What drives vibrational cooling in an excited stated?Period of Vibrational ModesExplaining the Relative Energies of Various Vibrational ModesRotational-vibrational spectroscopy of a moleculeDo different vibrational modes correspond to vibrational energy levels?Spectroscopy of excited state lithiumWhy does vibrational cooling occur much faster than sponteneous emission in the condensed phase?Notation for excited statesVibrational Self-Consistent FieldDo molecules change shape when they become vibrationally excited?Heisenberg uncertainty principle and lifetime of the excited state
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What drives vibrational cooling in an excited stated?
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What drives vibrational cooling in an excited stated?
Period of Vibrational ModesExplaining the Relative Energies of Various Vibrational ModesRotational-vibrational spectroscopy of a moleculeDo different vibrational modes correspond to vibrational energy levels?Spectroscopy of excited state lithiumWhy does vibrational cooling occur much faster than sponteneous emission in the condensed phase?Notation for excited statesVibrational Self-Consistent FieldDo molecules change shape when they become vibrationally excited?Heisenberg uncertainty principle and lifetime of the excited state
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As we excite a molecule from its ground state, $S_0,v=0$, to some excited state in a higher vibrational state, i.e. $S_1,v'=3$, what drives vibrational cooling within that excited state manifold (such as $S_1,v'=0 leftarrow S_1,v'=3 $?) This is a fast process, typically on the order of hundreds of femtoseconds, and it occurs prior to any photorelaxation (as per Kasha's rule). Due to the vibrational density of states being quite high near the first excited state, is it the anharmonic coupling to these states that drives vibrational cooling? Furthermore, is there a quantum mechanistic/mathematical way of depicting the probability/rate of this process (as in, what would the relevant perturbation term $hatH'$ for dealing with this problem in the context of Fermi's golden rule)?
physical-chemistry quantum-chemistry spectroscopy
asked 9 hours ago
kallekalle
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As we excite a molecule from its ground state, $S_0,v=0$, to some excited state in a higher vibrational state, i.e. $S_1,v'=3$, what drives vibrational cooling within that excited state manifold (such as $S_1,v'=0 leftarrow S_1,v'=3 $?) This is a fast process, typically on the order of hundreds of femtoseconds, and it occurs prior to any photorelaxation (as per Kasha's rule). Due to the vibrational density of states being quite high near the first excited state, is it the anharmonic coupling to these states that drives vibrational cooling? Furthermore, is there a quantum mechanistic/mathematical way of depicting the probability/rate of this process (as in, what would the relevant perturbation term $hatH'$ for dealing with this problem in the context of Fermi's golden rule)?
physical-chemistry quantum-chemistry spectroscopy
asked 9 hours ago
kallekalle
383 bronze badges
New contributor
kalle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Where are you, in a condensed phase or in a dilute gas? A compact lattice has no space for local vibrations.
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– Karl
2 hours ago
$begingroup$
Good point - I guess the form of vibrational cooling depends on the nature of the state. I'm more interested in the condensed phase, where solvent/bath interactions or possible. My interest mainly lies in the crystalline phase.
$endgroup$
– kalle
1 hour ago
add a comment |
$begingroup$
As we excite a molecule from its ground state, $S_0,v=0$, to some excited state in a higher vibrational state, i.e. $S_1,v'=3$, what drives vibrational cooling within that excited state manifold (such as $S_1,v'=0 leftarrow S_1,v'=3 $?) This is a fast process, typically on the order of hundreds of femtoseconds, and it occurs prior to any photorelaxation (as per Kasha's rule). Due to the vibrational density of states being quite high near the first excited state, is it the anharmonic coupling to these states that drives vibrational cooling? Furthermore, is there a quantum mechanistic/mathematical way of depicting the probability/rate of this process (as in, what would the relevant perturbation term $hatH'$ for dealing with this problem in the context of Fermi's golden rule)?
physical-chemistry quantum-chemistry spectroscopy
asked 9 hours ago
kallekalle
383 bronze badges
New contributor
kalle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
As we excite a molecule from its ground state, $S_0,v=0$, to some excited state in a higher vibrational state, i.e. $S_1,v'=3$, what drives vibrational cooling within that excited state manifold (such as $S_1,v'=0 leftarrow S_1,v'=3 $?) This is a fast process, typically on the order of hundreds of femtoseconds, and it occurs prior to any photorelaxation (as per Kasha's rule). Due to the vibrational density of states being quite high near the first excited state, is it the anharmonic coupling to these states that drives vibrational cooling? Furthermore, is there a quantum mechanistic/mathematical way of depicting the probability/rate of this process (as in, what would the relevant perturbation term $hatH'$ for dealing with this problem in the context of Fermi's golden rule)?
physical-chemistry quantum-chemistry spectroscopy
physical-chemistry quantum-chemistry spectroscopy
asked 9 hours ago
kallekalle
383 bronze badges
New contributor
kalle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 9 hours ago
kallekalle
383 bronze badges
New contributor
kalle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 9 hours ago
kallekalle
383 bronze badges
New contributor
kalle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 9 hours ago
kallekalle
383 bronze badges
asked 9 hours ago
kallekalle
383 bronze badges
383 bronze badges
New contributor
kalle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
kalle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Where are you, in a condensed phase or in a dilute gas? A compact lattice has no space for local vibrations.
