The quantum mechanical harmonic oscillator

A general intermolecular potential V(Q), where Q represents the nuclear coordinate, can be expressed as an expansion in powers of Q. This leads to a series of terms linear, quadratic etc. in Q. Mathmatically, we write

where Q₀ is the equilibrium nuclear position. The significance of this expansion is that we are taking an arbitrary potential V(Q) and expressing it as a polynomial. At equilibrium (i.e. when Q = Q₀) the first derivative ¶V/¶Q = 0. This is equivalent to saying that the force on the nuclei is zero since F = -¶V/¶Q. If we ignore terms higher than quadratic then we have expressed the potential in terms of a quadratic function. The potential V(Q) and the linear and quadratic terms are shown below.

By the way V(Q₀) is an arbitrary vertical offset. In the above Figure V(Q₀) = 0 whereas in the Figure below V(Q₀) = -1000.

The point of these figures is that we can approximate the true potential (the red curve) using a harmonic approximation (the blue curve). It is important to bear in mind that this approximation may not be very good for some molecules and some vibrations. However, the use of this approximation is widespread and it has lead to great insight into molecular vibrations. If we assume that V(Q₀) = 0 and that we can equate the second derivative (i.e. the curvature of V(Q) at Q₀ with a force constant, k then we can write

This is the equation for the blue parabola in the Figures above.

The classical harmonic harmonic oscillator

The historic reason for using a force constant k comes from Hooke's law. The components of the classical harmonic oscillator are illustrated in the Figure below.

A classical parabolic potential is derived from the force equation F = -kx. In this equation F is the restoring force for a deviation of a spring from the position x = 0.

Hooke's law states that the restoring force is linearly proportional to the displacement. The constant of proportionality is k. Instead of simply returning to its equilibrium position the mass shoots past it and continues until it compresses the spring with a displacement equal to the initial extension. Of course, for a classical system this is an ideal view that ignores friction and non-ideality of the spring etc. The motion is called harmonic motion and it is well represented by a sinuisoidal function. To see this let us use F = ma to solve for the motion of the mass. Here m is the mass of the object and a is the acceleration after we let it go from its extended position. Thus F = ma = -kx.

and therefore

You verify the solution by taking second derivative of cos(wt) and showing that it is equal to -w²cos(wt). We have introduced the definition

We call w the angular frequency and it is related to the frequency n by w = 2pn. Note the relation k = mw² that will be useful below.

Quantum mechanical harmonic oscillator

One might imagine that the quantum mechanical harmonic oscillator will behave similarly, however, there is a difference. Even if we use the same potential function kx²/2 we will find that the defined harmonic motion is not consistent with the Uncertainty Principle. For example, molecules vibrate in a crystal at T = 0 K. If we could define their motion using a sinusoidal function, we would know both their position and momentum to arbitrary accuracy. To see this we consider the quantum mechanical hamiltonian for the harmonic oscillator.

is a general expression where the equilibrium position is x = 0. In other words x replaces Q - Q₀ above. Why use x sometimes and Q other times? This is a good question. Q is often used to denote nuclear coordinates as distinct from electronic coordinates q. Since the harmonic oscillator problem reduces to a one-dimensional problem it is convenient to emphasize this by using the variable x instead, and this done in most books. As we have seen V(x) = kx²/2 so

The Schrodinger equation is HY = EY and thus we set this up to solve for E and Y. We will state the solutions and then demonstrate that they are in fact solutions. The general solution is

Hermite polynomials are obtained by taking the derivative of a Gaussian function.

Some of the Hermite polynomials are listed below

H₀(y) = 1

H₁(y) = 2y

H₂(y) = 4y² - 2

H₃(y) = 8y³ - 12y

H₄(y) = 16y⁴ - 48y² + 12

The energies are

The constant a is

The vibrational wave functions and quantized energy levels are represented in the Figure below.

Example: show that Y = exp{-ax²/2} is a solution to the harmonic oscillator Schrödinger equation. Is this wavefunction normalized?

Solution:

The terms in x² cancel since k = mw². Thus, the solution is correct provided that

This is in fact the energy of the lowest vibrational energy level. This energy is often called the zero point energy. That is because it is the energy in a vibrational mode when only the lowest or v = 0 vibrational state is occupied at absolute zero.

You can verify this for the solution v = 1 as an exercise.

Next we consider normalization. If the function were normalized then the integral

Here the wave function is real so Y* = Y. Thus,

Clearly, the wave function is not normalized. Note that we have solved a Gaussian integral to obtain this result.

The normalization constant N is found by insisting that

Here this means that

Interpretation of the quantum mechanical solution

What does the Gaussian function mean? The interpretation of Y*Y is that it is a probability distribution. Thus, the Gaussian function indicates the nuclear probability amplitude. In the lowest energy state, v = 0, that is a Gaussian function centered on x = 0. However, for higher levels the probability of being at x = 0 diminishes and the probability of being near the classical turning point increases. Thus, the quantum mechanical motion has a certain similarity with the classical. In the classical solution the particle has zero velocity only at the turning points and it spends more time near the turning points than elsewhere because it must slow down and change direction. This is reflected in the larger probability near the turning points for higher quantum numbers.

The lobes shown in the Figure below represent the probability for finding the nuclei in a given region of space.