**Required math: calculus **

**Required physics: harmonic oscillator**

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Sec 2.3.

When we solve the SchrÃ¶dinger equation for the harmonic oscillator we find that we can peel off an exponential factor and what remains to be solved is the differential equation:

We can solve this equation using a power series, and we find that in order for the solution to be normalizable, the power series must contain only a finite number of terms. The energy of the oscillator is determined by the number of terms we allow the series to have, and the energy levels for the harmonic oscillator turn out to be:

where

The stationary states for the oscillator are then

where was introduced as a shorthand variable:

and the coefficients satisfy the recursion relation

where is another shorthand variable for the energy:

Once we decide on how many terms the series will have, this fixes the energy since all terms beyond a particular point in the recursion formula must vanish. Since the recursion formula relates every *second* coefficient, and for any given stationary state the energy must have only one value, the series can contain only even terms or only odd terms, but not both. That is, once we have chosen an energy level, the corresponding stationary state wave function will be either even or odd.

Let’s have a look at the first few stationary states that are generated by this recursion relation. The ground state corresponds to a polynomial that contains only the zeroth degree term, so is non-zero and all higher coefficients are zero. From the recursion relation 5 with :

In this case, the complete wave function for the stationary ground state is

The next state is found by taking to eliminate all the even terms in the series, and then requiring that is the single term in the series. So we have

with corresponding stationary state

In both these cases, we must determine and by normalizing the wave functions as usual. So in the case of we must have

Notice that the variable we must integrate over is the original but we have the exponential expressed in terms of the convenience variable , so we need to transform the differential. Since both limits are infinite, they remain the same after the transformation.

The integral is the Gaussian integral, and its value is known to be , so we get

For the other stationary states, we can show that equation 1 has as its solutions the Hermite polynomials . Furthermore, these polynomials form an orthogonal set of functions when weighted by the factor , in the sense that

When , then, normalizing the stationary states requires that

Changing the variable of integration from to and using equation 20, we get

The final form for the stationary states of the harmonic oscillator is therefore

In the last line we have replaced the convenience variable by its original form in terms of the position variable .

The general solution of the time dependent SchrÃ¶dinger equation for the harmonic oscillator is then a linear combination of the stationary states multiplied by the complex energy term, as usual

Note that the sum starts at , unlike most potentials where the sum starts at .

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