2
$\begingroup$

I am finding that the eigenfunctions of my Hermitian differential operator are not orthogonal and I do not know why. Consider the differential operator

$$ \mathcal{L} = x^2 \frac{d^2}{dx^2} + 2x \frac{d}{dx} + c $$

where $c$ is a constant. This is an Hermitian operator with respect to the inner product

$$ \langle \psi , \phi \rangle =\int_{- \infty}^\infty dx\psi^*\phi$$

We have

$$ \langle \psi , \mathcal{L} \phi \rangle =\int_{- \infty}^\infty dx\psi^*\left( x^2 \frac{d^2 \phi}{dx^2} + 2x \frac{d \phi}{dx} + c \phi\right) \\ = \int_{- \infty}^\infty dx \left(\frac{d^2}{dx^2} \left( x^2 \psi^* \right) \phi - \frac{d}{dx}\left( 2x \psi^* \right) \phi + c \psi^* \phi \right) \\ = \int_{- \infty}^\infty dx\left( 2\psi^* + 4x \frac{d \psi^*}{dx} + x^2 \frac{d^2 \psi^*}{dx^2} - 2\psi^* - 2x \frac{d \psi^*}{dx} + c\psi^* \phi\right) \\ = \langle \mathcal{L} \psi , \phi \rangle $$

where I have assumed my solutions vanish at $\pm \infty$ so the boundary terms vanish when I integrate by parts. So my operator is Hermitian and I expect my eigenfunctions to be orthogonal. Consdider the eigenvalue equation $\mathcal{L} \psi = \lambda \psi$, this yields the differential equation

$$ \quad x^2 \psi''(x) + 2x \psi'(x) + (c - \lambda)\psi = 0$$

The eigenvalue equation is therefore an Euler differential equation. If we take a trial solution $\psi(x) = x^n$, then substituting this in yields the quadratic equation

$$ n^2 + n + (c- \lambda) = 0 \quad \Rightarrow \quad n=-\frac{1}{2} \pm \frac{1}{2} \sqrt{1- 4(c-\lambda)}$$

Suppose we took the special case where the eigenvalues are negative and of the form $\lambda = -E^2$, for some $E$, and let $ c = \frac{1}{4}$, then we have $ n = -\frac{1}{2} \pm i E$ and the solutions will be given by

$$ \psi_\pm(x) = \frac{1}{\sqrt{x}} x^{\pm i E}$$

My problem is that these solutions do not appear to be orthogonal for different eigenvalues. If we take the solutions whose eigenvalues are $\lambda $ and $\lambda'$, then the inner product would be

$$ \langle \psi , \psi' \rangle = \int_{-\infty}^\infty dx \frac{1}{x} x^{\pm i (E'-E)} $$

which according to Wolfram is divergent. I am not sure why my solutions for different eigenvalues are not orthogonal. Any hints would be greatly appreciated.

$\endgroup$
6
  • 3
    $\begingroup$ Are you sure you're supposed to work on the whole real line? $\endgroup$ Commented Oct 20, 2020 at 21:16
  • 3
    $\begingroup$ The spectral theorem only applies to self-adjoint operators. Being Hermitian is necessary but not sufficient to be self-adjoint. $\endgroup$ Commented Oct 20, 2020 at 21:28
  • $\begingroup$ @CameronWilliams Yes $\endgroup$ Commented Oct 20, 2020 at 21:48
  • 3
    $\begingroup$ It's only for eigenfunctions in $L^2$ that the usual proof that eigenfunctions with different eigenvalues are orthogonal goes through. The eigenfunctions you wrote down aren't in $L^2$ for any value of $E$. Also you need to be a bit more careful about what $\sqrt{x}$ means if $x$ is negative. $\endgroup$ Commented Oct 20, 2020 at 21:52
  • 2
    $\begingroup$ @QiaochuYuan's comment is exactly why I asked if you're supposed to be working on $\mathbb{R}$, OP. These are generalized eigenfunctions in a sense, but even then it's messy. Presumably a sort of quantization on the eigenvalues comes in to make sure things are "well" defined re: making sense of powers of $x$. $\endgroup$ Commented Oct 20, 2020 at 23:37

2 Answers 2

2
$\begingroup$

The function of yours that you have named $\psi_{\pm}$ are not both in $L^2(0,\infty)$ or both in $L^2(-\infty,0)$. Are they? Orthogonality has no meaning when it comes to functions that are not in the same inner product space.

$\endgroup$
2
  • $\begingroup$ I don’t understand why they both aren’t valid in $ (0,\infty)$? They can be written as $\psi_\pm (x) = \frac{1}{\sqrt{x}} e^{\pm i E \ln x}$ and that is valid for both $\pm$ surely? $\endgroup$ Commented Oct 23, 2020 at 8:47
  • $\begingroup$ @Matt0410 : Are these functions square-integrable? If not, they are not in the space, and you can't form inner products with them. $e^{isx}$ is not in $L^2(\mathbb{R})$, for example. $\endgroup$ Commented Oct 24, 2020 at 1:50
1
$\begingroup$

It is also worth realizing that, although in happy/lucky situations, an operator that is "apparently" symmetric/hermitian/whatever behaves as we'd expect from simpler (finite-dimensional?) linear algebra, things are more complicated for unbounded operators on (infinite-dimensional) Hilbert spaces.

Namely, a given "symmetric/hermitian/whatever" unbounded operator $T$ defined on a dense subspace (otherwise re-set-up the question) can have none, or infinitely-many extensions which are truly "self-adjoint" in a sense that is faithful to our expectations. In particular, when the operators are differential operators, solving the obvious differential equation (even for $L^2$ solutions...) often inadvertently asks for eigenvectors for the adjoint $T^*$, when $T$ (as initially described), while apparently "symmetric/hermitian/whatever" is not "fully self-adjoint", in the sense that $T^*$ is NOT even "symmetric/hermitian/whatever"... So its eigenvectors (even in $L^2$) do not necessarily have real eigenvalues.

A tangible example (occurred somewhere on MSE, but I can't find it just now) is the (as they say) "formally self-adjoint" $T=x^3\circ {d\over dx}+{d\over dx}\circ x^3$, where "$x^3$" means the multiplication operator. The corresponding differential equation for eigenvectors/eigenvalues only has $L^2({\mathbb R})$ solutions for eigenvalues in (I forget which) either upper or lower complex half-plane. What?!? None for real eigenvalues. :)

In fact, this sort of craziness (which I thought years ago was just a hobby for specialists) was laid out very clearly in vonNeumann's and in Stone's work on such operators, c. 1930. That is, there is a parametrization of all possible "genuinely self-adjoint" extensions of a symmetric/hermitian/whatever operator...

One should also be aware of the anecdote that, supposedly, when Dirac (one of my heroes) was informed that vonNeumann had clarified the distinction between symmetric/hermitian/whatever and self-adjoint operators, he said "there's a difference???" And I'm sympathetic to that attitude, because, indeed, in many "practical" situations, the natural symmetric/hermitian/whatever operator is "essentially self-adjoint", in the sense that it has a unique truly-self-adjoint extension. Laplacians on (complete?) Riemannian manifolds have this property, though it was only proven in the 1950s. ... BUT c. 1981, Y. Colin de Verdiere made use of the weirdness of the situation to give (really following Lax-Phillips) a new proof of the meromorphic continuation of Eisenstein series... AND to set up some situations that suggested a connection between such operators and the Riemann Hypothesis. :)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .