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 $$V(x) = \frac{1}{2}kx^2  $$ is non-zero  and can theoretically range from $$x =[-\infty \text{, } \infty] $$.

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:

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

With:

$$v = \text{Quantum number} $$

$$N_v = \text{Normalization factor}$$

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

The first four Hermite polynomials are:

$$\begin{align} H_0(x) &= 1, \\ H_1(x) &= 2x, \\ H_2(x) &= 4x^2 - 2, \\ H_3(x) &= 8x^3 - 12x. \\ \end{align} $$

$$m = \text{Mass of particle} $$

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

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

$$x=\text{Position}  $$

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:

$$\int_{-\infty}^{\infty} \mid \psi (x) \mid^2 dx =1 $$

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.

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

$$N_{v}^{2} \int_{-\infty}^{\infty} H_{v}^{2} \left( \sqrt{\frac{mw}{\hbar}}x \right) \cdot \exp\left( \frac{-mw}{\hbar}x^2 \right) dx = $$

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

$$N_{v}^{2} \int_{-\infty}^{\infty} H_{v}(y) H_{v}(y) \cdot \exp\left( -y^2 \right) \sqrt{\frac{\hbar}{mw}} dy = $$

$$N_{v}^{2} \sqrt{\frac{\hbar}{mw}} (-1)^{v} \int_{-\infty}^{\infty} H_{v}(y) \frac{d^{v}}{dy^{v}} \exp\left( -y^2 \right) dy = $$

$$N_{v}^{2} \sqrt{\frac{\hbar}{mw}} \int_{-\infty}^{\infty}\exp(-y^{2}) \frac{d^{v}}{dy^{v}} H_{v}(y) dy = $$

$$N_{v}^{2} \sqrt{\frac{\hbar}{mw}} 2^{v}v!\sqrt{\pi} = $$

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

Sub in v=1

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

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 $$\alpha = \left( \sqrt{\frac{k\mu}{\hbar^{2}}} \right)  $$

$$\int_{-\infty}^{\infty} \mid \psi_{1}\mid^2 dx = \left( \frac{4 \alpha^3 }{\pi} \right)^{1/2} \int_{-\infty}^{\infty} x^2 \exp^{-\alpha x^2} dx = \left( \frac{4 \alpha^3 }{\pi} \right)^{1/2} \left( \frac{1}{2 \alpha} \left( \frac{\pi}{\alpha} \right)^{1/2} \right) = 1 $$