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I How are Mixed State Ammonia Molecules Separated by an E-Field?

  1. May 2, 2016 #1
    I am working on a paper about Ammonia masers. It looks like Ammonia molecules are usually found in a superposition of even and odd parity states that are eigenstates of the inversion potential. That is the double well potential of the Nitrogen to tunnel through the Hydrogen plane. If it punches through the centerline of this Hydrogen triad, it can be simplified to a 1-D Schrödinger equation. The ground state is an even parity function that looks similar to a sine wave, and the first excited state is anti-symmetrical with the same number of nodes as the ground state with opposite parity. These two states are very nearly degenerate. Apparently, they can be put through a Stern-Gerlock type of device and be sorted. Normally, the superposition of this system causes the Nitrogen to tunnel through from ##\lvert + >## to ##\rvert - >## and oscillate back and forth. These are not eigenstates, but a superposition of the even and odd states that are stationary. The whole thing is explained well in James Binney's QM book online:

    http://www-thphys.physics.ox.ac.uk/people/JamesBinney/qb.pdf

    This is/was a free ebook with his open courseware QM class. It is not bootleg, notice where it is linked to. All of this is discussed from page 81-84.

    It is not clear to me the mechanism by which the excited states are attracted to the stronger portion of the inhomogeneous electric field. There is an explanation in this text, but it seems a little hand wavy to me (maybe I am just dense and missing something). I am aware this is also treated in the Feynman Lectures, but the bit about sorting is blown off in that article. This doesn't seem obvious and I would be greatly indebted to someone who could clarify this matter.

    Thanks,
    KQ6UP
     
  2. jcsd
  3. May 2, 2016 #2

    blue_leaf77

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    When an electric field is applied to these ammonia molecules, the ground and excited states will be polarized. These new states are basically the eigensvectors of matrix (5.26). Thus the dipole moment, which is proportional to ##\langle r \rangle##, will have non-zero value (before the E field was applied, the eigenstates of the pure ammonia Hamiltonian have definite parities and hence must have vanishing dipole moments). If furthermore, the applied field is a function of space, i.e. not uniform, a certain force will be exerted to the new eigenstates having non-zero dipole moment. This way, they will separate in the direction of the gradient of the applied field.
     
  4. May 2, 2016 #3
    Does the parity of the eigenstate cause the dipole moments to be aligned and the other to be anti-aligned to the E field? Where could I find a reference for this behavior. I have looked for this in reference to the Ammonia molecule, but have not been successful. A general statement of parity and polarity would be wonderful.

    Thanks,
    KQ6UP
     
  5. May 3, 2016 #4

    blue_leaf77

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    The argument goes like the following. Before there is any E field, the eigenstates of the ammonia are (I will omit normalization constant)
    $$
    |e\rangle = -|-\rangle + |+\rangle \\
    |g\rangle = |-\rangle + |+\rangle
    $$
    where ##|-\rangle ## and ##|+\rangle## are the states of the ammonian when most electrons are in the bottom and upper side of the hydrogen plane, respectively. These are the eigenvectors of matrix (5.24). If you further assume that the double potential well is along z direction and then calculate ##\langle z \rangle## for both ##|e\rangle## and ##|g\rangle## states, you will get zero. Since the dipole moment is proportional to ##z##, ##\langle p \rangle \propto \langle z \rangle##, the dipole moments for the original eigenstates are also zero. But you want to separate the molecules using E field, this cannot happen unless your molecules have non-vanishing dipole moment. Therefore, you apply an electric field to this beam of ammonia to modify the Hamiltonian such that it takes the form of matrix (5.26).

