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In Section 2.3 of the second edition of Modern Quantum Mechanics (which discusses the harmonic oscillator), Sakurai derives the relation $$Na\left|n\right> = (n-1)a\left|n\right>,$$ and states that

this implies that $a\left|n\right>$ and $\left|n-1\right>$ are the same up to a multiplicative constant.

To my sensibilities, this is only implied if the $\lambda$-eigenspace of the number operator $N:=a^{\dagger}a$ corresponding to $\lambda=n-1$ is one-dimensional. If it is multidimensional, then we cannot say that $a\left|n\right>$ and $\left|n-1\right>$ are proportional. So (unless I've made some fundamental error) how do we know that the $\lambda$ eigenspaces of $N$ are one dimensional?

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1 Answers1

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OP wrote(v1):

So (unless I've made some fundamental error) how do we know that the $\lambda$-eigenspaces of $N$ are one dimensional?

Yes, OP is right. In general, we cannot know. There exist reducible unitary representations of the Heisenberg algebra $[a,a^\dagger]=1$, where the eigenvalues of $N$ are degenerated.

However, if one assumes that the ket Hilbert space is a non-trivial irreducible unitary representation of the Heisenberg algebra, then one may show that the eigenvalues of $N$ must be non-degenerated. See e.g. this Phys.SE answer.

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