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import mmf_setup; mmf_setup.nbinit()
import os
from pathlib import Path
FIG_DIR = Path(mmf_setup.ROOT) / '../Docs/_build/figures/'
os.makedirs(FIG_DIR, exist_ok=True)
import logging; logging.getLogger("matplotlib").setLevel(logging.CRITICAL)
%matplotlib inline
import numpy as np, matplotlib.pyplot as plt
try: from myst_nb import glue
except: glue = None

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8. Sturm-Liouville Theory

Sturm-Liouville theory describes the general properties of equations like the time-independent Schrödinger equation, establishing the rigorous connection with linear algebra for wavefunctions. Here I will couch things in terms of the language of quantum mechanics.

Inserting a complete set of states in the position basis

\[\begin{gather*} \op{1} = \int\d{y}\ket{y}\bra{y} \end{gather*}\]

we have

\[\begin{gather*} \int\braket{x|\op{H}|y} \underbrace{\braket{y|\psi}}_{\psi(y)}\d{y} = \underbrace{\braket{x|\psi}}_{\psi(x)}E. \end{gather*}\]

The operators under consideration here are local in the sense that

\[\begin{gather*} \braket{x|\op{H}|y} = \delta(x-y)H\left(\diff{}{x}, x\right) \end{gather*}\]

so the equation can be expressed as in the book:

\[\begin{gather*} \mathcal{L}(x)\psi(x) = \psi(x)\underbrace{\lambda}_{E}. \end{gather*}\]

The idea is to find the spectrum of equations with the form

\[\begin{gather*} \op{H}\ket{\psi} = \ket{\psi}E. \end{gather*}\]

The basic idea is to make \(\op{H}\) self-adjoint, so that has a complete set of independent eigenvectors \(\ket{n}\) with real eigenvalues \(E_n\). Complications include:

1. The bare operator may not be self-adjoint, requiring some modification or special boundary conditions.

2. The spectrum might have regions with continuous eigenvalues \(E_{k}\) where \(k\) is part of a continuum and the corresponding states \(\ket{k}\) are not normalizable. The appropriate mathematical context for this is called a rigged Hilbert space (see e.g. [Madrid, 2005] or [Ballentine, 2014] for details.)

We summarize the key results:

• To check if \(\op{H}\) is self-adjoint, first see if it is hermitian. If so, then check if it is self-adjoint by integrating by parts. This may introduce boundary terms which spoil self-adjointness.

• Sometimes self-adjointness can be restored by adding a weight function:

  \[\begin{gather*} w(x)\mathcal{L}(x)\psi(x) = w(x)\psi(x)E. \end{gather*}\]

  This does not affect the eigenvalues or eigenfunctions, but, if needed, then this must appear in the inner product, and the eigenfunctions will only be orthogonal with this weight.

• Once a complete set of eigenstates is found, any function can be expressed as a linear combination:

  \[\begin{gather*} \psi(x) = \sum_{n}c_n\psi_n(x). \end{gather*}\]

All of this can be expressed in the following general formalism, if properly interpreted:

\[\begin{gather*} \op{H}\ket{n} = \ket{n}E_n, \tag{eigenvalue equation}\\ \braket{m|n} = \delta_{mn}, \tag{orthonormality}\\ \sum_{n}\ket{n}\bra{n} = \op{1}, \tag{completness}\\ \ket{\psi} = \sum_{n}\ket{n}\underbrace{\braket{n|\psi}}_{c_n}. \tag{expansion} \end{gather*}\]

1. If a weight \(w(x)\) is needed to make \(\op{H}\) self-adjoint, then it should be incorporated into the inner product:

   \[\begin{gather*} \braket{f|g} = \int f^*(x)g(x) w(x)\d{x}. \end{gather*}\]

In physics, the convention is to keep the factors of \(2\pi\) with the momentum integrals and momentum delta-functions. In mathematics, things tend to be more symmetric with factors of \(\sqrt{2\pi}\) appearing everywhere.

2. If the spectrum is continuous, then \(\delta_{mn}\) should become a Dirac delta function and the sums \(\sum_{n}\) should become integrals. One must treat the normalization with a bit of care. For example, the following are common in physics for position and momentum eigenstates:

   \[\begin{gather*} \int\d{x}\; \ket{x}\bra{x} = \int\frac{\d{x}}{2\pi} \ket{k}\bra{k} = \op{1}, \tag{completeness}\\ \braket{x|y} = \delta(x-y), \qquad \braket{k|q} = (2\pi)\delta(k-q), \tag{orthogonality}\\ \braket{x|k} = e^{\I k x}. \end{gather*}\]

The general trick is to insert the appropriate factors of \(\op{1}\). For example, the compute the Fourier transform \(\psi_k = \braket{k|\psi}\) from the wavefunction \(\psi(x) = \braket{x|\psi}\) we simply insert \(\op{1}\) in the appropriate form:

\[\begin{gather*} \psi_k = \braket{k|\psi} = \braket{k|\op{1}|\psi} = \int\d{x}\;\underbrace{\braket{k|x}}_{\braket{x|k}^*}\braket{x|\psi} = \int\d{x}\;e^{-\I k x}\psi(x). \end{gather*}\]

