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

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:

${\displaystyle \psi _{v}=N_{v}\cdot H_{v}\left({\sqrt {\frac {mw}{\hbar }}}x\right)\cdot \exp \left({\frac {-mw}{2\hbar }}x^{2}\right)}$

With:

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

${\displaystyle N_{v}={\text{Normalization factor}}}$

${\displaystyle H_{v}=v{\text{-th Hermite polynomial}}}$

The first four Hermite polynomials are:

{\displaystyle {\begin{aligned}H_{0}(x)&=1,\\H_{1}(x)&=2x,\\H_{2}(x)&=4x^{2}-2,\\H_{3}(x)&=8x^{3}-12x.\\\end{aligned}}}

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

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

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

${\displaystyle x={\text{Position}}}$

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:

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

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.

${\displaystyle \psi _{v}=N_{v}\cdot H_{v}\left({\sqrt {\frac {mw}{\hbar }}}x\right)\cdot \exp \left({\frac {-mw}{2\hbar }}x^{2}\right)}$

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

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

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

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

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

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

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

Sub in v=1

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

Check

Prove that ${\displaystyle \psi _{1}}$ is normalized (using notation from class)

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

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