quantum mechanics - Non-Degeneracy of Eigenvalues of Number Operator for Simple Harmonic Oscillator

Possible Duplicate:
Proof that the One-Dimensional Simple Harmonic Oscillator is Non-Degenerate?

I'm trying to convince myself that the eigenvalues $n$ of the number operator $N=a^{\dagger}a$ for the quantum simple harmonic oscillator are non-degenerate.

I can't see a way to do this just given the operator algebra for creation and annihilation operators. Is there an easy way to show this, or does it depend on something deeper? I'd appreciate any detailed argument or insight! Many thanks in advance.

Answer

Recall $ \hat{H} = \left( \hat{N} + \frac{1}{2} \right) $ and $ \left[ \hat{a}, \hat{a}^\dagger \right] = 1 $ (dropping $\hbar$ and $\omega$).

1. 
   

   Assume the ground state $\left|0\right>$ is non-degenerate. You can prove this by solving $\left=0$ in position representation, but I don't know how to do it algebraically. The rest of the proof is algebraic.

   
   


2. 
   

   Let the first excited state be $k$-fold degenerate: $\left|1i\right>$, $i=1,\ldots,k$, where $\left|1i\right>$ orthonormal. Then, by the algebra we have $$ \hat{a} \left|1i\right> = \left|0\right> $$ and $$ \hat{a}^\dagger \left|0\right> = \sum_i c_i \left|1i\right> $$ where $ \sum_i c_i^\star c_i = 1 $.

   
   


3. 
   

   Now, for these states to be eigenstates of $\hat{H}$ with energy $\frac{3}{2}$ they must be eigenvalues of $\hat{N}$ with eigenvalue 1. This requires

   
   
   

$$ \begin{matrix} \hat{N}\left|1i\right> &=& \hat{a}^\dagger \hat{a}\left|1i\right>\\ &=& \hat{a}^\dagger \left|0\right> \\ \left|1i\right> &=& \sum_j c_j \left|1j\right> \end{matrix}$$

This must hold for all $i$, which leads to an immediate contradiction (no solution for the $c_i$) unless $k=1$.

Induction proves non-degeneracy for the higher states.