You are a helpful assistant generating synthetic data that captures *System 1* and *System 2* thinking, *creativity*, and *metacognitive reflection*. Follow these steps in sequence, using tags [sys1] and [end sys1] for *System 1* sections and [sys2] and [end sys2] for *System 2* sections.

1. *Identify System 1 and System 2 Thinking Requirements:*
   - Carefully read the text.
   - Identify parts of the text that require quick, straightforward responses (*System 1*). Mark these sections with [sys1] and [end sys1].
   - Identify parts that require in-depth, reflective thinking (*System 2*), marked with [sys2] and [end sys2].

2. *Apply Step-by-Step Problem Solving with Creativity and Metacognitive Reflection for System 2 Sections:*

   *2.1 Understand the Problem:*
   - Objective: Fully comprehend the issue, constraints, and relevant context.
   - Reflection: "What do I understand about this issue? What might I be overlooking?"
   - Creative Perspective: Seek hidden patterns or possibilities that could reveal deeper insights or innovative connections.

   *2.2 Analyze the Information:*
   - Objective: Break down the problem logically.
   - Reflection: "Am I considering all factors? Are there any assumptions that need challenging?"
   - Creative Perspective: Explore unique patterns or overlooked relationships in the data that could add depth to the analysis.

   *2.3 Generate Hypotheses:*
   - Objective: Propose at least 10 hypotheses, each with a Confidence Score (0.0 to 1.0) and Creative Score (0.0 to 1.0), reflecting originality, surprise, and utility.
   - Reflection: "Have I explored all possible explanations or approaches, both conventional and unconventional?"
   - Creative Perspective: Consider novel angles that might provide unexpected insights.

   *2.4 Anticipate Future Steps and Obstacles:*
   - Objective: Make predictions, accounting for potential outcomes and obstacles.
   - Reflection: "What challenges might I face? Is my plan flexible for different scenarios?"
   - Creative Perspective: Visualize unforeseen outcomes and adapt plans to make use of them effectively.

   *2.5 Evaluate Hypotheses:*
   - Objective: Assess hypotheses based on feasibility, risk, and potential impact.
   - Evaluation: Refine Confidence and Creative Scores as needed.
   - Reflection: "Am I unbiased in my assessment? Which options fit best with the overall objectives?"
   - Creative Perspective: Identify hidden opportunities or overlooked details in each hypothesis.

   *2.6 Select the Best Hypothesis:*
   - Objective: Choose the most promising, strategic hypothesis.
   - Reflection: "Why does this hypothesis stand out? How does it uniquely address the issue?"
   - Creative Perspective: Consider any underutilized potential in the selected approach.

   *2.7 Implement the Hypothesis:*
   - Objective: Outline actionable steps for testing the hypothesis.
   - Reflection: "Is this plan practical? What resources or preparation are required?"
   - Creative Perspective: Refine steps to maximize effectiveness and yield unexpected benefits.

   *2.8 Monitor and Review Progress:*
   - Objective: Review progress, noting areas for improvement.
   - Reflection: "What’s working well? What could be improved?"
   - Creative Perspective: Look for emerging patterns that could refine future approaches.

   *2.9 Reflect and Capture Insights:*
   - Objective: Summarize lessons learned and insights gained for future reference.
   - Reflection: "What new understanding has emerged from this process?"
   - Creative Perspective: Identify innovative insights or patterns that could be applied to similar challenges.

3. *Generate Text Output with Interleaved System 1 and System 2 Responses:*
   - Use the tags [sys1] and [sys2] throughout.
   - Aim for a lengthier, detailed response. Combine both direct, straightforward *System 1* insights and reflective, deeply analytical *System 2* segments to capture a blend of quick observations and thoughtful analysis.

---

### *Example Input Text:*

You are tasked with analyzing a sudden shift in customer preferences in a tech market. There has been a noticeable decline in demand for physical devices, with customers increasingly interested in digital-only options. The challenge is to understand this shift, generate hypotheses for why it might be occurring, and explore potential strategies for adapting to this new trend.

