The Simple Harmonic Oscillator

This chapter will introduce a system that is fundamental to our understanding of more physical phenomena than any other. Although the “simple” harmonic oscillator seems to be only the combination of the most mundane components, the formalism developed to explain the behavior of a mass, spring, and damper is used to describe systems that range in size from atoms to oceans. Our investigation goes beyond the “traditional” treatments found in the elementary physics textbooks. For example, the introduction of damping will open a two-way street: a damping element (i.e., a mechanical resistance,Rm) will dissipate the oscillator’s energy, reducing the amplitudes of successive oscillations, but it will also connect the oscillator to the surrounding environment that will return thermal energy to the oscillator. The excitation of a harmonic oscillator by an externally applied force, displacement, or combination of the two will result in a response that is critically dependent upon the relationship between the frequency of excitation and the natural frequency of the oscillator and will introduce the critical concepts of mechanical impedance, resonance, and quality factor. Finally, the harmonic oscillator model will be extended to coupled oscillators that are represented by combinations of several masses and several springs.


Schrödinger's Equation and the Ground State Wave Function
The classical equation of motion for a one-dimensional simple harmonic oscillator with a particle of mass m attached to a spring having spring constant C is The total energy is clearly constant in time.It is often useful to picture the time-development of a system in phase space, in this case a two-dimensional plot with position on the x-axis, momentum on the y-axis.For dimensional consistency, we would plot mω x against p.It is evident that in these variables, the point representing the system in phase space moves clockwise around a circle of radius 2mE centered at the origin.
Note that in the classical problem, we could choose any point x, p, place the system there and it would then move in a circle about the origin.In the quantum problem, on the other hand, we cannot specify the initial x, p precisely, because of the uncertainly principle.
The best we can do is to place the system initially in a small cell in phase space, of size .In fact, we shall find that in quantum mechanics, phase space is always divided into cells of essentially this size for each pair of variables./ 2 x p ∆ ⋅∆ = From the expression for total energy above, the Schrödinger equation for the quantum oscillator follows in standard fashion: What will the solutions to this Schrödinger equation look like?Since the potential ½mω 2 x 2 increases without limit on going away from x = 0, no matter how much kinetic energy the particle has, if it gets far enough from the origin the potential energy dominates, and the (bound state) wavefunction will decay increasing rapidly as x increases further.(Obviously, for a real physical oscillator there is a limit on the height of the potential-we will assume that limit is much greater than the energies of interest in our problem.) We know that when a particle penetrates a barrier of constant height V 0 (greater than the particle's kinetic energy) the wave function decreases exponentially into the barrier, as , where But, in contrast to this constant height barrier, the "height" of the simple harmonic oscillator potential continues to increase as the particle penetrates to larger x.Obviously, in this situation the decay will be faster than exponential.If we (rather naïvely) assume it is more or less locally exponential, but with a local α varying with V 0 , neglecting E relative to V 0 in the expression for α suggests that α itself is proportional to x, so maybe the wavefunction decays as e To check this idea, we insert The ψ(x) is just a factor here, and it is never zero, so can be cancelled out.This leaves a quadratic expression which must have the same coefficients of x 0 , x 2 on the two sides, that is, the coefficient of x 2 on the left hand side must be zero: This fixes the wave function.Equating the constant terms fixes the energy: So the conjectured form for the wave function is in fact the exact solution for the lowest energy state!(It's the lowest state because it has no nodes.)Also note that even in this ground state the energy is nonzero, just as it was for the square well.The central part of the wave function must have some curvature to join together the decreasing wave function on the left to that on the right.This "zero point energy" is sufficient in one physical case to melt the lattice-helium is liquid even down to absolute zero temperature (checked down to microkelvins!) because the wave function spread destabilizes the solid lattice that will form with sufficient external pressure.

