Radiation from a Harmonic Oscillator

Consider an electron in a one-dimensional harmonic oscillator potential aligned along the $x$-axis. According to Sect. 5.8, the unperturbed energy eigenvalues of the system are

$$E_n = (n+1/2) \hbar \omega_0,$$ (1135)

where $\omega_0$ is the frequency of the corresponding classical oscillator. Here, the quantum number $n$ takes the values $0,1,2,\cdots$. Let the $\psi_n(x)$ be the (real) properly normalized unperturbed eigenstates of the system.

Suppose that the electron is initially in an excited state: i.e., $n>0$. In principle, the electron can decay to a lower energy state via the spontaneous emission of a photon of the appropriate frequency. Let us investigate this effect. Now, according to Eq. (1131), the system can only make a spontaneous transition from an energy state corresponding to the quantum number $n$ to one corresponding to the quantum number $n'$ if the associated electric dipole moment

$$(d_x)_{n,n'} = \langle n\vert e x\vert n'\rangle = e\int_{-\infty}^{\infty} \psi_n(x) x \psi_{n'}(x) dx$$ (1136)

is non-zero [since $d_{if}\equiv (d_x)_{n,n'}^{ 2}$ for the case in hand]. However, according to Eq. (416),

$$\int_{-\infty}^\infty \psi_n x \psi_{n'} dx =\sqrt{\frac... ...sqrt{n} \delta_{n,n'+1} + \sqrt{n'} \delta_{n,n'-1}\right).$$ (1137)

Since we are dealing with emission, we must have $n>n'$. Hence, we obtain

$$(d_x)_{n,n'} = e \sqrt{\frac{\hbar n}{2 m_e \omega_0}} \delta_{n,n'+1}.$$ (1138)

It is clear that (in the electric dipole approximation) we can only have spontaneous emission between states whose quantum numbers differ by unity. Thus, the frequency of the photon emitted when the $n$th excited state decays is

$$\omega_{n,n-1} = \frac{E_n - E_{n-1}}{\hbar} = \omega_0.$$ (1139)

Hence, we conclude that, no matter which state decays, the emitted photon always has the same frequency as the classical oscillator.

According to Eq. (1131), the decay rate of the $n$th excited state is given by

$$w_n = \frac{\omega_{n,n-1}^{ 3} (d_x)_{n,n-1}^{ 2}}{3\pi \epsilon_0 \hbar c^3}.$$ (1140)

It follows that

$$w_n = \frac{n e^2 \omega_0^{ 2}}{6\pi \epsilon_0 m_e c^3}.$$ (1141)

The mean radiated power is simply

$$P_n = \hbar \omega_0 w_n = \frac{e^2 \omega_0^{ 2}}{6\pi \epsilon_0 m_e c^3} [E_n -(1/2) \hbar \omega_0].$$ (1142)

Classically, an electron in a one-dimensional oscillator potential radiates at the oscillation frequency $\omega_0$ with the mean power

$$P= \frac{e^2 \omega_0^{ 2}}{6\pi \epsilon_0 m_e c^3} E,$$ (1143)

where $E$ is the oscillator energy. It can be seen that a quantum oscillator radiates in an almost exactly analogous manner to the equivalent classical oscillator. The only difference is the factor $(1/2) \hbar \omega_0$ in Eq. (1142)--this is needed to ensure that the ground-state of the quantum oscillator does not radiate.