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:

    1. 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).
    2. 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.


2 Answers 2


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

$$\frac{\hat{H}}{\hbar\omega} ~:=~ \hat{N}+\nu{\bf 1}, \qquad\qquad \nu\in\mathbb{R},\tag{1}$$ $$ \hat{N}~:=~\hat{a}^{\dagger}\hat{a}, \tag{2}$$ $$ [\hat{a},\hat{a}^{\dagger}]~=~{\bf 1}, \qquad\qquad[{\bf 1}, \cdot]~=~0.\tag{3}$$

(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,

$$ {\cal A}~:=~ \text{associative algebra generated by $\hat{a}$, $\hat{a}^{\dagger}$, and ${\bf 1}$}.\tag{4}$$

The spectrum of a semi-positive operator $\hat{N}=\hat{a}^{\dagger}\hat{a}$ is always non-negative,

$$ {\rm Spec}(\hat{N})~\subseteq~ [0,\infty[.\tag{5}$$

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

$$ |n_0- 1\rangle,\quad |n_0- 2\rangle, \quad\ldots \tag{6}$$

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

$$ \hat{a}|n_0 - m\rangle~=~0.\tag{7}$$

Assume that $m$ is the smallest of such integers. The norm is

$$\begin{align} 0 ~=~& || ~\hat{a}|n_0 - m\rangle ~||^2 \cr ~=~& \langle n_0 - m|\hat{N}|n_0 - m\rangle \cr ~=~& ( n_0 - m) \underbrace{||~|n_0 - m\rangle~||^2}_{>0}, \end{align}\tag{8}$$

so the original eigenvalue is an integer

$$ n_0 ~=~ m\in\mathbb{N}_0,\tag{9}$$

and eq. (7) becomes

$$ \hat{a}|0\rangle ~=~0,\qquad\qquad \langle 0 |0\rangle ~\neq~0.\tag{10}$$

We can next use the raising ladder (creation) operator $\hat{a}^{\dagger}$ repeatedly to define new eigenstates

$$ |1\rangle,\quad |2\rangle,\quad \ldots.\tag{11}$$

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,

$$ \begin{align} || ~\hat{a}^{\dagger}|n\rangle ~||^2 ~=~& \langle n|\hat{a}\hat{a}^{\dagger}|n\rangle\cr ~=~& \langle n|(\hat{N}+1)|n\rangle\cr ~=~& (n+1) ~\langle n|n\rangle~>~0. \end{align}\tag{12}$$

So $V$ contains at least one full copy of a standard Fock space

$${\rm span}_{\mathbb{C}}\left\{|n\rangle \mid n\in\mathbb{N}_0\right\}. \tag{13}$$

Notice that the Fock space (13) is invariant under the action of the Heisenberg algebra (4), i.e. it is a representation thereof.

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). In this case the normalized eigenstates $|n\rangle$ are unique up to phase factors.

III) Finally, if $V$ is not irreducible, then the operator $\hat{N}$ has an eigenstate outside the Fock space (13). We can then repeat the argument in section II to find a linearly independent copy of a standard Fock space inside $V$, i.e. $V$ becomes a direct sum of several Fock spaces. In this latter case, the ground state energy-level is degenerate.

  • 4
    $\begingroup$ In particular, case III) means that one needs the representation in form of a differential operator to conclude nondegeneracy. (In the abstract case, one instead assumes irreducibility.) $\endgroup$ Commented Mar 30, 2012 at 13:31
  • $\begingroup$ @ArnoldNeumaier. With this answer and your comment, it seems I would conclude the following. Given the irreducibility of the set $\{ P, Q \}$ and the definitions of $a$ and $a^\dagger$ in terms of $P$ and $Q$, the assumed irreducibility condition above would now follow and thus guarantee uniqueness. I have seen a proof of the irreducibility of $\{ P, Q \}$ which relies on the differential operator representation. Perhaps the next question is if there is an alternative proof which does not invoke this representation? $\endgroup$ Commented Apr 7, 2012 at 3:35
  • $\begingroup$ @Evan: There cannot be a proof without the differential operator representation. For the tensor product of several irreducible representations $j=1,...,N$ produces a reducible representation for $P=\sum P_j$, $Q=\sum Q_j/N$, the center of mass motion. $\endgroup$ Commented Apr 12, 2012 at 13:27
  • $\begingroup$ I know the post is old, but it is not clear to me what guarantees that $N$ is positive semidefinite $\endgroup$
    – lucabtz
    Commented Mar 9, 2022 at 12:15
  • 1
    $\begingroup$ Hi ekardnam_ Thanks for the feedback. Hint: $\langle \psi | \hat{N}|\psi\rangle = || \hat{a}|\psi\rangle ||^2\geq 0$. $\endgroup$
    – Qmechanic
    Commented Mar 9, 2022 at 12:24

Every one dimensional potential system has a unique vacuum because it is the minimum of the following functional

$$ \int |\psi'|^2 + V(x) |\psi|^2 dx$$

Which is minimized by a real positive (nodeless) wavefunction. If there are two different minima (if there is a degeneracy), then a linear combination of the two wavefunctions have a node, and this is inconsistent with a regular potential.

The only way to have degenerate ground states (or two separate independent ground states) is for the potential V to have an infinite hard wall separating different regions. Otherwise, the ground state wavefunction is everywhere nonvanishing and the linear combination/node argument above works.

For the harmonic oscillator, it's a little more trivial to prove than this general stuff, because the ground state is annihilated by the annihilation operator $x+ip$, and this is a first order differential equation with exactly one solution up to rescaling, which is the ground state Gaussian.


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