or we can just use the projection operator
( 1
√
2
√^1
2)
( 1
√
2
√^1
2)
=
( 1
21
1 2
21
2)
.
( 1
21
1 2
21
2)
N
2
(
0
1
)
=
N
4
( 1
√
2
√^1
2)
18.9 Other Two State Systems*
18.9.1 The Ammonia Molecule (Maser)
The Feynman Lectures (Volume III, chapters 8 and 9) makes a complete study of the two ground
states of the Ammonia Molecule. Feynman’s discussion is very instructive. Feynman starts with two
states, one with the Nitrogen atom above the plane defined by the three Hydrogen atoms, and the
other with the Nitrogen below the plane. There is clearly symmetry between these two states. They
have identical properties. This is an example of an SU(2) symmetry,like that in angular momentum
(and the weak interactions). We just have two states which are different but completely symmetric.
Since the Nitrogen atom can tunnel from one side of the molecule to the other, there are cross terms
in the Hamiltonian (limiting ourselves to the two symmetric ground states).
〈ψabove|H|ψabove〉=〈ψbelow|H|ψbelow〉=E 0〈ψabove|H|ψbelow〉=−AH=
(
E 0 −A
−A E 0
)
We can adjust the phases of the above and below states to makeAreal.
The energy eigenvalues can be found from the usual equation.
∣
∣
∣
∣
E 0 −E −A
−A E 0 −E
∣
∣
∣
∣ = 0
(E 0 −E)^2 = A^2
E−E 0 = ±A
E = E 0 ±A
Now find the eigenvectors.
Hψ=Eψ
(
E 0 −A
−A E 0)(
a
b)
= (E 0 ±A)
(
a
b)
(
E 0 a−Ab
E 0 b−Aa)
=
(
(E 0 ±A)a
(E 0 ±A)b)
These are solved ifb=∓a. Substituting auspiciously, we get.
(
E 0 a±Aa
E 0 b±Ab
)
=
(
(E 0 ±A)a
(E 0 ±A)b