The Harmonic Oscillator in One Dimension

Next we solve for the energy eigenstates of the harmonic oscillator potential \bgroup\color{black}$V(x)={1\over 2}kx^2={1\over 2}m\omega^2x^2$\egroup, where we have eliminated the spring constant \bgroup\color{black}$k$\egroup by using the classical oscillator frequency \bgroup\color{black}$\omega=\sqrt{k\over m}$\egroup. The energy eigenvalues are

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

The energy eigenstates turn out to be a polynomial (in \bgroup\color{black}$x$\egroup) of degree \bgroup\color{black}$n$\egroup times \bgroup\color{black}$e^{-m\omega x^2/\hbar}$\egroup. So the ground state, properly normalized, is just

\begin{displaymath}\bgroup\color{black}u_0(x)=\left({m\omega\over\pi\hbar}\right)^{1\over 4} e^{-m\omega x^2/\hbar}.\egroup\end{displaymath}

We will later return the harmonic oscillator to solve the problem by operator methods.



Jim Branson 2013-04-22