Quantum Chemistry/Example 2

The correspondence principle for the particle in a 1D box

Question

Starting with the wave equation of a 1D box:

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

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:

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

Where $L$ is the length of the 1D box, $n$ is the principle quantum number, and $x$ 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 $\Psi ^{*}$ 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:

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

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

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

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:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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.