Quantum Harmonic Oscillator Number & Ladder Operators

A quantum harmonic oscillator is similar in many ways to a classical one; it is simply a particle that undergoes repetitive motion, bound by a potential with an equilibrium point. The major difference between quantum and classical harmonic oscillators, however, is that in quantum systems, particles can occupy only certain discrete energy levels. Here, the number operator and ladder operators — operators concerned with energy levels — of quantum harmonic oscillators are derived.

Number Operator

In quantum physics, the energy levels of a system are numbered by quantum numbers, $n$, which in the case of harmonic oscillators are $n\in \{0,1,2,...\}$. The number operator, $N$, for quantum harmonic oscillators is an operator which returns the quantum number, $n$, for particles of the corresponding energy level. In other words, $N\lvert n \rangle = n \lvert n\rangle$, where $\lvert n \rangle$ is an eigenstate with an eigenvalue of quantum number $n$. To derive the number operator, start with the one-dimensional Hamiltonian, $\hat{H}$, for harmonic oscillators (derived here in this post on zero-point energy):

\begin{equation}
\hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2\hat{x}^2
\end{equation}

where $m$ is the mass of the particle, $\omega$ is the angular frequency of the particle's wave function, and $\hat{p}$ and $\hat{x}$ are respectively the one-dimensional momentum operator and the one-dimensional position operator. Quantum wave functions share many of the same properties as electromagnetic waves. For electromagnetic waves, the energy of a photon, a single discrete quanta of electromagnetic radiation, is given by $\hbar\omega$, where $\omega$ is the angular frequency of the electromagnetic wave. Similarly, for a quantum harmonic oscillator, the energy increment between energy levels is also given by $\hbar\omega$, where $\omega$ is now the angular frequency of the particle's wave function. Because the number operator returns the quantum number of the particle, it follows that $\hat{H}=\hbar\omega(N + 1/2)$, such that when $n=0$, the particle has the zero-point energy $E_0 = \hbar\omega/2$.

\begin{equation}
\hbar\omega\left(N + \frac{1}{2}\right) = \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2\hat{x}^2
\end{equation}
\begin{equation}
N = \frac{\hat{p}^2}{2m\hbar\omega} + \frac{m\omega\hat{x}^2}{2\hbar} - \frac{1}{2}
\end{equation}
\begin{equation}
N = \frac{1}{2} \left( \sqrt{\frac{m\omega}{\hbar}}\hat{x} - i \sqrt{\frac{1}{m\hbar\omega }}\hat{p}\right) \left( \sqrt{\frac{m\omega }{\hbar}}\hat{x} + i\sqrt{\frac{1}{m\hbar\omega }}\hat{p} \right) - \frac{i}{2\hbar}\left[\hat{x},\hat{p}\right] - \frac{1}{2}
\end{equation}
\begin{equation}
N = \frac{m\omega}{2\hbar} \left( \hat{x} - \frac{i}{m\omega}\hat{p}\right) \left( \hat{x} + \frac{i}{m\omega}\hat{p} \right)
\end{equation}

Ladder Operators

Ladder operators refer to the lowering and raising operators which, when applied to eigenstates, respectively lower and raise the eigenvalue of some other operator, in the case of harmonic oscillators, the number operator. These lowering and raising operators are also respectively known as the annihilation and creation operators. To be able to lower the eigenvalue of the number operator by one, the lowering operator, $a$, must satisfy the following equation:

\begin{equation}
Na\lvert n \rangle = (n - 1)a\lvert n \rangle
\end{equation}
\begin{equation}
Na\lvert n \rangle = aN\lvert n \rangle - a\lvert n \rangle
\end{equation}
\begin{equation}
[N,a] = -a
\end{equation}

Looking back at equation $(3)$, note that instead of equation $(4)$, it can also be written that:

\begin{equation}
N = \frac{1}{2} \left( \sqrt{\frac{m\omega}{\hbar}}\hat{x} + i \sqrt{\frac{1}{m\hbar\omega}}\hat{p}\right) \left( \sqrt{\frac{m\omega}{\hbar}}\hat{x} - i\sqrt{\frac{1}{m\hbar\omega}}\hat{p} \right) - \frac{i}{2\hbar}\left[\hat{p},\hat{x}\right] - \frac{1}{2}
\end{equation}
\begin{equation}
N = \frac{m\omega}{2\hbar} \left( \hat{x} + \frac{i}{m\omega}\hat{p}\right) \left( \hat{x} - \frac{i}{m\omega}\hat{p} \right) - 1
\end{equation}