    Now, put numbers for matrix in equation (5.26), e.g. ##\bar{E} = 4##, ##qEs = 3##, and ##A=1##. The new eigenstates for this matrix will be
    $$
    |e'\rangle = -6.2|-\rangle + |+\rangle \\
    |g'\rangle = 0.16|-\rangle + |+\rangle
    $$
    In these states ##\langle e' |z| e' \rangle## will be negative and ##\langle g' |z| g' \rangle## will be positive. In other words, in these states ##\langle e'|P|e' \rangle## and ##\langle g'|P|g' \rangle## will be nonzero (polarized) and have opposite signs. If furthermore the field is non-uniform, ##\partial_xE## will be nonzero and there will be force exerted onto the polarized molecules, which is equation (5.30), whose direction depends on the signs of ##\langle P \rangle##.
     
    Last edited: May 3, 2016
  6. May 3, 2016 #5

    DrDu

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    You can also look at this like this: The parity eigenstates have a vanishing dipole moment, but, due to their near degeneracy, a very high polarisability. If you do perturbation theory with the electric field as a perturbation, you will find that the lower state is strongly paramagnetic, while the upper state shows a strong diamagnetic behaviour.
    What I also want to remark: The splitting is only observable when the nuclear spins form a chiral configuration, so that the original ammonia molecule and it's inverted mirror image become distinguishable.
    You can find lots of interesting (but somewhat highbrow) stuff on the ammonia maser in the book by Bunker and Jensen, Molecular Symmetry and Spectroscopy.
     
  7. May 4, 2016 #6
    Thanks guys, both of those responses helps. Was also talking to my professor, and he also made an argument using the expectation value for z.

    The book reference was also useful, as I can reference something with this extra info.

    Thanks,
    Chris KQ6UP
     
  8. May 4, 2016 #7
    I followed all of your argument save this bit:

    Are the coefficients in front of the ##\lvert ->## kets only affecting that ket, or should there be parenthesis covering the whole quantity ##(\lvert -> +\lvert +>)##? Also, Please show how you got the new coefficients from your example above. I am thinking those are new coefficients for the eigenvectors like the ##a_i##'s in:
    $$\psi=\sum _{ i }^{ N }{ a_i \phi_i } $$
    Studying Ammonia is the first time I have used the matrix form of the Schrödinger's equation, so I am a bit of a novice here.

    Thanks,
    Chris KQ6UP
     
  9. May 4, 2016 #8

    blue_leaf77

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    No, it's just the coefficient of one ket.
    By solving matrix (5.26) using the numbers I wrote there. Ok, first before there was any E field, the Hamiltonian of your ammonina molecule is given by matrix (5.24). Try to give any number you wish to this matrix, no matter what number you give this matrix will always have eigenvectors of the form
    $$
    |e\rangle = -c|-\rangle + c|+\rangle \\
    |g\rangle = c|-\rangle + c|+\rangle
    $$
    The above form implies that the electrons in the ammonia in the states ##|e\rangle## and ##|g\rangle## can be found above and below the hydrogen plane with the same probability, therefore the expectation value of ##z## is zero.

    Now if you give E field, the Hamiltonian of your ammonia, in the first order of perturbation, will be given by matrix (5.26). Put any number you like, e.g. like what I suggested before. Then the eigenvectors of this new matrix will be
    $$
    |e'\rangle = -6.2|-\rangle + |+\rangle \\
    |g'\rangle = 0.16|-\rangle + |+\rangle
    $$
    For the calculation, see e.g. this calculator. When the ammonia is in ##|e'\rangle## state, its electron will be more often below the hydrogen plane because the absolute value of the coefficient of the state ##|-\rangle## is bigger than ##|+\rangle##. Consquenctly, ##\langle z \rangle < 0## and ##\langle P \rangle = |\langle P \rangle|\hat{z}##. In ##|g'\rangle ## state, you more likely find the electron in the upper part of the hydrogen plane, and thus ##\langle z \rangle > 0## and ##\langle P \rangle = -|\langle P \rangle|\hat{z}##.
     
  10. May 7, 2016 #9
    Thank you for helping me with this. It looks good. I am working on a powerpoint presentation on for the paper right now, and I will be digging into this again in a day or so.

    Regards,
    KQ6UP
     
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