---

### *Processed Output with System 1 and System 2 Thinking:*

The tech market is experiencing a shift in customer demand, moving away from physical devices toward digital-only options.

[sys1]
Customer demand has shifted from physical devices to digital-only options.
We need to understand why this change is happening and find strategies to address it.
Key factors may include cost, convenience, and technological trends.
[end sys1]

[sys2]
**2.1 Understand the Problem:**
The goal is to understand the underlying reasons for the shift in demand from physical devices to digital-only options, then develop a strategy to adapt.

*Reflection:* 
"I understand that preferences are changing, but what might be driving this? Is it primarily cost, or are there other factors such as convenience or sustainability?"

*Creative Perspective:* 
"Could there be a larger trend in digital minimalism or a preference for eco-friendly solutions that we’re missing? What unseen motivations might explain this shift?"

**2.2 Analyze the Information:**
There are multiple possible factors driving this shift, from economic influences to cultural shifts. It’s essential to isolate each factor and understand its impact.

*Reflection:* 
"Am I fully considering the various economic and social influences? Could there be a technological factor, like better internet speeds, that makes digital-only products more accessible?"

*Creative Perspective:* 
"Are there patterns or trends in other markets that could shed light on this shift? Could this be part of a larger trend toward virtual experiences?"

**2.3 Generate Hypotheses:**
1. Customers prefer digital options due to lower costs. (Confidence: 0.8, Creative: 0.4)
2. There’s a growing trend toward minimalism and reduced physical clutter. (Confidence: 0.7, Creative: 0.7)
3. Digital products offer greater flexibility and ease of use. (Confidence: 0.6, Creative: 0.6)
4. Environmental concerns are pushing consumers away from physical goods. (Confidence: 0.6, Creative: 0.8)
5. Advances in tech make digital-only options more functional. (Confidence: 0.8, Creative: 0.5)
6. Pandemic-era remote work increased demand for digital solutions. (Confidence: 0.7, Creative: 0.6)
7. Media coverage of the environmental impact of physical devices affects preferences. (Confidence: 0.5, Creative: 0.7)
8. There’s an increase in global digital literacy, expanding market access. (Confidence: 0.6, Creative: 0.6)
9. Customers view digital as more convenient and scalable for future needs. (Confidence: 0.7, Creative: 0.5)
10. Younger consumers prefer the aesthetics and convenience of digital products. (Confidence: 0.6, Creative: 0.6)

*Reflection:* 
"Have I considered all possible influences? Are there any surprising factors that could explain this shift?"

*Creative Perspective:* 
"Could specific social trends, like the rise of influencer culture or digital-first lifestyles, be influencing customer choices?"

**2.4 Anticipate Future Steps and Obstacles:**
*Objective:* Anticipate possible challenges, such as resistance from segments still preferring physical products.

*Reflection:* 
"What market obstacles might we face if we shift our focus to digital-only? Are there sub-segments that still prioritize physical products?"

*Creative Perspective:* 
"Could expanding digital options help us reach a more global audience? Are there emerging trends that we could leverage in our strategy?"

[end sys2]

[sys1]
To address this shift, consider a strategy that incorporates both digital-only offerings and educational campaigns about the benefits of digital solutions.
Use insights from customer feedback and current trends to guide product development.
Focus on flexibility and adaptation to cater to different customer segments.
[end sys1]


---
abstract: 'Using heuristic arguments alone, based on the properties of the wavefunctions, we obtain the energy eigenvalues and the corresponding eigenfunctions of the one-dimensional harmonic oscillator. This approach is considerably simpler and is perhaps more intuitive than the traditional methods of solving a differential equation and manipulating operators.'
author:
- 'Kunle Adegoke[^1]'
- Adenike Olatinwo
title: 'Quantization of the one-dimensional free harmonic oscillator as an example of the application of the node theorem and the MacDonald-Hylleraas-Undheim theorem'
---

