Thank you for the video. Small question. Which parameters should we keep the same calculating entire system and isolated atoms? For instance, if we use smearing for entire system, should we use exact the same degauss for isolated atoms? Or we can choose smaller degauss, if it is applicable, to decrease smearing contribution?
Nice tutorial!!! Can you explain why we use 12 angstroms lattice parameter for isolated atom calculation what will heppen in case if I increase this lattice parameter to 20 angstroms or otherwise decrease it to 7 angstroms ?
I think I already mentioned this in the video. But the key thing is that Quantum ESPRESSO inherently applies periodic boundary conditions to your system in 3 dimensions. This means that whatever system you define it will essentially be repeated with the periodicity given by the lattice parameters. For cohesive energy, you require the energy of an isolated atom. But Quantum ESPRESSO will repeat the atom I have in my system periodically. So if I have a small lattice parameter, then the atom will have periodic images that are nearer. If the lattice parameter is large, then the periodic images would be further. If the images are nearer then they will interact. You wouldn't call an atom interacting with other similar atoms as isolated. So this wouldn't work. If the periodic images are farther, then after a certain threshold, the interactions between them would become negligible. Then you can consider your atom to be isolated. You can find out the threshold or the lattice parameter size after which the interactions become negligible yourself, by running SCF for different lattice parameters. Once the energy stops changing beyond a threshold, you can consider the atom to be effectively isolated.
Sir I have a small question. when the electronic configuration is like [Ar] 4s² 3d¹ what nspin value I have to take? As 4s is filled and it is the outer shell so nspin = 1. Is this correct?
The 3d electron is also a valence electron. Since there is one unpaired electron the value of nspin should be 2. Also you can always verify these things. You can run two calculations: one with nspin=1 and nspin=2. Then you can compare the energies from the two calculations. The one with the lower energy would be the correct option.
I'm trying to compute the cohesive energy of a perovskite. I have modeled the unit cell structure using vesta and did convergence testing to find the optimal ecut and k-points. Now, this is a silly question, but for the isolated atoms, should I use single atom structures expanded with a large amount of lattice constant, or should I use the standard conventional structure? My thinking is that, what I should use is the previous, but some guy told me to use the "standard conventional" for the isolated atoms. It did not make sense to me because then that structure will have its own cohesive energy, but, what do you think?
You're completely right. For the isolated atoms the unit cell size should be large enough to prevent inter-image interaction s. The atom is isolated after all. Also, one must use an orthorhombic cell ( with different lattice constants in different directions) instead of a cubic cell to remove any symmetry in the wave function as that can also affect the energies. I think I forgot this and used a cubic cell in this tutorial.
In that case both npin=1 and nspin=2 should give the same energy. You should still verify that. Also, the magnetization should be zero even with nspin=2 if there are no unpaired electrons in the atom.
The energy of Sc, using your input file is not equal with and without spin polarization. I'm getting a value of -156.63416812 Ry without spoin polarization and -156.64404984 Ry with spin polarization. Ry is a big unit of energy. 1 Ry= 13.6 eV. Please note, I use a higher value of Ecutwfc= and Ecutrho, compared to what you used: ecutrho = 4.42315e+02 ecutwfc = 4.91461e+01 Also, I am getting a cohesive energy of -4.2 eV/atom. What are you getting and what is the expected value?
@@ManasSharma07 what is the formula used for this purpose... pls elaborate. the value 156 I am also getting for Al I am getting 39.2 Ry. and for ScAl2 this is around 1888 ry. what is the formula used?
@@sankhasubhramukhopadhyay8005 This is the formula I used www.wolframalpha.com/input?i=-1888.553770%2F24+-+%288*-156.64404984+%2B+16*-39.24939399%29%2F24
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Very helpful. Thanks!!!
Good work
Thank you for the video. Small question. Which parameters should we keep the same calculating entire system and isolated atoms? For instance, if we use smearing for entire system, should we use exact the same degauss for isolated atoms? Or we can choose smaller degauss, if it is applicable, to decrease smearing contribution?
Hello sir, very nice lecture. Can you also make a tutorial on how to calculate attachment energy of a surface using QE? Appreciate your efforts!!
Nice tutorial!!! Can you explain why we use 12 angstroms lattice parameter for isolated atom calculation what will heppen in case if I increase this lattice parameter to 20 angstroms or otherwise decrease it to 7 angstroms ?
