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:
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.
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.