Quantum Chemistry/Example 29

Quantum Mechanical Harmonic Oscillator Wavefunction

A system undergoing harmonic motion around an equilibrium is known as a harmonic oscillator.

In quantum chemistry, the harmonic oscillator refers to a simplified model often used to describe how a diatomic molecule vibrates. This is because it behaves like two masses on a spring with a potential energy that depends on the displacement from the equilibrium, but the energy levels are quantized and equally spaced. The potential energy $V(x)=\frac{1}{2}k{x}^2$ is non-zero and can theoretically range from $x=[-\mathrm{\infty }\text{ , }\mathrm{\infty }]$.

The wavefunction for the quantum harmonic oscillator is given by the Hermite polynomials multiplied by the Gaussian function. The general form of the wavefunction is:

${\psi }_v=N_v\cdot H_v\left(\sqrt{\frac{mw}{\hslash }}x\right)\cdot \exp \left(\frac{-mw}{2\hslash }{x}^2\right)$

With:

$v=\text{Quantum number}$

$N_v=\text{Normalization factor}$

$H_v=v\text{-th Hermite polynomial}$

The first four Hermite polynomials are:

$\begin{array}{cc}{H}_0(x)& =1,\\ H_1(x)& =2x,\\ H_2(x)& =4x^2-2,\\ H_3(x)& =8x^3-12x.\\ \end{array}$

$m=\text{Mass of particle}$

$w=\text{Angular frequency of the oscillator}$

$\hslash =\text{Reduced Planck's constant}$

$x=\text{Position}$

Normalization of a Wavefunction

The probability of finding the particle in any state is given by the square of the wavefunction. Therefore normalizing a wavefunction in quantum mechanics means ensuring that the total probability of finding a particle in all possible positions is equal to 1. A normalized wavefunction is one that satisfies the normalization condition:

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\mid \psi (x){\mid }^{2}dx=1$

Normalization is important because it ensures that the probability of finding the particle somewhere in space is 100%.

Example

Show the derivation of the normalization factor of the v=1 state of the harmonic oscillator beginning from the unnormalized wavefunction.

${\psi }_v=N_v\cdot H_v\left(\sqrt{\frac{mw}{\hslash }}x\right)\cdot \exp \left(\frac{-mw}{2\hslash }{x}^2\right)$

$N_v^2\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{H}_v^2\left(\sqrt{\frac{mw}{\hslash }}x\right)\cdot \exp \left(\frac{-mw}{\hslash }{x}^2\right)dx=$

$\text{Let }y=\left(\sqrt{\frac{mw}{\hslash }}\right)x$

$N_v^2\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{H}_v(y)H_v(y)\cdot \exp (-y^2)\sqrt{\frac{\hslash }{mw}}dy=$

$N_v^2\sqrt{\frac{\hslash }{mw}}(-1{)}^v\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{H}_v(y)\frac{d^v}{d{y}^v}\exp (-y^2)dy=$

$N_v^2\sqrt{\frac{\hslash }{mw}}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\exp (-y^2)\frac{d^v}{d{y}^v}{H}_v(y)dy=$

$N_v^2\sqrt{\frac{\hslash }{mw}}{2}^{v}v!\sqrt{\pi }=$

$N_v=\frac{1}{\sqrt{2^{v}v!}}{\left(\frac{mw}{\pi \hslash }\right)}^{1/4}$

Sub in v=1

$N_1=\frac{1}{\sqrt{2^{1}1!}}{\left(\frac{mw}{\pi \hslash }\right)}^{1/4}=\frac{1}{\sqrt{2}}{\left(\frac{mw}{\pi \hslash }\right)}^{1/4}$

Check

Prove that ${\psi }_1$ is normalized (using notation from class)

${\psi }_1={\left(\frac{4{\alpha }^3}{\pi }\right)}^{1/4}x\exp \left(\frac{-\alpha x^2}{2}\right)$, where $\alpha =\left(\sqrt{\frac{k\mu }{{\hslash }^2}}\right)$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\mid {\psi }_1{\mid }^{2}dx={\left(\frac{4{\alpha }^3}{\pi }\right)}^{1/2}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^2{\exp }^{-\alpha x^2}dx={\left(\frac{4{\alpha }^3}{\pi }\right)}^{1/2}\left(\frac{1}{2\alpha }{\left(\frac{\pi }{\alpha }\right)}^{1/2}\right)=1$