Inversion applicable to three-dimensional lattices only
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I have just started my first course in solid state physics and while studying symmetries, Inversion is defined as "A point operation which is applicable to three-dimensional lattices only. This symmetry implies that each point located at r relative to a lattice point has an identical point located at -r relative to same lattice point."
Now if I have a two dimensional lattice with a square as the unit cell and a single atom as a basis, relative to a fixed point O(taken as origin in the fig attached), i can find a point B and B'; C and C' at x ,-x and y,-y respectively.
So, is there an additional criteria for inversion? Where I am getting it wrong?
solid-state-physics symmetry crystals
asked 6 hours ago
Harshdeep SinghHarshdeep Singh
767
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add a comment |
$begingroup$
I have just started my first course in solid state physics and while studying symmetries, Inversion is defined as "A point operation which is applicable to three-dimensional lattices only. This symmetry implies that each point located at r relative to a lattice point has an identical point located at -r relative to same lattice point."
Now if I have a two dimensional lattice with a square as the unit cell and a single atom as a basis, relative to a fixed point O(taken as origin in the fig attached), i can find a point B and B'; C and C' at x ,-x and y,-y respectively.
So, is there an additional criteria for inversion? Where I am getting it wrong?
solid-state-physics symmetry crystals
asked 6 hours ago
Harshdeep SinghHarshdeep Singh
767
$endgroup$
add a comment |
$begingroup$
I have just started my first course in solid state physics and while studying symmetries, Inversion is defined as "A point operation which is applicable to three-dimensional lattices only. This symmetry implies that each point located at r relative to a lattice point has an identical point located at -r relative to same lattice point."
Now if I have a two dimensional lattice with a square as the unit cell and a single atom as a basis, relative to a fixed point O(taken as origin in the fig attached), i can find a point B and B'; C and C' at x ,-x and y,-y respectively.
So, is there an additional criteria for inversion? Where I am getting it wrong?
solid-state-physics symmetry crystals
asked 6 hours ago
Harshdeep SinghHarshdeep Singh
767
$endgroup$
I have just started my first course in solid state physics and while studying symmetries, Inversion is defined as "A point operation which is applicable to three-dimensional lattices only. This symmetry implies that each point located at r relative to a lattice point has an identical point located at -r relative to same lattice point."
Now if I have a two dimensional lattice with a square as the unit cell and a single atom as a basis, relative to a fixed point O(taken as origin in the fig attached), i can find a point B and B'; C and C' at x ,-x and y,-y respectively.
So, is there an additional criteria for inversion? Where I am getting it wrong?
solid-state-physics symmetry crystals
solid-state-physics symmetry crystals
asked 6 hours ago
Harshdeep SinghHarshdeep Singh
767
asked 6 hours ago
Harshdeep SinghHarshdeep Singh
767
asked 6 hours ago
Harshdeep SinghHarshdeep Singh
767
asked 6 hours ago
Harshdeep SinghHarshdeep Singh
767
asked 6 hours ago
Harshdeep SinghHarshdeep Singh
767
767
add a comment |
add a comment |
1 Answer
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I've actually never seen inversion symmetry defined to only apply to three dimensional lattices before. It's usually nothing more than the operation taking $mathbf{r}$ to $-mathbf{r}$, when $mathbf{r}$ is measured relative to some center of inversion - regardless of dimension. (It's convenient to choose this point as the origin, but not necessary.)
However, there is a clear and crucial difference between inversion in 2D and 3D one should keep in mind. In two spatial dimensions, inversion about the origin is simply equivalent to a $180^circ$ rotation about an axis going through the origin. In three dimensions the inversion operation involves both a $180^circ$ rotation and a reflection. This additional reflection actually changes the orientation. To see this, draw the usual $x$, $y$, $z$ coordinate axes, and then draw what you'd get after an inversion operation: the $-x$, $-y$, and $-z$ axes. One of the coordinate system will be right-handed, and the other left-handed. Perhaps the book you're using considers this orientation change a crucial part of inversion symmetry, or perhaps its authors wanted to avoid the redundancy between $C_2$ rotation and inversion in 2D.
Side note: The above distinction generalizes to odd- or even-dimensional lattices, see Wikipedia.
answered 5 hours ago
AnyonAnyon
48939
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1
$begingroup$
Excellent point about the inversions playing a different role in 2D and 3D. If you view the symmetries of the lattice as a transformation group, this distinction becomes quite important.
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– Michael Seifert
5 hours ago
add a comment |
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$begingroup$
I've actually never seen inversion symmetry defined to only apply to three dimensional lattices before. It's usually nothing more than the operation taking $mathbf{r}$ to $-mathbf{r}$, when $mathbf{r}$ is measured relative to some center of inversion - regardless of dimension. (It's convenient to choose this point as the origin, but not necessary.)
However, there is a clear and crucial difference between inversion in 2D and 3D one should keep in mind. In two spatial dimensions, inversion about the origin is simply equivalent to a $180^circ$ rotation about an axis going through the origin. In three dimensions the inversion operation involves both a $180^circ$ rotation and a reflection. This additional reflection actually changes the orientation. To see this, draw the usual $x$, $y$, $z$ coordinate axes, and then draw what you'd get after an inversion operation: the $-x$, $-y$, and $-z$ axes. One of the coordinate system will be right-handed, and the other left-handed. Perhaps the book you're using considers this orientation change a crucial part of inversion symmetry, or perhaps its authors wanted to avoid the redundancy between $C_2$ rotation and inversion in 2D.
Side note: The above distinction generalizes to odd- or even-dimensional lattices, see Wikipedia.
answered 5 hours ago
AnyonAnyon
48939
$endgroup$
1
$begingroup$
Excellent point about the inversions playing a different role in 2D and 3D. If you view the symmetries of the lattice as a transformation group, this distinction becomes quite important.
$endgroup$
– Michael Seifert
5 hours ago
add a comment |
$begingroup$
I've actually never seen inversion symmetry defined to only apply to three dimensional lattices before. It's usually nothing more than the operation taking $mathbf{r}$ to $-mathbf{r}$, when $mathbf{r}$ is measured relative to some center of inversion - regardless of dimension. (It's convenient to choose this point as the origin, but not necessary.)
However, there is a clear and crucial difference between inversion in 2D and 3D one should keep in mind. In two spatial dimensions, inversion about the origin is simply equivalent to a $180^circ$ rotation about an axis going through the origin. In three dimensions the inversion operation involves both a $180^circ$ rotation and a reflection. This additional reflection actually changes the orientation. To see this, draw the usual $x$, $y$, $z$ coordinate axes, and then draw what you'd get after an inversion operation: the $-x$, $-y$, and $-z$ axes. One of the coordinate system will be right-handed, and the other left-handed. Perhaps the book you're using considers this orientation change a crucial part of inversion symmetry, or perhaps its authors wanted to avoid the redundancy between $C_2$ rotation and inversion in 2D.
Side note: The above distinction generalizes to odd- or even-dimensional lattices, see Wikipedia.
answered 5 hours ago
AnyonAnyon
48939
$endgroup$
1
$begingroup$
Excellent point about the inversions playing a different role in 2D and 3D. If you view the symmetries of the lattice as a transformation group, this distinction becomes quite important.
$endgroup$
– Michael Seifert
5 hours ago
add a comment |
$begingroup$
I've actually never seen inversion symmetry defined to only apply to three dimensional lattices before. It's usually nothing more than the operation taking $mathbf{r}$ to $-mathbf{r}$, when $mathbf{r}$ is measured relative to some center of inversion - regardless of dimension. (It's convenient to choose this point as the origin, but not necessary.)
However, there is a clear and crucial difference between inversion in 2D and 3D one should keep in mind. In two spatial dimensions, inversion about the origin is simply equivalent to a $180^circ$ rotation about an axis going through the origin. In three dimensions the inversion operation involves both a $180^circ$ rotation and a reflection. This additional reflection actually changes the orientation. To see this, draw the usual $x$, $y$, $z$ coordinate axes, and then draw what you'd get after an inversion operation: the $-x$, $-y$, and $-z$ axes. One of the coordinate system will be right-handed, and the other left-handed. Perhaps the book you're using considers this orientation change a crucial part of inversion symmetry, or perhaps its authors wanted to avoid the redundancy between $C_2$ rotation and inversion in 2D.
Side note: The above distinction generalizes to odd- or even-dimensional lattices, see Wikipedia.
answered 5 hours ago
AnyonAnyon
48939
$endgroup$
I've actually never seen inversion symmetry defined to only apply to three dimensional lattices before. It's usually nothing more than the operation taking $mathbf{r}$ to $-mathbf{r}$, when $mathbf{r}$ is measured relative to some center of inversion - regardless of dimension. (It's convenient to choose this point as the origin, but not necessary.)
However, there is a clear and crucial difference between inversion in 2D and 3D one should keep in mind. In two spatial dimensions, inversion about the origin is simply equivalent to a $180^circ$ rotation about an axis going through the origin. In three dimensions the inversion operation involves both a $180^circ$ rotation and a reflection. This additional reflection actually changes the orientation. To see this, draw the usual $x$, $y$, $z$ coordinate axes, and then draw what you'd get after an inversion operation: the $-x$, $-y$, and $-z$ axes. One of the coordinate system will be right-handed, and the other left-handed. Perhaps the book you're using considers this orientation change a crucial part of inversion symmetry, or perhaps its authors wanted to avoid the redundancy between $C_2$ rotation and inversion in 2D.
Side note: The above distinction generalizes to odd- or even-dimensional lattices, see Wikipedia.
answered 5 hours ago
AnyonAnyon
48939
edited 5 hours ago
answered 5 hours ago
AnyonAnyon
48939
answered 5 hours ago
AnyonAnyon
48939
answered 5 hours ago
AnyonAnyon
48939
48939
1
$begingroup$
Excellent point about the inversions playing a different role in 2D and 3D. If you view the symmetries of the lattice as a transformation group, this distinction becomes quite important.
$endgroup$
– Michael Seifert
5 hours ago
add a comment |
1
$begingroup$
Excellent point about the inversions playing a different role in 2D and 3D. If you view the symmetries of the lattice as a transformation group, this distinction becomes quite important.
$endgroup$
– Michael Seifert
5 hours ago
1
1
$begingroup$
Excellent point about the inversions playing a different role in 2D and 3D. If you view the symmetries of the lattice as a transformation group, this distinction becomes quite important.
$endgroup$
– Michael Seifert
5 hours ago
$begingroup$
Excellent point about the inversions playing a different role in 2D and 3D. If you view the symmetries of the lattice as a transformation group, this distinction becomes quite important.
$endgroup$
– Michael Seifert
5 hours ago
add a comment |
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