# The role of the harmonic oscillator eigenfunctions in quantum optics

In quantum optics we quantize the electromagnetic field and describe it using the harmonic oscillator model and the formalism of annihilation and creation operators. For the electric field operator we find for example for a single mode $$E(z,t) \propto \sin(kz)\cdot q(t)$$ and a similar expression for the magnetic field which has a time dependence $$p(t)$$. From the operators $$q$$ and $$p$$ we define the annihilation and creation operators and the quadrature operators.

However, I have been wondering about the role of the eigenfunctions of the harmonic oscillators for this model.

In the classical harmonic oscillator $$p$$ and $$q$$ correspond to the momentum and position of the particle over time, whereas they are only operators in quantum optics. The states of the electromagnetic field, for example $$|n \rangle$$. "We" never actually looked at the wave function that would correspond to that state. Is there a physical meaning to the wave functions?

• Indeed, the space of an electromagnetic field is just a label in field quantization and so distinctly different than its dynamical variables quantized. A field is basically an infinity of coupled oscillators so their notional q,p are abstract formal variables, bearing no relationship to your z, for example, beyond being somehow labeled by it. Jan 13, 2019 at 0:19

$$\newcommand{\Ket}{\left|#1\right>}$$ $$\newcommand{\Bra}{\left<#1\right|}$$

The number states, or Fock states, are eigenstates of both the Hamiltonian of the quantum harmonic oscillator and the number operator $$\hat{n}=\hat{a}^\dagger \hat{a}$$. They are thus states of definite energy and definite number of photons.

Now, let's try to calculate the mean electric field and variance of our number state $$\Ket{n}$$. You can do that by promoting the electric field to an operator $$\begin{equation} \hat{E}=E_0(\hat{a}e^{i\phi}+c.c.) \end{equation}$$ On a Fock state you get $$\begin{equation} \langle E\rangle=\langle n| \hat{E}|n\rangle=0 \\ (\Delta E)^2=\langle E^2\rangle-\langle E\rangle^2=E_0^2(2n+1) \end{equation}$$ We can see that we are far away from the classical regime. Indeed the mean of the electric field vanishes and a measurement at any time may lead to any outcome, determined by $$\Delta E$$. This reflects the fact that a Fock state lacks any information about the phase of the field.

To recover a classical oscillation behavior we may want to introduce coherent states, that are called "the most classical states of light", that are eigenstates of $$\hat{a}$$ such that $$\begin{equation} \hat{a}\Ket{\alpha}=a\Ket{\alpha} \end{equation}$$ with complex numbers such that $$\alpha=|\alpha|e^{i\theta}$$. Now the mean electric field reads $$\begin{equation} \langle E\rangle=\langle \alpha| \hat{E}|\alpha\rangle=2E_0|\alpha|cos(\theta+\phi) \end{equation}$$ which reflects the sinusoidal behavior of the classical electric field.

In the picture you can see the behavior of the electric field for the vacuum state $$\Ket{0}$$ and a coherent state $$\Ket{\alpha}$$ (unfortunately I don't have the source of the picture, I saved it on my computer in my Quantum Optics course folder). I hope I got the point of your question.

• That comes quite close to what I was asking but it is not the "full answer"... Jan 13, 2019 at 10:53
• Clarify then what the full answer should include and if I have the knowledge I will update my answer. What else are you asking specifically? Jan 13, 2019 at 13:47