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# 12.9: Radiation from Harmonic Oscillator

Consider an electron in a one-dimensional harmonic oscillator potential aligned along the $$x$$-axis. According to Section [sosc], the unperturbed energy eigenvalues of the system are $E_n = (n+1/2)\,\hbar\,\omega_0,$ 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: that is, $$n>0$$. 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 Equation ([e3.115]), 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 $(d_x)_{n,n'} = \langle n|e\,x|n'\rangle = e\int_{-\infty}^{\infty} \psi_n(x)\,x\,\psi_{n'}(x)\,dx$ is non-zero [because $$d_{if}\equiv (d_x)_{n,n'}^{\,2}$$ for the case in hand]. However, according to Equation ([e5.xxx]), $\int_{-\infty}^\infty \psi_n\,x\,\psi_{n'}\,dx =\sqrt{\frac{\hbar}{2\,m_e\,\omega_0}}\left(\sqrt{n}\,\delta_{n,n'+1} + \sqrt{n'}\,\delta_{n,n'-1}\right).$ Because we are dealing with emission, we must have $$n>n'$$. Hence, we obtain $(d_x)_{n,n'} = e\,\sqrt{\frac{\hbar\,n}{2\,m_e\,\omega_0}}\,\delta_{n,n'+1}.$ 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 $\omega_{n,n-1} = \frac{E_n - E_{n-1}}{\hbar} = \omega_0.$ Hence, we conclude that, no matter which state decays, the emitted photon always has the same frequency as the classical oscillator.

According to Equation ([e3.115]), the decay rate of the $$n$$th excited state is given by $w_n = \frac{\omega_{n,n-1}^{\,3}\,(d_x)_{n,n-1}^{\,2}}{3\pi\,\epsilon_0\,\hbar\,c^{\,3}}.$ It follows that $w_n = \frac{n\,e^{\,2}\,\omega_0^{\,2}}{6\pi\,\epsilon_0\,m_e\,c^{\,3}}.$ The mean radiated power is simply $\label{e13.126} P_n = \hbar\,\omega_0\,w_n = \frac{e^{\,2}\,\omega_0^{\,2}}{6\pi\,\epsilon_0\,m_e\,c^{\,3}}\,[E_n -(1/2)\,\hbar\,\omega_0].$ Classically, an electron in a one-dimensional oscillator potential radiates at the oscillation frequency $$\omega_0$$ with the mean power $P= \frac{e^{\,2}\,\omega_0^{\,2}}{6\pi\,\epsilon_0\,m_e\,c^{\,3}}\,E,$ 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 Equation ([e13.126])—this is needed to ensure that the ground-state of the quantum oscillator does not radiate.
