Some clarifications about Point Groups rapresentations for molecules
$begingroup$
This is what I've understood until now about this subject, please confirm if what I write is true or not.
Choosen one molecule, there is a certain number of symmetry operations which this molecule satisfies (for example $C_{2}-i-S_{4}^{3}$...). The set of these operations forms the point group of the molecule. I can also choose a metric, for example the internal coordinates of the molecule (that is a vector with dimension $3N-6$), the cartesian displacment of the atoms (that is a vector with dimension $3N$) or the atoms (that is a vector with dimension $N$). Choosen one metric, I can build a rapresentation of the point group of the molecule, by looking at the way in which the vector of the metric is transformed by each symmetry operation. In this way, I can build for each operations a square matrix which has the same dimension of the metric and which is the rapresentation of that symmetry operation. If the group I'm considering has order $h$ (number of different operations) I will have at most $h$ differents matrices, because there can be two or more operations with the same matrix. I now have let's say $c$ differents matrices with $c<m$, and I make a change of basis in the vectorial space I'm considering. For one of the $c$ matrices, which I've called $A$ I get:
$$Y=AX rightarrow X'=TX and Y'=TYrightarrow Y'=TAT^{-1}X' $$
So in this new basis I've obtained $c$ new matrices where each matrix is similar to the previous one. I've read that exists a particular change of basis for which all the new matrices are in block diagonal form. Unfortunately they don't specify exactly how these new matrices are. I mean, are all in their respective Jordan normal form? If so, there must exist a specific matrix $T$ of generalized eigenvectors which must be common to all the matrices. Similarly, if all my matrices are diagonalisable, the new $c$ matrices are the corrispective diagonal form of each one? Also in this case there must exist a specific matrix $T$ of eigenvectors which must be common to all the matrices. Someone could explain me this?
Thanks a lot
group-theory
asked Dec 15 '18 at 16:35
LandauLandau
447
$endgroup$
add a comment |
$begingroup$
This is what I've understood until now about this subject, please confirm if what I write is true or not.
Choosen one molecule, there is a certain number of symmetry operations which this molecule satisfies (for example $C_{2}-i-S_{4}^{3}$...). The set of these operations forms the point group of the molecule. I can also choose a metric, for example the internal coordinates of the molecule (that is a vector with dimension $3N-6$), the cartesian displacment of the atoms (that is a vector with dimension $3N$) or the atoms (that is a vector with dimension $N$). Choosen one metric, I can build a rapresentation of the point group of the molecule, by looking at the way in which the vector of the metric is transformed by each symmetry operation. In this way, I can build for each operations a square matrix which has the same dimension of the metric and which is the rapresentation of that symmetry operation. If the group I'm considering has order $h$ (number of different operations) I will have at most $h$ differents matrices, because there can be two or more operations with the same matrix. I now have let's say $c$ differents matrices with $c<m$, and I make a change of basis in the vectorial space I'm considering. For one of the $c$ matrices, which I've called $A$ I get:
$$Y=AX rightarrow X'=TX and Y'=TYrightarrow Y'=TAT^{-1}X' $$
So in this new basis I've obtained $c$ new matrices where each matrix is similar to the previous one. I've read that exists a particular change of basis for which all the new matrices are in block diagonal form. Unfortunately they don't specify exactly how these new matrices are. I mean, are all in their respective Jordan normal form? If so, there must exist a specific matrix $T$ of generalized eigenvectors which must be common to all the matrices. Similarly, if all my matrices are diagonalisable, the new $c$ matrices are the corrispective diagonal form of each one? Also in this case there must exist a specific matrix $T$ of eigenvectors which must be common to all the matrices. Someone could explain me this?
Thanks a lot
group-theory
asked Dec 15 '18 at 16:35
LandauLandau
447
$endgroup$
1
$begingroup$
Have you looked at some small examples? The matrices will generally not be in Jordan normal form.
$endgroup$
– Tobias Kildetoft
Dec 15 '18 at 22:01
$begingroup$
Yes, unfortunately in the example I've seen the matrices are all put in diagonal form. However I thought that the only diagonal block matrix to which a generic matrix could be similar was its Jordan/diagonal form. You say it is not true? @Tobias Kildetoft
$endgroup$
– Landau
Dec 15 '18 at 22:39
add a comment |
$begingroup$
This is what I've understood until now about this subject, please confirm if what I write is true or not.
Choosen one molecule, there is a certain number of symmetry operations which this molecule satisfies (for example $C_{2}-i-S_{4}^{3}$...). The set of these operations forms the point group of the molecule. I can also choose a metric, for example the internal coordinates of the molecule (that is a vector with dimension $3N-6$), the cartesian displacment of the atoms (that is a vector with dimension $3N$) or the atoms (that is a vector with dimension $N$). Choosen one metric, I can build a rapresentation of the point group of the molecule, by looking at the way in which the vector of the metric is transformed by each symmetry operation. In this way, I can build for each operations a square matrix which has the same dimension of the metric and which is the rapresentation of that symmetry operation. If the group I'm considering has order $h$ (number of different operations) I will have at most $h$ differents matrices, because there can be two or more operations with the same matrix. I now have let's say $c$ differents matrices with $c<m$, and I make a change of basis in the vectorial space I'm considering. For one of the $c$ matrices, which I've called $A$ I get:
$$Y=AX rightarrow X'=TX and Y'=TYrightarrow Y'=TAT^{-1}X' $$
So in this new basis I've obtained $c$ new matrices where each matrix is similar to the previous one. I've read that exists a particular change of basis for which all the new matrices are in block diagonal form. Unfortunately they don't specify exactly how these new matrices are. I mean, are all in their respective Jordan normal form? If so, there must exist a specific matrix $T$ of generalized eigenvectors which must be common to all the matrices. Similarly, if all my matrices are diagonalisable, the new $c$ matrices are the corrispective diagonal form of each one? Also in this case there must exist a specific matrix $T$ of eigenvectors which must be common to all the matrices. Someone could explain me this?
