this question might seem ridiculous to you, but I will really appreciate it if you could clarify the difference between a harmonic oscillator mode and a cavity mode. Below is my depiction.

For a harmonic oscillator, using annihilation and creation operator, we can write the Hamiltonian as $$H=\omega(a^\dagger a+\frac{1}{2}).$$ And its wave function corresponds to $|2\rangle$ is $$\psi(x)=\frac{1}{2\sqrt{2}} \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}e^{-\frac{m\omega x^2}{2\hbar}}H_2(\sqrt{\frac{m\omega}{\hbar}}x).$$ It will only take one argument to define the state of the system.

For a eigenmode of a cavity with length $L$, say $a_k$ mode, naively, the mode distribution is $$\phi_k(x)=\sqrt{\frac{2}{L}}\sin kx.$$ in which $k=\frac{2n\pi}{L}$. Its Hamiltonian can also be written as $$H=\omega_k(a^\dagger_k a_k+\frac{1}{2}).$$ Now if we have two photons that occupy this same mode. The wave function now would be $$\Phi(x_1,x_2)=\sqrt{2}\phi_k(x_1)\phi_k(x_2).$$ It takes two argument to define the wave function. So what is the difference between the two.

My thought is that in a cavity, we are dealing with a field, even though we write it as $a_k$, it is actually composed of infinite many oscillators (in real space). While the harmonic oscillator is just one oscillator. But I am uncertain about the details. Is there anyone could give me some explanation about the difference?

  • 2
    $\begingroup$ Indeed, each cavity mode behaves like a harmonic oscillator. And you can specify which cavity mode you're talking about by adding the subscript $k$. $\endgroup$ – knzhou Nov 6 '18 at 14:11
  • $\begingroup$ Each time you canonically quantize a field, the total Hamiltonian, i.e the spatial integration of Hamiltonian density reduces to that form with creation and annihilation operator. As the Hamiltonian comes to be kind of same so people bring the analogy with the quantum harmonic oscillator. $\endgroup$ – Galilean Nov 6 '18 at 16:50
  • $\begingroup$ @knzhou, But why for harmonic oscillator, it only takes one argument to define the wavefunction (even there are two excitations), but for cavity mode, it will take two arguments to define the wave function (for two excitations)? What is the difference between that? $\endgroup$ – yangcs11 Nov 7 '18 at 2:46
  • $\begingroup$ @yangcs11 A cavity mode is just a set of harmonic oscillators, so your statement is: "when one harmonic oscillator has two excitations, we can define the state by just the number "$2$". But when multiple harmonic oscillators have two excitations in total, to define the state we need to say where the excitations are." $\endgroup$ – knzhou Nov 7 '18 at 6:11
  • $\begingroup$ Could you explain what the "wavefunction" in the last equation is supposed to be? I'd be surprised if there was any physical meaning to that function. $\endgroup$ – Wolpertinger Dec 18 '18 at 15:46

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