Raman transition and IR transition
$begingroup$
In a character table one can find the reducible representation of vibration of a molecule. In such a table we also see things like $xy,z,x...$. If this corresponds with an irrep we say that IR transitions can occur if $x,y,z$ and quadratic terms give Raman transitions. This is because the dipole moment and polarizability are non-zero and can change in that case. My question is: where does this come from? I prefer a physical or mathematical answer involving quantum mechanics since I am from physics.
spectroscopy ir-spectroscopy
asked 7 hours ago
DaniDani
1684
$endgroup$
add a comment |
$begingroup$
In a character table one can find the reducible representation of vibration of a molecule. In such a table we also see things like $xy,z,x...$. If this corresponds with an irrep we say that IR transitions can occur if $x,y,z$ and quadratic terms give Raman transitions. This is because the dipole moment and polarizability are non-zero and can change in that case. My question is: where does this come from? I prefer a physical or mathematical answer involving quantum mechanics since I am from physics.
spectroscopy ir-spectroscopy
asked 7 hours ago
DaniDani
1684
$endgroup$
add a comment |
$begingroup$
In a character table one can find the reducible representation of vibration of a molecule. In such a table we also see things like $xy,z,x...$. If this corresponds with an irrep we say that IR transitions can occur if $x,y,z$ and quadratic terms give Raman transitions. This is because the dipole moment and polarizability are non-zero and can change in that case. My question is: where does this come from? I prefer a physical or mathematical answer involving quantum mechanics since I am from physics.
spectroscopy ir-spectroscopy
asked 7 hours ago
DaniDani
1684
$endgroup$
In a character table one can find the reducible representation of vibration of a molecule. In such a table we also see things like $xy,z,x...$. If this corresponds with an irrep we say that IR transitions can occur if $x,y,z$ and quadratic terms give Raman transitions. This is because the dipole moment and polarizability are non-zero and can change in that case. My question is: where does this come from? I prefer a physical or mathematical answer involving quantum mechanics since I am from physics.
spectroscopy ir-spectroscopy
spectroscopy ir-spectroscopy
asked 7 hours ago
DaniDani
1684
asked 7 hours ago
DaniDani
1684
asked 7 hours ago
DaniDani
1684
asked 7 hours ago
DaniDani
1684
asked 7 hours ago
DaniDani
1684
1684
add a comment |
add a comment |
1 Answer
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$begingroup$
To simplify slightly both Raman and IR spectroscopy show the vibrational modes of a molecule (though the techniques used to reveal these are very different).
IR spectroscopy relies on coupling between the electromagnetic field of light passing through a sample and the electric dipole of the molecule. But that absorption is only possible is the vibration in the molecule changes the molecules dipole. Consider, for example, carbon dioxide, a linear molecule with no net dipole moment. Symmetric stretches (ie both oxygens moving the same amount in opposite directions) don't change the dipole so can't be detected by IR. But asymmetric stretches or bending will change the dipole and can be detected by IR.
Raman spectroscopy (simplifying a little) relies on detecting changes to the molecules polarisability not the dipole. So the symmetric vibrations of carbon dioxide are detectable because they do change the polarisability of the molecule.
In both cases the symmetry tables tell you whether the symmetry of the vibrational mode changes the electric dipole or the polarisability of the molecule. In simple molecules physical intuition can usually tell the same thing but symmetry tables are more reliable.
A full understanding would involve a quantum mechanical description of the interaction of EM radiation with the possible vibrational modes of a molecule considering the allowed and forbidden transitions. But the physical intuition above captures the essentials without getting too mathematical.
answered 4 hours ago
matt_blackmatt_black
18.6k352106
$endgroup$
$begingroup$
It might be worth mentioning in addition that polarisability involves a change in the shape or size of the 'electron cloud' and so the em field see the projection of this i.e. a 2D shape change hence quadratic terms in the point group.
$endgroup$
– porphyrin
2 mins ago
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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active
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votes
$begingroup$
To simplify slightly both Raman and IR spectroscopy show the vibrational modes of a molecule (though the techniques used to reveal these are very different).
IR spectroscopy relies on coupling between the electromagnetic field of light passing through a sample and the electric dipole of the molecule. But that absorption is only possible is the vibration in the molecule changes the molecules dipole. Consider, for example, carbon dioxide, a linear molecule with no net dipole moment. Symmetric stretches (ie both oxygens moving the same amount in opposite directions) don't change the dipole so can't be detected by IR. But asymmetric stretches or bending will change the dipole and can be detected by IR.
Raman spectroscopy (simplifying a little) relies on detecting changes to the molecules polarisability not the dipole. So the symmetric vibrations of carbon dioxide are detectable because they do change the polarisability of the molecule.
In both cases the symmetry tables tell you whether the symmetry of the vibrational mode changes the electric dipole or the polarisability of the molecule. In simple molecules physical intuition can usually tell the same thing but symmetry tables are more reliable.
A full understanding would involve a quantum mechanical description of the interaction of EM radiation with the possible vibrational modes of a molecule considering the allowed and forbidden transitions. But the physical intuition above captures the essentials without getting too mathematical.
answered 4 hours ago
matt_blackmatt_black
18.6k352106
$endgroup$
$begingroup$
It might be worth mentioning in addition that polarisability involves a change in the shape or size of the 'electron cloud' and so the em field see the projection of this i.e. a 2D shape change hence quadratic terms in the point group.
