Analytical Method for Solving the Simple Harmonic Oscillator

In contrast to the elegant method described above to solve the harmonic oscillator, there is another "brute force" method to find out the eigenvalues and eigenfunctions. This method uses exapansion of the wavefunction in a power series.

Let us start with the Schrodinger equation:

$-{\frac {\hbar ^{2}}{2m}}{\frac {\mathrm {d} ^{2}\psi }{\mathrm {d} x^{2}}}+{\frac {1}{2}}m\omega ^{2}x^{2}\psi =E\psi$

or, ${\frac {\mathrm {d} ^{2}\psi }{\mathrm {d} x^{2}}}=(\xi ^{2}x^{2}-k)\psi$ where $k={\frac {2mE}{\hbar ^{2}}}$, $\xi ^{2}={\frac {m^{2}\omega ^{2}}{\hbar ^{2}}}$

We shall look at the asymptotic behavior

At large x, ${\frac {\mathrm {d} ^{2}\psi }{\mathrm {d} x^{2}}}\approx \xi ^{2}x^{2}\psi$

To find its solution, let us make the following ansatz:

$\psi (x)=e^{kx^{2}}$

Substituting this in the asymptotic equation, we get

$4k^{2}x^{2}+2K=\xi ^{2}x^{2}\!$

or in the large x limit,

$k=\pm \xi /2$

with this value of k,

$\psi (x)=Ae^{-{\xi x^{2}/2}}+Be^{\xi x^{2}/2}$

For $\psi (x)$ to remain finite at the origin, $B=0\!$.

So $\psi (x)=Ae^{-{\xi x^{2}/2}}$ for large $x$

Now that we have separated out the asymptotic behavior, we shall postulate that the complete solution, valid everywhere, can be written as:

$\psi (x)=h(x)e^{-{\xi x^{2}/2}}$ where $h(x)$ is some polynomial. It is clear that $h(x)$ must diverge slower than the rate at which $e^{-{\xi x^{2}/2}}$ converges for large $x$.

$\psi '=[h'(x)-\xi xh(x)]e^{-{\xi x^{2}/2}}$

$\psi ''=[h''(x)-2\xi xh'(x)+(\xi ^{2}x^{2}-\xi )h(x)]e^{-{\xi x^{2}/2}}$

Putting this back in the differential equation, we get

$h''(x)-2\xi xh'(x)+(k-\xi )h(x)=0\!$

let us try a series solution for $h(x)$

$h(x)=\sum _{j=0}^{\infty }a_{j}x^{j}$ $h'(x)=\sum _{j=0}^{\infty }ja_{j}x^{j-1}$ $h''(x)=\sum _{j=0}^{\infty }j(j-1)a_{j}x^{j-2}$

Substituting and equating coefficient of each power on both sides, we get the recursion relation

$\sum _{j=0}^{\infty }[(j+1)(j+2)a_{j+2}-(2j\xi +\xi -k)]=0$

$a_{j+2}={\frac {2j\xi +\xi -k}{(j+1)(j+2)}}a_{j}$x

unless this terminates after a finite number of terms, the whole solution will blow up at $x=\pm \infty .$ So

$2n\xi +\xi -k=0\!$

where

$n$ is a non-negative integer.

$k=\xi (2n+1)\!$

Since $k$ depends on the energy, we get

$E=\hbar \omega \left(n+{\frac {1}{2}}\right)$ $n=0,1,2,3,...$.

Once $k$ is constrained as above, we have

$a_{j+2}={\frac {-2\xi (n-j)}{(j+1)(j+2)}}a_{j}$.

Hence the series starts with either $a_{0}$ or $a_{1}$, and will be even or odd, respectively. These are called Hermite polynomials. The properly normalized eigenfunctions are ( see 'Introduction to Quantum Mechanics, 2nd ed. , by D. J. Griffiths for reference)

$\psi _{n}(x)=\left({\frac {m\omega }{\pi \hbar }}\right)^{1/4}{\frac {1}{\sqrt {2^{n}n!}}}H_{n}(x)e^{-{\xi x^{2}/2}}$

sample problem