I think I already mentioned this in the video. But the key thing is that Quantum ESPRESSO inherently applies periodic boundary conditions to your system in 3 dimensions. This means that whatever system you define it will essentially be repeated with the periodicity given by the lattice parameters. For cohesive energy, you require the energy of an isolated atom. But Quantum ESPRESSO will repeat the atom I have in my system periodically. So if I have a small lattice parameter, then the atom will have periodic images that are nearer. If the lattice parameter is large, then the periodic images would be further. If the images are nearer then they will interact. You wouldn't call an atom interacting with other similar atoms as isolated. So this wouldn't work. If the periodic images are farther, then after a certain threshold, the interactions between them would become negligible. Then you can consider your atom to be isolated.
You can find out the threshold or the lattice parameter size after which the interactions become negligible yourself, by running SCF for different lattice parameters. Once the energy stops changing beyond a threshold, you can consider the atom to be effectively isolated.
Sir I have a small question. when the electronic configuration is like [Ar] 4s² 3d¹ what nspin value I have to take? As 4s is filled and it is the outer shell so nspin = 1. Is this correct?
The 3d electron is also a valence electron. Since there is one unpaired electron the value of nspin should be 2.
Also you can always verify these things. You can run two calculations: one with nspin=1 and nspin=2. Then you can compare the energies from the two calculations. The one with the lower energy would be the correct option.
@@PhysWhiz thank you sir
No need to call me "sir".
I'm trying to compute the cohesive energy of a perovskite. I have modeled the unit cell structure using vesta and did convergence testing to find the optimal ecut and k-points. Now, this is a silly question, but for the isolated atoms, should I use single atom structures expanded with a large amount of lattice constant, or should I use the standard conventional structure?
My thinking is that, what I should use is the previous, but some guy told me to use the "standard conventional" for the isolated atoms. It did not make sense to me because then that structure will have its own cohesive energy, but, what do you think?
You're completely right. For the isolated atoms the unit cell size should be large enough to prevent inter-image interaction s. The atom is isolated after all.
Also, one must use an orthorhombic cell ( with different lattice constants in different directions) instead of a cubic cell to remove any symmetry in the wave function as that can also affect the energies. I think I forgot this and used a cubic cell in this tutorial.
@@PhysWhiz big help!
If last orbit is fully filled and there are no unpaired electrons then also I have to put nspin = 2 ?
In that case both npin=1 and nspin=2 should give the same energy. You should still verify that.
Also, the magnetization should be zero even with nspin=2 if there are no unpaired electrons in the atom.
I am trying to calculate cohesive energy of ScAl2. ans coming out is much much larger than actual value. Another thing is that I am trying with nspin=2, nspin=1 for Sc, In both cases energy is coming as almost equal. Where I am going wrong. pls give me suggestions . I am giving my input script below, for ScAl2, Sc, and Al
ScAl2
&CONTROL
calculation = 'scf'
outdir = './out/'
prefix = 'ScAl2'
pseudo_dir = '.'
verbosity = 'high'
/
&SYSTEM
degauss = 1d-02
ecutrho = 400d0
ecutwfc = 40d0
ibrav = 0
nat = 24
ntyp = 2
occupations = 'smearing'
smearing = 'gaussian'
/
&ELECTRONS
conv_thr = 1.2000000000d-09
electron_maxstep = 150
mixing_beta = 4.0000000000d-01
startingpot = 'atomic'
startingwfc = 'atomic+random'
/
ATOMIC_SPECIES
Sc 44.9559 Sc.pbe-spn-kjpaw_psl.1.0.0.UPF
Al 26.9815 Al.pbe-nl-kjpaw_psl.1.0.0.UPF
ATOMIC_POSITIONS {crystal}
Sc 0.00000000000000 0.00000000000000 0.50000000000000
Sc 0.25000000000000 0.25000000000000 0.75000000000000
Sc 0.00000000000000 0.50000000000000 0.00000000000000
Sc 0.25000000000000 0.75000000000000 0.25000000000000
Sc 0.50000000000000 0.00000000000000 0.00000000000000
Sc 0.75000000000000 0.25000000000000 0.25000000000000
Sc 0.50000000000000 0.50000000000000 0.50000000000000
Sc 0.75000000000000 0.75000000000000 0.75000000000000
Al 0.12500000000000 0.12500000000000 0.12500000000000
Al 0.62500000000000 0.37500000000000 0.87500000000000
Al 0.37500000000000 0.12500000000000 0.37500000000000
Al 0.37500000000000 0.37500000000000 0.12500000000000
Al 0.12500000000000 0.62500000000000 0.62500000000000
Al 0.62500000000000 0.87500000000000 0.37500000000000
Al 0.37500000000000 0.62500000000000 0.87500000000000
Al 0.37500000000000 0.87500000000000 0.62500000000000
Al 0.62500000000000 0.12500000000000 0.62500000000000
Al 0.12500000000000 0.37500000000000 0.37500000000000
Al 0.87500000000000 0.12500000000000 0.87500000000000
Al 0.87500000000000 0.37500000000000 0.62500000000000
Al 0.62500000000000 0.62500000000000 0.12500000000000
Al 0.12500000000000 0.87500000000000 0.87500000000000
Al 0.87500000000000 0.62500000000000 0.37500000000000
Al 0.87500000000000 0.87500000000000 0.12500000000000
K_POINTS {automatic}
8 8 8 0 0 0
CELL_PARAMETERS {angstrom}
7.55530537 0.00000000 0.00000000
0.00000000 7.55530537 0.00000000
0.00000000 0.00000000 7.55530537
Sc
&CONTROL
calculation = 'scf'
outdir = './out/'
prefix = 'Sc'
pseudo_dir = '.'
