Confusion in Quantum Harmonic Oscillator

Question

I am confused with the meaning of the particle number of a quantum harmonic oscillator. Classically, the Hamiltonian of harmonic oscillator in phase space is defined as follows: $$H = \frac{p^{2}}{2m} + \frac{1}{2} m \omega^{2} x^{2}$$

This Hamiltonian describes the total energy of an object in simple harmonic motion. By constructing the annihilation operator $\hat{a}$, we can diagonalize the Hamiltonian and it becomes the quantum Harmonic oscillator with Hamiltonian $\hat{H}$
$$\hat{H} = \hbar \omega \left( \hat{a}^{\dagger} \hat{a} + \frac{1}{2} \right)$$

We know that $\hat{n} = \hat{a}^{\dagger} \hat{a} $ is the number operator which counts the particle number in the system and eigenvalues of $n = 0,1,2,\ldots$. My confusion is that starting from a classical equation, the Hamiltonian describes the one-body total energy. However, when we diagonalize it and promote the Hamiltonian to quantum operator $\hat{H}$, the particle number of the system can be $n = 0,1,2,\ldots$ but not equal 1. Therefore, I want to know what is the exact meaning of particle number in quantum harmonic oscillator. Why it differs from the classical picture of Harmonic oscillator which describes one-particle total energy? I appreciate any comment.

Answer 1

There are two answers to this question. One of them corresponds to the fact that explicitly for a quantum harmonic oscillator, the eigenvalues $n$ of the number operator give you the number of phonons inside the oscillator, which is a fancy way of saying that it has been excited to the $n^{th}$ energy level.

That is to say that if the energy of your $1-D$ oscillator is $3\hbar \omega/2$, you say that if it was previously in the ground state, it has just absorbed one phonon of energy $\hbar \omega$.

The second one is more advanced, and is related to the quantisation of an electromagnetic field, which is generally taught at the introductory quantum optics level. The point is that electromagnetic field is quantised such that photons are modeled as quantum harmonic oscillators. In that case, the number of photons in the beam are given by $a^\dagger a$, each having energy $\hbar \omega$

In fact, a coherent beam of light with a fixed frequency is known as the coherent state in quantum optics, with its explicit functional form as:

$$|\alpha\rangle=\exp|-\alpha|^2/2 \sum_{n=0}^{\infty}\dfrac{\alpha^n}{n!} |n\rangle$$

where $\alpha$ is an eigenvalue of the annihilation operator $a$ with eigenvalue equation being $a|\alpha\rangle=\alpha|\alpha\rangle$ and $|n\rangle$ is the $n^{th}$ stationary state of the quantum harmonic oscillator.

Answer 2

Thank you for the solutions above. I have reviewed some literature and realise that the the number operator counts for the energy quanta( excitation modes) but not the classical one-body in SHO motion. In Ben Simons "Condensed Matter Field Theory"(p.22), it says that SHO of course is a single-particle problem. We can interpret the energy states $E_{n} = \hbar \omega \big( n + \frac{1}{2} \big)$ as an accumulation of $n$ elementary entities, or quasi-particles having energy $\hbar \omega$