Thanks a lot
group-theory
asked Dec 15 '18 at 16:35
LandauLandau
447
$endgroup$
This is what I've understood until now about this subject, please confirm if what I write is true or not.
Choosen one molecule, there is a certain number of symmetry operations which this molecule satisfies (for example $C_{2}-i-S_{4}^{3}$...). The set of these operations forms the point group of the molecule. I can also choose a metric, for example the internal coordinates of the molecule (that is a vector with dimension $3N-6$), the cartesian displacment of the atoms (that is a vector with dimension $3N$) or the atoms (that is a vector with dimension $N$). Choosen one metric, I can build a rapresentation of the point group of the molecule, by looking at the way in which the vector of the metric is transformed by each symmetry operation. In this way, I can build for each operations a square matrix which has the same dimension of the metric and which is the rapresentation of that symmetry operation. If the group I'm considering has order $h$ (number of different operations) I will have at most $h$ differents matrices, because there can be two or more operations with the same matrix. I now have let's say $c$ differents matrices with $c<m$, and I make a change of basis in the vectorial space I'm considering. For one of the $c$ matrices, which I've called $A$ I get:
$$Y=AX rightarrow X'=TX and Y'=TYrightarrow Y'=TAT^{-1}X' $$
So in this new basis I've obtained $c$ new matrices where each matrix is similar to the previous one. I've read that exists a particular change of basis for which all the new matrices are in block diagonal form. Unfortunately they don't specify exactly how these new matrices are. I mean, are all in their respective Jordan normal form? If so, there must exist a specific matrix $T$ of generalized eigenvectors which must be common to all the matrices. Similarly, if all my matrices are diagonalisable, the new $c$ matrices are the corrispective diagonal form of each one? Also in this case there must exist a specific matrix $T$ of eigenvectors which must be common to all the matrices. Someone could explain me this?
Thanks a lot
group-theory
group-theory
asked Dec 15 '18 at 16:35
LandauLandau
447
asked Dec 15 '18 at 16:35
LandauLandau
447
asked Dec 15 '18 at 16:35
LandauLandau
447
asked Dec 15 '18 at 16:35
LandauLandau
447
asked Dec 15 '18 at 16:35
LandauLandau
447
447
1
$begingroup$
Have you looked at some small examples? The matrices will generally not be in Jordan normal form.
$endgroup$
– Tobias Kildetoft
Dec 15 '18 at 22:01
$begingroup$
Yes, unfortunately in the example I've seen the matrices are all put in diagonal form. However I thought that the only diagonal block matrix to which a generic matrix could be similar was its Jordan/diagonal form. You say it is not true? @Tobias Kildetoft
$endgroup$
– Landau
Dec 15 '18 at 22:39
add a comment |
1
$begingroup$
Have you looked at some small examples? The matrices will generally not be in Jordan normal form.
$endgroup$
– Tobias Kildetoft
Dec 15 '18 at 22:01
$begingroup$
Yes, unfortunately in the example I've seen the matrices are all put in diagonal form. However I thought that the only diagonal block matrix to which a generic matrix could be similar was its Jordan/diagonal form. You say it is not true? @Tobias Kildetoft
$endgroup$
– Landau
Dec 15 '18 at 22:39
1
1
$begingroup$
Have you looked at some small examples? The matrices will generally not be in Jordan normal form.
$endgroup$
– Tobias Kildetoft
Dec 15 '18 at 22:01
$begingroup$
Have you looked at some small examples? The matrices will generally not be in Jordan normal form.
$endgroup$
– Tobias Kildetoft
Dec 15 '18 at 22:01
$begingroup$
Yes, unfortunately in the example I've seen the matrices are all put in diagonal form. However I thought that the only diagonal block matrix to which a generic matrix could be similar was its Jordan/diagonal form. You say it is not true? @Tobias Kildetoft
$endgroup$
– Landau
Dec 15 '18 at 22:39
$begingroup$
Yes, unfortunately in the example I've seen the matrices are all put in diagonal form. However I thought that the only diagonal block matrix to which a generic matrix could be similar was its Jordan/diagonal form. You say it is not true? @Tobias Kildetoft
$endgroup$
– Landau
Dec 15 '18 at 22:39
add a comment |
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$begingroup$
Have you looked at some small examples? The matrices will generally not be in Jordan normal form.
$endgroup$
– Tobias Kildetoft
Dec 15 '18 at 22:01
$begingroup$
Yes, unfortunately in the example I've seen the matrices are all put in diagonal form. However I thought that the only diagonal block matrix to which a generic matrix could be similar was its Jordan/diagonal form. You say it is not true? @Tobias Kildetoft
$endgroup$
– Landau
Dec 15 '18 at 22:39