Simple Harmonic Oscillator (SHO) Quantum Physics II Recommended Reading: Harris: chapter 4 section 8

The Simple Harmonic Oscillator (Classical) (1) One of the most fundamental problems of physics is the simple harmonic oscillator. This problem occurs in areas as diverse as molecular vibrations, lasers, quantum descriptions of the electromagnetic field, phonon propagation, calssical mechanics ( pendulum, mass on spring) Consider a particle of mass m subject to a restoring force F = -kx. Classically (Newton 2nd Law) the equation of motion is where (2)  0 is the natural frequency of the oscillator, it depends on the mass m of the particle and the strength of the spring k A general solution to equation (1) is

Quantum S.H.O. The Schrodinger equation uses the Potential rather than the force Recall that force and potential are related Knowing F we can calculate U Hooke’s Law Where k = spring constant N.m -1. (3) In quantum mechanics solving the simple harmonic oscillator reduces to solving the Schrodinger equation for the potential

Schrodinger Equation for S H O The Schrodinger equation for the potential given in equation (3) is (4) Difficult differential equation to solve. So we start by guessing its solutions and require that they satisfy boundary conditions. The wave-function must approach zero as x    and in this limit the solutions of eqn (4) behave like exponentials of -x 2. and comparing with the solutions for the infinite potential well, we expect just one peak in the lowest energy wave function.  x

(5) Where A and b are constants to be determined by requiring that this wave-function satisfies equation (4). Take first and second derivative of (5) and plug into (4) (6) (7) Since the potential is symmetric about the origin expect the wave- functions to have alternating even and odd parity So, for the ground state we try a solution of the form differentiate again collect terms together

Now plug equations (5) and (7) for  (x) and (d 2  / dx 2 ) into equation (4) and see if there is a solution. (8) Equation (8) must hold for all values of x  the coefficients of each power of x must be equal to zero, this then gives and So we can now solve for b and E. From (9) we find (9) (10) and from eqn (10)(11)(12) The constant A must be found from the Normalisation Condition collect terms together

The NORMALISATION Condition isNormalisation So the constant A is therefore Putting this together with equation (5) we find Equation (12) gives the quantised lowest energy level while equation (14) gives the lowest wavefunction of a particle of mass m confined in a potential kx 2 /2. This is the ground state solution of the oscillator (Lowest energy) = ZERO POINT ENERGY (13) (14) (12)

Ground State (n=0) wavefunction Note that the oscillator extends beyond the classical turning points of the oscillator. (Compare this to the finite potential well where the wavefunction penetrated into the classically forbidden region. classically forbiddn regions

Complete Solution The full solution of the Schrodinger equation for Harmonic Oscillator potential (equation 4) is difficult. To find the wavefunctions and allowed energies, we assume that the general solution is of the form where H n (x) are polynomial functions of x in which the highest power of x is x n. That is We then find these functions by substituting equation 16 into the TISE and solving for the coefficients a n, a lot of algebra (15) (16) The polynomials that satisfy the Schrodinger equation for the Harmonic Oscillator potential are called the HERMITE Polynomials Recall and define then the first few Hermite Polynomials are:

Hermite Polynomials The wave functions all have the form of a Hermite Polynomial H n (x) multiplied by a Gaussian exp(-bx 2 ) function. So we first look at the shape of the Hermite polynomials H 4 (x) H 0 (x) H 1 (x) H 2 (x) H 3 (x) H 5 (x), H 6 (x)...

Hermite Polynomials The first five Hermite polynomials are defined as where N n is the Normalisation Constant which is determined from the normalisation condition. In general the normalisation constant for the n th wavefunction is given by What do the wavefunctions look like??

Ground State Wave function All of the SHO wave functions have a Gaussian term exp(-z 2 /2) which looks like this Normalisation Factor Which for the first 5 wave functions is Putting everything together we find that the wavefunctions look like

SHO Wave Functions n=0 n=1 n=2 n=3 n=4 n=5

SHO Wave Functions Note PARITY of wavefunctions: Alternating even and odd parity n = 0: even n = 2: even n = 4: even n = 1: odd n = 3: odd

SHO Wavefunctions First four wavefunctions and corresponding probability distributions for the SHO potential.

SHO Energy Levels We have already seen above that the ground state solution of the oscillator (Lowest energy = ZERO POINT ENERGY) is given by (12) What about the energies of the higher states. When we solve the Schrodinger equation we find that the energies are quantised (just like the particle in an infinite potential well). However, the energy levels (spectrum) of the SHO are given by n=0 n=5 n=4 n=6 n=3 n=2 n=1 Note that the energy levels are evenly spaced

Time r eq Atom oscilating backwards and forwards Application: Molecular Vibration

Chemical bonds are not rigid, the atoms in a molecule vibrate about an equilibrium position. The force required to stretch or compress the bond length to r is proportional to (r-r eq )  The energy required to stretch the bond is Just like a spring which obeys Hooke’s Law. where x = (r-r eq ) k = force constant of bond, units N.m -1  r eq r m1m1 m2m2 r Can consider molecule as a single oscillating particle with an effective (reduced) mass

In a diatomic molecule both masses oscillate relative to the centre of mass of the molecule. centre of mass. m1m1 m2m2 This id equivalent to a single oscillating particle of mass  (reduced mass) 

Potential Energy Curve Connection between shape of potential energy curve and strength of bond k. Potential energy curve is approximately parabolic and the force constant k is related to the potential by Big  steep potential  large k Small  shallow potential  small k HCl, k = 516 Nm -1,  weak. CO, k = 1902 Nm -1,  strong.

Energy Levels in a Molecule Example: 1 H 35 Cl has a force constant of 519 cm -1 find its oscillation frequency 0. And therefore We can excite the 1 H 35 Cl vibration with radiation of this wavelength  infra-red First we calculate the reduced mass of the molecule we can then calculate the vibrational frequency of the molecule This gives rise to the whole area of Infra red (IR) Spectroscopy of molecules

Hz n=0 n=2 n=1 n=3 n=4 Selection Rules:  n =  1 +1  absorption of photon -1  emission of photon Vibrational Spectrum The energy of a photon required to cause a transition from the n = 0 to the n = 1 state is Spectrum has only one line. Usually in the infra red region all transitions will occur at the same frequency. Do you see why?

Schrodinger Equation Where  is the effective mass of the molecule H n (y) = Hermite Polynomials, and vibrational quantum number, n = 0, 1, 2, 3, … See Applet at following website: