Lecture 5: Harmonic oscillator, Morse Oscillator, 1D Rigid Rotor

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1 Lecture 5: Harmonic oscillator, Morse Oscillator, 1D Rigid Rotor It turns out that the boundary condition of the wavefunction going to zero at infinity is sufficient to quantize the value of energy that are allowed. ( E v = v + 1 ) ω v = 0,1,2,... 2 where ω = k/m is the angular frequency of the oscillator. Thus the energy levels are equally spaced starting from 1/2 ω. The corresponding wavefunctions are given by ψ v (x) = N v H v (y)e y2 /2 y = x α 2 2 = α km In the above solution, H v (y) is a Hermite Polynomial function and N v is the normalization constant. The Hermite polynomials that satisfy a differential equation H v 2yH v + 2vH v = 0 and they can be derived using the power series method. The first few polynomials are given by H 0 (y) = 1 H 1 (y) = 2y H 2 (y) = 4y 2 2 H 3 (y) = 8y 3 12y Notice that the odd polynomials contain only powers of y and are odd functions and the even polynomials are even functions. This is a general feature of the Hermite polynomials and some other polynomials that we will encounter. Consider v = 0. We have E 0 = 1 2 ω 1

2 the ground state energy or the zero point energy. The wavefunction is given by ψ 0 (x) = N 0 e x2 /2α 2 which is simply a Gaussian function. The normalization of the Gaussian integrals can be carried out using + e x2 dx = π and it turns out that N 2 0 = 1/α π. The wavefunction for the first excited state v = 1 is given by 2x /2α2 ψ 1 (x) = N 1 e x2 α This function is odd and has a node at x = 0. We plot the first few wavefunctions and the squares of the wavefunctions, i.e. the probability densities as a function of x. The quantum harmonic oscillator shows a finite probability in classically forbidden regions as described below. Consider the v = 0 state wherein the total energy is 1/2 ω. We see that the total energy E is equal to the potential energy V when 1 2 ω = 1 2 kx2 m which leads to x m = ±α, the maximum allowed displacement. Classically, an oscillator of energy E 0 will oscillate harmonically between x = α and x = α. However, the quantum mechanical oscillator has a nonzero probability of being in the region beyond x = ±α. This phenomenon is referred to as tunneling. The Harmonic oscillator is a model for studying vibrations of molecules. In fact, the vibrational energy is modeled using the Harmonic oscillator model and can closely be matched with Infra-red spectroscopy experiments. However, it is not really sufficient to describe the potential energy of bonds, since 2

3 U(x) Figure 1: Energy Levels of a Harmonic Oscillator shown in the potential energy well ψ ψ 0 x ψ 1 ψ 2 x x Figure 2: The first few wavefunctions of a harmonic osciallator we would like to have a breaking of bonds when the bond is stretched. In this regard, another model potential is used called the Morse potential given by V (x) = D e ( 1 e ax ) 2 3

4 where the constant a is given by ( k a = 2D e ) 1/2 We just mention this potential here for completion. It can be solved exactly and the energy levels are closely related to the harmonic oscillator for lower values of v but get closer as v gets larger. 0 R R e D e Figure 3: The potential energy of a Morse oscillator showing the energy levels 4

5 1 The Rigid Rotor Model We now go on to describe rotational motion of a particle. We first consider the 1D rigid rotor which corresponds to a particle freely moving on a ring or a rod freely spinning about an axis perpendicular to its length (Z-axis). The Hamiltonian operator for this system is given by Ĥ = ˆL 2 z 2I where L z refers to the Z-component of the angular momentum and I refers to the momoent of inertia. Note that I = µr 2 where µ is the reduced mass and R is the radius of the orbit for circular motion. We can write the operator for Z-component of the angular momentum as ( ˆL z = ˆxˆp y ŷˆp x = i x y y ) x This is most conveniently expressed in spherical polar coordinates using the transformations x = r sin θ cos φ y = r sin θ sin φ z = r cos θ Using elementary methods in partial derivatives, we can show that ˆL z = i φ which leads to the expression for the Hamiltonian in terms of φ as Ĥ = 2 2 2I φ 2 which leads to the Schrödinger equation which is solved by 2 2 ψ(φ) = Eψ(φ) 2I φ 2 ψ(φ) = Ae im lφ ( 2IE m l = ± 2 ) 1/2 5

6 Invoking the periodicity condition ψ(φ) = ψ(φ + 2π), we get e im l2π = 1 or m l = 0, ±1, ±2,... and the energy values are given by E ml = m2 l 2 2I and the normalized wavefunctions are given by ψ(φ) = 1 2π e im lφ Notice that except the ground states, all states are doubly degenerate. Also, notice that the energy eigenstates are also eigenfunctions of the angular momentum operator with eigenvalue L z = m l. 6