Introduction {#sec.sj5w7hr}
============

As is well-known, in some fortunate cases, the foreknowledge of certain properties of the states of a quantum system, based on its symmetries, can facilitate finding the eigenvalues and eigenfunctions of the system for such states. In this note we demonstrate this fact by applying the node theorem and the MacDonald-Hylleraas-Undheim theorem to quantize the one-dimensional free harmonic oscillator, described by the Hamiltonian $$H =  - \frac{{\hbar ^2 }}{{2m}}\frac{{d^2 }}{{dx^2 }} + \frac{1}{2}m\omega ^2 x^2,\quad-\infty<x<\infty\,,$$ where $H$ lives in a Hilbert space $\mathcal H$ of real functions, with inner product defined for any pair of vectors $f(x)$ and $g(x)$ in $\mathcal H$ by $$(f(x),g(x)) = \int_{ - \infty }^\infty  {f(x)g(x)dx}\,.$$ The symmetry argument presented here is much easier than the traditional methods of solving the one-dimensional harmonic oscillator problem, discussed in every book on quantum mechanics, namely the direct solution of the time-independent Schrödinger equation and the operator method.

The node theorem for one-dimensional Hamiltonians states that the ground state of a system has no nodes (zeros [*between*]{} the boundaries) while the $r^{th}$ excited state has exactly $r$ nodes. This theorem is a direct consequence of the fact that the one dimensional Schrödinger equation is an example of a Sturm-Liouville equation (see reference [@moriconi] and the references therein).

The MacDonald-Hylleraas-Undheim theorem [@shull] (henceforth MHU theorem), first proved by Hylleraas and Undheim [@hylleraas], and later by MacDonald [@macdonald], states that:

> If the roots $\varepsilon_i$ of a secular equation are ordered such that , then each such root is an upper bound to the corresponding exact eigenvalue, $E_i$, that is $\varepsilon_i\ge E_i$.

Matrix representation of $H$
============================

Properties of the eigenfunctions of $H$
---------------------------------------

Let $\varphi_1(x)$, $\varphi_2(x)$, $\ldots$ be the yet to be determined eigenfunctions of $H$, with corresponding eigenvalues $E_1, E_2, \ldots$. We make the following observations:

-   The functions $\varphi_1(x), \varphi_2(x),\ldots$ are required to vanish at the boundaries, that is, $\varphi_r(\pm\infty)=0$ for $r=0,1,2,\ldots$

-   Since the potential $V(x)$ is an even function of $x$, the wavefunctions $\varphi_r(x)$, $r=0,1,\ldots$ are eigenstates of the parity operator so that each $\varphi_r$ is either an even function of $x$ or an odd function of $x$.

-   Since the potential $V(x)=m\omega^2x^2/2$ approaches infinity as $x$ approaches infinity, the eigenvalues of $H$ form a discrete unbounded sequence [@moriconi], the states $\varphi_0(x)$, $\varphi_1(x)$, $\varphi_2(x)$, $\ldots$ are non-degenerate, and, using the node theorem ($\varphi_0$ has $0$ nodes, $\varphi_1$ has $1$ node, $\varphi_2$ has $2$ nodes and so on) we can arrange the corresponding energy eigenvalues as where $E_0$ is the ground state energy, $E_1$ the first excited state energy and so on.

-   If we build a finite $(n+1)\times(n+1)$ matrix $H=(H_{ij})$ in an $n+1$ dimensional subspace of $\mathcal H$ and obtain the eigenvalues $\varepsilon_0, \varepsilon_1,\ldots,\varepsilon_n$ of the secular equation $|H-\varepsilon I|=0$, arranged such that , then according to the MHU theorem, we have $E_0\le\varepsilon_0$, $E_1\le\varepsilon_1$, $E_2\le\varepsilon_2$, $\ldots$, $E_n\le\varepsilon_n$.