Likewise, the inverse Fourier transform is

\[\begin{gather*} \psi(x) = \braket{x|\psi} = \braket{x|\op{1}|\psi} = \int\frac{\d{k}}{2\pi}\braket{x|k}\braket{k|\psi} = \int\frac{\d{k}}{2\pi} e^{\I k x}\psi_k. \end{gather*}\]

Self-Adjoint vs Hermitian

A hermitian operator \(\op{H}\) is simply one whose matrix elements are hermitian

\[\begin{gather*} H_{mn} = \braket{m|\op{H}|n}, \qquad H_{mn} = H_{nm}^*, \qquad \mat{H} = \mat{H}^\dagger. \end{gather*}\]

A self-adjoint operator is one that is its own adjoint, meaning

\[\begin{gather*} \braket{\op{H}m|n} = \braket{m|\op{H}n} \end{gather*}\]

for all states \(\ket{m}\) and \(\ket{n}\). In a finite-dimensional vector space, all hermitian matrices are self-adjoint, but the same need not be true in infinite-dimensional Hilbert spaces.

For example, consider the momentum operator \(\op{H} = \op{p}\) for a particle in a box from \(x\in [a, b]\). The position space representation is

\[\begin{gather*} \braket{x|\op{p}\psi} = -\I\hbar \psi'(x). \end{gather*}\]

To be self-adjoint, we must have

\[\begin{gather*} \braket{\op{p}f|g} = \braket{f|\op{p}g},\\ \int_{a}^{b} [-\I\hbar f'(x)]^*g(x)\d{x} = \int_{a}^{b} f^*(x)[-\I\hbar g'(x)]\d{x},\\ \int_{a}^{b} f'(x)^*g(x)\d{x} = -\int_{a}^{b} f^*(x)g'(x)\d{x},\\ \int_{a}^{b} [f^*g]' \d{x} = f^*(a)g(a) - f^*(b)g(b) = 0, \end{gather*}\]

for all functions \(f(x)\) and \(g(x)\). This is integration by parts, and clearly this will not be satisfied if \(f(x)\) and \(g(x)\) do not appropriately vanish at the endpoints of the interval.

In this case, self-adjointness can be ensured by imposing Dirichlet boundary conditions \(f(a) = f(b)\) on our functions, as we do when we consider a particle in an infinite square well.

Another Example of a Self-Adjoint Extension

Another interesting example of a problem that is not self-adjoint concerns the quantum mechanics of a particle on the semi-infinite interval \(r \in [0, \infty)\) with the unbounded potential \(V(r) = -m\alpha r^4 = -\hbar^2 a r^4/2m\). This is not self-adjoint for different reasons in that the solutions are unbounded, but there is a way to render this well-behaved.

The key comes from considering the classical trajectories:

\[\begin{gather*} \ddot{q} = \diff{\dot{q}}{q}\dot{q} = 4 \alpha q^3,\\ \frac{\dot{q}^2-\dot{q}_0^2}{2} = \alpha (q^4 - q_0^4). \end{gather*}\]

As the particle falls away from the origin, we expect the velocity (and thus momentum) to behave as

\[\begin{gather*} \dot{q} = \sqrt{\dot{q}_0^2 + 2\alpha(q^4 - q_0^4)} \sim \sqrt{2\alpha}q^2. \end{gather*}\]

The probability of finding a particle in some region \(\d{q}\) is proportional to the amount of time it spends there, and hence, inversely proportional to the velocity. We thus expect

\[\begin{gather*} \abs{\psi(r)}^2 \propto \frac{1}{\dot{q}} \propto \frac{1}{r^2}. \end{gather*}\]

Now notice that the probability in the tails is finite:

\[\begin{gather*} \int_{r_0}^{\infty}\frac{1}{r^2}\d{r} = \frac{1}{r_0}. \end{gather*}\]

This suggests that we might be able to express our problem in a space of \(L_2\) functions with an appropriate boundary condition at infinity.

Equivalently, consider a particle starting (t=0) at rest \(\dot{q}_0 = 0\) at position \(q=q_0\). The previous condition of normalizing the wavefunction is equivalent to noting that the particle gets to \(q=\infty\) in finite time!:

\[\begin{gather*} t = \int_{q_0}^{\infty} \frac{1}{\sqrt{2\alpha}\sqrt{q^4 - q_0^4}}\d{q} = \frac{1}{q_0\sqrt{2\alpha}}\int_{1}^{\infty} \frac{1}{\sqrt{x^4 - 1}}\d{x}\\ = \frac{1}{q_0\sqrt{2\alpha}}\frac{\sqrt{\pi}\Gamma(5/4)}{\Gamma(3/4)} = \frac{1}{q_0\sqrt{\alpha}}0.927\dots \end{gather*}\]

(Quick dimensions check: \([\sqrt{\alpha}] = \sqrt{E/MD^4} = 1/DT\), so \([q_0\sqrt{\alpha}] = 1/T\).)

One way to do this is to consider a limiting case of a hard wall

\[\begin{gather*} V(r) = \begin{cases} -m\alpha r^4 & r < R\\ \infty & R \geq R. \end{cases} \end{gather*}\]

For each \(R\), the Sturm-Liouville problem will be self-adjoint because of the physical boundary condition \(\psi(R) = 0\). The limit of \(R\rightarrow \infty\) is interesting because, as \(R\) gets larger, the potential admits more and more bound states, but the nature of these bound states for \(r \ll R\) becomes universal. This provides an example of a renormalization-group limit cycle closely related to the Efimov effect.