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Solutions for ammonia as two-state system

  1. Oct 2, 2014 #1
    The state of an ammonia molecule approximated by a two-state system as in the Feynman Lectures can be written [itex]|\psi\rangle = C_{1}|1\rangle + C_{2}|2\rangle[/itex], where the general solution for the coefficients is

    C_{1} = \frac{a}{2}e^{i(E_{0} - A)t/\hbar} + \frac{b}{2}e^{i(E_{0} + A)/\hbar}$$


    C_{2} = \frac{a}{2}e^{i(E_{0} - A)t/\hbar} - \frac{b}{2}e^{i(E_{0} + A)/\hbar}$$

    By taking [itex]a = \sqrt{2}[/itex] and [itex]b = 0[/itex], we get a stationary state [itex]|I\rangle[/itex] with energy [itex]E_{0} - A[/itex]. By taking [itex]a = 0[/itex] and [itex]b= \sqrt{2}[/itex], we get a stationary state [itex]|II\rangle[/itex] with energy [itex]E_{0} + A[/itex]. For each of these states, we have [itex]\Delta E =0[/itex].

    By taking [itex]a = 1[/itex] and [itex]b = 1[/itex], the molecule is in state [itex]|1\rangle[/itex] at [itex]t = 0[/itex] and in state [itex]|2\rangle[/itex] at [itex]t = \frac{\pi \hbar}{2A}[/itex]. The probability oscillates between states [itex]|1\rangle[/itex] and [itex]|2\rangle[/itex], and the system is in a constant superposition of states [itex]|I\rangle[/itex] and [itex]|II\rangle[/itex], so we have [itex]\Delta E = A[/itex] for this state.

    Now, what about the other possible solutions? For example, what does the solution with [itex]a = \sqrt{1.5}[/itex] and [itex]b = \sqrt{0.5}[/itex] stand for? What to make of the fact that the probability to find the molecule in state [itex]|1\rangle[/itex] would never be 1 in this state? Is there a connection to be made with the time-energy uncertainty relation?
  2. jcsd
  3. Oct 3, 2014 #2
    After discussing with a colleague, maybe I found the answer to my own question. We can imagine the molecule to be in a state for which [itex]a = \sqrt{1.5}[/itex] and [itex]b = \sqrt{0.5}[/itex], but after any measurement of the position of the nitrogen atom (over or below the plane of hydrogens) or of the energy ([itex]E_{0} - A[/itex] or [itex]E_{0} + A[/itex]), the molecule could only be left in one of the states [itex]|1\rangle[/itex], [itex]|2\rangle[/itex], [itex]|I\rangle[/itex] or [itex]|II\rangle[/itex], so these are the only ones that are worth discussing. Does that make sense?
  4. Oct 3, 2014 #3


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    You could ask the same question about the states ##\vert I \rangle## and ##\vert II \rangle##. The probability to find the molecule in state ##\vert 1 \rangle## is never 1 in those states either, right?

    More precisely, they're the only ones worth discussing if the measurements you mention are the only ones you are interested in. If you were interested in other measurements, you might find other states that were worth discussing.
  5. Oct 3, 2014 #4
    Right, but the probability to find the molecule in ##\vert 1 \rangle## is always 1/2 in a stationary state, while it could vary with time between 0.3 and 0.7 for another state.

    I thought I could relate this to the uncertainty principle, since ##\Delta E \lt A## for such a state.
  6. Oct 3, 2014 #5
    Hi. Complex numbers a and b where |a|^2+|b|^2=1 show the mixing of two energy eigen states |I> and |II>.
    Energy expextation value of the state is expressed by a and b as

    We should relate |a|^2 and |b|^2, e^-(E_0+A)/kT and e^-(E_0-A)/kT in statistical mechanics.
    Last edited: Oct 3, 2014
  7. Oct 3, 2014 #6


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    You can, but the first thing is to realize that the uncertainty principle applies to all states, including stationary states. The uncertainty principle, at least in the version that's applicable here, says (roughly) that ##\Delta x \Delta p \ge \hbar##. Strictly speaking, this relates the uncertainty in position to the uncertainty in momentum, not energy, but for an object with known mass (which the ammonia molecule is), you can relate the energy and momentum, so you can assume that if you know ##\Delta E##, you know ##\Delta p##; one is just a constant times the other. In particular, if ##\Delta E = 0##, then ##\Delta p = 0##.

    This means that, in the stationary states, ##\vert I \rangle## and ##\vert II \rangle##, since ##\Delta E = 0##, we must have ##\Delta x = \infty##; in other words, the particle has an equal probability to be in any position. (I'm speaking loosely, because the underlying model we are using here doesn't actually have a configuration space with an infinite number of possible positions; but we're just being heuristic here.) If you look at the amplitudes for the possible positions of the particle in these states, you will see that this is indeed true.

    The other extreme, so to speak, is the position eigenstates, i.e., the states ##\vert 1 \rangle## and ##\vert 2 \rangle##. These have ##\Delta x = 0##, so they must have ##\Delta E = \infty## (because they must have ##\Delta p = \infty##). In other words, these states have an equal probability of having any energy. One way to see this is to change basis, so that the basis vectors are the energy eigenstates, ##\vert I \rangle## and ##\vert II \rangle##; in this basis, the amplitudes for the two position eigenstates give equal probabilities for both.

    Btw, this observation about change of basis, along with my note above about speaking loosely, is relevant to the answer you gave to your original question, about these states being the only ones worth discussing. The model of the ammonia molecule that we are using is a limited model; it doesn't cover all possible states of one nitrogen atom and three hydrogen atoms. It only covers the states that are useful in modeling this particular molecule, in which the atoms are all combined into a single bound state, and we are only considering the internal configuration space, so to speak, of that molecule. We aren't considering how the molecule as a whole might be moving--is it part of a gas? A liquid? At high temperature? Low temperature? We aren't considering how the molecule might interact with other molecules or atoms or particles nearby. We're only looking at the molecule as an isolated system, and the two pairs of states we've been discussing--the position eigenstates (1 and 2) and the energy eigenstates (I and II)--each form a basis of the state space for that isolated system, and each basis is a basis of eigenstates for a particular operator (position or energy) of interest.

    Having said all that, we can how see how states other than the four we've discussed work. Consider the state with ##a = b = 1## that you mention in the OP. That state has ##\Delta E = A##. What is its ##\Delta x##? It will be something like ##\Delta x = C \hbar / A## (where ##C## is a constant that depends on how, specifically, ##\Delta E## and ##\Delta p## are related). In other words, states with finite but nonzero uncertainty in energy will also have finite but nonzero uncertainty in position, and the specific relationship between the two is given by the uncertainty principle, but can be verified by looking at the amplitudes in each basis (the energy basis and the position basis).

    Btw, I know you actually asked about the time-energy uncertainty principle, but that actually doesn't work quite the same, because time is not an operator in QM, and the uncertainty principle relates operators. So I think it's actually better to think about position-momentum uncertainty and rely on the fact that energy and momentum are related, so that energy eigenstates are also momentum eigenstates. For more on the issues that come into play if you actually try to think about time-energy uncertainty, see here:

    Last edited: Oct 3, 2014
  8. Oct 3, 2014 #7
    Thank you PeterDonis!
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