$endgroup$
– Karl
2 hours ago
$begingroup$
Good point - I guess the form of vibrational cooling depends on the nature of the state. I'm more interested in the condensed phase, where solvent/bath interactions or possible. My interest mainly lies in the crystalline phase.
$endgroup$
– kalle
1 hour ago
add a comment |
$begingroup$
Where are you, in a condensed phase or in a dilute gas? A compact lattice has no space for local vibrations.
$endgroup$
– Karl
2 hours ago
$begingroup$
Good point - I guess the form of vibrational cooling depends on the nature of the state. I'm more interested in the condensed phase, where solvent/bath interactions or possible. My interest mainly lies in the crystalline phase.
$endgroup$
– kalle
1 hour ago
$begingroup$
Where are you, in a condensed phase or in a dilute gas? A compact lattice has no space for local vibrations.
$endgroup$
– Karl
2 hours ago
$begingroup$
Where are you, in a condensed phase or in a dilute gas? A compact lattice has no space for local vibrations.
$endgroup$
– Karl
2 hours ago
$begingroup$
Good point - I guess the form of vibrational cooling depends on the nature of the state. I'm more interested in the condensed phase, where solvent/bath interactions or possible. My interest mainly lies in the crystalline phase.
$endgroup$
– kalle
1 hour ago
$begingroup$
Good point - I guess the form of vibrational cooling depends on the nature of the state. I'm more interested in the condensed phase, where solvent/bath interactions or possible. My interest mainly lies in the crystalline phase.
$endgroup$
– kalle
1 hour ago
add a comment |
1 Answer
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Normal modes are technically orthogonal so no energy could flow from one to another, but of course this is just a mathematical construct to make life simple for us, and in reality 'anharmonicity' (physical and electronic) will allow the energy to move about. This is to say that these modes are able to couple to one another. Fermi golden rule is the usual starting point although the density of states may not be that high at low vibrational levels. The calculation involves calculating the Franck-Condon density of states from the initial to final level. The process is called IVR or intramolecular vibrational redistribution.
In many types molecules, when isolated in the vapour phase, the initially excited level can still remain excited for many nanoseconds and the fluorescence lifetime and spectrum from these levels can easily be measured, e.g. naphthalene, pyrene, benzanthracene etc. This is particularly the case when the FC excited mode has a distinctly different frequency and displacement to other modes so that the FC factors to other vibrational levels is small. An interesting case is that of many types planar aromatic dyes which show persistent wavepacket behaviour of the FC excited mode even in solution at room temperature.
The theory of radiationless transitions has been thoroughly worked out mainly for intersystem crossing but the theory is essentially the same with different operators. Look for papers from the 1970's by S. H. Lin et al, W. Siebrand et al, J. Jortner et al. and K. Freed et al.
answered 6 hours ago
porphyrinporphyrin
19.1k33 silver badges58 bronze badges
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$begingroup$
Normal modes are technically orthogonal so no energy could flow from one to another, but of course this is just a mathematical construct to make life simple for us, and in reality 'anharmonicity' (physical and electronic) will allow the energy to move about. This is to say that these modes are able to couple to one another. Fermi golden rule is the usual starting point although the density of states may not be that high at low vibrational levels. The calculation involves calculating the Franck-Condon density of states from the initial to final level. The process is called IVR or intramolecular vibrational redistribution.
In many types molecules, when isolated in the vapour phase, the initially excited level can still remain excited for many nanoseconds and the fluorescence lifetime and spectrum from these levels can easily be measured, e.g. naphthalene, pyrene, benzanthracene etc. This is particularly the case when the FC excited mode has a distinctly different frequency and displacement to other modes so that the FC factors to other vibrational levels is small. An interesting case is that of many types planar aromatic dyes which show persistent wavepacket behaviour of the FC excited mode even in solution at room temperature.
