$ by $g$ and obtain the matrix $E=gL$ with $2M$ positive elements on the diagonal (when the matrix $D$ is Hermitian and positive-definite). The Hamiltonian gets to

$H = B^{\dagger} E B$.

Finally, one expands the matrix products to arrive at the Hamiltonian in the form (omitting constant terms and $\hbar = 1$)

$H = \sum_k \omega_{\beta} (k) b_{\beta}^{\dagger} (k) b_{\beta} (k),$

where index $\beta$ runs from 1 to $M$ and count magnon modes. This transition from the matrix form to the "serial" form involves recombination of the "particle" and "hole" parts of $E(k)$ and $E(-k)$ to obtain the full solution of the magnon problem at the level of LSWT. Note that the "particle" and "hole" are the same physical excitations in the context of the bosonic problem (see OKK).

There are more details about it in the paper about Magnopy, I will add a reference here when it appears.

Best,
Andrey

[XiankangTang]

I agree with your analysis here. However, this approach might limit Magnopy to merely being a software for simulating magnon spectra in the future, making it difficult to expand further.

[adrybakov]

Dear Xiankang,

At the version of 0.4.3 of Magnopy the eigenenergies in the "particle-hole" picture (i. e. by "splitting" each magnon eigenstate in two equal parts) can be obtained as (assume that the k-point is given in absolute values of the reciprocal space)

import magnopy
import numpy as np

# Some spin Hamiltonian
spinham = ...
# Set of spin directions that define the classical "vacuum state"
spin_directions = ...
# Some k-point in reciprocal space given in absolute coordinates
k = ...

lswt = magnopy.LSWT(spinham=spinham, spin_directions=spin_directions)

E, G = magnopy.solve_via_colpa(D = lswt.GDM(k = k))

E = np.diag(E)

N = E.shape[0] // 2
g = np.diag(np.concatenate((np.ones(N), -np.ones(N))))
L = g @ E

Here E is $2M\times 2M$ diagonal matrix with $2M$ positive diagonal elements; L is a $2M\times 2M$ diagonal matrix with first $M$ diagonal elements being positive and second $M$ being negative; and G is a $2M\times 2M$ transformation matrix that transforms old into new boson vector operators as $B = G A$.

Magnopy will be developed based on (i) interests and needs of the community and (ii) focus of its authors.

Best,
Andrey