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 16 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 16 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 16 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 16 hours ago
DaniDani
1684
asked 16 hours ago
DaniDani
1684
asked 16 hours ago
DaniDani
1684
asked 16 hours ago
DaniDani
1684
asked 16 hours ago
DaniDani
1684
1684
add a comment |
add a comment |
1 Answer
1
active
oldest
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 14 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
9 hours ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "431"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
var $window = $(window),
onScroll = function(e) {
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom)) {
StackExchange.using('gps', function() { StackExchange.gps.track('embedded_signup_form.view', { location: 'question_page' }); });
$window.unbind('scroll', onScroll);
}
};
$window.on('scroll', onScroll);
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f109191%2framan-transition-and-ir-transition%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
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 14 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
9 hours 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 14 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
9 hours 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 14 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 14 hours ago
matt_blackmatt_black
18.6k352106
edited 9 hours ago
answered 14 hours ago
matt_blackmatt_black
18.6k352106
answered 14 hours ago
matt_blackmatt_black
18.6k352106
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
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
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
add a comment |
Thanks for contributing an answer to Chemistry Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
var $window = $(window),
onScroll = function(e) {
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom)) {
StackExchange.using('gps', function() { StackExchange.gps.track('embedded_signup_form.view', { location: 'question_page' }); });
$window.unbind('scroll', onScroll);
}
};
$window.on('scroll', onScroll);
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f109191%2framan-transition-and-ir-transition%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
var $window = $(window),
onScroll = function(e) {
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom)) {
StackExchange.using('gps', function() { StackExchange.gps.track('embedded_signup_form.view', { location: 'question_page' }); });
$window.unbind('scroll', onScroll);
}
};
$window.on('scroll', onScroll);
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
var $window = $(window),
onScroll = function(e) {
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom)) {
StackExchange.using('gps', function() { StackExchange.gps.track('embedded_signup_form.view', { location: 'question_page' }); });
$window.unbind('scroll', onScroll);
}
};
$window.on('scroll', onScroll);
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
var $window = $(window),
onScroll = function(e) {
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom)) {
StackExchange.using('gps', function() { StackExchange.gps.track('embedded_signup_form.view', { location: 'question_page' }); });
$window.unbind('scroll', onScroll);
}
};
$window.on('scroll', onScroll);
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
