Quantum mechanical harmonic oscillator - where does the number operator come from?

Question

I have recently been learning about the quantum harmonic oscillator and how it is described in the language of ladder operators. At the moment the logic behind the number operator seems incomplete to me. As I understand it, the Hamiltonian can easily be shown to be

$$ \hat{H} = \hbar\omega(\hat{a}^\dagger\hat{a} + \frac{1}{2}), $$

and then by comparison with the well-known energy spectrum of a harmonic oscillator, this suggests that the eigenvalues of the number operator $\hat{N} = \hat{a}^\dagger\hat{a}$ give the (integer) energy level of the oscillator. But in actually proving that $\hat{N}$ gives the appropriate eigenvalues, the general approach seems to be to use the normalised ladder operator relations:

$$ \hat{a}|n\rangle = \sqrt{n}|n-1\rangle \\ \hat{a}^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle. $$

However, to my knowledge these normalising prefactors themselves come from applying the number operator when attempting to normalise the result of an application of a ladder operator!

I would like to know if there is a method of acquiring these normalisations on the ladder operators without assuming the number operator, or alternatively if there is a way of showing the number operator works that doesn't rely on already knowing the normalisations.

Answer 1

You've got the logic of the derivation wrong, I think. I recommend re-reading it, but here's an outline of the process, and you'll see that it's not circular:

1. Show that $\hat{H}$ can be factored as $\hat{a}^{\dagger}\hat{a}+\hat{I}/2$, where the raising and lowering operators are defined in terms of $\hat{x}$ and $\hat{p}$ as usual. Note that the number operator and $\hat{H}$ clearly commute, and so they share an eigenbasis.

2. Derive a set of commutation relations using the commutation relations for $\hat{x}$ and $\hat{p}$. The results are $[\hat{a},\hat{a}^{\dagger}]=1$, $[\hat{a},\hat{a}^{\dagger}\hat{a}]=\hat{a}$, $[\hat{a}^{\dagger},\hat{a}^{\dagger}\hat{a}]=-\hat{a}^{\dagger}$, etc.

3. Using the commutation relations directly, show that $\hat{a}|n\rangle$ is an eigenvector of $\hat{a}^{\dagger}\hat{a}$ with eigenvalue $n-1$. We don't know yet that $n$ is an integer, and (crucially, to show that we are not making a circular argument as the OP suggests) we don't know yet what the constant of proportionality is. That is, we know that $\hat{a}|n\rangle = C_n|n-1\rangle$, but we don't know what $C_n$ is.

4. Argue that there must be a unique ground state $|G\rangle$ that is annihilated by $\hat{a}$, and use this to show that the eigenvalue is zero, i.e., $\hat{a}^{\dagger}\hat{a}|G\rangle = 0$. From this, we can conclude that the $n$'s are non-negative integers.

5. Assuming that by $|n\rangle$ we mean the normalized eigenstates, calculate $\langle n | \hat{a}^{\dagger}\hat{a} |n \rangle$ in two ways to determine the value of $C_n$. Similarly, we can figure out the action of $\hat{a}^{\dagger}$ on $|0\rangle$.

Answer 2

First of all, I stress that the ladder operators are not defined as $$ \hat{a}|n\rangle = \sqrt{n}|n-1\rangle \\ \hat{a}^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle. $$ Rhey are instead defined in terms of the position and momentum operators with suitable normalization factors in order to find the canonical commutation relations $$[a,a^\dagger]= I\:.$$

However, in order to dispel any doubts about the existence of any logical loop involving ladder operators, I point out an independent way to find the eigenvalues $n= 0,1,2,\ldots$ of the number operator $N$. This procedure does not pass through the ladder operators (and their normalization).

Consider the Hermite function (see Applications here: Hermite polynomials) $$\psi_n : \mathbb{R} \to \mathbb{C}\:,$$ where $$\psi_n(x) := (-1)^n(2^n n!\sqrt{\pi})^{-1/2}e^{x^2/2}\frac{d^n}{dx^n} e^{-x^2/2}\:,\quad n=0,1,2,\ldots\:.$$

These functions have the following nice properties.

1. They satisfy $L^2(\mathbb{R},dx)$ orthogonality relations $$\int_{\mathbb{R}} \overline{\psi_n(x)}\psi_n(x) dx = \delta_{mn}\:.$$

2. $\{\psi_n\}_{n \in \mathbb{N}}$ is a Hilbert basis of $L^2(\mathbb{R},dx)$

3. They satisfy the Hermite equation which I re-arrange suitably: $$-\frac{1}{2}\frac{d^2\psi_n}{dx^2} + \frac{1}{2}x^2 \psi_n(x) - \frac{1}{2}\psi_n(x)= n\psi_n(x)\:.$$

As an immediate consequence of 3, we see that, defining $$N := -\frac{1}{2}\frac{d^2}{dx^2} + \frac{1}{2}x^2 - \frac{1}{2}\:,\tag{1}$$ for instance on the Schwartz space, this operator turns out to be symmetric and the $\psi_n$ are a Hilbert basis of eigenvectors with eigenvalues $n\in \mathbb{N}$. This is sufficient to prove (by the Nelson theorem and the uniqueness of spectral decomposition) that $N$ is (essentially) selfadjoint on that domain and its spectrum is a point spectrum which exactly coincides with $\mathbb{N}$.

It is clear that $N$ defined in (1) is the number operator when $\hbar=\omega=1$. The general case can be treated similarly with easy re-adaptations.