The standard treatment of the one-dimensional quantum simple harmonic oscillator (SHO) using the raising and lowering operators arrives at the countable basis of eigenstates $\{\vert n \rangle\}_{n = 0}^{\infty}$ each with corresponding eigenvalue $E_n = \omega \left(n + \frac{1}{2}\right)$. Refer to this construction as the abstract solution.
How does the abstract solution also prove uniqueness? Why is there only one unique sequence of countable eigenstates? In particular, can one prove the state $\vert 0\rangle$ is the unique ground state without resorting to coordinate representation? (It would then follow that the set $\{\vert n \rangle\}_{n = 0}^{\infty}$ is also unique.)
The uniqueness condition is obvious if one solves the problem in coordinate representation since then one works in the realm of differential equations where uniqueness theorems abound. Most textbooks ignore this detail (especially since they often solve the problem both in coordinate representation and abstractly), however I have found two exceptions:
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Shankar appeals to a theorem which proves one-dimensional systems are
non-degenerate, however this is unsatisfactory for two reasons:- Not every one-dimensional system is non-degenerate, however a general result can be proven for a large class of potentials (the SHO potential is in such a class).
- The proof requires a departure from the abstract solution since it classifies the potentials according to their functional properties.
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Griffiths addresses this concern in a footnote stating that the equation $a \vert 0\rangle = 0$ uniquely determines the state $\vert 0\rangle$. Perhaps this follows from the abstract solution, however I do not see how.
Best Answer
I) It depends on how abstract OP wants it to be. Say that we discard any reference to 1D geometry, and position and momentum operators $\hat{q}$ and $\hat{p}$. Say that we only know that
$$\tag{1}\frac{\hat{H}}{\hbar\omega} ~:=~ \hat{N}+\nu{\bf 1}, \qquad\qquad \nu\in\mathbb{R},$$ $$\tag{2} \hat{N}~:=~\hat{a}^{\dagger}\hat{a}, $$ $$\tag{3} [\hat{a},\hat{a}^{\dagger}]~=~{\bf 1}, \qquad\qquad[{\bf 1}, \cdot]~=~0.$$
(Since we have cut any reference to geometry, there is no longer any reason why $\nu$ should be a half, so we have generalized it to an arbitrary real number $\nu\in\mathbb{R}$.)
II) Next assume that the physical states live in an inner product space $(V,\langle \cdot,\cdot \rangle )$, and that $V$ form a non-trivial irreducible unitary representation of the Heisenberg algebra,
$$\tag{4} {\cal A}~:=~ \text{associative algebra generated by $\hat{a}$, $\hat{a}^{\dagger}$, and ${\bf 1}$}.$$
The spectrum of a semi-positive operator $\hat{N}=\hat{a}^{\dagger}\hat{a}$ is always non-negative,
$$\tag{5} {\rm Spec}(\hat{N})~\subseteq~ [0,\infty[.$$
In particular, the spectrum ${\rm Spec}(\hat{N})$ is bounded from below. Since the operator $\hat{N}$ commutes with the Hamiltonian $\hat{H}$, we can use $\hat{N}$ to classify the physical states. Let us sketch how the standard argument goes. Say that $|n_0\rangle\neq 0$ is a normalized eigenstate for $\hat{N}$ with eigenvalue $n_0\in[0,\infty[$. We can use the lowering ladder (annihilation) operator $\hat{a}$ repeatedly to define new eigenstates
$$\tag{6} |n_0- 1\rangle,\quad |n_0- 2\rangle, \quad\ldots$$
which however could have zero norm. Since the spectrum ${\rm Spec}(\hat{N})$ is bounded from below, this lowering procedure (6) must stop in finite many steps. There must exists an integer $m\in\mathbb{N}_0$ such that zero-norm occurs
$$\tag{7} \hat{a}|n_0 - m\rangle~=~0.$$
Assume that $m$ is the smallest of such integers. The norm is
$$\tag{8} 0 ~=~ || ~\hat{a}|n_0 - m\rangle ~||^2 ~=~ \langle n_0 - m|\hat{N}|n_0 - m\rangle ~=~ ( n_0 - m) \underbrace{||~|n_0 - m\rangle~||^2}_{>0},$$
so the original eigenvalue is an integer
$$\tag{9} n_0 ~=~ m\in\mathbb{N}_0,$$
and eq. (7) becomes
$$\tag{10} \hat{a}|0\rangle ~=~0,\qquad\qquad \langle 0 |0\rangle ~\neq~0.$$
We can next use the raising ladder (creation) operator $\hat{a}^{\dagger}$ repeatedly to define new eigenstates
$$\tag{11} |1\rangle,\quad |2\rangle,\quad \ldots.$$
By a similar norm argument, one may see that this raising procedure (11) cannot eventually create a zero-norm state, and hence it goes on forever/doesn't stop. Inductively, at stage $n\in\mathbb{N}_0$, the norm remains non-zero,
$$\tag{12} || ~\hat{a}^{\dagger}|n\rangle ~||^2 ~=~ \langle n|\hat{a}\hat{a}^{\dagger}|n\rangle~=~ \langle n|(\hat{N}+1)|n\rangle ~=~ (n+1) ~\langle n|n\rangle~>~0. $$
So $V$ contains at least one full copy of the standard Fock space. On the other hand, by the irreducibility assumption, the vector space $V$ cannot be bigger, and $V$ is hence just a standard Fock space (up to isomorphism).
III) Finally, if $V$ is not irreducible, then $V$ could be a direct sum of several Fock spaces. In the latter case, the ground state energy-level is degenerate.