3

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.

Community
  • 1
Edward Hughes
  • 7,876
  • You know that Fock states $|n\rangle$ generated by applying the creation operator $n$ times to the vacuum have different energies, so if you can show that these states form a complete basis then you're done. – Mark Mitchison Dec 12 '12 at 01:25
  • @MarkMitchison: I'm a little rusty on my foundational quantum mechanics - could you expand on the relevance of the Hermite polynomials (possibly in the context of a full answer). Many thanks! – Edward Hughes Dec 12 '12 at 01:28
  • The Hermite polynomials are the wavefunctions corresponding to the eigenstates of the harmonic oscillator. You can find them recursively from the ground state by using the creation and annihilation operators in the position representation. For example, the vacuum (ground) state is defined by $$ \hat{a}|0\rangle = \sqrt{\frac{m\omega}{2\hbar}}\left(\hat{x} + \frac{i}{m\omega}\hat{p}\right)|0\rangle = 0 $$ $$ \Rightarrow \sqrt{\frac{m\omega}{2\hbar}}\left(x + \frac{\hbar}{m\omega}\frac{\partial}{\partial x}\right) \psi_0(x) = 0 $$ $$ \Rightarrow \psi_0(x) \sim e^{-\frac{m\omega x^2}{2\hbar}} $$ – Mark Mitchison Dec 12 '12 at 11:51
  • @MarkMitchison: okay, so showing that the Hermite polynomials span is sufficient, because each Hermite polynomial has a different $N$ eigenvalue, and if they span then there couldn't possibly be degeneracy for any eigenvalue. But proving that they span is standard functional analysis, so we're done. Am I correct? In that case the key is moving to explicit wavefunctions in the position representation, since the algebra is insufficient to get this result. – Edward Hughes Dec 13 '12 at 13:09
  • Yes, that was my argument. I'm not sure it's correct to say that you cannot do it with algebra; I seem to remember the last time I did this with creation and annihilation operators I could satisfy myself that nothing apart from Fock states could be eigenstates of the SHO Hamiltonian. Maybe I was being lazy though... In any case, I'm sure there is a way to do it without invoking the position representation: there is nothing special about the position representation! – Mark Mitchison Dec 13 '12 at 15:37
  • @MarkMitchison: but surely there is something you gain from going to the position representation, namely more information about what the states look like. Once we know concretely that the states are the Hermite polynomials and the Hilbert space is the space of normalisable wavefunctions then we are done. But isn't it precisely the position representation which gives this to us? I guess momentum space would work just as well, but am I wrong in thinking that we gain information by thinking about wavefunctions rather than just states? Thanks for all your help! – Edward Hughes Dec 13 '12 at 23:48
  • Yes, but you just gain information about position! In the momentum basis you gain information about the momentum distribution (the wavefunctions are the same in either basis for the harmonic oscillator because the Hamiltonian is symmetric under $x \leftrightarrow p$ in an appropriate system of units). Any basis will do, position has no special status. The ladder operators are just a canonical rotation of position and momentum operators after all, so you could equally go to the coherent state basis; the Hamiltonian doesn't care what basis you use. – Mark Mitchison Dec 14 '12 at 00:39
  • However, you may have a point that projecting onto a basis gives you something useful. My intuition is that it does not really give you anything more than the algebra does, since the form of the position space (say) wavefunctions ultimately depends on the algebraic relationships between the position operator and the Hamiltonian. Going to the position basis might be the quickest way of doing it though. – Mark Mitchison Dec 14 '12 at 00:47
  • Possible duplicate: http://physics.stackexchange.com/q/23028/2451 , https://physics.stackexchange.com/q/39307/2451 and links therein. – Qmechanic Jan 11 '13 at 09:26

2 Answers2

1

I don't just use operator algebra here, so you probably already know the following, but just in case it helps:

  1. Eigenstates of the number operator are eigenstates of the Hamiltonian, since $\hat{H}=\hbar\omega\left(\hat{N}+\frac{1}{2}\right)$.
  2. Bound states in one dimension are non-degenerate (see e.g. http://arxiv.org/abs/0706.1135 for conditions). The proof given in that link hinges on choosing two bound states with the same energy and using the Schrodinger equation to easily show that the Wronskian $W=\psi_2\psi'_1-\psi_1\psi'_2$ is constant. By evaluating $W$ at $\infty$, the constant must be zero (with conditions), so that the states must be proportional, and therefore the same after normalisation.
Ramashalanka
  • 1,274
  • 10
  • 13
1

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 $\left<x\right|\hat{a}\left|0\right>=0$ 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.

Michael
  • 16,512
  • 2
  • 49
  • 69