$endgroup$
– porphyrin
2 mins ago
add a comment |
$begingroup$
To simplify slightly both Raman and IR spectroscopy show the vibrational modes of a molecule (though the techniques used to reveal these are very different).
IR spectroscopy relies on coupling between the electromagnetic field of light passing through a sample and the electric dipole of the molecule. But that absorption is only possible is the vibration in the molecule changes the molecules dipole. Consider, for example, carbon dioxide, a linear molecule with no net dipole moment. Symmetric stretches (ie both oxygens moving the same amount in opposite directions) don't change the dipole so can't be detected by IR. But asymmetric stretches or bending will change the dipole and can be detected by IR.
Raman spectroscopy (simplifying a little) relies on detecting changes to the molecules polarisability not the dipole. So the symmetric vibrations of carbon dioxide are detectable because they do change the polarisability of the molecule.
In both cases the symmetry tables tell you whether the symmetry of the vibrational mode changes the electric dipole or the polarisability of the molecule. In simple molecules physical intuition can usually tell the same thing but symmetry tables are more reliable.
A full understanding would involve a quantum mechanical description of the interaction of EM radiation with the possible vibrational modes of a molecule considering the allowed and forbidden transitions. But the physical intuition above captures the essentials without getting too mathematical.
answered 4 hours ago
matt_blackmatt_black
18.6k352106
$endgroup$
$begingroup$
It might be worth mentioning in addition that polarisability involves a change in the shape or size of the 'electron cloud' and so the em field see the projection of this i.e. a 2D shape change hence quadratic terms in the point group.
$endgroup$
– porphyrin
2 mins ago
add a comment |
$begingroup$
To simplify slightly both Raman and IR spectroscopy show the vibrational modes of a molecule (though the techniques used to reveal these are very different).
IR spectroscopy relies on coupling between the electromagnetic field of light passing through a sample and the electric dipole of the molecule. But that absorption is only possible is the vibration in the molecule changes the molecules dipole. Consider, for example, carbon dioxide, a linear molecule with no net dipole moment. Symmetric stretches (ie both oxygens moving the same amount in opposite directions) don't change the dipole so can't be detected by IR. But asymmetric stretches or bending will change the dipole and can be detected by IR.
Raman spectroscopy (simplifying a little) relies on detecting changes to the molecules polarisability not the dipole. So the symmetric vibrations of carbon dioxide are detectable because they do change the polarisability of the molecule.
In both cases the symmetry tables tell you whether the symmetry of the vibrational mode changes the electric dipole or the polarisability of the molecule. In simple molecules physical intuition can usually tell the same thing but symmetry tables are more reliable.
A full understanding would involve a quantum mechanical description of the interaction of EM radiation with the possible vibrational modes of a molecule considering the allowed and forbidden transitions. But the physical intuition above captures the essentials without getting too mathematical.
answered 4 hours ago
matt_blackmatt_black
18.6k352106
$endgroup$
To simplify slightly both Raman and IR spectroscopy show the vibrational modes of a molecule (though the techniques used to reveal these are very different).
IR spectroscopy relies on coupling between the electromagnetic field of light passing through a sample and the electric dipole of the molecule. But that absorption is only possible is the vibration in the molecule changes the molecules dipole. Consider, for example, carbon dioxide, a linear molecule with no net dipole moment. Symmetric stretches (ie both oxygens moving the same amount in opposite directions) don't change the dipole so can't be detected by IR. But asymmetric stretches or bending will change the dipole and can be detected by IR.
Raman spectroscopy (simplifying a little) relies on detecting changes to the molecules polarisability not the dipole. So the symmetric vibrations of carbon dioxide are detectable because they do change the polarisability of the molecule.
In both cases the symmetry tables tell you whether the symmetry of the vibrational mode changes the electric dipole or the polarisability of the molecule. In simple molecules physical intuition can usually tell the same thing but symmetry tables are more reliable.
A full understanding would involve a quantum mechanical description of the interaction of EM radiation with the possible vibrational modes of a molecule considering the allowed and forbidden transitions. But the physical intuition above captures the essentials without getting too mathematical.
answered 4 hours ago
matt_blackmatt_black
18.6k352106
edited 32 mins ago
answered 4 hours ago
matt_blackmatt_black
18.6k352106
answered 4 hours ago
matt_blackmatt_black
18.6k352106
answered 4 hours ago
matt_blackmatt_black
18.6k352106
18.6k352106
$begingroup$
It might be worth mentioning in addition that polarisability involves a change in the shape or size of the 'electron cloud' and so the em field see the projection of this i.e. a 2D shape change hence quadratic terms in the point group.
$endgroup$
– porphyrin
2 mins ago
add a comment |
$begingroup$
It might be worth mentioning in addition that polarisability involves a change in the shape or size of the 'electron cloud' and so the em field see the projection of this i.e. a 2D shape change hence quadratic terms in the point group.
$endgroup$
– porphyrin
2 mins ago
$begingroup$
It might be worth mentioning in addition that polarisability involves a change in the shape or size of the 'electron cloud' and so the em field see the projection of this i.e. a 2D shape change hence quadratic terms in the point group.
$endgroup$
– porphyrin
2 mins ago
$begingroup$
It might be worth mentioning in addition that polarisability involves a change in the shape or size of the 'electron cloud' and so the em field see the projection of this i.e. a 2D shape change hence quadratic terms in the point group.
$endgroup$
– porphyrin
2 mins ago
add a comment |
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