Harmonic Oscillator Eigenket Notation

Question

I'm reading the $3^{\mathrm{rd}}$ edition of Sakurai and Napolitano's Modern Quantum Mechanics, and I have a brief question about the notation used to describe the eigenstates of the harmonic oscillator.

The authors define the usual creation ($a^{\dagger}$) and annihilation ($a$) operators, and define the operator $N = a^{\dagger}a$.

They then label the eigenvectors of $N$ by their corresponding eigenvalue $n$, i.e., $N|n\rangle = n|n\rangle$.

They go on to note that

\begin{equation*} Na|n\rangle = (n-1)a|n\rangle, \end{equation*}

(this is clear) and that this implies that $a|n\rangle$ and $|n-1\rangle$ are the same up to a multiplicative constant.

What is the best way to see this? If the eigenvectors of $N$ are labeled by their corresponding eigenvalue, is it true that

\begin{equation*} (n-1)a|n\rangle = a(n-1)|n\rangle = a|n-1\rangle = Na|n\rangle? \end{equation*}

Even in that case, how does it follow that $a|n\rangle$ and $|n-1\rangle$ are the same up to a multiplicative constant? I suspect this question might be deeper than just notation, but I'm not sure enough to be certain.

Edit: in response to the first (and second) downvote, how can I improve the question? From a mathematical perspective, it's not clear to me how the result I'm asking about holds. I come from a background of math and not physics, so I apologize for any foregone conclusions I might be missing here.

Answer 1

You may check that, since the 1D oscillator spectrum is non-degenerate, the eigenstates of N are unique, and normalized as $\langle n| n\rangle$.

As invited in the comment, define $$ |n\rangle = \frac{(a^\dagger)^n}{\sqrt{n!}} |0\rangle ~~~~ \leadsto ~~~~N|n\rangle = n|n\rangle \\ \langle n | aa^\dagger | n \rangle = \langle n|\left([a, a^\dagger] + a^\dagger a\right)| n \rangle = \langle n|(N + 1)|n\rangle = n + 1 \\ \Rightarrow a^\dagger | n\rangle = \sqrt{n + 1} | n + 1\rangle \\ \Rightarrow|n\rangle = \frac{a^\dagger}{\sqrt{n}} | n - 1 \rangle = \frac{(a^\dagger)^2}{\sqrt{n(n - 1)}} | n - 2 \rangle = \cdots = \frac{(a^\dagger)^n}{\sqrt{n!}}|0\rangle. $$

Answer 2

$[N,a] = -a$

then, $Na|n\rangle = aN|n\rangle - a|n\rangle = an|n\rangle - a|n\rangle = (n-1)a|n\rangle$

We can think of $N$ as operator, $a|n\rangle$ as eigenstate and $(n-1)$ as eigenvalue.

In other words, it means the 'lowering operator' lowers the number $n$ of eigenstate (oscillator).

In conclusion, $a|n\rangle \propto |n-1\rangle$

Answer 3

We label the eigenstates of the Harmonic Oscillator by the respective eigenvalues of the number operator $N=a^{\dagger}a$. So the state $|n\rangle$ means that $N|n\rangle = n |n\rangle$ etc. Similarly $N|n-1\rangle = (n-1) |n-1\rangle$. Since the eigenvalues of the harmonic oscillator (and therefore the number operators) are non-degenerate it means that any state $|\psi\rangle$ that maintains $N|\psi\rangle = (n-1)|\psi\rangle$ must be the state $|n-1\rangle$, up to some constant multiplicative factor. So if we have $N|\psi\rangle = (n-1)|\psi\rangle$ we can conclude that $|\psi\rangle = C |n-1\rangle$ with $C$ some complex number.

As was shown in other answers, the commutation relations imply that $Na|n\rangle = (n-1)a|n\rangle$ meaning that $a|n\rangle$ is the eigenstate of $N$ with eigenvalue $(n-1)$ up to some constant multiplicative factor. The only missing piece is that this state is unique, in other words that the spectrum is non-degenerate, and a proof for that you can see in the linked answer.