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SU(2) symmetry of the ammonia molecule?

  1. Mar 15, 2013 #1
    Hi,
    I'm an undergraduate taking the basic quantum classes and on my own, I'm trying to wrap my mind around how symmetry and group theory applies in Q.M. and theoretical physics in general; it's coming along slowly but surely!

    Can someone please explain why the ammonia molecule is said to exhibit SU(2) symmetry? I can't imagine how the two symmetric states of ammonia have anything to do with the group of 2x2 complex unitary matrices of det=1.

    Thanks
     
  2. jcsd
  3. Mar 15, 2013 #2

    Vanadium 50

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    Where did you hear that?
     
  4. Mar 15, 2013 #3

    Bill_K

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    The ammonia molecule has two states, the first one |1> in which the N is above the three H's, the other one |2> where the N is below them. Both states have the same energy E (two-fold degeneracy).

    Moreover, any orthogonal linear combinations of |1> and |2> will also have the same energy E. The linear combinations can have complex coefficients, but must preserve the norm of the state. In other words, the most general state with energy E is given by U|ψ> where U is a 2x2 matrix and <ψ|UU|ψ> = 1. That is, U is unitary. But the U(2) symmetry is actually SU(2) since one of the transformations just adds a phase factor, which doesn't count!
     
    Last edited: Mar 15, 2013
  5. Mar 15, 2013 #4
    Ah, so why is that considered a "symmetry?" I always thought a symmetry was of the form: "_____ is left invariant upon action by ____" What is being left invariant? Just the amplitudes?
     
  6. Mar 15, 2013 #5

    Vanadium 50

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    I ask again - where did you see that?

    You're asking us to explain something that we can't see. This is not going to go well. Bill has a very good answer, but I could just as easily argue that the symmetry group is Z2. We need context.
     
  7. Mar 15, 2013 #6
    When I think of the symmetry of ammonia, I think [itex] C_{3v} [/itex].

    Clarification would certainly help.
     
  8. Mar 15, 2013 #7
  9. Mar 16, 2013 #8

    DrDu

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    That's only the approximate point group symmetry of the ammonia molecule which would hold if it were rigid. However, as there is an inversion mode where the N can swing through the plane formed by the H atoms one has to consider the full molecular symmetry group which is a subgroup of the complete nuclear permutation and inversion group (CNPI). In the context of the OP this is not too relevant. It is all about the approximate degeneracy of the symmetric and antisymmetric eigenstates of the inversion mode. If these where really degenerate, there would be an additional SU(2) symmetry, as Bill_K already explained. It is a very old discussion why superpositions of configurations are seldomly observed, especially of enantiomeric states: look out for "Hund's paradox".
     
  10. Mar 16, 2013 #9
    this is probably treated in feynman lectures quantum mechanics,where he has chosen the above ones as two base states.
     
  11. Mar 17, 2013 #10

    DrDu

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    This inversion mode of the ammonia molecule has been studied intensively as it forms the basis of the ammonia maser. One interesting point is that the inversion mode only relates two different states for certain values of the sum of the spins of the three hydrogen atoms (S=1/2). This spin state can be obtained in two different ways, either coupling atom 1 and two into a singlet or coupling 1 and 3 into a singlet (other possibilities can be expressed as a linear combination of the two). The inversion will interconvert these two states
     
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