In harmonic oscillator, as vibration quantum number increases then does the distance of oscillation of particle from the mean position where there is minimum potential energy increases? If yes, then the expression of potential energy $(kx^2/2)$ which we derived from Taylor expansion of potential energy at mean position should not hold as we neglect high powers terms $(x-x_0)^3$ etc.but as x increases so we can't neglect high power terms. Then the expression of quantized energy $(n+1/2)\hbar\omega$ which we get by solving Schrödinger equation also doesn't hold.

The wave function diagram also shows that the distance from mean position increases as vibration quantum number increases.

Please tell me where is flaw in my argument.


2 Answers 2


Classically speaking, the (idealised) harmonic oscillator is defined by Hooke's law. Quoting from Wikipedia:

In classical mechanics, a harmonic oscillator is a system that, when displaced from its equilibrium position, experiences a restoring force $F$ proportional to the displacement $x$:

$$\vec{F} = -k\vec{x}$$

where $k$ is a positive constant. If $F$ is the only force acting on the system, the system is called a simple harmonic oscillator.

In one dimension we can drop the vectors and write this as $F = -kx$. But the force is also the (negative) derivative of the potential, i.e. $F = -\mathrm dV/\mathrm dx$, and integrating this to get $V = kx^2/2 + C$ is trivial. The constant of integration $C$ is arbitrary, but it doesn't fundamentally change the physics (it just moves the energy scale), so we can ditch it by setting it to zero.

If you had a potential energy with higher-order terms in $x$ (i.e. $x^3$ and above), then the derivative with respect to $x$ would contain terms that are $x^2$ and above. This directly contradicts the definition of the harmonic oscillator. It could be some other kind of oscillator, but its motion would not be called simple harmonic.

Consequently, for this particular oscillator, trying to perform a Taylor expansion is a little pointless: there are no terms in $x^3$ or above, so all the higher-order derivatives $\mathrm d^3V/\mathrm dx^3, \cdots$ are all zero, and all those terms in the Taylor expansion are zero. Of course, in the general case, Taylor analysis of a different oscillator could be very useful. Just not here.

The potential energy of the quantum harmonic oscillator is directly inherited from the classical simple harmonic oscillator, by replacing the classical quantities $x$ and $p~ (= mv)$ with their quantum analogues $\hat{x}$ and $\hat{p} ~(= -\mathrm i\hbar(\partial/\partial x))$. So given that the classical potential energy is $V = kx^2/2$, it's not too much of a stretch to get the quantum potential energy $V = kx^2/2$.


In a strictly harmonic oscillator then the energy and potential are exactly as you state. This is not the case in any real molecule because all real molecule dissociate at some energy, and so higher terms in $x$ should be included. The energy levels are changed in accordance with how the potential energy in the Schroedinger equation is changed. A common potential is the Morse potential of the form $(1-\exp(-bx)^2$ as this allows the Schroedinger equation to be solved exactly, see Wikipedia for details and graphs. (Very often, in practice, the potential is derived from experiment as it is very hard to calculate the potential accurately using MO methods. )

In your second paragraph, for the harmonic potential the average internuclear distance is the same for all energy levels. The average value is $\langle x \rangle =\int_{-\infty}^\infty \psi(x)_n x\psi(x)_n dx$. This is an 'odd function' which means that the integral over all space (or any symmetrical limits) is exactly zero. The wavefunction multiplied by itself must be symmetrical $\psi(x)^2 = \psi(-x)^2$ and multiplying by $x$ makes the integral odd meaning that if the function is $F$ then $F(x)=-F(-x)$.

The square of the average value is not zero, $\langle x^2 \rangle =\int_{-\infty}^\infty \psi(x)_n x^2\psi(x)_n dx$ as this is an 'even function'.


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