Classically, the molecule’s moment of inertia is the same for ori-
entations a/1 and a/2, so we would expect there to be two states
with the same energy. We can always add an arbitrary constant to
the energies, so if they’re the same, we can just say they’re both
zero. Does this mean that quantum-mechanically, we simply have
Hˆ = 0? That would be boring. But this cannot be true, for the
following reason. According to the Schr ̈odinger equation, a state of
definite energy is a state that has a definite frequency, so it lasts
forever, just twirling its phase angle around in the complex plane at
a rateω=E/~. So if state 1 were a state of definite energy, then ac-
cording to the Schr ̈odinger equation if we initially put the molecule
in state 1 it would stay in that state forever. But this cannot be
the case, because we know it is possible for the molecule to switch
from state 1 to state 2 by turning itself inside out like an umbrella
caught by a gust of wind. The possibility of this type of inversion is
not just an optional thing. Vibrations that flex the shape will exist
due to zero-point motion (p. 963). Even if inversion requires a lot
of energy, and the molecule doesn’t have that much energy, there is
at least some probability of having quantum-mechanical tunneling
from 1 to 2. If we prepare the molecule in state 1, and then observe
it at some later time, there is some nonzero probability of finding
it in state 2. This is a contradiction, so our assumption ofHˆ = 0
must have been false.
So the Hamiltonian is not zero, but we already know the full
variety of forms that the Hamiltonian of a two-state system can have.
We only have a couple of parameters to play with, the numbersε
andf. We haveε= 0 by symmetry, so the only possible form for
the Hamiltonian is this:
Hˆ| 1 〉=f| 2 〉
Hˆ| 2 〉=f∗| 1 〉.Because we can define the states| 1 〉and| 2 〉with any phases we like,
we are free to takefto be real,f∗=f, although this implies a certain
relationshipbetweenthe phases of| 1 〉and| 2 〉. If we visualize these
states as bell-shaped functions of anx coordinate describing the
position of the nitrogen relative to the plane of the hydrogens, then
it would be nice to have a phase convention such that where the tails
of the wavefunctions overlap, inside the barrier, they have the same
phase. This turns out to be the case whenfis real and negative,
so we will assume that from now on. Recycling our previous result
for the energies, we haveE=±
√
ε^2 +f^2 =±f. If the tunneling
probability approaches zero, then we expectfto go to zero, and
the energy splitting approaches zero, as we had expected classically.
are there two states, or only one? The analysis in this case is rather intricate,
and depends on the Pauli exclusion principle and the fact that the hydrogen
atoms are all identical, that there are three of them, and that their nuclei are
fermions. See Townes and Schawlow, Microwave Spectroscopy, 1955, pp. 69-71.Section 14.7 Applications to the two-state system 993