Nearly everything we interact with in ordinary life is made of molecules.  Molecules of any size can be thought of as nuclei held together by electrons in much the same way as masses would be connected by springs.  As such, molecules vibrate just as masses attached by springs do.  [br][br]One difference between macroscopic and microscopic oscillators is that we only tend to notice the quantization of energy levels in microscopic ones.  Another difference as we'll see when we numerically calculate the wave functions for the QHO is that such molecular oscillators can extend to lengths that would be classically forbidden and lead to negative kinetic energies.  This is much like the barrier tunneling in that there is an exponentially decaying tail on the wave function in such regions where U(x)>E.[br][br]The fact that the energy levels of the QHO are quantized should not be surprising since it was Max Planck that suggested it.  He suggested that oscillators can not just vibrate at any amplitude, but must contain an integer multiple of quanta, or a total energy of [math]E=\mathbb{N}_1hf[/math], which we'll see was actually incorrect, but close enough to explain blackbody radiation.  [br][br]Albert Einstein suggested that light itself must be seen as coming as packets of energy (or field quanta) of the same amount of energy, E=hf.  I hope you see that these facts are all related.  Molecules oscillate with quantized energies, and they give off photons as they transition from higher to lower quantized energy states.   

A molecular oscillator is just like a mass/spring system from first semester.  The big and important difference is that the mass/spring system had one fixed (and assumed unmovable) support on one end and a mass on the other.  In such cases, as you probably recall, the angular frequency of the vibration is just [math]\omega=\sqrt{\frac{\kappa}{m}[/math] where I use [math]\kappa[/math] (kappa) for the elastic constant rather than k here so it doesn't get confused with the wave number.[br][br]The difference between that mass/spring system and a vibrating molecular oscillator is that all the atoms in a molecule are free to move.  In the simplest case of a diatomic molecule we have to account for the fact that both atoms move. To do this, in the expression of angular frequency, we use what's called the [b]reduced mass [/b]of the system.  The reduced mass accounts for the fact that with a movable mass on either end, it will be twice as easy for a given spring to draw them together - since they both move rather than just one of them.  It also allows us to find a value to use for mass when the two atoms are of different mass such as CO (carbon monoxide).  The expression for reduced mass is [br][br][center][math]m_{reduced}=\frac{m_1m_2}{m_1+m_2}.[/math][/center]So in the case of CO, for instance, the value in grams per mole (which would need to be converted to kg) is just [math]m_{reduced}=\frac{12(16)}{12+16}=6.85.[/math]

As a general rule, molecules will oscillate at many different frequencies all at the same time.  The bigger the molecule, the larger that number will be.  For any polyatomic molecule that is non-linear in structure the number of frequencies (some may be redundant or degenerate due to symmetry) is 3N-6, where N is the number of atoms in the molecule.  For linear molecules it is 3N-5.  [br][br]Thus for a diatomic molecule like the diatomic nitrogen that makes up much of the air we breathe, we have 3(2)-5 = 1 vibration which occurs when the molecule stretches the "spring" between the two atoms.  With a molecule like water, which is non-linear due to its bend, we have 3(3)-6 = 3 frequencies.  [br][br]For caffeine, one of my favorite molecules, there are 24 atoms which contribute to 3(24)-6 = 66 frequencies of vibration.  As you can see the number grows rapidly.  Any chunk of bulk material has very, very many frequencies of vibration that may be essentially thought of as a continuum of values.  [br][br]Whatever the case may be, those frequencies have quantized energies.  If we know the frequency of the vibration, we know that the corresponding energy of such a QHO will be[br][br][center][math]E_N=(N+\frac{1}{2})hf=(N+\frac{1}{2})\hbar\omega.[/math][/center]These results come from solving Schrödinger's equation using the same potential energy that we derived in first semester for the mass on a spring, or [math]U(x)=\frac{1}{2}\kappa x^2.[/math]  While the addition of this potential energy term to Schrödinger's equation isn't hard to understand, the solution of the equation becomes very difficult.  It requires sophisticated techniques and is usually written in terms of special functions called Hermite polynomials developed exactly for an equation of this type.  We will not go into the analytic solution, but using GeoGebra we can rather easily solve Schrödinger's equation numerically.[br][br]Before doing that, however, I want to point out that this result should be contrasted with Max Planck's E=Nhf.  We find that the QHO actually has a non-zero ground state energy of [math]E=\frac{1}{2}hf[/math].  This is called the zero point energy, and indicates that molecules can never stop moving such that they cease to contain vibrational energy, even at absolute zero.  Notice that while this result is different from that of Max Planck, that the energy level steps are identical in both equations, and that's why his hypothesis still predicted correctly the blackbody spectrum of an object. 

