# SIMPLE HARMONIC OSCILLATOR

## Introduction

The simple harmonic oscillator (SHO) is a model for molecular vibration. It represents the relative motion of atoms in a diatomic molecule or the simultaneous motion of atoms in a polyatomic molecule along an "normal mode" of vibration.

## Classical Oscillator

The classical harmonic oscillator is described by the following second order differential equation: . Here x is the displacement of the oscillating particle from its equilibrium position (x=0), t is the time, k is the Hooke's law force constant, and m is the particle mass. There are two integration constants in the general solution: x(t) = Acos(wt) + Bsin(wt). This solution describes an oscillatory motion with angular frequency w; the total energy of the oscillation is E = k(A2+B2)/2. Initial conditions are used to evaluate the constants. Thus, if the initial displacement and velocity are given as x0 and v0, the values are A=x0, B=v0/w. The total energy of this system is conserved and oscillates between kinetic and potential as seen from K(t) = (k/2)[A2sin2wt - 2ABcoswtsinwt + B2cos2wt] and V(t) = (k/2)[A2cos2wt + 2ABcoswtsinwt + B2sin2wt]. The extremes of displacement, ±xmax, are called the classical turning points and are given by xmax2 = 2E/k.

Notice in this solution that the frequency is determined by the system parameters k and m. Further, the allowed energies of the oscillator form a continuum 0 < E < ¥.

## Quantum Oscillator

For the quantum mechanical description, we use the Hamiltonian operator, , and Schroedinger's time independent equation, Hy=Ey. The solutions of this equation supply the allowed energy levels and corresponding energy eigenstate wave functions. Namely, we must solve the second order DE: subject to the usual conditions on y (finite, continuous, single valued, differentiable).

Here we exhibit the traditional solution in terms of Hermite polynomials. The solution is conveniently divided into three steps.

### Step one, Dimensionless form and natural units.

Scale the coordinate by a natural length unit for the system: x = ay, where the size of a is yet to be determined, and let u(y) = y(ay). Then the SHO Schroedinger equation becomes . If the natural length unit is chosen to be then we find , where    and  . The dimensionless form of the Schroedinger equation contains no units; its wave function u(y) is continuous, finite, and differentiable. These boundary conditions will require the energy levels e to be quantized. Dimensions can be restored to the solution using E = eEnatural, and y(x) = (a)-1/2u(x/a).

### Step two, Asymptotic solution.

In the limit of very large y, the eigenvalue is negligible compared to the potential. Namely, we have the limiting form , whose solutions, valid in the limit of large y, are u(y) = exp(±y2/2). The upper sign is unacceptable since then u is no longer finite in the limit of increasing y. Hence the asymptotic large variable limit of the solution is u(y) ~ exp(-y2/2).

As a result, we introduce the unknown form of the wave function at small y: u(y) = H(y) exp(-y2/2). When this is substituted into the dimensionless Schroedinger equation, we find the DE for H: H''-2yH'+(2e-1)H=0. This result is known as Hermite's differential equation. (Don't confuse the function H(y) with the hamiltonian operator, H.)

### Step three, Hermite Polynomials.

Solutions to Hermite's differential equation, are found in standard texts (e.g., H. Margenau and G. M. Murphy, The Mathematics of Physics and Chemistry, [van Nostrand, 1956]). One approach is the series method in which a power series expansion is assumed and recurrence relations are found to evaluate the coefficients of all powers. It is found in this way that the series diverges at large y unless 2e-1 is an even integer and that such solutions are finite series, i.e. polynomials. Hence, we have an acceptable solution if 2e-1 = 2n and n = 0, 1, ... .

In dimensionless form, the SHO energy levels are given by e   = n + 1/2. When the units are restored, E = Enatural(n + 1/2) = h_bar w (n + 1/2).

Hermite polynomials are one of the commonly used "special functions" and are tabulated in standard references (e.g., M. Abramowitz and I. A. Stegun, Handbook of Mathematical Functions [Dover, 1972]). Some modern mathematics software tools contain these functions too.

The complete wave function for the SHO has the form u(y) = NnHn(y) exp(-y2/2) where Nn is a normalizing factor. Or, when units are restored, the wave function is y(x) = (a)-1/2NnHn(x/a) exp(-x2/2a2).

Another analytical method of solution factors the hamiltonian into the product of two first-order differential operators, the creation and the annihilation operators or the step-up and step-down operators (see e.g., A. Das & A.C. Melissinos, Quantum Mechanics, [Gordon and Breach, 1986]; p. 364 ff). This factorization method

## Problems(see references for more exercises)

1. Derive expressions for the integration constants A and B of the classical oscillator in terms of the total energy and the initial displacement.
2. An alternate parameterization of the classical solution is sometimes used. Namely, x(t) = Dcos(wt+f) in which D is the amplitude of the displacement and f is the phase angle. Derive the formulas that express D and f in terms of A and B. Also derive the inverse relations for A and B in terms of D and f.
3. Many useful properties apply to the Hermite polynomials, Hn(x). Among these is the recursion formula, Hn+1 -2xHn +2nHn-1=0. Use this relation and the first two members, H0 = 1, H1=2x, to derive the following
H0 = 1
H1 = 2x
H2 = 4x2 -2
H3 = 8x3 - 12x
H4 = 16x4 - 48 x2 + 12
H5 = 32x5 - 160x3 + 120x
H6 = 64x6 - 480x4 + 720x2 - 120
4. For diatomic molecules, the SHO model fits reasonably well to the observed vibrational spectrum (as vibrational energy increases the fit is less good). For such systems, the appropriate mass is the "reduced mass", m = mAmB/(mA+mB), where mA and mB are atomic masses of the AB molecule. Hydrogen chloride in the gas phase shows strong absorptions in the infrared region of the electromagnetic spectrum at wave lengths 3446.5 nm, 1764.3 nm, 1198.0 nm, 915.45 nm, and 746.46 nm. These are assigned to the vibrational transitions 0 --> 1, 0 --> 2, etc. Use these data to
(a) Determine the vibrational frequency of HCl.
(b) Find the force constant of the HCl bond (assume the spectrum belongs to the common isotopes 1H and 35Cl).
(c) Predict the vibrational spectrum of 1H37Cl.
5. Compute the probability to find the quantum mechanical SHO beyond the classical turning points. Do this for the ground state by any method. [Hint: the error function and complementary error function are requied.]

## References

1. P. W. Atkins and R.S. Frieman, Molecular Quantum Mechanics [Oxford University Press, 1997], Ch. 2.
2. Almost any physical chemistry textbook will have a treatment of the SHO. For example, K.J. Laidler and J. H. Meiser, Physical Chemistry [Houghton Mifflin Company, 1995].
3. Introductory physics quantum mechanics textbooks also treat the SHO. For example, A. Das and A. C. Melissinos, Quantum Mechanics [Gordon and Breach, 1986].