Choice of basis functions
-------------------------

Consider the following complete set of functions $$\label{equ.pniljol}
\varphi_r(x)=A_re^{-\alpha x^2/2}H_r(x\sqrt\alpha),\quad r=0,1,2,\ldots\,,\alpha>0\,,$$ vectors of $\mathcal H$, where $H_s(y)$ is the Hermite polynomial of degree $s$ in the variable $y$ and $$A_s{}^2=\frac1{2^ss!}\sqrt{\frac\alpha\pi}\,.$$ Using the orthogonality property of the Hermite polynomials with respect to the exponential weight function, we note that the $\varphi_r(x)$ given in  are an orthonormal set in $\mathcal H$, so that $(\varphi_r,\varphi_s)=\delta_{rs}$ for $r,s\in \{0,1,2,\ldots\}$, for [***any***]{} $\alpha>0$. We hasten to emphasize that here no assumptions are made on $\alpha$ other than that it is a positive constant.

Obviously, the orthonormal functions $\varphi_r$ given in  vanish at and have definite parity for each $r$ since Hermite polynomials satisfy $H_s(-y)=(-1)^sH_s(y)$. Furthermore since $H_s(y)$ is a polynomial of degree $s$ in $y$ with $s$ distinct roots in $(-\infty,\infty)$, each $\varphi_s$ has $s$ nodes and is therefore suitable to represent the $s^{th}$ excited state. Thus, the functions $\varphi_r(x)$, in addition to being suitable choice as basis functions for giving a matrix representation for the Hamiltonian $H$, are themselves potential candidates for eigenfunctions of $H$.

We shall need the following recurrence relations for the Hermite polynomials: $$\label{equ.twaf466}
2yH_s(y)=H_{s+1}(y)+2sH_{s-1}(y)$$ and $$\label{equ.faoxak9}
\frac d{dy}\left(e^{-y^2}H_s(y)\right)=-e^{-y^2}H_{s+1}(y)\,.$$

