Harmonic Oscillator Raising Operator

We wish to find the matrix representing the 1D harmonic oscillator raising operator.

We use the raising operator equation for an energy eigenstate.

\begin{displaymath}\bgroup\color{black}A^\dag u_n=\sqrt{n+1}u_{n+1}\egroup\end{displaymath}

Now simply compute the matrix element.

\begin{displaymath}\bgroup\color{black}A^\dag _{ij}=\langle i\vert A^\dag \vert j\rangle=\sqrt{j+1}\delta_{i(j+1)}\egroup\end{displaymath}

Now this Kronecker delta puts us one off the diagonal. As we have it set up, i gives the row and j gives the column. Remember that in the Harmonic Oscillator we start counting at 0. For i=0, there is no allowed value of j so the first row is all 0. For i=1, j=0, so we have an entry for \bgroup\color{black}$A^\dag _{10}$\egroup in the second row and first column. All he entries will be on a diagonal from that one.


\begin{displaymath}\bgroup\color{black}A^\dag =\left(\matrix{
0&0&0&0&...\cr
...
...r
0&0&0&\sqrt{4}&...\cr
...&...&...&...&... }\right)\egroup\end{displaymath}

Jim Branson 2013-04-22