Question About Harmonic Oscillator in Griffiths - E=0 or E=1/2hw?

[adartsesirhc]

I'm reading Griffiths', and I had a question about the harmonic oscillator.

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

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

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

 

[CompuChip]

Watch out! The statement a_- \psi_0 = 0 just says that when you apply the ladder operator to this state, you get nothing, i.e. the vacuum. Physicists have a tendency to use "natural" notation, and you will often see things like a_- |0\rangle = 0. Note that here the |0\rangle on the left hand side is a state (wavevector) and the 0 on the right hand side is the zero vector, which is not a state. Griffiths is just kind enough to call it \psi_0 instead of |0\rangle so you don't get completely confused

You say that a_- \psi_0 is a solution to the Schrodinger equation: that's right. But it isn't a physical solution. It's just the solution \psi = 0. The energy of \psi_0 is \frac12 \hbar \omega (formula 2.61) which can be seen by applying the Hamiltonian in the form (2.56) to this state and using that a_- annihilates it.

Hope that makes it more clear.

 

[adartsesirhc]

Thanks, CompuChip!