Harmonic Oscillator Solution with Operators

We can solve the harmonic oscillator problem using operator methods. We write the Hamiltonian in terms of the operator

$$\bgroup\color{black}A\equiv \left(\sqrt{m\omega\over2\hbar}x+i{p\over\sqrt{2m\hbar\omega}}\right)\egroup$$

.

$$\bgroup\color{black}H={p^2\over 2m}+{1\over 2}m\omega^2x^2=\hbar\omega(A^\dagger A+{1\over 2})\egroup$$

We compute the commutators

$$\bgroup\color{black}[A,A^\dagger]={i\over 2\hbar}(-[x,p]+[p,x])=1 \egroup$$

$$\bgroup\color{black}[H,A]=\hbar\omega[A^\dagger A,A]=\hbar\omega[A^\dagger,A]A=-\hbar\omega A\egroup$$

$$\bgroup\color{black}[H,A^\dagger]=\hbar\omega[A^\dagger A,A^\... ...\hbar\omega A^\dagger[A,A^\dagger]=\hbar\omega A^\dagger\egroup$$

If we apply the the commutator $[H,A]$ to the eigenfunction $u_n$, we get $[H,A]u_n=-\hbar\omega Au_n$ which rearranges to the eigenvalue equation

$$\bgroup\color{black}H(Au_n)=(E_n-\hbar\omega)(Au_n).\egroup$$

This says that $(Au_n)$ is an eigenfunction of $H$ with eigenvalue $(E_n-\hbar\omega)$ so it lowers the energy by $\hbar\omega$. Since the energy must be positive for this Hamiltonian, the lowering must stop somewhere, at the ground state, where we will have

$$\bgroup\color{black}Au_0=0.\egroup$$

This allows us to compute the ground state energy like this

$$\bgroup\color{black}Hu_0=\hbar\omega(A^\dagger A+{1\over 2})u_0={1\over 2}\hbar\omega u_0\egroup$$

showing that the ground state energy is ${1\over 2}\hbar\omega$. Similarly, $A^\dagger$ raises the energy by $\hbar\omega$. We can travel up and down the energy ladder using $A^\dagger$ and $A$, always in steps of $\hbar\omega$. The energy eigenvalues are therefore

$$\bgroup\color{black}E_n=\left(n+{1\over 2}\right)\hbar\omega.\egroup$$

A little more computation shows that

$$\bgroup\color{black}Au_n=\sqrt{n}u_{n-1}\egroup$$

and that

$$\bgroup\color{black}A^\dagger u_n=\sqrt{n+1}u_{n+1}.\egroup$$

These formulas are useful for all kinds of computations within the important harmonic oscillator system. Both $p$ and $x$ can be written in terms of $A$ and $A^\dagger$.

$$\bgroup\color{black} x=\sqrt{\hbar\over 2m\omega}(A+A^\dagger) \egroup$$

$$\bgroup\color{black} p=-i\sqrt{m\hbar\omega\over 2}(A-A^\dagger) \egroup$$