I am calculating lambda using two different methods implemented in EPW.
First one is obtained by activating anisotropic Eliashberg equation related tags.
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eliashberg = .true.
laniso = .true.
limag = .true.
lpade = .true.
The result is as below:
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===================================================================
Solve anisotropic Eliashberg equations
===================================================================
Finish reading .freq file
Fermi level (eV) = 1.4490810867E+00
DOS(states/spin/eV/Unit Cell) = 7.4601791612E-01
Electron smearing (eV) = 1.0000000000E-01
Fermi window (eV) = 4.0000000000E-01
Nr irreducible k-points within the Fermi shell = 134 out of 231
2 bands within the Fermi window
Finish reading .egnv file
Max nr of q-points = 466
Finish reading .ikmap files
Start reading .ephmat files
Finish reading .ephmat files
lambda_max = 1.3814131 lambda_k_max = 0.7463819
Electron-phonon coupling strength = 0.5890385
Estimated Allen-Dynes Tc = 1.0488626 K for muc = 0.16000
Estimated BCS superconducting gap = 0.0001591 eV
and the second one is obtained by activating a2f tags and deactivating anisotropic Eliashberg tags.
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a2f = .true.
phonselfen = .true.
The result is as below:
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===================================================================
Eliashberg Spectral Function in the Migdal Approximation
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lambda : 0.5689812
lambda_tr : 0.7333977
Estimated Allen-Dynes Tc
logavg = 0.0006900 l_a2F = 0.5735213
mu = 0.10 Tc = 2.164145346702 K
mu = 0.12 Tc = 1.691229225724 K
mu = 0.14 Tc = 1.276401502717 K
mu = 0.16 Tc = 0.922975116778 K
mu = 0.18 Tc = 0.632737937327 K
mu = 0.20 Tc = 0.405375457882 K
The lambda obtained by the first one is 0.5890385, while the lambda from the second method is 0.5689812. and T_c is also slightly different.
Could it be understood it is because of the effect of anisotropic nature of the system? or some bugs in the code or numerical inaccuracy? I've calculated the second part by re-loading previously obtained files that required solving the Eliashberg equation. Any comments would be helpful and appreciate.
Best regards,
Jun-Ho