# Quantum Chemistry/Example 2

The correspondence principle for the particle in a 1D box

## Question

Starting with the wave equation of a 1D box:

${\displaystyle \Psi \left(x\right)={\sqrt {\frac {2}{L}}}sin\left({\frac {n\pi }{L}}x\right)}$

Show that the average probability of the particle in a 1D box follows the correspondence principle given that the average probability according to classical mechanics is:

${\displaystyle P\left(x\right)_{classical}={\frac {1}{L}}}$

Where ${\displaystyle L}$ is the length of the 1D box, ${\displaystyle n}$ is the principle quantum number, and ${\displaystyle x}$ is the position of the particle in a 1D box.

## Solution

The probability distribution of a particle in a 1D box is represented by the wavefunction multiplied by it's complex conjugate ${\displaystyle \Psi ^{*}}$ over the full length of the box. The probability of finding a particle in a specific range is determined by integrating the wavefunction multiplied by it's complex conjugate over a distance between two given values:

${\displaystyle P\left(x\right)=\int _{a}^{b}\Psi ^{*}(x)\Psi (x)}$

${\displaystyle \left\vert \Psi ^{2}\right\vert ={\sqrt {\frac {2}{L}}}\sin \left({\frac {n\pi }{L}}x\right)\cdot {\sqrt {\frac {2}{L}}}\sin \left({\frac {n\pi }{L}}x\right)}$

${\displaystyle \left\vert \Psi ^{2}\right\vert ={\frac {2}{L}}sin^{2}\left({\frac {n\pi }{L}}x\right)}$

The correspondence principle states that as quantum numbers become large, quantum mechanics reproduces expected results from classical mechanics. Therefore, the average probability of finding a particle in a 1D box for quantum mechanics should match classical mechanics as the quantum number reaches infinity according to the correspondence principle.

The average value of an integrand is given by the formula:

${\displaystyle f_{avg}\left(x\right)={\frac {1}{b-a}}\int _{a}^{b}f\left(x\right)dx}$

In this example, the function to be integrated is a ${\displaystyle \sin ^{2}x}$ function, which is a function with continuously repeating cycles from ${\displaystyle 0}$ to ${\displaystyle \pi }$. Therefore, determining the average over one cycle determines the average over an infinite amount of cycles, going to infinity represents the principle quantum ${\displaystyle n}$ going to very large numbers. So, the average of ${\displaystyle \sin ^{2}x}$ as ${\displaystyle n}$ goes to infinity is determined between one cycle from ${\displaystyle 0}$ to ${\displaystyle \pi }$.

${\displaystyle {\overline {\sin ^{2}\left(x\right)}}={\frac {1}{\pi -0}}\int _{0}^{\pi }\sin ^{2}\left(x\right)dx}$, as ${\displaystyle {x}\rightarrow \infty }$

Using the trigonometric relationships below, the solution to the original integral becomes trivial.

${\displaystyle \cos ^{2}x=1-\sin ^{2}x}$

${\displaystyle \cos {2x}=\cos ^{2}x-\sin ^{2}x}$

${\displaystyle \sin ^{2}x={\frac {1}{2}}-{\frac {\cos {2x}}{2}}}$

The new value of ${\displaystyle \sin ^{2}x}$ is inserted into the integral and solved for.

${\displaystyle {\overline {\sin ^{2}{\left(x\right)}}}={\frac {1}{\pi }}\int _{0}^{\pi }\left[{\frac {1}{2}}-{\frac {\cos {2x}}{2}}\right]dx}$ , as ${\displaystyle {x}\rightarrow \infty }$

${\displaystyle {\overline {\sin ^{2}{\left(x\right)}}}={\frac {1}{\pi }}\left({\frac {x}{2}}-{\frac {\sin {2x}}{4}}\right){\Big |}_{0}^{\pi }}$, as ${\displaystyle {x}\rightarrow \infty }$

${\displaystyle {\overline {\sin ^{2}{\left(x\right)}}}={\frac {1}{\pi }}\left({\frac {\pi }{2}}-0\right)}$, as ${\displaystyle {x}\rightarrow \infty }$

${\displaystyle {\overline {\sin ^{2}{\left(x\right)}}}={\frac {1}{2}}}$, as ${\displaystyle {x}\rightarrow \infty }$

Thus, the average value of ${\displaystyle \sin ^{2}x}$ as the principle quantum number goes to infinity is equal to ${\displaystyle {\frac {1}{2}}}$. By plugging that value into the probability distribution formula for a particle in a 1D box, the average probability becomes ${\displaystyle {\frac {1}{L}}}$.

${\displaystyle \left\vert \Psi ^{2}\left(x\right)\right\vert ={\frac {2}{L}}\sin ^{2}\left({\frac {n\pi }{L}}x\right)}$, as ${\displaystyle {n}\rightarrow \infty }$

${\displaystyle \left\vert \Psi ^{2}\left(x\right)\right\vert ={\frac {2}{L}}\cdot {\frac {1}{2}}}$

${\displaystyle \left\vert \Psi ^{2}\left(x\right)\right\vert ={\frac {1}{L}}}$

This matches the average probability of a particle in a 1D box for classical mechanics as given in the question and demonstrates the correspondence principle using the 1D box model.