Proportionality of states in quantum harmonic oscillator

Question

What is the justification for $a_{\pm} \psi_{n}$ being proportional to $\psi_{n\pm1}$ in a quantum harmonic oscillator?

Here $a_{\pm}$ is the raising/lowering ladder operator.

Answer 1

In short, we say that $\hat a|n⟩$ is a multiple of $|n-1⟩$ because it's an eigenstate of $\hat n=\hat a^\dagger \hat a$ with eigenvalue $n$. That is, $$ \hat n\left(\hat a|n⟩\right)=\hat a^\dagger \hat a\hat a|n⟩=\left(\hat a\hat a^\dagger-1\right)\hat a|n⟩=\hat a\left(\hat a^\dagger\hat a-1\right)|n⟩ =\hat a(n-1)|n⟩=(n-1)\left(\hat a|n⟩\right), $$ where we've used the commutator $[\hat a,\hat a^\dagger]=\hat a\hat a^\dagger-\hat a^\dagger \hat a=1$ to switch the two operators.

For a simple harmonic oscillator the eigenvalues are nondegenerate, so any two eigenstates with the same eigenvalue give must be proportional.

An exactly analogous argument works for $\hat a^\dagger |n⟩$.

(On the other hand, you may wonder why we know the eigenvalues are nondegenerate. This is pretty much the assumption you bring in when you say 'simple' harmonic oscillator, but it's got some maths behind it. Specifically, the same ladder argument I showed above allows you to reduce the degeneracy of each eigenvalue to the degeneracy of the ground state, and to form unambiguous chains of $\hat n$ eigenstates that correspond to a specific ground state. The ground state might indeed be degenerate, if e.g. there are other coordinates, particles, or spin degrees of freedom involved, but if you find a way to tag the different ground states, then you can unambiguously tag all the other eigenstates, so you're left with a collection of separate simple oscillators.)

Answer 2

(incomplete answer...)

Mathematically one tries to build a representation of an algebra generated by a unit element $1$ and the two ladder operator which satisfy their usual commutation relations.

The Hamiltonian is an element of the algebra as a sum of product of the $a, a^{\dagger}$ and the unit. So is the "Number of particle operator N" and if I remember well the $\psi_n$ are defined as a normalized eigenvector of N for the eigenvalue n.

If one can proove that each eigenspace is only one dimensional, then if one finds an eigenvector for the eigenvalue $n+1$ it is necessarily proportional to $\psi_{n+1}$. I guess that your lecture notes showed that $a^{\pm}\psi_n$ is an eigenvector of N with eigenvalue $n\pm 1$

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More concretely, representation of an algebra means that one wants to understand each element of the algebra as a matrix (or the infinite dimension version: operator) acting on some vector space. When one knows that an operator is self-adjoint (e.g. N) then one can find a basis of the vector space composed entirely with the eigenvectors of the operator.

Just to be very explicit: the $a$ and $a^{\dagger}$ are completely determined if one knows their action on the vectors of a basis (one can write its associated "matrix"), which the $\psi_n$ are as eigenvectors of a self-adjoint operator.