a/The ammonia molecule,
in states that are inverted relative
to one another.
Here the constantfis a complex number with units of energy. The
interpretation is thatεtells us how much energy splitting we would
have had if the magnetic field had not had anyxorycomponents,
whilefbrings in the effect of those components. We could go ahead
and work out the eigenvalues of this operator by writing down the
eigenvalue equation and solving it by brute force, but the result
is likely to seem less mysterious if we instead apply the following
physical argument.
Althoughflies at some point in the complex plane with some
phase angle argf, such phase angles in quantum mechanics are not
directly observable. Since energiesareobservable, it follows that the
two eigenvalues of energy can only depend on the magnitude off,
not on its phase. By rotational invariance (sec. 3.4.2, p. 195), we also
know that these energies can only depend on|B|=√
B^2 x+By^2 +Bz^2 ,
and in fact when the direction of the field is fixed they must be
proportional to|B|(not to, e.g., the cube of the field). We have
already interpretedεas being essentiallyBz, except for a constant
of proportionality, so it follows from units that the energies must be
of the formE=±√
ε^2 + (...)|f|^2 =±√
ε^2 + (...)f∗f, where (...)
represents a universal unitless constant, which turns out to be 1.
We therefore have for the energies the result
E=±√
ε^2 +f∗f.
Note that our earlier result ofE=±εis recovered whenf= 0.14.7.2 The ammonia molecule
I chose the example of the proton in a magnetic field in the
preceding section for ease of computation, but the treatment of the
general case wheref 6 = 0 may not have seemed especially compelling,
since we would always have the freedom to align ourzaxis with the
field, givingf= 0. But our results from that analysis are of much
greater generality. They do not depend on any facts about the
system other than the fact that it is a system with two states. To
see the full power and generality of this approach, we will apply it
to the ammonia molecule, NH 3 , shown in figure a.
At ordinary temperatures, this molecule is likely to be rotating,
and its angular momentum will have some component about its
symmetry axis (the left-right axis in the diagram). Let’s say, for
example, that the angular momentum vector points to the right,
which we’ll say is the positivexdirection. Then the two orientations
of the molecule shown in figure a are distinguishable. In one, the
electric dipole vector (example 6, p. 586) points in the same direction
as the angular momentum vector, and in the other they point in
opposite directions.^8 For a fixed angular momentum, we have a
two-state system, as in section 14.7.1.(^8) This argument shows that whenLz 6 = 0 we have two distinguishable states,
but it does not necessarily tell us anything about the converse. WhenLz= 0,
992 Chapter 14 Additional Topics in Quantum Physics