Solution of the Differential Equation

Now we actually want to solve (4.5). We introduce dimensionless quantities

$$q:=\sqrt{\frac{m\omega}{\hbar}}x,\quad \varepsilon :=\frac{E}{\hbar \omega},\quad \phi(q):=\psi(x).$$ (230)

Then, (4.5) becomes

$$\phi''(q)+(2\varepsilon -q^2)\phi(q)=0.$$ (231)

For large $q\to \infty$, one can neglect the term $\propto \varepsilon$. This yields the asymptotic behaviour of $\phi(q\to\pm\infty)$,

$$\phi(q\to\pm\infty) \propto e^{\pm q^2/2},$$ (232)

which you can check by differentiating

$$\phi'(q\to\pm\infty) \propto \pm qe^{\pm q^2/2},\quad \phi''(q\to\pm\infty) \propto \pm e^{\pm q^2/2} + q^2 e^{\pm q^2/2} \to q^2 e^{\pm q^2/2}.$$ (233)

This roughly is an example of how an asymptotic analysis of a differential equation is performed; if you are interested for more mathematical details of the interesting theory of asymptotic analysis have a look at the book by Bender and Orszag.

We obviously have two different solutions: one grows to infinity as $q\to\pm\infty$, while the other goes to zero. Wave functions have to be normalized which is impossible for the solution that grows to infinity. We exclude that solution and write $\phi(q)$ as

$$\phi(q)=e^{-q^2/2}h(q),$$ (234)

which is an ANSATZ with an up to now unknown function $h(q)$ that we wish to determine. To do so, we plug it into our differential equation (4.7) and use

$$\phi'(q) = -q e^{-q^2/2}h(q) + e^{-q^2/2}h'(q)$$

$$\phi''(q) = -e^{-q^2/2}h(q) +q^2 e^{-q^2/2}h(q) - 2 qe^{-q^2/2}h'(q)+ e^{-q^2/2}h''(q),$$ (235)

which leads to

$$h''(q)-2qh'(q)+(2\varepsilon -1) h(q) =0.$$ (236)

We try to solve this by a power series

$$h(q) = \sum_{k=0}^{\infty} a_kq^k,\quad h'(q)=\sum_{k=0}^{\infty} ka_kq^{k-1}$$

$$h''(q) = \sum_{k=2}^{\infty} k(k-1)a_{k}q^{k-2}=\sum_{k=0}^{\infty} (k+1)(k+2)a_{k+2}q^{k}.$$ (237)

We insert these series into (4.12),

$$\sum_{k=0}^{\infty}\left[ (k+1)(k+2)a_{k+2} -2k a_k +(2\varepsilon -1)a_k \right]q^k=0.$$ (238)

The left side of the equation must be zero for any value of $q$ which can only be true if all the coefficients in $[...]$ vanish. The powers $q^k$ form an infinite basis $q^0,q^1,q^2,q^3,...$; if any function expanded in this basis is zero, all expansion coefficients must be zero. From $[...]=0$ in (4.14), we therefore have

$$a_{k+2}=\frac{2k - 2\varepsilon +1}{(k+1)(k+2)}a_k.$$ (239)

This is a recursion relation for the coefficients $a_k$. For large $k\to \infty$, one has

$$a_{k+2}\approx \frac{2}{k}{a_k},\quad k\to\infty,$$ (240)

unless the $a_k$ become zero above some $k=n$. The infinite power series $h(q)$ becomes asymptotical equal to the exponential function $e^{q^2}$ for large $q$: consider

$$e^{q^2}=\sum_{j}\frac{q^{2j}}{j!}\leadsto \quad [k=2j ] \frac{a_{k+2}}{a_k}=\frac{(k/2)!}{(k/2+1)!}=\frac{1}{k/2+1}\to \frac{2}{k},\quad k\to \infty.$$ (241)

Now, this is obviously not what we had intended with our Ansatz, because this would mean that the wave function $\phi(q)= e^{-q^2/2}h(q)\to e^{q^2/2}$ which means that it is no longer normalizable. The only possibility for a solution $\phi(q)$ that vanishes as $q\to\pm\infty$ therefore is obtained by demanding that the $a_k$ become zero above some $n=k$ whence $h(q)$ becomes a polynomial of finite degree. For this to be the case, the numerator in (4.15) has to vanish for some $k=n$ which means

$$2\varepsilon = 2n+1\leadsto \varepsilon\equiv \varepsilon_n=n+\frac{1}{2}.$$ (242)

The possible energy values $E$ are therefore

$$E\equiv E_n = \hbar \omega \left( n+\frac{1}{2}\right),\quad n=0,1,2,3,...$$ (243)

This is the famous quantization of the energy of the harmonic oscillator, which Planck had postulated to explain the blackbody radiation in 1900!

For each non-negative integer $n$ we obtain one energy eigenvalue and the corresponding eigenfunction $\phi_n(q)=h_n(q)e^{-q^2/2}$ from the finite recursion formula (4.15) for the polynomial $h_n(q)$. Here, we already use the index $n$ to denote the $n$-th solution. The polynomials $h(q)$ fulfill the differential equation (4.12) with $2\varepsilon=n$, that is

$$h''(q)-2qh'(q)+ 2 n h(q) =0.$$ (244)

Figure: Lowest eigenstate wave functions $\psi_n(x)$, Eq.(4.21) of the one-dimensional harmonic oscillator potential $V(x)=(1/2)m\omega^2x^2$ (black curve). Wave functions are in units $(m\omega/\pi\hbar)^{1/4}$. The curves have an offset for clarity.

The polynomials $h_n(q)$ that fulfill (4.20) are called Hermite polynomials $H_n(q)$ if they are normalized such that the wavefunctions $\psi_n(x)=\phi_n(q)$ are normalized: the result for the normalized eigenfunctions $\psi_n(x)$ with eigenenergy $E_n$, that is the solutions of (4.5), is

$$\fbox{$ \begin{array}{rcl} \displaystyle \psi_n(x)&=&\left(\frac{... ...2\hbar}x^2}\\ H_n(q)&=&(-1)^n e^{q^2}\frac{d^n}{dq^n}e^{-q^2}. \end{array}$\ }$$ (245)

We do not prove the explicit form of the Hermite polynomials here; in the next section we will learn an alternative method to calculate the $E_n$ and the $\psi_n(x)$ anyway. Here, we calculate $H_n(x)$ for the first $n$, using (4.21) (denote $q$ by $x$ here)

$$H_0(x) = 1$$

$$H_1(x) = 2x$$

$$H_2(x) = 4x^2 -2$$

$$H_3(x) = 8x^3-12 x$$

$$H_4(x) = 16x^4-48 x^2 + 12.$$ (246)

The lowest eigenfuncions $\psi_n(x)$ are shown in Fig. (4.1).