7.5 The quantum harmonic oscillator  (Page 2/4)

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The quantum harmonic oscillator

One problem with this classical formulation is that it is not general. We cannot use it, for example, to describe vibrations of diatomic molecules, where quantum effects are important. A first step toward a quantum formulation is to use the classical expression $k=m\text{ }{\omega }^{\text{ }2}$ to limit mention of a “spring” constant between the atoms. In this way the potential energy function can be written in a more general form,

$U\left(x\right)=\frac{1}{2}m{\omega }^{\text{ }2}{x}^{\text{ }2}.$

Combining this expression with the time-independent Schrӧdinger equation gives

$-\frac{\hslash }{2m}\phantom{\rule{0.2em}{0ex}}\frac{{d}^{\text{ }2}\psi \left(x\right)}{d\text{ }{x}^{2}}+\frac{1}{2}m{\omega }^{\text{ }2}{x}^{\text{ }2}\psi \left(x\right)=E\psi \left(x\right).$

To solve [link] —that is, to find the allowed energies E and their corresponding wave functions $\psi \left(x\right)$ —we require the wave functions to be symmetric about $x=0$ (the bottom of the potential well) and to be normalizable. These conditions ensure that the probability density $|\psi \left(x\right){|}^{\text{ }2}$ must be finite when integrated over the entire range of x from $\text{−}\infty$ to $+\infty$ . How to solve [link] is the subject of a more advanced course in quantum mechanics; here, we simply cite the results. The allowed energies are

${E}_{n}=\left(n+\frac{1}{2}\right)\hslash \omega =\frac{2n+1}{2}\hslash \omega ,\phantom{\rule{0.5em}{0ex}}n=0,1,2,3,...$

The wave functions that correspond to these energies (the stationary states or states of definite energy) are

${\psi }_{n}\left(x\right)={N}_{n}{e}^{\text{ }-{\beta }^{\text{ }2}{x}^{\text{ }2}\text{/}2}{H}_{n}\left(\beta x\right),\phantom{\rule{0.5em}{0ex}}n=0,1,2,3,...$

where $\beta =\sqrt{m\text{ }\text{ }\omega \text{/}\hslash }$ , ${N}_{n}$ is the normalization constant, and ${H}_{n}\left(y\right)$ is a polynomial of degree n called a Hermite polynomial . The first four Hermite polynomials are

$\begin{array}{c}{H}_{0}\left(y\right)=1\hfill \\ {H}_{1}\left(y\right)=2y\hfill \\ {H}_{2}\left(y\right)=4{y}^{2}-2\hfill \\ {H}_{3}\left(y\right)=8{y}^{3}-12y.\hfill \end{array}$

A few sample wave functions are given in [link] . As the value of the principal number increases, the solutions alternate between even functions and odd functions about $x=0$ .

Classical region of harmonic oscillations

Find the amplitude A of oscillations for a classical oscillator with energy equal to the energy of a quantum oscillator in the quantum state n .

Strategy

To determine the amplitude A , we set the classical energy $E=k{x}^{2}\text{/}2=m\text{ }{\omega }^{2}{A}^{2}\text{/}2$ equal to ${E}_{n}$ given by [link] .

Solution

We obtain

${E}_{n}=m\text{ }{\omega }^{2}{A}_{n}^{\text{ }2}\text{/}2\phantom{\rule{0.5em}{0ex}}⇒\phantom{\rule{0.5em}{0ex}}{A}_{n}=\sqrt{\frac{2}{m\text{ }{\omega }^{\text{ }2}}{E}_{n}}=\sqrt{\frac{2}{m\text{ }{\omega }^{\text{ }2}}\phantom{\rule{0.2em}{0ex}}\frac{2n+1}{2}\hslash \omega }=\sqrt{\left(2n+1\right)\frac{\hslash }{m\text{ }\omega }}.$

Significance

As the quantum number n increases, the energy of the oscillator and therefore the amplitude of oscillation increases (for a fixed natural angular frequency. For large n , the amplitude is approximately proportional to the square root of the quantum number.

Several interesting features appear in this solution. Unlike a classical oscillator, the measured energies of a quantum oscillator can have only energy values given by [link] . Moreover, unlike the case for a quantum particle in a box, the allowable energy levels are evenly spaced,

$\text{Δ}E={E}_{n+1}-{E}_{n}=\frac{2\left(n+1\right)+1}{2}\hslash \omega -\frac{2n+1}{2}\hslash \omega =\hslash \omega =h\text{ }f.$

When a particle bound to such a system makes a transition from a higher-energy state to a lower-energy state, the smallest-energy quantum carried by the emitted photon is necessarily hf . Similarly, when the particle makes a transition from a lower-energy state to a higher-energy state, the smallest-energy quantum that can be absorbed by the particle is hf . A quantum oscillator can absorb or emit energy only in multiples of this smallest-energy quantum. This is consistent with Planck’s hypothesis for the energy exchanges between radiation and the cavity walls in the blackbody radiation problem.

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