Comparing the above equation to equation $(5)$, we thus have:

\begin{equation}
\frac{m\omega}{2\hbar} \left( \hat{x} - \frac{i}{m\omega}\hat{p}\right) \left( \hat{x} + \frac{i}{m\omega}\hat{p} \right) - \frac{m\omega}{2\hbar} \left( \hat{x} + \frac{i}{m\omega}\hat{p}\right) \left( \hat{x} - \frac{i}{m\omega}\hat{p} \right)= - 1
\end{equation}
\begin{equation}
\sqrt{\frac{m\omega}{2\hbar}}\left( \hat{x} - \frac{i}{m\omega}\hat{p}\right) \sqrt{\frac{m\omega}{2\hbar}} \left( \hat{x} + \frac{i}{m\omega}\hat{p} \right)a - \sqrt{\frac{m\omega}{2\hbar}} \left( \hat{x} + \frac{i}{m\omega}\hat{p}\right) \sqrt{\frac{m\omega}{2\hbar}}\left( \hat{x} - \frac{i}{m\omega}\hat{p} \right)a= - a
\end{equation}

It now becomes apparent, considering equation $(5)$ and the above equation, that to satisfy equation $(8)$, $a$ should be defined as:

\begin{equation}
a = \sqrt{\frac{m\omega}{2\hbar}} \left( \hat{x} + \frac{i}{m\omega}\hat{p}\right)
\end{equation}

Notice that since $N$ is a Hermitian operator, taking the Hermitian adjoint of equation $(8)$ gives:

\begin{equation}
[N,{a}^\dagger] = {a}^\dagger
\end{equation}
\begin{equation}
N {a}^\dagger \lvert n\rangle = {a}^\dagger N\lvert n\rangle + {a}^\dagger\lvert n\rangle
\end{equation}
\begin{equation}
N {a}^\dagger \lvert n\rangle = (n+1){a}^\dagger \lvert n\rangle
\end{equation}

Interestingly, $a^\dagger$, the Hermitian adjoint of $a$, is found to be the raising operator. Thus, the raising operator is defined as:

\begin{equation}
a^\dagger = \sqrt{\frac{m\omega}{2\hbar}} \left( \hat{x} - \frac{i}{m\omega}\hat{p}\right)
\end{equation}

Applying the Ladder Operators

By equations $(5)$ and $(10)$, the number operator can be written in terms of the ladder operators like so:

\begin{align}
N &=a^\dagger a \\
N &=aa^\dagger -1
\end{align}

The above relations suggest that applying $a$ and $a^\dagger$ respectively to an eigenstate not only lowers and raises the eigenstate's energy level, but also multiplies the eigenstate by some constant, $C$ and $D$ respectively.

\begin{align}
a\lvert n \rangle &= C \lvert n-1 \rangle \\
a^\dagger\lvert n \rangle &= D \lvert n+1 \rangle
\end{align}
\begin{align}
\langle n \rvert a^\dagger a\lvert n \rangle &= \langle n-1 \rvert|C|^2\lvert n-1 \rangle \\
\langle n \rvert a a^\dagger\lvert n \rangle &= \langle n+1 \rvert|D|^2\lvert n+1 \rangle
\end{align}
\begin{align}
\langle n \rvert N \lvert n \rangle &= \langle n-1 \rvert|C|^2\lvert n-1 \rangle \\
\langle n \rvert (N + 1) \lvert n \rangle &= \langle n+1 \rvert|D|^2\lvert n+1 \rangle
\end{align}
\begin{align}
n \langle n \mid n \rangle &= |C|^2 \langle n-1 \mid n-1 \rangle \\
(n+1) \langle n \mid n \rangle &= |D|^2 \langle n+1 \mid n+1 \rangle
\end{align}
\begin{align}
n &= |C|^2 \\
n+1 &= |D|^2
\end{align}
\begin{align}
C &= \sqrt{n} \\
D &= \sqrt{n+1}
\end{align}

Therefore, it is derived that the applications of the ladder operators to state vector $\rvert n\rangle$ give:

\begin{align}
a\rvert n \rangle &= \sqrt{n}\rvert n-1 \rangle \\
a^\dagger\rvert n \rangle &= \sqrt{n+1}\rvert n+1 \rangle
\end{align}