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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.

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    $\begingroup$ Related: physics.stackexchange.com/q/23028/2451 , physics.stackexchange.com/q/54691/2451 and links therein, $\endgroup$
    – Qmechanic
    Commented Dec 25, 2016 at 6:42
  • $\begingroup$ We do not need to derive the energy eigenvalues using ladder operators; rather, the ladder operators represent a convenient way to arrive at solutions of the simple harmonic oscillator based on its quantization. Comparatively look at how we find wavefunctions and eigenvalues using the series solution method (google.ca/url?sa=t&source=web&rct=j&url=http://…) and try to consider how the forms of the solutions come to be. Then relate it to the ladder operators. $\endgroup$
    – bleuofblue
    Commented Dec 25, 2016 at 6:49

3 Answers 3

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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.

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  • $\begingroup$ 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? $\endgroup$ Commented Dec 26, 2016 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. $\endgroup$ Commented Dec 26, 2016 at 8:08
  • $\begingroup$ Dear Valter, I have question. If you have time, I'd appreciate a reply. You say: in particular no continuous spectrum takes place. Could you elaborate why this follows exactly? Can't there be points of the continuous spectrum which are limit points of the point spectrum in general? Thanks in advance. $\endgroup$ Commented Apr 11, 2023 at 20:42
  • $\begingroup$ Is it because no sequence of these eigenvalues can converge, actually? $\endgroup$ Commented Apr 11, 2023 at 21:08
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    $\begingroup$ In the considered case the point spectrum has no accumulation points and the eigenvectors form a Hilbert basis. This is sufficient to argue that the spectrum is made of eigenvalues only. (When the eigenvectors form a Hilbert basis, the only possibility for some points in the continuous part is that these points are accumulation points of the point spectrum.) $\endgroup$ Commented Apr 11, 2023 at 21:12
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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

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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.
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  • $\begingroup$ 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. $\endgroup$ Commented Dec 25, 2016 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
    Commented Dec 25, 2016 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. $\endgroup$ Commented Dec 25, 2016 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? $\endgroup$ Commented Dec 26, 2016 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. $\endgroup$ Commented Dec 26, 2016 at 8:23

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