Raman transition and IR transition












3












$begingroup$


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







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$endgroup$

















    3












    $begingroup$


    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







    share|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      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







      share|improve this question









      $endgroup$




      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






      share|improve this question













      share|improve this question











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      share|improve this question










      asked 16 hours ago









      DaniDani

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      1684






















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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.






          share|improve this answer











          $endgroup$













          • $begingroup$
            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
            9 hours ago











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          1 Answer
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          active

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          6












          $begingroup$

          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.






          share|improve this answer











          $endgroup$













          • $begingroup$
            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
            9 hours ago
















          6












          $begingroup$

          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.






          share|improve this answer











          $endgroup$













          • $begingroup$
            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
            9 hours ago














          6












          6








          6





          $begingroup$

          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.






          share|improve this answer











          $endgroup$



          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.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 9 hours ago

























          answered 14 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
            9 hours 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
            9 hours 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
          9 hours 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
          9 hours ago


















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