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Consider standard quantum harmonic oscillator, $H = \frac{1}{2m}P^2 + \frac{1}{2}m\omega^2Q^2$.

We can solve this problem by defining the ladder operators $a$ and $a^{\dagger}$. One can show that there is a unique "ground state" eigenvector $\psi_0$ with $H\psi_0 = \frac{1}{2}\hbar\omega\psi_0$ and furthermore that given any eigenvector $\psi$ of $H$ with eigenvalue $E$, the vector $a^{\dagger}\psi$ is also an eigenvector of $H$ with eigenvalue $E + \hbar\omega$.

However, it is usually stated that we now have all eigenvectors of $H$ by considering all vectors of the form $(a^{\dagger})^n\psi_0$.

How do we know that we have not missed any eigenvectors by this process? e.g. how do we know that eigenvalues are only of the form $E_n = (n+\frac{1}{2})\hbar\omega$?

Also a slightly more technical question, how do we know that the continuous spectrum of $H$ is empty?

The technical details I am operating with are that $\mathcal{H} = L^2(\mathbb{R})$ and all operators ($H, P, Q$) are defined on Schwartz space, so that they are essentially self-adjoint with their unique self-adjoint extensions corresponding to the actual observables.

user140223
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3 Answers3

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It is sufficient to prove that the vectors $|n\rangle$ form a Hilbert basis of $L^2(\mathbb R)$. This fact cannot be completely established by using the ladder operators. To prove that the span of the afore-mentioned vectors is dense in the Hilbert space, one should write down the explicit expression of the wavefunctions of the said vectors recognizing that they are the well-known Hilbert basis of Hermite functions. Since the vectors $|n\rangle$ are a Hilbert basis, from standard results of spectral theory, the operator $$\sum_n \hbar \omega(n +1/2 ) |n\rangle \langle n | \tag{1}$$ (using the strong operator topology which defines the domain of this operator implicitly) is self-adjoint and its spectrum is a pure point spectrum made of the numbers $\hbar \omega(n +1/2 ) $ with $n$ natural. This fact proves that the initial symmetric Hamiltonian operator you described in your post and defined on the Schwartz space admits at least one self-adjoint extension with the said spectrum (in particular no continuous spectrum takes place). To prove that it is the unique self-adjoint extesion, i.e., that the initial symmetric operator is essentially self-adjoint, the shortest way is to observe that the vectors $|n\rangle$ are necessarily analytic vectors of the initial Hamiltonian (notice that all the afore-mentioned vectors stay in the Schwartz space which is the initial domain) because they are eigenvectors. Since they are a Hilbert basis, their span is dense. Under these hypotheses, a celebrated theorem by Nelson implies that the initial symmetric Hamiltonian operator is essentially self-adjoint and thus (1) is the only self-adjoint extension of the initial symmetric Hamiltonian operator. As a final comment, it is interesting to remark that (1) is not a differential operator differently from the naive initial Hamiltonian which is a differential operator but it is not self-adjoint.

Valter Moretti
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I am no expert, but I can think of two ways to think about why there are no more eigenvalues than those you mentioned.

1) the eigenstates asossiated with the mentioned eigenvalues form a complete basis, and so any addition of a new eigenvalue would imply the addition of an additional eigenvector, orthogonal to all the rest but nonzero, which is absurd.

2) more concretely, if you had an eigenvalue different from those, you could apply the anihilation operator enough times to get an eigenvalue forbidden by the constraints of the hamiltonian

On the continuum part of the spectrum, I don't have an answer and I would like to read one

Popeye
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A neat argument is given in E.E.Commins' book, section 6.13.1. I'll give a brief sketch here:

  • All eigenvalues $n$ of $N=a^{\dagger}a$ are $\geq 0$.
  • There is a unique ket $\lvert 0\rangle$ s.t. $\;a\lvert 0\rangle=0$.
  • If $\lvert n\rangle$ is an eigenket of $N$ with eigenvalue $n$, then $\lvert n-1\rangle$ is an eigenket with eigenvalue $n-1$.
  • Hence all eigenvalues of $N$ are integers $\geq 0$. If there was a positive eigenvalue $k \notin \mathbb{Z}^+$, we could apply $a$ to the corresponding eigenket until we found an eigenket with negative eigenvalue (effectively "jumping over" $\lvert 0\rangle$), contradicting the first statement.
user28400
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