A Particle in a Symmetric Box: With Application to Hydrogen Chain
Copyright (c) RDMCHEM LLC 2019

Overview


While early theories that quantized energies of light and matter were able to reproduce results of these early experiments, they did not explain why energy was quantized. De Broglie provided a glimpse into the source of quantization in describing the waveparticle duality of light and matter. If matter were to behave like a wave, then there must be a waveequation for matter! This activity introduces the waveequation for matter, known as Schrodinger equation, and the simplest ideal system: a particle in symmetric box. In this chapter, you will explore two hallmarks of quantum mechanics: discrete allowed energy levels and the probabilistic motion of matter, in stark contrast to classical mechanics with its continuous energies and deterministic trajectories. How can these seemingly contradictory behaviors of matter both be true? To answer this question, you will explore the concept of the Bohr correspondence principle, which says that as system size or energy increases, a quantum system will approach classical behavior. To illustrate this concept, consider the simplest of all quantum systems: a particle of mass m in a onedimensional symmetrical box of length L.
The potential energy term for this system is described by
V(x) = $\left\{\begin{array}{cc}0& \frac{L}{2}<x<\frac{L}{2}\\ \infty & \mathbf{else}\end{array}\right.\,$
and the timeindependent Schrodinger equation is given by
$\frac{{\mathrm{\ℏ}}^{2}}{2m}\frac{{{d}}^{2}}{{\mathrm{dx}}^{2}}\mathrm{psi;}\left(x\right)plus;V\left(x\right)\mathrm{psi;}\left(x\right)equals;\mathrm{Epsi;}\left(x\right)period;$
The solutions to this equation are given by
$En\=\frac{{n}^{2}{h}^{2}}{8{\mathrm{mL}}^{2}}$
and
$\mathrm{\ψ}n\left(x\right)\=\sqrt{\frac{2}{L}}\mathrm{sin}\left(\frac{\mathrm{n\πx}}{L}\+\frac{\mathrm{n\π}}{2}\right)\.$
In the Interactive Concepts section below, you will explore the behavior of E_{n} and $\mathrm{\psi}n\left(x\right)$as m, L, and n increase. Despite its simplicity, this model can be used to describe translational motion of ideal gases, for example. In the Applications in Chemistry section, you will also see that the particle in a box can be used to model πelectrons in conjugated compounds!


Interactive Concepts



Discrete Energy Levels, Density of States, and the Bohr Correspondence Principle


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In this section, you will explore the energies of the particle in a 1D Box, which are given by
$En\=\frac{{n}^{2}{h}^{2}}{8{\mathrm{mL}}^{2}}$ (1)
where n = 1, 2, 3, .... In particular, you will see how E_{n} depends on the length of the box, L, and the particle mass, m. From the Equation (1), it is clear that as L increases, E_{n} decreases. Likewise for m.
It may be more instructive to see how the energy levels change. For this, you will make use of the plot function in Maple to visualize the density of states as a function of L and m. The density of states is related to the number of allowed energy levels that exist in a given energy range and is important in how energy is distributed throughout the system. For classical systems with a continuum of energy levels, the density of states is essentially infinite, meaning any energy is allowed in a given energy range. However, for quantum systems, such as atoms and molecules with a discrete energy manifold, the density of states is finite. This begs the question: when does a particle behave classically or quantum mechanically?
To answer this question, consider the simplest quantum particle, the particle in a box, and begin by looking at the effect of increasing m and L. In the Maple Input provided below, enter any positive real value for m and for L to see how the energy levels of the particle in a box are affected. [Note: To avoid the hassle of thinking in such small magnitudes , we use 'atomic units', whose magnitudes are on the order of unity. In atomic units, Z = 1, the unit for length is the Bohr radius, a_{0}, and the unit for mass is the mass of the electron, m_{e}. Therefore, a length of L = 2 is actually 2 a_{0}, or 1.058×10^{31 }m. Likewise, a mass of m = 2 is actually 2 ⋅ (9.11×10^{31}kg), or 1.822×10^{30} kg The unit for energy is the Hartree, E_{H}.]
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$\mathrm{with}\left(\mathrm{plots}\right)\:\mathrm{with}\left(\mathrm{LinearAlgebra}\right)colon;\mathrm{with}\left(\mathrm{ScientificConstants}\right)colon;$

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$L\u22541\;num;\mathrm{Enter\; length\; of\; box\; in\; atomic\; units}$

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$m\u22541\;num;\mathrm{Enter\; mass\; of\; particle\; in\; atomic\; units}$

