3.2 Solving the Quantum Harmonic Oscillator

The first step is change our variable from $x$ to some dimensionless variable $y$. This will simplify the notation somewhat, which makes it easier to see what is going on.

We have three physical constants, $\mathrm{\hslash }$, $m$, and $\omega$ in equation 3.9. These have units of $\mathrm{kg}{\mathrm{m}}^2{\mathrm{s}}^{-1}$, $\mathrm{kg}$, and ${\mathrm{s}}^{-1}$ respectively. If we combine these in the form $\frac{\mathrm{\hslash }}{m\omega }$, this has units of ${\mathrm{m}}^2$. Therefore, if we define

this is a natural length scale for the problem, so we can define our new dimensionless length parameter $y$ as

Note that $x_0$ has nothing to do with equilibrium position. For a given oscillator, this is a constant. Substituting $x=y{x}_0$ into equation 3.9 and expanding the definition of $x_0$, we get

where we have defined

as the dimensionless energy parameter.

As $y\to \pm \mathrm{\infty }$, we can neglect $\epsilon u$ compared to $y^{2}u$ as the former will be negligible. Equation 3.15 then becomes

If we guess a solution of the form $A{e}^{-\frac{y^2}{2}}$, then we get

where in the last part we have neglected terms that are small when $y\to \pm \mathrm{\infty }$. Therefore, this function is a solution to the TISE for large $y$. Note that $B{e}^{+\frac{y^2}{2}}$ is also a valid solution, but is does not have a finite limit as $y\to \pm \mathrm{\infty }$ and is therefore not normalisable, so we will throw it out.

So the behaviour of the full solution to the TISE must be Gaussian for large $y$, but what about the rest of the function? We now suppose that it takes the form

and substitute this into equation 3.15 to see what we get.

Taking derivatives, we find

Substituting these in, we get

If we solve this differential equation for $h(y)$, we get the full solution for equation 3.15. Luckily, this equation is a well-known equation in mathematics called “Hermite’s Equation” (with $2\lambda =\epsilon -1$)! We will now go through the solution via a series method.

Assuming that $h(y)$ can be written as an infinite polynomial (a power series), we can substitute the following into equation 3.26:

What we find is a recurrence relation for the coefficients:

The limiting behaviour for the ratio of the subsequent coefficients is

which diverges more quickly than the Gaussian converges. This means that for the solutions to be valid wavefunctions, meaning for them to be normalisable, the power series must terminate. For each solution with a series terminating at $j=1$, $j=2$, $j=3$, etc., we obtain one valid eigenstate.

For some $j$ to be the highest non-vanishing coefficient, we must have the numerator of the recurrence relation be zero, i.e.

If we substitute back in $\epsilon =\frac{2E}{\mathrm{\hslash }\omega }$, this gives us the energy eigenvalues of the quantum harmonic oscillator:

where we have replaced $j$ with $n$. Note that in contrast to the infinite square well, the label for the energy eigenvalues starts at 0 instead of 1. This is just a matter of preference for how the equations look and doesn’t mean anything physically.

The ground state energy is $E_0=\frac{1}{2}\mathrm{\hslash }\omega$, the first excited state is $E_1=\frac{3}{2}\mathrm{\hslash }\omega$, and so on. The energy levels are equally spaced, increasing in single units of $\mathrm{\hslash }\omega$.

So what are the polynomials $h_n(y)$? To determine the coefficients, we need $a_0$ to calculate all the even coefficients and $a_1$ to calculate all the odd coefficients. Both the even and odd coefficients must terminate for the whole series to terminate, but the value of $E$ only allows us to terminate one or the other. Therefore, valid eigenstates will either have $a_0=0$ and only odd powers of $x$ or $a_1=0$ and only even power of $x$ in $h(y)$.

For the ground state, only $a_0\ne 0$, so $h(y)=a_0$ and the ground state eigenfunction is

For the first excited state, only $a_1\ne 0$, so $h(y)=a_{1}y$ and the eigenfunction is therefore

For the second excited state, $a_0$ and $a_2$ are nonzero so $h(y)=a_0+a_2{x}^2$. The dimensionless energy has the value

so the recurrence relation gives

Hence the eigenfunction takes the form

The process carries on like this. The polynomials generated are called Hermite polynomials, and denoted $h_n(y)$. Note that it is conventional to introduce a minus sign to the normalisation constant $a_0$ or $a_1$ so that the leading-order terms in the polynomial have a positive sign. The first few Hermite polynomials are

These coefficients are slightly different to the ones calculated above, but it doesn’t matter since we have to normalise the whole wavefunction anyway.