← Exercise library ← All exercises for this theme

Theme 3 — Postulates and Applications

The ammonia molecule: a two-level system

Keywords: two-level system · ammonia · tunnelling · oscillations · beats · Bohr frequency · time evolution

Exercise 1 : The ammonia molecule: a two-level system

In the ammonia molecule $\mathrm{NH}_3$ , the three hydrogen atoms form a triangle, and the nitrogen atom can lie on either side of the plane of this triangle. The molecule is modelled as a two-level system with orthonormal basis $(\ket G,\ket D)$ , where $\ket G$ (respectively $\ket D$ ) represents the nitrogen atom located to the left (respectively to the right) of the plane. In this basis, the Hamiltonian is

\begin{aligned} \hat H=\begin{pmatrix}E_0&-A\\-A&E_0\end{pmatrix}, \end{aligned}

where $E_0$ is real and $A>0$ represents the coupling due to the nitrogen atom tunnelling from one side to the other. Let $d$ denote the electric dipole moment of the molecule, which is $+d$ in the state $\ket G$ and $-d$ in the state $\ket D$ ; this quantity is represented by the observable $\hat D=d\bigl(\ket G\bra G-\ket D\bra D\bigr)$ .

Question 1

Determine the eigenenergies and eigenstates of

\hat H

. Are the states

\ket G

and

\ket D

stationary?

Solution

We write

\hat H=E_0\mathbf{1}-A\sigma_x

, where

\sigma_x

is the Pauli matrix. The eigenvalues of

\sigma_x

are

\pm1

, with eigenvectors

(\ket G\pm\ket D)/\sqrt2

. Hence the eigenstates of

\hat H

are

\ket S=\frac{\ket G+\ket D}{\sqrt2},\quad E_S=E_0-A, \qquad \ket{AS}=\frac{\ket G-\ket D}{\sqrt2},\quad E_{AS}=E_0+A .

This can be checked directly: $\hat H\ket S=\frac{1}{\sqrt2}\bigl[(E_0-A)\ket G+(-A+E_0)\ket D\bigr]=(E_0-A)\ket S$ . The ground level is the symmetric state $\ket S$ , and the separation between the two levels is $2A$ . The states $\ket G$ and $\ket D$ , in which the nitrogen atom is localised on one side, are not eigenstates of $\hat H$ : they are not stationary. Without tunnelling coupling ( $A=0$ ), they would be degenerate with energy $E_0$ ; the coupling lifts this degeneracy.

Question 2

At time

t=0

, the molecule is prepared in the state

\ket G

. Determine

\ket{\psi(t)}

, then the probability

P_D(t)

of finding the nitrogen atom on the right at time

t

Solution

We decompose the initial state in the eigenbasis of

\hat H

\ket G=(\ket S+\ket{AS})/\sqrt2

. Each eigenstate evolves by acquiring the phase

e^{-iEt/\hbar}

corresponding to its energy:

\ket{\psi(t)}=\frac{1}{\sqrt2}\Bigl(e^{-i(E_0-A)t/\hbar}\ket S+e^{-i(E_0+A)t/\hbar}\ket{AS}\Bigr) =\frac{e^{-iE_0t/\hbar}}{\sqrt2}\Bigl(e^{iAt/\hbar}\ket S+e^{-iAt/\hbar}\ket{AS}\Bigr).

Returning to the basis $(\ket G,\ket D)$ by substituting the expressions for $\ket S$ and $\ket{AS}$ gives

\begin{aligned} \ket{\psi(t)}&=e^{-iE_0t/\hbar}\left[\frac{e^{iAt/\hbar}+e^{-iAt/\hbar}}{2}\ket G+\frac{e^{iAt/\hbar}-e^{-iAt/\hbar}}{2}\ket D\right]\\ &=e^{-iE_0t/\hbar}\Bigl[\cos\frac{At}{\hbar}\ket G+i\sin\frac{At}{\hbar}\ket D\Bigr]. \end{aligned}

The probability of finding the nitrogen atom on the right is therefore

P_D(t)=\sin^2\frac{At}{\hbar}=\frac12\Bigl(1-\cos\frac{2At}{\hbar}\Bigr).

The nitrogen atom oscillates from one side of the plane of the hydrogen atoms to the other. The transfer is complete after a time $\pi\hbar/(2A)$ , and the period of the probability oscillations is $\pi\hbar/A$ .

Question 3

Calculate the expectation value

\langle\hat D\rangle(t)

of the dipole moment. At what frequency does it oscillate? What is the probability of measuring each of the two energies at time

t

Solution

We have

\langle\hat D\rangle=d\,(P_G-P_D)=d\bigl(\cos^2\frac{At}{\hbar}-\sin^2\frac{At}{\hbar}\bigr)=d\cos\frac{2At}{\hbar}

. The mean dipole moment oscillates at the angular frequency

\omega=2A/\hbar=(E_{AS}-E_S)/\hbar

, which is the Bohr frequency associated with the two levels. By contrast, the probabilities of measuring the two energies are

P(E_S)=|\braket{S}{\psi(t)}|^2=\frac12,\qquad P(E_{AS})=\frac12,

independent of time: this is the conservation of the energy distribution established in Lesson 1. The oscillations appear only for an observable, such as $\hat D$ , which does not commute with $\hat H$ and connects the two eigenstates: this is a quantum beat.

Question 4

The ammonia inversion frequency measured by microwave spectroscopy is

\nu\simeq23{,}9

GHz. Deduce the separation

2A

between the two levels, in electronvolts, as well as the time required for the nitrogen atom to pass from left to right. Take

h\simeq4{,}14\times10^{-15}

eV s.

Solution

The dipole oscillation frequency is

\nu=\omega/(2\pi)=2A/h

. It follows that

2A=h\nu\simeq4{,}14\times10^{-15}\times2{,}39\times10^{10}\simeq9{,}9\times10^{-5}\ \mathrm{eV}.

This separation is very small compared with electronic energies, which are of the order of an electronvolt, reflecting the weakness of the tunnelling coupling through the barrier formed by the plane of the hydrogen atoms. The time required to pass from left to right is half the period of the dipole oscillations:

\frac{\pi\hbar}{2A}=\frac{h}{4A}=\frac{1}{2\nu}\simeq\frac{1}{2\times2{,}39\times10^{10}}\ \mathrm{s}\simeq2{,}1\times10^{-11}\ \mathrm{s}.

This transition between $\ket S$ and $\ket{AS}$ is the one used to build the first maser, in 1954.