verbosity = 'high'
/
&SYSTEM
degauss = 1d-02
angle1(1) = 0.00000d+00
angle2(1) = 0.00000d+00
constrained_magnetization = 'atomic'
ecutrho = 400d0
ecutwfc = 40d0
ibrav = 0
nat = 1
ntyp = 1
nspin = 2
occupations = 'smearing'
smearing = 'gaussian'
starting_magnetization(1) = 1.00000d-01
/
&ELECTRONS
conv_thr = 1.2000000000d-09
electron_maxstep = 150
mixing_beta = 4.0000000000d-01
startingpot = 'atomic'
startingwfc = 'atomic+random'
/
ATOMIC_SPECIES
Sc 44.9559 Sc.pbe-spn-kjpaw_psl.1.0.0.UPF
ATOMIC_POSITIONS {crystal}
Sc 0.50000000000000 0.50000000000000 0.50000000000000
K_POINTS {gamma}
CELL_PARAMETERS {angstrom}
14.00000000 0.00000000 0.00000000
0.00000000 14.00000000 0.00000000
0.00000000 0.00000000 14.00000000
Al
&CONTROL
calculation = 'scf'
outdir = './out/'
prefix = 'Al'
pseudo_dir = '.'
verbosity = 'high'
/
&SYSTEM
degauss = 1d-02
angle1(1) = 0.00000e+00
angle2(1) = 0.00000e+00
constrained_magnetization = "atomic"
ecutrho = 400d0
ecutwfc = 40d0
ibrav = 0
nat = 1
ntyp = 1
nspin = 2
occupations = 'smearing'
smearing = 'gaussian'
starting_magnetization(1) = 1.00000e-01
/
&ELECTRONS
conv_thr = 1.2000000000d-09
electron_maxstep = 150
mixing_beta = 4.0000000000d-01
startingpot = 'atomic'
startingwfc = 'atomic+random'
/
ATOMIC_SPECIES
Al 26.9815 Al.pbe-nl-kjpaw_psl.1.0.0.UPF
ATOMIC_POSITIONS {crystal}
Al 0.87500000000000 0.87500000000000 0.12500000000000
K_POINTS {gamma}
CELL_PARAMETERS {angstrom}
14.00 0.00000000 0.00000000
0.00000000 14.00 0.00000000
0.00000000 0.00000000 14.00
The energy of Sc, using your input file is not equal with and without spin polarization. I'm getting a value of -156.63416812 Ry without spoin polarization and -156.64404984 Ry with spin polarization. Ry is a big unit of energy. 1 Ry= 13.6 eV.
Please note, I use a higher value of Ecutwfc= and Ecutrho, compared to what you used:
ecutrho = 4.42315e+02
ecutwfc = 4.91461e+01
Also, I am getting a cohesive energy of -4.2 eV/atom. What are you getting and what is the expected value?
@@ManasSharma07 what is the formula used for this purpose... pls elaborate. the value 156 I am also getting for Al I am getting 39.2 Ry. and for ScAl2 this is around 1888 ry. what is the formula used?
@@sankhasubhramukhopadhyay5633 You didn't answer my question in my comment:
What are you getting and what is the expected value?
@ManasSharma07 I am getting value around 117 Ry. And actual ans is 4.16 eV ..your calculation is right but what formula I have to use
@@sankhasubhramukhopadhyay8005 This is the formula I used
www.wolframalpha.com/input?i=-1888.553770%2F24+-+%288*-156.64404984+%2B+16*-39.24939399%29%2F24