The Harmonic Oscillator

The harmonic oscillator for which the restoring force F is proportional to the deviation of x from the equilibrium position can be described by the potential energy V(x). Since F ~ −^d/_dxV(x) the potential energy can be written as follows:

And the Hamiltonian operator and the Schroedinger equation:

where µ is the reduced mass. Since we want to be as brief as possible, we will introduce new variables z² = x²(^kµ/_h²)^½ and ε = ^E/_h(^µ/_k)^½ = ^E/_hw with [w  =  (^k/_µ)^½] and finally obtain (^d²/_dx²  =  (^kµ/_h²)^½d²/_dz²):

One who is not interested in the mathematical derication can go to the link here to see the final result.

Now we want to factorize the term (−^d²/_dz² + z²) as for instance −a² + b² = (−a + b) (a + b). It's not so easy for the operator because we must pay attention to the precise consequence. And we produce for check
 

where    ½ (−^d²/_dz² + z²)  =  H     and     Hy  =  e y   one can obtain:

Now it's the equation for the harmonic oscillator.Let's now try to calculate the permutation relation between a₋ and a₊:

Applying a₋ to our equation a₊a₋ y  =  (ε −½)y :

By applying the operator a₋ to our wavefunction ywe obtain the wavefunction (a₋y) that is enough for our harmonic oscillator equation a₊a_-y = (ε − ½) y, for the eigenvalue which is smaller than about 1: (e- ½)  →  (e-³/₂).  The operator a₋  decreases the eigenvalue on 1 and therefore it's also called the annihilation operator.

Since ε is the measure of energy E (ε = ^E/_hw) then it should be the lowest state y_o. If we apply the annihilation operator to y_o then we cannot obtain more deep state. This is not possible! And finally it's considered for the lowest state

The lowest energy state has the energy ½hw. And calculating the corresponding wavefunction one can obtain:

C_o can be obtained from normalization condition  _-¥ò^+∞y_o* y_o dz = 1 in the following way:

And now applying a₊ to our equation a₊a_-y = (ε − ½) y:

The application of a₊ to y increases the eigenvalue on about 1. Therefore a₊ is also called the production operator. If we apply a₊ to the lowest state y_o then we will obtain the wavefunction y₁ that is the eigenfunction closer to the higher eigenvalue:
 

and so on.

Generally one can obtain for any n-wavefunction y_n

Result:

In general one can write down the following formula for possible energy values:
 

The ground state energy level E_o = ½ hw is the zero point of energy. Generally one can obtain for the wavefunction of n-state
 

where the constant C_n can be obtained from the normalization condition _-¥ò^+∞y_n* y_n dz = 1:
 

and the function H_n(z) is the Hermitian polynomial:
 

All other polynomials can be obtained from the recursion formula
 

for instance in order to obtain H₃(z) one must do the following H_n−1(z) ® H₂(z) and H_n−2(z) ® H₁(z):

H₃(z)  =  2z · (4z²−2)− 2(2) (2z)  =  8z³ − 12z
 

Fig. 1: Four lowest wavefunctions and its probability density for the harmonic oscillator.

The highest detection probability in the ground state is in the equilibrium position (see Fig. 1 for n=0). When increasing vibrations the quantity |y_n|² would be higher for classical turning points E = V(x); see figure to the left for n=5, 10, 20.

How much does the probability density for the classical oscillator in the range [x, x+dx]?

Classical oscillator:          x  =  x_m sinwt               x_m = maximum deviation

In the time interval dt (in which one observes the particle) the deviation dx will be as follows:

The quantity 1/w(x_m²−x²)^½ shows the probability density to find particle in the range [x, x+dx]; i.e. the highest chance to detect particle is in the classical turning points x_m. 

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