... | @@ -12,7 +12,7 @@ It returns a dictionary with all the wyckoff positions, and for each of them how |
... | @@ -12,7 +12,7 @@ It returns a dictionary with all the wyckoff positions, and for each of them how |
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**Detailed description:**
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**Detailed description:**
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We want to get rid of the atom labels and replace them with something independent, that is intrinsic, that no matter if you have formula like MgCu3 and FeTi3. These two might have the same prototype (let's assume that they have). If you have Fe at position a and 2Ti at position b and one Ti at position c, we cannot directly compare and see that it is equal to this. To choose we introduce `x_1` and `x_2` notation that is solving the problem.
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We want to get rid of the atom labels and replace them with something independent, that is intrinsic, that no matter if you have formula like MgCu3 and FeTi3. These two might have the same prototype (let's assume that they have). If you have Fe at position *a* and 2Ti at position *b* and one Ti at position *c*, we cannot directly compare and see that it is equal to this. To choose we introduce `x_1` and `x_2` notation that is solving the problem.
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we replace with new label and we get `x_2x_13` and `x_2x_13` in this way we can compare that formula and see that is equal and atom at position a is atom `x_2` and other is `x_1`
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we replace with new label and we get `x_2x_13` and `x_2x_13` in this way we can compare that formula and see that is equal and atom at position a is atom `x_2` and other is `x_1`
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Problem is when 2 atoms have exactly the same numerical count. e.g. Fe2O2 - how to decide which one you call `x_1` and which `x_2`. So what we do if they are equal: we begin to look at the first Wyckoff position and we decide if one is present more that the other, than we choose that one. And then we look at other Wyckoff position within that Wyckoff position and we look again (because it could be that both `x_1` and `x_2` here, Fe and O and one cannot decide). If we cannot yet decide because they are both here we look at the next one in alphabetical order until we can decide or if you cannot decide - it is always the same it means they are equivalent - the two atoms are equivalent so it does not matter cause we cannot ever decide, you can choose one is `x_1` and one `x_2 or vice versa - it will work the same.
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Problem is when 2 atoms have exactly the same numerical count. e.g. Fe2O2 - how to decide which one you call `x_1` and which `x_2`. So what we do if they are equal: we begin to look at the first Wyckoff position and we decide if one is present more that the other, than we choose that one. And then we look at other Wyckoff position within that Wyckoff position and we look again (because it could be that both `x_1` and `x_2` here, Fe and O and one cannot decide). If we cannot yet decide because they are both here we look at the next one in alphabetical order until we can decide or if you cannot decide - it is always the same it means they are equivalent - the two atoms are equivalent so it does not matter cause we cannot ever decide, you can choose one is `x_1` and one `x_2 or vice versa - it will work the same.
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