I'm reading Griffiths', and I had a question about the harmonic oscillator.(adsbygoogle = window.adsbygoogle || []).push({});

Griffiths solves the Schrodinger equation using ladder operators, and he then says that there must be a "lowest rung," or [tex]\psi_{0}[/tex], such that a_[tex]\psi_{0}[/tex] = 0. I'm guessing this also means that E = 0 for a_[tex]\psi_{0}[/tex], which is why it wouldn't be allowable.

Because a_[tex]\psi_{0}[/tex] is (mathematically, at least) a solution to the Schrodinger equation, it corresponds to the energy ([tex]E_{0}[/tex] - [tex]\hbar\omega[/tex]), which equals zero. Doesn't this mean that [tex]E_{0}[/tex] = [tex]\hbar\omega[/tex]?

But then this conflicts with Griffith's statement that [tex]E_{0}[/tex] = [tex]\frac{1}{2}\hbar\omega[/tex]. HELP!

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Harmonic Oscillator

**Physics Forums | Science Articles, Homework Help, Discussion**