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).

\begin{displaymath}\bgroup\color{black}\langle\psi_{above}\vert H\vert\psi_{abov...
...e=\langle\psi_{below}\vert H\vert\psi_{below}\rangle=E_0\egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black}\langle\psi_{above}\vert H\vert\psi_{below}\rangle=-A\egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black}H=\left(\matrix{E_0 & -A \cr -A & E_0}\right)\egroup\end{displaymath}

We can adjust the phases of the above and below states to make \bgroup\color{black}$A$\egroup real.

The energy eigenvalues can be found from the usual equation.

\begin{eqnarray*}
\left\vert\matrix{E_0-E & -A \cr -A & E_0-E}\right\vert & = & ...
...
(E_0-E)^2 & = & A^2 \\
E-E_0 & = & \pm A \\
E & = & E_0\pm A
\end{eqnarray*}


Now find the eigenvectors.

\begin{eqnarray*}
H\psi=E\psi \\
\left(\matrix{E_0 & -A \cr -A & E_0}\right)\le...
..._0b-Aa}\right)=\left(\matrix{(E_0\pm A)a \cr (E_0\pm A)b}\right)
\end{eqnarray*}


These are solved if \bgroup\color{black}$b=\mp a$\egroup. Substituting auspiciously, we get.

\begin{displaymath}\bgroup\color{black}\left(\matrix{E_0a\pm Aa \cr E_0b\pm Ab}\right)=\left(\matrix{(E_0\pm A)a \cr (E_0\pm A)b}\right)\egroup\end{displaymath}

So the eigenstates are

\begin{eqnarray*}
E=E_0-A &\qquad & \left(\matrix{{1\over\sqrt{2}} \cr {1\over\s...
...d & \left(\matrix{{1\over\sqrt{2}} \cr -{1\over\sqrt{2}}}\right)
\end{eqnarray*}


The states are split by the interaction term.

Feynman goes on to further split the states by putting the molecules in an electric field. This makes the diagonal terms of the Hamiltonian slightly different, like a magnetic field does in the case of spin.

Finally, Feynman studies the effect of Ammonia in an oscillating Electric field, the Ammonia Maser.

Jim Branson 2013-04-22