The theory of radiationless transitions has been thoroughly worked out mainly for intersystem crossing but the theory is essentially the same with different operators. Look for papers from the 1970's by S. H. Lin et al, W. Siebrand et al, J. Jortner et al. and K. Freed et al.
answered 6 hours ago
porphyrinporphyrin
19.1k33 silver badges58 bronze badges
$endgroup$
add a comment |
$begingroup$
Normal modes are technically orthogonal so no energy could flow from one to another, but of course this is just a mathematical construct to make life simple for us, and in reality 'anharmonicity' (physical and electronic) will allow the energy to move about. This is to say that these modes are able to couple to one another. Fermi golden rule is the usual starting point although the density of states may not be that high at low vibrational levels. The calculation involves calculating the Franck-Condon density of states from the initial to final level. The process is called IVR or intramolecular vibrational redistribution.
In many types molecules, when isolated in the vapour phase, the initially excited level can still remain excited for many nanoseconds and the fluorescence lifetime and spectrum from these levels can easily be measured, e.g. naphthalene, pyrene, benzanthracene etc. This is particularly the case when the FC excited mode has a distinctly different frequency and displacement to other modes so that the FC factors to other vibrational levels is small. An interesting case is that of many types planar aromatic dyes which show persistent wavepacket behaviour of the FC excited mode even in solution at room temperature.
The theory of radiationless transitions has been thoroughly worked out mainly for intersystem crossing but the theory is essentially the same with different operators. Look for papers from the 1970's by S. H. Lin et al, W. Siebrand et al, J. Jortner et al. and K. Freed et al.
answered 6 hours ago
porphyrinporphyrin
19.1k33 silver badges58 bronze badges
$endgroup$
add a comment |
$begingroup$
Normal modes are technically orthogonal so no energy could flow from one to another, but of course this is just a mathematical construct to make life simple for us, and in reality 'anharmonicity' (physical and electronic) will allow the energy to move about. This is to say that these modes are able to couple to one another. Fermi golden rule is the usual starting point although the density of states may not be that high at low vibrational levels. The calculation involves calculating the Franck-Condon density of states from the initial to final level. The process is called IVR or intramolecular vibrational redistribution.
In many types molecules, when isolated in the vapour phase, the initially excited level can still remain excited for many nanoseconds and the fluorescence lifetime and spectrum from these levels can easily be measured, e.g. naphthalene, pyrene, benzanthracene etc. This is particularly the case when the FC excited mode has a distinctly different frequency and displacement to other modes so that the FC factors to other vibrational levels is small. An interesting case is that of many types planar aromatic dyes which show persistent wavepacket behaviour of the FC excited mode even in solution at room temperature.
The theory of radiationless transitions has been thoroughly worked out mainly for intersystem crossing but the theory is essentially the same with different operators. Look for papers from the 1970's by S. H. Lin et al, W. Siebrand et al, J. Jortner et al. and K. Freed et al.
answered 6 hours ago
porphyrinporphyrin
19.1k33 silver badges58 bronze badges
$endgroup$
Normal modes are technically orthogonal so no energy could flow from one to another, but of course this is just a mathematical construct to make life simple for us, and in reality 'anharmonicity' (physical and electronic) will allow the energy to move about. This is to say that these modes are able to couple to one another. Fermi golden rule is the usual starting point although the density of states may not be that high at low vibrational levels. The calculation involves calculating the Franck-Condon density of states from the initial to final level. The process is called IVR or intramolecular vibrational redistribution.
In many types molecules, when isolated in the vapour phase, the initially excited level can still remain excited for many nanoseconds and the fluorescence lifetime and spectrum from these levels can easily be measured, e.g. naphthalene, pyrene, benzanthracene etc. This is particularly the case when the FC excited mode has a distinctly different frequency and displacement to other modes so that the FC factors to other vibrational levels is small. An interesting case is that of many types planar aromatic dyes which show persistent wavepacket behaviour of the FC excited mode even in solution at room temperature.
The theory of radiationless transitions has been thoroughly worked out mainly for intersystem crossing but the theory is essentially the same with different operators. Look for papers from the 1970's by S. H. Lin et al, W. Siebrand et al, J. Jortner et al. and K. Freed et al.
answered 6 hours ago
porphyrinporphyrin
19.1k33 silver badges58 bronze badges
answered 6 hours ago
porphyrinporphyrin
19.1k33 silver badges58 bronze badges
answered 6 hours ago
porphyrinporphyrin
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answered 6 hours ago
porphyrinporphyrin
19.1k33 silver badges58 bronze badges
19.1k33 silver badges58 bronze badges
add a comment |
add a comment |
kalle is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Where are you, in a condensed phase or in a dilute gas? A compact lattice has no space for local vibrations.
$endgroup$
– Karl
2 hours ago
$begingroup$
Good point - I guess the form of vibrational cooling depends on the nature of the state. I'm more interested in the condensed phase, where solvent/bath interactions or possible. My interest mainly lies in the crystalline phase.
$endgroup$
– kalle
1 hour ago