Now, the next set few lines of Maple input will create a plot of the energies (in atomic units) of the particle in a box up to a maximum energy:
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$E\left(n\,L\right)\u2254\frac{{n}^{2}}{8\cdot m\cdot {L}^{2}}semi;\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\mathrm{Emax}\u225410semi;\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\mathbf{for}n\mathbf{from}1\mathbf{by}1\mathbf{while}E\left(ncomma;L\right)\le \mathrm{Emax}\mathbf{do}\mathbf{end}\mathbf{do}colon;\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\phantom{\rule[0.0ex]{0.0em}{0.0ex}}n\u2254n1colon;\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\mathrm{ME}\u2254\mathrm{Vector}\left(ncomma;1comma;\left(n\right)\to E\left(ncomma;L\right)\right)colon;\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\mathrm{ML}\u2254\mathrm{Vector}\left(ncomma;1comma;\left(n\right)\to L\right)colon;\phantom{\rule[0.0ex]{0.0em}{0.0ex}}A\u2254\mathrm{plot}\left(\mathrm{Emax}comma;xequals;0..2\cdot Lcomma;yequals;0..\mathrm{Emax}comma;\mathrm{tickmarks}equals;\left[0comma;0\right]\right)colon;\phantom{\rule[0.0ex]{0.0em}{0.0ex}}B\u2254\mathrm{plot}\left(\mathrm{ML}comma;\mathrm{ME}comma;\mathrm{symbolsize}equals;15comma;\mathrm{style}equals;\mathrm{point}comma;\mathrm{symbol}equals;\mathrm{circle}comma;\mathrm{color}equals;''Blue''\right)colon;\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\mathrm{plots}\left[\mathrm{display}\right]\left(Acomma;B\right)$

${E}{\u2254}\left({n}{\,}{L}\right){\mapsto}\frac{{{n}}^{{2}}}{{8}{\cdot}{m}{\cdot}{{L}}^{{2}}}$
 
${\mathrm{Emax}}{\u2254}{10}$
 
Now increase L and m in the Maple input above to see how the density of states changes as each increases. What do you notice? What happens to the density of states as m and L increase?

Answer


What we see is a demonstration of the Bohr correspondence principle, which says that as the size or energy of a quantum system increases, it approaches the behavior of a classical system. For example, for a small m and L, the spacing between energy levels is relatively large, and the discrete nature of the quantum system is clearly evident. But as m and L increase, the spacing between adjacent energy levels decreases, and for large enough m or L, the system appears to be more classical with 'all' energies being possible!



Wavefunctions for the Particle in a Box


In the section above, you saw a terrific illustration of the Bohr correspondence principle in seeing the density of states for the particle in a box increase with increasing m and L. Another hallmark of quantum mechanics is the interpretation of the wavefunction as a probability amplitude for finding the particle at a given value of x. Now, you will again consider the correspondence principle as you look at the wavefunctions and the resulting probability density for a particle in a 1D symmetrical box. For the potential
V(x) = $\left\{\begin{array}{cc}0& \frac{L}{2}<x<\frac{L}{2}\\ \infty & \mathbf{else}\end{array}\right.\,$
the resulting wavefunctions are given by
$\mathrm{\ψ}\left(x\right)\=\sqrt{\frac{2}{L}}\mathrm{sin}\left(\frac{\mathrm{n\πx}}{L}\+\frac{\mathrm{n\π}}{2}\right)comma;$
where n = 1, 2, 3, .... The corresponding probability density then is given by
$\mathrm{\ψ}\*\left(x\right)\mathrm{\ψ}\left(x\right)\=\frac{2}{L}{\mathrm{sin}}^{2}\left(\frac{\mathrm{npi;x}}{L}plus;\frac{\mathrm{npi;}}{2}\right)$
In the Maple input below, you will plot the ψ*ψ for a particle in a box with L = 1. Use the slide bar to change the value of n and answer the discussion questions.
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$\mathrm{restart}\:\mathrm{with}\left(\mathrm{plots}\right)\:$

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$\mathrm{psi}\left(n\,x\right)\u2254\mathrm{sqrt}\left(2\right)\cdot \mathrm{sin}\left(\mathrm{n}\cdot \mathrm{Pi}\cdot \mathrm{x}plus;\frac{\mathrm{n}\cdot \mathrm{Pi}}{2}\right)semi;$

