14
$\begingroup$

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.

$\endgroup$
12
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The ladder operators, expressed in the position basis, actually generate the recursion relation for the Hermite polynomials, and their repeated action is closely related to the Rodrigues formula for Hermite polynomials. Moreover, up to normalization, the annihilation operator acting on the ground state produces a differential equation with Gaussian solution. Thus wouldn't you say that using ladder operators is completely equivalent to using actual basis functions in the position representation? $\endgroup$ – ZeroTheHero Dec 26 '16 at 2:03
  • $\begingroup$ What you cannot prove only using ladder operators is the fact that the span of the vectors $|n\rangle$ is dense in $L^2$. Orthogonality can proved easily with the ladder operator formalism instead. $\endgroup$ – Valter Moretti Dec 26 '16 at 8:08
4
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.
$\endgroup$
  • $\begingroup$ The point with this sort of "proofs" is that, thiugh they include some interesting idea, every issue concerning domains is disregarded so that their status is dubious. $\endgroup$ – Valter Moretti Dec 25 '16 at 14:41
  • $\begingroup$ Can you elaborate? Admittedly I'm not a mathematician and I don't know what you mean with "issues concerning domains". It would be helpful if you could point out explicitly where this sort of elementary textbook proof lacks in mathematical rigor. $\endgroup$ – user28400 Dec 25 '16 at 14:56
  • $\begingroup$ Well, first of all the operator constructed out of creation and annihilation operators is not self-adjoint but only symmetric. By construction its domain is a dense subdomain of Schwartz space. Since it is not self-adjoint the spectral theorem is not valid in principle and any discussion on its spectrum dealing with that domain is, in principle, doubtful. Therefore one must first prove that the considered operator admits some self-adjoint extension, possibly unique, on a larger domain, then with this extension any question about completeness of the spectrum can be tackled. $\endgroup$ – Valter Moretti Dec 25 '16 at 15:41
  • $\begingroup$ Not sur I follow when you say the operators constructed out of creation and destruction operators are not self-adjoint. Agreed that while the creation destruction operators are not themselves self-adjoint, their product N surely is. Yes there are issues with domains but do such issues arise for continuous and differentiable potentials such as the harmonic oscillator? $\endgroup$ – ZeroTheHero Dec 26 '16 at 2:09
  • $\begingroup$ Well, the picture is quite complicated if one wishes to pay attention to every mathematical details. For instance, $a^\dagger$ is not the adjoint of $a$but only a restriction... Here you have operators defined in the Schwartz space and you look for self adjoint extension of some of them. $\endgroup$ – Valter Moretti Dec 26 '16 at 8:23

protected by Qmechanic Dec 25 '16 at 14:54

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.