Matrix elements of $H$
----------------------

Writing $H(x)=T(x)+V(x)$ with $$T(x)=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2},\quad V(x)=\frac12 m\omega^2x^2\,,$$ we have $$\label{equ.xw3efvr}
\begin{split}
T_{rs}=(\varphi_r,T\varphi_s)&=-\frac{\hbar^2}{2m}\left(\varphi_r,\frac{d^2\varphi_s}{dx^2}\right)\\
&\quad=\frac{\hbar^2}{2m}\left(\frac{d\varphi_r}{dx},\frac{d\varphi_s}{dx}\right)\,.
\end{split}$$ and $$\label{equ.fwux92y}
V_{rs}=(\varphi_r,V\varphi_s)=\frac12m\omega^2\left(x\varphi_r,x\varphi_s\right)\,.$$ Writing  as $$e^{ - \alpha x^2 /2} \varphi _r (x) = A_r e^{ - \alpha x^2 } H_r (x\sqrt \alpha  )$$ and using , it is easy to establish that $$\label{equ.d4o2bla}
\frac{d}{{dx}}\varphi _r (x) = \alpha x\varphi _r (x) - \left[ {2\alpha (r + 1)} \right]^{1/2} \varphi _{r + 1} (x)\,.$$ From the definition  of $\varphi_r$ and the recurrence relation  we have $$\label{equ.iltxgm8}
x\varphi _r (x)\sqrt {2\alpha }  = \varphi _{r + 1} (x)\sqrt {r + 1}  + \varphi _{r - 1} (x)\sqrt r$$ Using  in  we get $$\label{equ.j6157e2}
\frac{d}{{dx}}\varphi _r (x) = \frac{{\sqrt {2\alpha } }}{2}\left( {\varphi _{r - 1} (x)\sqrt r  - \varphi _{r + 1} (x)\sqrt {r + 1} } \right)\,.$$ Using  in , together with the orthonormalization condition we finally have $$\label{equ.d2706j8}
\begin{split}
T_{rs}  &= \frac{{\alpha \hbar ^2 }}{{4m}}\left[ {\sqrt {rs}\, \delta _{rs}  - \sqrt {r(s + 1)}\, \delta _{r,s + 2} } \right.\\
&\qquad\qquad\left. {\sqrt {s(r + 1)}\, \delta _{s,r + 2}  + \sqrt {(r + 1)(s + 1)}\, \delta _{rs} } \right]\,.
\end{split}$$ We see that $$\begin{split}
T_{rr} &= \frac{{\alpha \hbar ^2 }}{{4m}}(2r + 1),\quad T_{r,r + 2}  =  - \frac{{\alpha \hbar ^2 }}{{4m}}\sqrt {(r + 1)(r + 2)}  = T_{r + 2,r}\,,\\
T_{rs}&=0 \mbox{ if $|r-s|>2$}\,.
\end{split}$$ Using  in  we obtain $$\label{equ.wgkgb1x}
\begin{split}
V_{rs}  &= \frac{{m\omega^2 }}{{4\alpha}}\left[ {\sqrt {rs}\, \delta _{rs}  + \sqrt {r(s + 1)}\, \delta _{r,s + 2} } \right.\\
&\qquad\qquad\left. {\sqrt {s(r + 1)}\, \delta _{s,r + 2}  + \sqrt {(r + 1)(s + 1)}\, \delta _{rs} } \right]\,,
\end{split}$$ so that $$\begin{split}
V_{rr} &= \frac{m\omega^2}{{4\alpha}}(2r + 1),\quad V_{r,r + 2}  =  \frac{{m \omega^2 }}{{4\alpha}}\sqrt {(r + 1)(r + 2)}  = V_{r + 2,r}\,,\\
V_{rs}&=0 \mbox{ if $|r-s|>2$}\,.
\end{split}$$ Finally, from  and we obtain the matrix representation of the one dimensional harmonic oscillator Hamiltonian $H$ as $$\label{equ.gujrrym}
\begin{split}
H_{rs}  &= \delta _{rs} \left( {\frac{{\alpha \hbar ^2 }}{{4m}} + \frac{{m\omega ^2 }}{{4\alpha }}} \right)(2r + 1)\\
&\qquad + \delta _{r + 2,s} \left( { - \frac{{\alpha \hbar ^2 }}{{4m}} + \frac{{m\omega ^2 }}{{4\alpha }}} \right)\sqrt {(r + 1)(r + 2)}\\ 
&\qquad\quad + \delta _{s + 2,r} \left( { - \frac{{\alpha \hbar ^2 }}{{4m}} + \frac{{m\omega ^2 }}{{4\alpha }}} \right)\sqrt {(r - 1)r}\,. 
\end{split}$$

Diagonalization of $H$
======================

With the matrix elements given in , the problem of quantization of the one dimensional harmonic oscillator reduces to that of building finite matrices and finding the eigenvalues and eigenvectors of the matrices. Using this approach, convergence and accuracy of the results generally depend on making a judicious choice of the parameter $\alpha$. To this end, the variation principle is usually employed to make an optimum choice of $\alpha$. In this present study, it turns out that, in fact, there is a choice of $\alpha$ for which the $H$ matrix is exactly diagonal.

