VI The harmonic oscillator

The classical harmonic oscillator describes a particle subject to a restoring force $F=-m{\omega }^{2}x$ proportional to the distance from an equilibrium position $x=0$. Newton’s equation $m\ddot{x}=F$ results in an oscillatory motion $x(t)=x_0\cos \omega t+(v_0/\omega )\sin \omega t$, where $\omega =2\pi /T$ and $T$ is the oscillation period. In this solution, $x_0=x(0)$ is the initial position and $v_0=\dot{x}(0)$ is the initial velocity of the particle. According to $F=-V^{\prime }$, the force $F=-m{\omega }^{2}x$ corresponds to a parabolic potential energy

$$V(x)=\frac{1}{2}m{\omega }^2{x}^2.$$

(91)

In order to solve this problem quantum mechanically, we follow our standard steps. The Schrödinger equation of the harmonic oscillator is given by

$$E\psi (x)=-\frac{{\mathrm{\hslash }}^2}{2m}{\psi }^{\prime \prime }(x)+\frac{1}{2}m{\omega }^2{x}^2\psi (x).$$

(92)

This equation is again a linear differential equation of second order, but now one coefficient is position dependent. From our general considerations we already can anticipate the following:

• •

  The solutions $\psi (x)$ are continuous.

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  The derivatives ${\psi }^{\prime }(x)$ of the solutions are also continuous.

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  As $V(x)\to \mathrm{\infty }$ for $|x|\to \mathrm{\infty }$, the particle cannot escape to infinity at finite energy $E$. This only permits bound states, which decay $\psi (x)\to 0$ as $|x|\to \mathrm{\infty }$.

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  The energies of the bound states are discrete (i.e., only at certain energies we can find valid solutions of the Schrödinger equation).

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  The ground state energy $E_0$ will be larger than the classical minimal energy: $E_0>0$.

Let us first focus on finding the mathematical solutions of the Schrödinger equation. We start by bringing the Schrödinger equation into a simpler, rescaled form. Denote $\psi (x)=\phi (x\sqrt{m\omega /\mathrm{\hslash }})$, which corresponds to a rescaled position $y=x\sqrt{m\omega /\mathrm{\hslash }}$. According to the chain rule, ${\psi }^{\prime }=\sqrt{m\omega /\mathrm{\hslash }}{\phi }^{\prime }$ and ${\psi }^{\prime \prime }=(m\omega /\mathrm{\hslash }){\phi }^{\prime \prime }$. We also write energy as $E=\frac{\mathrm{\hslash }\omega }{2}ϵ$. In terms of $\phi$, the Schrödinger equation then reads

$$\frac{\mathrm{\hslash }\omega }{2}ϵ\phi (y)=-\frac{\mathrm{\hslash }\omega }{2}{\phi }^{\prime \prime }(y)+\frac{\mathrm{\hslash }\omega }{2}{y}^2\phi (y),$$

(93)

or, after cancelling the common factor $\frac{\mathrm{\hslash }\omega }{2}$,

$$ϵ\phi (y)=-{\phi }^{\prime \prime }(y)+y^2\phi (y).$$

(94)

Next, we introduce the new function $\phi (y)=e^{-y^2/2}f(y)$. By applying the product rule, we then find the following standard differential equation:

$$ϵf=-f^{\prime \prime }+2y{f}^{\prime }+f.$$

(95)

When $\epsilon \ne 2n+1$, where $n=0,1,2,\mathrm{\dots }$ is an integer, the solutions of Equation (95) grow rapidly for $|y|\to \mathrm{\infty }$, even outgrowing the factor $\exp (-y^2/2)$ appearing in $\phi (y)=e^{-y^2/2}f(y)$. These solutions do not fulfil our boundary conditions for $|x|\to \mathrm{\infty }$ and hence have to be discarded.

For $\epsilon =2n+1$, $n=0,1,2,\mathrm{\dots }$, however, Eq. (95) has simple solutions $f(y)=H_n(y)$ which are just polynomials of degree $n$. These polynomials are known as the Hermite polynomials. The first two of these polynomials are simply $H_0=1$ and $H_1=2y$. The other polynomials can be calculated recursively, $H_{n+1}(y)=2y{H}_n(y)-2n{H}_{n-1}(y)$.

The first two solutions correspond to the ground state and the first excited state of the harmonic oscillator:

Ground state: $n=0$, $f=1$, which gives to $f^{\prime }=f^{\prime \prime }=0$. This is a solution of Eq. (95) if $ϵ=1$, which corresponds to an energy $E_0=\frac{1}{2}\mathrm{\hslash }\omega$. The wave function is given by

$${\psi }_0(x)=c_0\exp \left(-x^2\frac{m\omega }{2\mathrm{\hslash }}\right),$$

where $c_0$ can be determined from the normalisation condition ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|\psi (x)|}^2𝑑x=1$ (see below). This gives $c_0={\left(\frac{m\omega }{\pi \mathrm{\hslash }}\right)}^{1/4}$.

First excited state: $n=1$, $f=2y$, hence $f^{\prime }=2$ and $f^{\prime \prime }=0$. This is a solution of Eq. (95) if $ϵ=3$, which corresponds to $E_1=\frac{3}{2}\mathrm{\hslash }\omega$. The wave function is given by

$${\psi }_1(x)=c_{1}x\exp \left(-x^2\frac{m\omega }{2\mathrm{\hslash }}\right),$$

with normalisation constant $c_1={\left(4\frac{m^3{\omega }^3}{\pi {\mathrm{\hslash }}^3}\right)}^{1/4}$.

The ground state wave function of the harmonic oscillator provides us with a good occasion to practice once more the normalisation of the wave function. Since we want to interpret ${|{\psi }_0(x)|}^2=P(x)$ as the probability density for position, we require

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|{\psi }_0(x)|}^2𝑑x=1.$$

With ${\psi }_0(x)=c_0\exp (-x^2\frac{m\omega }{2\mathrm{\hslash }})$ this integral reads

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{c}_0^2\exp \left(-x^2\frac{m\omega }{\mathrm{\hslash }}\right)𝑑x.$$

We use the standard integral ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\exp (-a{x}^2)𝑑x=\sqrt{\pi /a}$, where we set $a=\frac{m\omega }{\mathrm{\hslash }}$:

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{c}_0^2\exp \left(-x^2\frac{m\omega }{\mathrm{\hslash }}\right)𝑑x=c_0^2\sqrt{\frac{\pi \mathrm{\hslash }}{m\omega }}.$$

The equates to $1$ if $c_0={\left(\frac{m\omega }{\pi \mathrm{\hslash }}\right)}^{1/4}$, in agreement with the value given in the previous section.

In general, the normalised bound-state wave functions are given by

$${\psi }_n=C_n{H}_n(x\alpha )e^{-x^2{\alpha }^2/2},$$

(96)

$n=0,1,2,3,\mathrm{\dots }$, $C_n={\alpha }^{1/2}{\pi }^{-1/4}{(n!2^n)}^{-1/2}$, where $\alpha =\sqrt{\frac{m\omega }{\mathrm{\hslash }}}$.

The associated bound-state energies are

$$E_n=\mathrm{\hslash }\omega \left(n+\frac{1}{2}\right),n=0,1,2,\mathrm{\dots }.$$

(97)

Note that the difference between two consecutive energies is given by Planck’s relation $E=\mathrm{\hslash }\omega$.

You could now draw diagrams of the potential, bound state energies, ground- and excited state wave functions, and the associated probability densities of the particle position.