Molecules vibrate. It was Max Planck who suggested that the vibrational energy levels of such molecules are multiples of E=hf. This allowed him to correctly predict the spectrum of light given off by black bodies. It was also the emitted radiation (photons) of those oscillators that allowed Einstein to show why the photoelectric effect experiments worked the way they did. In other words, when quantum oscillators drop to lower energy levels and lose energy equal to E=hf, the photons they emit carry away exactly that amount of energy.[br][br]While those two ideas contributed to the success of early quantum physics, it turns out Max Planck was wrong about the energy levels of quantum oscillators - the name given to vibrating molecules. It turns out he was correct about the energy [i]between [/i]the levels, but missed the fact that the lowest energy state is not E=hf, but is instead half that value. [br][br]In this section we will discuss the quantum harmonic oscillator (QHO), and derive the energy levels using the Wilson-Sommerfeld quantization condition.

Starting with the expression [math]\oint p\;dx = \mathbb{N}_1 h,[/math] we wish to find the energy levels of a QHO. This will be the first occasion where will will have to actually evaluate the integral rather than pulling momentum out as a constant. To do this, we need an expression for [math]p(x)[/math] in terms of energies - both kinetic and potential.[br][br]To do this we should write total energy is the sum of kinetic and potential energies, or [math]E=K+U.[/math] The kinetic energy is related to the momentum by [math]K=\frac{p^2}{2m}.[/math][br][br]The potential energy is the same one we derived in first semester for a mass/spring system, or [math]U=\frac{1}{2}kx^2,[/math] where [math]x[/math] represents displacement from equilibrium. [br][br]Combined, this allows us to write [math]E=\frac{p^2}{2m} + \frac{1}{2}kx^2.[/math] Solving for momentum, we get the expression that needs to go in the W-S integral:[br][br][center][math]p=\sqrt{2m(E-\frac{1}{2}kx^2)}. [/math][/center][br][br]Plugging into the W-S quantization condition gives us:[br][br][center][math]\oint p\;dx=\oint \sqrt{2m(E-\frac{1}{2}kx^2)}\;dx=h\mathbb{N}_1.[/math][/center] [br][br]Since the integral needs to be taken over a complete cycle of the oscillator, we need to find the end points (in x) of such an oscillator. At the end points, if we think of a classical oscillator, the kinetic energy is zero. This implies [math]E=U=\frac{1}{2}kx^2,[/math] which leads to [math]x=\pm\sqrt{\frac{2E}{k}}.[/math][br][br]The last step is applying these limits to the integral. Since we start where we finish, and since the potential is symmetric, we can double the integral and integrate only over half the cycle - a common practice from calculus. This give us a final integral that we need to evaluate:[br][br][center][math]2\int_{-\sqrt{\frac{2E}{k}}}^{\sqrt{\frac{2E}{k}}} \sqrt{2m(E-\frac{1}{2}kx^2)}\;dx=\mathbb{N}_1 h.[/math][/center]

There are several ways to solve this integral. The first and perhaps easiest way that came to mind for me was the following:[br][br][center][math][br]2\int_{-\sqrt{\frac{2E}{k}}}^{\sqrt{\frac{2E}{k}}} \sqrt{2m(E-\frac{1}{2}kx^2)}\;dx=\mathbb{N}_1 h \\[br]2\sqrt{2mE}\int_{-\sqrt{\frac{2E}{k}}}^{\sqrt{\frac{2E}{k}}} \sqrt{(1-\frac{k}{2E}x^2)}\;dx=\mathbb{N}_1 h \\[br]\text{Substitute \cos\theta = \sqrt{\frac{k}{2E}} x.}\\[br]\text{This gives dx=-\sin\theta\ \;d\theta\sqrt{\frac{2E}{k}}.}\\[br]\text{Looking at the expression for x, our limits become \pm\pi and 0.} \\[br]\text{Making all those substitutions turns our integral into:}\\[br]-2\sqrt{2mE}\sqrt{\frac{2E}{k}} \int_{\pi}^{0} \sqrt{1-\cos^2\theta}\sin\theta\;d\theta=\mathbb{N}_1 h. \\[br]-\frac{4E}{\omega} \int_{\pi}^{0} \sin^2\theta\;d\theta=\mathbb{N}_1 h \\[br]\text{From here, either a substitution using the double angle formula, Euler's formula, or memory, may be used.} \\[br]\text{Recall that we have looked at squared sine and cosine functions before. Over half cycles they always average 1/2.} \\[br]\text{That makes the integral evaluate to \pi/2. Plugging in gives:} \\[br]\frac{2 \pi E}{\omega}=h\mathbb{N}_1. \\[br]E_N = \frac{h}{2 \pi}\omega\mathbb{N}_1. \\[br]\text{It is common to define the reduced Planck's constant as \hbar = \frac{h}{2\pi}.} \\[br]E_N = \hbar\omega\mathbb{N}_1 \text{ or } E_N=hf\mathbb{N}_1.[br][/math][/center][br][br]All that work for such a simple result! This one is in error as compared with the more modern quantum mechanical treatment... unless we consider that waves sometimes reflect from boundaries with a phase inversion. That's what we'll consider next.[br]