Light typically provides the energy to allow a molecular oscillator like the molecules of air to be excited to energy levels above their ground states. In such cases there is what's called a [b]selection rule[/b] that limits the possibility of transitions. The rule is that the quantum number can only be raised or lowered by one, or [math]N\rightarrow N\pm 1.[/math] Given the fact that for any oscillator all the energy levels are separated by the same amount (hf), this means only a single wavelength of light may be absorbed or emitted. To find that wavelength, simply set the energy of a photon equal to [math]\Delta E.[/math] This gives:[br][br][center][math]E_{photon}=\Delta E = E_{N+1}-E_N = hf_{oscillator}. \\[br]\frac{hc}{\lambda}= hf_{oscillator}, \text{ or in terms of f } \\[br]hf_{light} = hf_{oscillator}. \\[br]\text{Here $f_{oscillator}$=\frac{\omega}{2\pi}, and \omega=\sqrt{k/m_{reduced}}.} \\[br]\text{Notice that the frequency of light absorbed is equal to the } \\[br]\text{frequency of the molecular oscillator. }[br][/math][/center][br][br]I should mention here that when light is absorbed by a molecule to excite it to a more excited vibrational state, that the molecule can simultaneously transition to a different rotational state. Therefore the spectrum is more complicated than this one frequency in actual practice, but in a way that can be easily predicted. Such spectra are called, rather appropriately, vibrational/rotational spectra. To read more about such spectra, please see this [url=https://en.wikipedia.org/wiki/Rotational%E2%80%93vibrational_spectroscopy]link[/url].

Solving equations like Schrödinger's equation numerically is rather similar to other equations we've solved numerically.  In the majority of cases we had second order differential equations to solve, which had to be written as a pair of first order equations.  This will still be the case since Schrödinger's equation is also second order.[br][br]Solving such an equation numerically is conceptually a simple process.  If we algebraically shuffle the differential equation such that we isolate the second derivative (curvature), we can think of that as a recipe.  Given a starting value of the wave function and a starting slope, that second derivative recipe will find each successive point on the function such that the curvature recipe is satisfied.  In tiny steps each additional point is found until what looks like a function is found. Keep in mind that this "function" obtained via numerical methods is really a series of discrete points rather than a true function.[br][br]The trouble with numerical solutions like this is that they will give us such a function for any choice of energy, and energy is what we want to find along with the wave function.  If you look at the shape of the wave functions that arise from arbitrary choices of energy, however, they will always diverge in the forbidden energy regions where U(x)>E, which we know is non-physical, just as we found we had to throw away the growing exponential term in our solution to the barrier tunneling problem.[br][br]If the energy, however, is chosen carefully, the solution will lead to an exponentially decaying tail in such regions, and when we see that convergence we know we have a valid energy value and corresponding wave function.  Due to numerical algorithms being very good, but not perfect solutions to equations, the convergence of that decaying tail may not be perfect, as any instability in the numerical solution will cause sudden divergence in the forbidden region even after the function looked like it was settling.  You'll see this in our lab exercise, and need not be concerned about it.

Schrödinger's equation for the QHO is written[br][br][center][math][br]-\frac{\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+U(x)\Psi(x)=E\Psi(x). \\[br]\text{Shuffling terms gives } \\[br]\frac{\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+(E-U(x))\Psi(x)=0 \\[br]\frac{d^2\Psi}{dx^2}+\frac{2m}{\hbar^2}(E-U(x))\Psi(x)=0 \\[br]\text{Plugging in the expression for potential energy using \kappa for spring constant gives } \\[br]\frac{d^2\Psi}{dx^2}+\frac{2m}{\hbar^2}(E-\frac{1}{2}\kappa x^2)\Psi(x)=0 \\[br]\text{Isolating the second derivative (the way GeoGebra needs it) gives: } \\[br]\frac{d^2\Psi}{dx^2}=-\frac{2m}{\hbar^2}(E-\frac{1}{2}\kappa x^2)\Psi(x)[br][/math][/center] [br][br]We need to arrange this equation as a pair of first order differential equations. I will use funny looking names like P for psi, and dPdx as a variable name to contain the derivative of the wave function since it's easy to understand. Thus we have to enter these two equations into GeoGebra:[br][code]P'(x,P,dPdx)=dPdx [/code] which simply defines the first derivative of the wave function, and [code]dPdx'(x,P,dPdx)=-2*(E-1/2*x^2)*P[/code] which defines the second derivative according to the equation we got above. In the second equation I assumed for simplicity that [math]m=\hbar=\kappa = 1.[/math] Doing this will give energies in units of [math]\hbar\omega.[/math] So if you do this calculation and find an energy of 2.0 leads to a proper wave function, understand that [math]E=2.0\hbar\omega.[/math][br][br]We will look into this solution further during your lab exercise. You'll see that it will be necessary to search for even and odd functions separately. This will be done by careful selection of boundary conditions. Below is an interactive graphic showing the even solutions to the QHO.[br][br]At least two things are instructive about the graphic. First you'll see that the wave function is wavy in the region where E>V, and decays in the region where E<V so long as you choose an energy that causes the solution to settle in that forbidden energy region. If not, the function will rapidly diverge in the forbidden region. The tendency to be wavy or oscillate where E>V is universal. Likewise, the decay of a wave function in forbidden energy regions is also universal.