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Ammonia molecule geometry and radiation

  1. May 27, 2015 #1
    What is the relationship between the nitrogen inversion (or "flip-flop" or "turning itself inside out") and the associated microwave radiation of the ammonia molecule ?
     
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  3. May 27, 2015 #2

    blue_leaf77

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    The molecular structure of ammonia is such that the three hydrogen atoms form the base of a tetrahedron with the vertex occupied by the nitrogen. The probability of position of this nitrogen atom is actually not fixed, it also has the possibility of going all the way down "trespassing" the three H's plane, in which case the nitrogen will be at the bottom (the tetrahedron is now flipped). This trespassing of the base of the tetrahedron might be modeled by a double well potential system, the left well corresponds to the "up" position of the N atom, while the right well is associated with the "down" position, or vice versa, the physical meaning will be unchanged anyway. You can try solving such system and you should find that the two lowest energy levels correspond to the symmetric, ##|S\rangle## and antisymmetric, ##|A\rangle##, wavefunctions, symmetric when there are hills in both wells, antisymmetric when one of well has a hill and the other has a valley. The energy difference between these two levels is what you observe as a microwave radiation.

    Now back to the up and down position of the N atom. Using our designation before, if in some situation you have high hill on the left well but shallow valley on the right, this means the N atom is more likely to be located in the "up" position. This situation might be represented as ##|up\rangle = 1/\sqrt{2}(|S\rangle - |A\rangle)##. The inverted problem where you have high hill on the right corresponds to the "down" position, whose wavefunction is ##|down\rangle = 1/\sqrt{2}(|S\rangle + |A\rangle)##. If you let one of them evolve in time, they will actually be oscillating one to another with oscillation frequency equal to, again, the microwave radiation.
     
  4. May 27, 2015 #3
    Does the higher energy level correspond to oscillation and the lower level to no oscillation between the two positions ?
     
  5. May 27, 2015 #4

    blue_leaf77

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    It's not like that, an eigenfunction of the Hamiltonian is stationary state, that is it will never change to another state. So both the ground state and the first excited state, which are ##|S\rangle## and ##|A\rangle## in our notation above, respectively, will not oscillate. However if you build a superposition state between those two energy eigenstates, for example ##|up\rangle## and let this new state evolve in time, which mathematically reads as ##|up , t\rangle = 1/\sqrt{2}(e^{-iE_S /\hbar t} |S\rangle - e^{-iE_A /\hbar t}|A\rangle) = 1/\sqrt{2}e^{-iE_S /\hbar t}( |S\rangle - e^{-i(E_A-E_S) /\hbar t}|A\rangle)##, you will see that this state will not stay as it is at t = 0. In particular when t is such that the complex exponential in the second term is equal to -1, this state will become ## |down\rangle## up to a phase factor. Let this state evolve further for the same amount of time it took to from t=0 to that when it becomes ##|down\rangle##, and you will obtain ##|up\rangle## again. You see, this evolution is an oscillation in time.
     
  6. May 28, 2015 #5
    Why not ? What is the physical difference between the two energy levels ?

    Stationary states have phases that vary at the same frequency. I would call that oscillation. However, I was not talking about the oscillation of the state.

    So you get that the probability of finding the molecule in one of it's position states changes with a frequency that happens to be the same as the frequency of the radiation associated with the energy states. Is there a reason why these frequencies are same ?
     
  7. May 28, 2015 #6

    blue_leaf77

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    I don't know what you meant by oscillation, if it is the phase factor ##e^{-iEt/\hbar}##, then yes it oscillates in that sense. However, the phase factor of each individual eigenstate is actually not really significant as you can always shift the energy scale by whatever value you want, what's important is the relative phase between two different states. Regarding comment #3 of yours, if you define the energy scale so that the ground state has zero energy, then it will conform with what you call by oscillation, the ground state doesn't oscillate because its phase factor is simply unity while all excited states will have time-dependent phase factor.
    Not with the same frequency, the frequency of the phase factor of each stationary state depends on its corresponding energy.
    I guess this is an example of coherence of a quantum system, which occurs because the eigenstates of the system keep accumulating phases over time with different rates.
    The phase factor of the second term in ##|up,t\rangle## is a function of the energy difference between the involved states.
     
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