# Confusion in Quantum Harmonic Oscillator

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.

• Yes, it is one-body Hamiltonian. But I am confused with that if the eigenvalues of number operator $\hat{n}$ is $0,1,2,\ldots,N$, why the Hamiltonian is called one particle Hamiltonian. I am new to condensed matter physics and I always got confused with the definition of many body Hamiltonian or single-particel Hamiltonian. Commented Jul 15, 2021 at 14:39
• Eigen values of the number operator denotes the excitation states, not the particle number Commented Jul 15, 2021 at 14:40
• Hi, @KP99, thanks for your comment. Yes, in the sense of simple quantum harmonic oscillator, I agree that the number operators counts for the excitation states. However, in QFT[Prof. D. Tong Notes (p.31) ](damtp.cam.ac.uk/user/tong/qft/qft.pdf), the number operator commutes with the Hamiltonian ensuring that particle number is conserved. That's why I am confused with the meaning of particle number in that simple model. Commented Jul 15, 2021 at 14:58
• In the comment after eqn (2.44), he mentions that ket $|p\rangle$ refers to eigenstate of single particle. He then carries over the same notations in multi-particle states and defines the number operator N which measures sum of excitation states for each particle for a given multi-particle state. Actually here N refers to the number of "virtual particles" and not the real particles. Commented Jul 15, 2021 at 15:30
• Thanks @KP99. I have the same point of view after reading M. Schwartz QFT. The number of particle in field theory sense means the total number of excitation modes. If we treat the excitation modes as the particle number, then $n$ really means the number operator which governs the particle number of the system, which is different from the classical sense. Commented Jul 15, 2021 at 15:35

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.

• Thanks for your comment @IndischerPhysiker. Yes, I do agree with your second answer by using EM field quantisation. When applying a creation operator, we actually create(excite) one photon from the vacuum. Commented Jul 15, 2021 at 15:15

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 $$\hbar \omega$$