${\mathrm{\psi}}{\u2254}\left({n}{\,}{x}\right){\mapsto}\sqrt{{2}}{\cdot}{\mathrm{sin}}{}\left({n}{\cdot}{\mathrm{\pi}}{\cdot}{x}{+}\frac{{1}}{{2}}{\cdot}{n}{\cdot}{\mathrm{\pi}}\right)$
 (2.2.1) 
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$\mathrm{Explore}\left(\mathrm{plot}\left({\left(\sqrt{2}\mathrm{sin}\left(\u230an\u2309\mathrm{\pi}x+\frac{\u230an\u2309\mathrm{\pi}}{2}\right)\right)}^{2}\,x\=0.5..0.5\,\mathrm{numpoints}\=500\,\mathrm{labels}\=\left[x\,''''\right]\right)\,'\mathrm{parameters}'=\left[n=1..100\right]\,'\mathrm{initialvalues}'=\left[n=1.00000000\right]\right)\;$


$\mathbf{n}$











For n = 1, what do you notice? Is the probability for finding the particle a constant, as in the classical analogue?

Answer


No, a maximum probability is seen in the center of the box! Very nonclassical behavior!

For n = 2, what do you notice?

Answer


There is a maxima at x = ± L/4 and a node in the center of the box where the probability for finding the particle is zero! That is, one will never find a particle in state n = 2 at the center of the box!! Even more nonclassical behavior!!

What happens as energy of the system increases even further?

Answer


n maxima and (n  1) nodes are evenly distributed throughout the box.

What happens as n goes to 10, 20, 30, 40, 50, ..., 100?

Answer


As n increases, we see essentially classical behavior with an equal probability for finding the particle at any x, in accord with the Bohr correspondence principle!!




Applications in Chemistry



Hydrogen Chain


As a simple, albeit somewhat contrived application, consider a linear chain of six H atoms, each atom 1 Angstrom away from its neighbors:
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$\mathrm{with}\left(\mathrm{QuantumChemistry}\right)\:\mathrm{Digits}\u225415\:$

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$\mathrm{molec}\u2254\left[\left[''H''\,0\,0\,0\right]\,\left[''H''\,0\,0\,1\right]\,\left[''H''\,0\,0\,2\right]\,\left[''H''\,0\,0\,3\right]\,\left[''H''\,0\,0\,4\right]\,\left[''H''\,0\,0\,5\right]\right]\;$

${\mathrm{molec}}{\u2254}\left[\left[{''H''}{\,}{0}{\,}{0}{\,}{0}\right]{\,}\left[{''H''}{\,}{0}{\,}{0}{\,}{1}\right]{\,}\left[{''H''}{\,}{0}{\,}{0}{\,}{2}\right]{\,}\left[{''H''}{\,}{0}{\,}{0}{\,}{3}\right]{\,}\left[{''H''}{\,}{0}{\,}{0}{\,}{4}\right]{\,}\left[{''H''}{\,}{0}{\,}{0}{\,}{5}\right]\right]$
 (3.1.1) 
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$\mathrm{PlotMolecule}\left(\mathrm{molec}\right)\;$

Rotate the plot to visualize the chain of Hatoms. (Or click on the plot and enter θ = 0, ϕ = 0, and ψ = 90.)
You will calculate the molecular orbitals, or wavefunctions, of the chain. A molecular orbital can be thought of a delocalized orbital across the entire molecule. So for a chain of Hatoms, the molecular orbitals should resemble particleinabox solutions! While the details of this calculation are beyond the scope of this activity, it is sufficient to note that six molecular orbitals will be constructed as a linear combination of the six 1s orbitals. According to the Pauli Principle, each molecular orbital may hold only two electrons with paired spins, so following Hund's rule, which orbital should be the HOMO?

Answer


For six electrons and two in each orbital, the HOMO should be the third molecular orbital. This can be checked by looking at the 'orbital occupation numbers', labeled in 'mo_occ' in the output below.

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$\mathrm{data}\u2254\mathrm{HartreeFock}\left(\mathrm{molec}\right)\;$

${\mathrm{table}}{}\left({\mathrm{\%id}}{\=}{18446744645440851006}\right)$
 (3.1.2) 
To visualize the calculated wavefunctions, use the Maple input below to plot the molecular orbitals.
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$\mathbf{for}i\mathbf{from}1\mathbf{by}1\mathbf{to}6\mathbf{do}\mathrm{DensityPlot3D}\left(\mathrm{molec}comma;\mathrm{data}comma;\mathrm{orbitalindex}equals;i\right)semi;\mathbf{end}\mathbf{do}semi;$

Again, click on each plot and orient the chain so that you can see the nodal pattern. Do the nodes correspond to what you expected for a particle in a box?
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