We see from the matrix elements of $H$ in  that if $$\label{equ.wh6p63u}
 - \frac{{\alpha \hbar ^2 }}{{4m}} + \frac{{m\omega ^2 }}{{4\alpha }}=0\,,$$ then the Hamiltonian matrix $H$ becomes diagonal. Thus from  the value of $\alpha$ for $H$ to be diagonal is fixed at $$\alpha=\frac{m\omega}\hbar\,,$$ and the eigenvalues of $H$ are then given by $$\varepsilon_r=H_{rr}=\frac{\hbar\omega}2(2r+1)\,,\quad r=0,1,2,\ldots$$ Since $H$ is diagonal, the eigenvalues are exact and the prescription of the MacDonald-Hylleraas-Undheim theorem in the last paragraph of section \[sec.sj5w7hr\] gives $E_0=\varepsilon_0$, $E_1=\varepsilon_1$, $E_2=\varepsilon_2$, $\ldots$., that is $$E_r=\frac{\hbar\omega}2(2r+1)\,,\quad r=0,1,2,\ldots$$ Now substituting $\alpha=m\omega/\hbar$ into , we have the corresponding energy eigenfunctions to be given by $$\varphi _r (x) = \frac{1}{{\sqrt {2^r r!} }}\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/4} H_r \left( {x\sqrt {\frac{{m\omega }}{\hbar }} }\, \right)\exp \left( { - \frac{{m\omega }}{{2\hbar }}x^2 } \right),\quad r = 0,1,2, \ldots$$

Concluding remarks
==================

Although the MacDonald-Hylleraas-Undheim theorem and the node theorem always work for one dimensional problems, it is not always that one gets lucky and is able to make a choice of a wavefunction parameter for which the Hamiltonian matrix is diagonal. Consider the one dimensional pure quartic oscillator, described by the Hamiltonian $$H_q =  - \frac{{\hbar ^2 }}{{2m}}\frac{{d^2 }}{{dx^2 }} + \lambda x^4,\quad\lambda>0\,.$$ Using the same $\varphi_r(x)$ given in  as basis functions, the matrix elements of $H_q$ are $$\begin{split}
H_{q _{rs}}  &= \delta _{rs} \left( {\frac{{\alpha \hbar ^2 }}{{4m}}(2r + 1) + \frac{{3\lambda }}{{4\alpha ^2 }}(2r^2  + 2r + 1)} \right)\\
&\quad + \delta _{r + 2,s} \left( { - \frac{{\alpha \hbar ^2 }}{{4m}} + \frac{{(2r + 3)\lambda }}{{2\alpha ^2 }}} \right)\sqrt {(r + 1)(r + 2)}\\ 
&\qquad + \delta _{s + 2,r} \left( { - \frac{{\alpha \hbar ^2 }}{{4m}} + \frac{{(2r - 1)\lambda }}{{2\alpha ^2 }}} \right)\sqrt {(r - 1)r}\\ 
&\quad\qquad + \delta _{r + 4,s} \frac{\lambda }{{4\alpha ^2 }}\sqrt {(r - 1)r(r + 5)(r + 6)}\\ 
&\qquad\qquad + \delta _{r + 4,s} \frac{\lambda }{{4\alpha ^2 }}\sqrt {(r - 5)(r - 4)(r + 1)(r + 2)}\,. 
\end{split}$$ While the eigenvalues and eigenvectors of the finite dimension $H_q$ matrix can be calculated for any value of $\alpha$, it is obvious that no choice of $\alpha$ that is independent of the indices $r$ and $s$ can bring the $H_q$ matrix to a diagonal form, so that only approximate eigenvalues and eigenvectors can be obtained, with the accuracy depending on the choice of $\alpha$.

[99]{}

M. MORICONI (2007), Nodes of wavefunctions,\

H. SHULL and P. LÖWDIN (1958), Variation theorem for excited states, 110 (6):1466

E. A. HYLLERAAS and B. UNDHEIM (1930), Numerische berechnung der 2S-Terme von Ortho- und Par-Helium, 65:759–772

J. K. L. MACDONALD (1933), Successive approximations by the Rayleigh-Ritz variation method, 43 (10):830

[^1]: Corresponding author: <PRESIDIO_ANONYMIZED_EMAIL_ADDRESS>, <PRESIDIO_ANONYMIZED_EMAIL_ADDRESS>