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Problem with molecular vibrations

  1. Jan 2, 2016 #1
    In IR spectroscopy, bonds vibrate and therefore there are moving charges and usually a changing dipole. If accelerating charges cause EM waves to be given off due to the changing electric field then shouldn't the vibrating bond be constantly emitting photons and therefore losing energy in a similar way to an electron moving around the nucleus? Doesn't this then make molecular vibrations impossible (well not impossible but inconsistent with them emitting single photons as they fall from a higher level to a lower level).

    I've also read that the infrared absorbed by molecules does not correspond to the frequency of the bond vibrations themselves but rather the difference in vibrational energy levels. If this is the case then what is the significance of the frequency of the molecular vibration when modeled as a spring?
     
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  3. Jan 3, 2016 #2

    blue_leaf77

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    In IR spectroscopy, as well as in any other spectroscopic techniques, the target molecule is initially in a stationary state, for absorption spectroscopy it's usually the vibrational ground state. Despite being assigned to a particular value of vibronic quantum number, this does not mean that the molecules is vibrating with the frequency associated with the vibronic quantum number the molecule is assigned to. The assignment to a quantum number only tells us that the quantum system in question is sitting calm in the corresponding energy state (in reality, an atom or molecule never stays in an excited state forever due to spontaneous emission) - there is no fluctuation of charge distribution in the system. Now when a perturbation comes, in this case in the form of EM wave in the IR spectrum, a pair of vibrational states can be coupled, depending on how close the IR frequency with the energy difference between the two states. It's this states coupling, or superposition, which causes the IR radiation to be emitted and subsequently measured.
    No, electrons around the nucleus will never lose its energy due to its movement. It will only lose energy when it interacts with external perturbation (stimulated emission and absorption) or internal (spontaneous emission).
    Yes, that's true.
     
  4. Jan 3, 2016 #3
    Thanks for the reply.

    Classically electrons moving around the nucleus are accelerating charge and therefore should lose energy in the same way classically a vibrating bond is accelerating charge and should also lose energy. No?

    I've seen illustrations showing that a passing IR wave basically drives the bond vibration due to the oscillating electric field passing through the molecule (when it is at the right frequency). Is it just a classical analogy and modeling it as a spring in this way allows solution of the shrodinger equation that produces results consistent with observations? Is this to say that all the different modes of vibration, symmetric and asymmetric stretching, bending etc. are just a classical means of looking at the system in order to assign a potential energy for the shrodinger equation and the molecule itself is not actually vibrating in any of these ways?
     
    Last edited: Jan 3, 2016
  5. Jan 3, 2016 #4
    By coupled here I assume you're not referring to rotational vibrational coupling and are just considering the transition between the pair of vibrational states. Also do you mean absorbed and not emitted?
     
  6. Jan 3, 2016 #5

    blue_leaf77

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    Classically yes. But unfortunately, the classical view cannot be applied to atoms or molecules.
    If the molecule stands alone in the universe and it is initially in the ground state, it won't be vibrating. The vibration is initiated by perturbation such as collision with other things or interaction with EM wave. Modelling molecule as a spring, for some reason, does yield a good approximate of the absorption spectrum in some frequency region. But it does not draw a connecting line at all with the solution of Schroedinger equation.
    Those vibrational modes as well as the specific way each of them vibrates can be derived using Schroedinger equation and these vibrations, again, must be initiated by certain perturbation, such as IR wave. There is perturbation, molecules vibrate, no perturbation, the molecules stand still.
    Yes, only between vibrational states.
     
  7. Jan 3, 2016 #6
    In your first message you say that there is no fluctuation of charge distribution in the molecule occupying excited states and then in your second message you say that the molecule vibrates in response to a perturbation as it transitions from say the ground state to some excited state. Doesn't the now vibrating molecule result in a fluctuation of charge distribution as the dipole moment changes? From your first post I understood that the molecules were not physically vibrating in any of the states as you said "sitting calm in the corresponding energy state" and therefore not necessarily vibrating with any particular frequency thereby resolving the issue of accelerating charge due to physical vibrations constantly needing to emit EM radiation as the molecule could just be said to be in the given state without actually vibrating. Your second post, however, suggests to me that actual physical vibrations do indeed take place, at least in the excited states, which leads me back to the initial problem I had with oscillating charges that should be continually emitting EM radiation as opposed to just losing energy in one go and emitting (usually) a single photon. Perhaps that's as a result of the classical view, which you said couldn't be applied, and in the quantum view there is some way for the molecule to physically vibrate and not lose energy until the transition down?