Higher Energy States
It is clear from the above discussion of the ground state that is the natural unit of length in this problem, and ω that of energy, so to investigate higher energy states we reformulate in dimensionless variables, Schrödinger's equation becomes Deep in the barrier, the ε term will become negligible, and just as for the ground state wave function, higher bound state wave functions will have behavior, multiplied by some more slowly varying factor.
We try solving this with a power series in ξ : Inserting this in the differential equation, and requiring that the coefficient of each power ξ n vanish identically, leads to a recurrence formula for the coefficients h n : Evidently, the series of odd powers and that of even powers are independent solutions to Schrödinger's equation.For large n >>ε, the recurrence relation simplifies to .
The series therefore tends to Ae Be So how do we find the nondiverging solutions?It is clear that the infinite power series must be stopped!The key is in the recurrence relation: if the energy satisfies 2ε = 2n + 1, with n an integer, h n+2 and all higher coefficients vanish.The remaining n th order polynomial is called a Hermite polynomial and written H n (ξ).
The standard normalization of the Hermite polynomials H n (ξ) is to take the coefficient of the highest power ξ n to be 2 n .The other coefficients are easy to find using the recurrence relation above, giving: So the bottom line is that the wavefunction for the n th excited state, having energy = n is a normalization constant to be determined in the next section.

Operator Approach to the Simple Harmonic Oscillator
Having scaled the position coordinate x to the dimensionless ξ by ξ = x/b, let us also scale the momentum from p to π = −i d/dξ (so ).The Hamiltonian is ( ) Dirac had the brilliant idea of factorizing this expression: the obvious thought − π isn't quite right, because it fails to take account of the noncommutativity of the operators, but the symmetrical version is fine, and we shall soon see that it leads to a very easy way of finding the eigenvalues and operator matrix elements for the oscillator, far simpler than using the wave functions we found above.Interestingly, Dirac's factorization here of a second-order differential operator into a product of first-order operators is close to the idea that led to his most famous achievement, the Dirac equation, the basis of the relativistic theory of electrons, protons, etc.
To continue, we define new operators by The solution, unnormalized, is 2 / 2 0 ( ) .Ce ξ ψ ξ − = (In fact, we've seen this equation and its solution before: this was the condition for the "least uncertain" wave function in the discussion of the Generalized Uncertainty Principle.) We denote the normalized set of eigenstates | 0

Now
and Therefore, if we take the set of orthonormal states | 0 as the basis in the Hilbert space, the only nonzero matrix elements of are (The column vectors in the space this matrix operates on have an infinite number of elements: the lowest energy, the ground state component, is the entry at the top of the infinite vector-so up the energy ladder is down the vector!) The adjoint For practical computations, we need to find the matrix elements of the position and momentum variables between the normalized eigenstates.
These matrices are, of course, Hermitian (not forgetting the i factor in p).
To find the matrix elements between eigenstates of any product of x's and p's, express all the x's and p's in terms of a's and 's, to give a sum of products of a's and a 's.Each product in this sum can be evaluated sequentially from the right, because each a or has only one nonzero matrix element when the product operates on one eigenstate.† a † † a

Normalizing the Eigenstates in x-space
The normalized ground state wave function is where we have gone back to the original x variable and normalized with the standard Gaussian result.
To find the normalized wave functions for the higher states, we first construct them using the creation operator acting on the ground state |0>, then write as a differential operator, acting on the ground state wave function given above.
We must now establish that this expression is the same as the Hermite polynomial wave function derived earlier, and to do that we need some further properties of the Hermite polynomials.

Hermite Polynomials
The Hermite polynomials are defined by 2 2 ( ) ( ) It follows immediately from the definition that the coefficient of the leading power is 2 n .
It is a straightforward exercise to check that H n is a solution of the differential equation We can transform ψ n (ξ) (from the end of the previous section) into this form by using the operator identity:

−
The standard approach to solving the general problem is to factor out the term, equation for h(ξ):

+
Actually we should have expected this-for a general value of the energy, the Schrödinger equation has the solution at large distances, and only at certain energies does the coefficient A vanish to give a normalizable bound state wavefunction.
indeed the same polynomials we found by the series solution of Schrödinger's equation earlier.
equivalence of the two approaches to Schrödinger's equation for the simple harmonic oscillator, and provides us with the overall normalization constants without doing cumbersome integrals.