The result above is the same one that Max Planck had proposed for the energy states of quantum oscillators so that he could predict the blackbody spectrum. The problem is that this result is at odds with later quantum theory for the ground state (or zero point) energy of the QHO. That zero point energy is supposed to be [math]E_0=\frac{1}{2}\hbar\omega, [/math] or only half of what this result predicts.[br][br]It is interesting to wonder why a real oscillator behaves differently than our results above predict. It could be that we are doing something fundamentally wrong and are therefore getting the wrong answer, but we do know that nature is full of waves, and we have treated our molecule as such. [b]The only thing we have not considered is whether our wave may be receiving a reflective phase change (inversion) from one or the other of the boundaries[/b]. [br][br]Using the potential energy function [math]U=\frac{1}{2}kx^2,[/math] there is symmetry and such a reflective phase change will either occur at neither boundary or at both, and from our early wave studies we know that in either case it won't make a difference. What we need to arrive at the correct zero point energy is only one reflective phase change. But why would that happen? [br][color=#1e84cc][br]NOTE: I should mention at this point that this material, as far as I am aware, has never been proposed before. In this sense read on and make of it what you may.[/color][br][br]It is both instructive to note and interesting to consider that in reality molecular bonds do not have a symmetric form. They are often modeled as either Morse potentials or Lennard-Jones potentials - both of which possess asymmetry. A link may be seen here: [url=https://en.wikipedia.org/wiki/Interatomic_potential]https://en.wikipedia.org/wiki/Interatomic_potential[/url][br][br]If you recall, when the speed of a wave changes as it travels through a medium, it will reflect. If it goes from high speed to low speed it inverts, and from low to high it does not. In the context of a potential function the impedance correlates with the curvature (2nd derivative) of the potential. In the case of the Morse and Lennard-Jones potentials, one side has an ever increasing value of second derivative and the other a decreasing one. This leads to the obvious conclusion that we should consider a reflective phase change in our calculation! Doing so will allow us to calculate the correct zero point energy for the QHO. Such a calculation was not considered by Wilson or Sommerfeld. Let's make that correction to see how it turns out.

If you go back to the original formulation of the Wilson-Sommerfeld quantization condition we quantified phase [math]\delta.[/math] That was assumed to be in multiples of [math]2\pi.[/math] What we were quantifying was total phase. If we go back to that, we should really have written:[br][br][center][math][br]\delta_{total}=\delta_{path-length}+\delta_{reflective} = \frac{2\pi}{h}\oint p\;dx \pm \pi=2\pi\mathbb{N}_1. \\[br]\delta_{total}=\frac{2\pi}{h}\oint p\;dx=2\pi\mathbb{N}_1\pm \pi. \\[br]\delta_{total}= \oint p\;dx=\hbar(\mathbb{N}_0\pm \frac{1}{2}). \\[br]\text{If you plug in the result of the integral from the first attempt at the QHO \\[br]and clean up the algebra, you get: } \\[br]E_N=(\mathbb{N}_0+\frac{1}{2})\hbar\omega. [br][/math][/center][br][br]These are the same values that one gets with much more cumbersome math using standard quantum theory. The only question is whether such a reflective phase change is justifiable in all cases, such as bonds deep within the crystal structure of a solid where bond symmetries are likely present. Certain symmetric modes of oscillation of a crystal may not exhibit such zero point energies due to atoms experiencing a symmetric potential due to the circumstances related to the motion of the surrounding atoms. Finding a way to test this experimentally would be interesting.