    Also it seems as if you're saying in the ground state there is no vibration but doesn't the zero point vibration associated with the ground state imply that the atoms are not completely at rest?
     
  8. Jan 3, 2016 #7

    blue_leaf77

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    Instead of
    I should probably have said: "If the molecule stands alone in the universe, for example in the ground state, it won't be vibrating."
    I used the ground state as an example because in reality, a molecule or atom in an excited state will not stay forever there, it will eventually go to the ground state by virtue of spontaneous emission. But, if only, spontaneous emission never existed, an atom or molecule in an excited state will not decay and it will not vibrate either because excited states as well as ground state are stationary states of the Hamiltonian.
    You must be referring to zero-point vibrational energy. It's a stationary state and therefore a molecule in this state will not vibrate.
     
  9. Jan 4, 2016 #8
    I think that we are both using the word vibration in a different way. I was using it to mean an actual physical vibration of something like a spring which is used to illustrate the mechanism of IR absorption though fallaciously as it would seem if nothing is actually vibrating that way. You, on the other hand, seem to be using vibration to mean a transition between states which is why I think you're mentioning standing alone in the universe i.e. without the presence of EM radiation or other molecules to collide with, and that whichever state it is in is not physically vibrating (my use of the word) or transitioning (your use of the term vibration) in the instance like you suggested were decay impossible.

    You say you need perturbation, in the form of say an IR wave, for vibration to take place. What I'm understanding from this is that the only time the dipole moment is changing or equivalently there is a fluctuation of charge density is during a transition, absorption or emission. All other times in between, the molecule is happy in whichever state it is in and not doing anything. What is physically different about the molecule between say the first level and the ground level? If the molecule is say a diatomic molecule (for simplicity's sake) then would it be correct to think of the ground state having one particular bond length and then when the molecule absorbs a photon the bond stretches, changing the dipole moment, and stays stretched until it decays emitting a photon and returning to its original length and dipole moment? This would be the vibration but every time the bond wanted to change lengths it would only be able to by discrete amounts and by absorbing or emitting a photon of appropriate energy. This is in contrast to my original picture of a bond constantly vibrating/oscillating with some amplitude when at a low vibrational level and then constantly vibrating/oscillating with some greater amplitude when at a higher vibrational level (this is with the same frequency at each level which also happened to be the same frequency of the IR wave that it could absorb). If this is the case then both our uses of the term "vibration" could be reconciled.

    Stationary state of the Hamiltonian means a time independent wavefunction which in turn leads to no vibration or movement? Surely the atoms cannot be standing still in the classical sense otherwise their momentum would be known exactly which would violate the uncertainty principle. Seeing as the wavefunction never ascribed a definite position or momentum in the first place, perhaps this is where the quantum view takes over. Is this what you mean by "not vibrating", maintaining the same wavefunction i.e. not vibrating in the quantum sense but not literally standing still in the classical sense? Do the details of how the stationary states avoid fluctuation of charge density or the need to continually emit photons then belong to some higher level topic in physics?
     
    Last edited: Jan 4, 2016
  10. Jan 4, 2016 #9

    blue_leaf77

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    For diatomic molecules, when it undergoes transition, it will vibrate in a similar manner as a spring does. I hope this statement reconcile our different view for the term vibration.
    Yes, that's right.
    It simply means it that the total energy of the system is higher in excited state than it is in ground state (just to note, that in molecules there are electronic, vibrational, and rotational energy levels).
    You can calculate the expectation value of interatomic distance for each eigenstate of the molecule, and it is time independent. For higher vibrational eigenstate, the interatomic distance should increase.
    I suggest that you start gaining the basic first by reading related literatures before making speculations, although some of them might be true. For introductory there is a book from Joachain and Bransden titled "Atomic and Molecular Physics", if you are more a chemist rather than physicist, may be "Molecular Quantum Mechanics" by Atkins and Friedman will suit you.
     
  11. Jan 7, 2016 #10

    DrDu

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    Maybe you should have a look at Landau and Lifshitz Vol. 3. Landau has a very clear description of how classical motion and transitions arise from quantum mechanics when one considers highly excited states.
     
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