Use Commutators to Derive HO Energies

We have computed the commutators

\begin{eqnarray*}[H,A]&=&-\hbar\omega A \\
{[H,A^\dagger]}&=&\hbar\omega A^\dagger \\

Apply \bgroup\color{black}$[H,A]$\egroup to the energy eigenfunction \bgroup\color{black}$u_n$\egroup.

\begin{eqnarray*}[H,A]u_n=-\hbar\omega Au_n \\
HAu_n-AHu_n=-\hbar\omega Au_n \\...
...(Au_n)=-\hbar\omega Au_n \\
H(Au_n)=(E_n-\hbar\omega)(Au_n) \\

This equation shows that \bgroup\color{black}$Au_n$\egroup is an eigenfunction of \bgroup\color{black}$H$\egroup with eigenvalue \bgroup\color{black}$E_n-\hbar\omega$\egroup. Therefore, \bgroup\color{black}$A$\egroup lowers the energy by \bgroup\color{black}$\hbar\omega$\egroup.

Now, apply \bgroup\color{black}$[H,A^\dagger]$\egroup to the energy eigenfunction \bgroup\color{black}$u_n$\egroup.

\begin{eqnarray*}[H,A^\dagger]u_n=\hbar\omega A^\dagger u_n \\
HA^\dagger u_n-A...
...r u_n) \\
H(A^\dagger u_n)=(E_n+\hbar\omega)(A^\dagger u_n) \\

\bgroup\color{black}$A^\dagger u_n$\egroup is an eigenfunction of \bgroup\color{black}$H$\egroup with eigenvalue \bgroup\color{black}$E_n+\hbar\omega$\egroup. \bgroup\color{black}$A^\dagger$\egroup raises the energy by \bgroup\color{black}$\hbar\omega$\egroup.

We cannot keep lowering the energy because the HO energy cannot go below zero.

\begin{displaymath}\bgroup\color{black} \langle \psi\vert H\vert\psi\rangle=
{1\... 2}m\omega^2\langle x\;\psi\vert x\;\psi\rangle\geq 0 \egroup\end{displaymath}

The only way to stop the lowering operator from taking the energy negative, is for the lowering to give zero for the wave function. Because this will be at the lowest energy, this must happen for the ground state. When we lower the ground state, we must get zero.

\begin{displaymath}\bgroup\color{black} Au_0=0 \egroup\end{displaymath}

Since the Hamiltonian contains \bgroup\color{black}$A$\egroup in a convenient place, we can deduce the ground state energy.

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

The ground state energy is \bgroup\color{black}$E_0={1\over 2}\hbar\omega$\egroup and the states in general have energies

\begin{displaymath}\bgroup\color{black} E=\left(n+{1\over 2}\right)\hbar\omega \egroup\end{displaymath}

since we have shown raising and lowering in steps of \bgroup\color{black}$\hbar\omega$\egroup. Only a state with energy \bgroup\color{black}$E_0={1\over 2}\hbar\omega$\egroup can stop the lowering so the only energies allowed are

\begin{displaymath}\bgroup\color{black} E=\left(n+{1\over 2}\right)\hbar\omega .\egroup\end{displaymath}

It is interesting to note that we have a number operator for \bgroup\color{black}$n$\egroup

H&=&\left(A^\dagger A+{1\over 2}\right)\hbar\omega \\
N_{op}&=&A^\dagger A \\
H&=&(N_{op}+{1\over 2})\hbar\omega \\

Jim Branson 2013-04-22