# Why do we nondimensionalize the Schrödinger equation when solving the quantum harmonic oscillator?

So if we aren't adding information to the equations, then what's the point of making these random-looking substitutions?

Those substitutions are quite helpful. Not because of the addition of new information (as you mentioned, we don't learn anything which we didn't know before), but because of the way they let us see what we already have. They also let you eyeball the system, as I like to call it, which essentially means that you can guess a lot more without resorting to computational tools.

Essentially, after making these substitutions,

1. You can easily think about what would happen if you change the proportions between the input values (this is a general advantage of nondimensionalization)

2. You can 'guess' possible solutions easily

Here's an explanation:

You didn't go through the substitutions, so here's a brief overview of the process of solving it. You can check each stage to make sure that you see what exactly is happening to the dimensions of the equation. Start off with the Schrödinger equation, $$\left(\frac{P^2}{2m}+\frac{1}{2}\omega^2x^2\right)\psi(x, t)=E\psi(x, t)$$ (Note that the LHS is the Hamiltonian, $$\mathscr{H}=T+V=\left(\frac{P^2}{2m}+\frac{1}{2}\omega^2x^2\right)$$, and this is all clearly dimensionally correct)

You're familiar with the concept of non-dimensionalizing, so I'll skim over that: take $$\epsilon=\frac{E}{\hbar\omega}$$, and get (after expanding your momentum operator and stuff) $$\frac{m\omega}{2\hbar}x^2\psi(x, t)-\frac{\hbar}{2m\omega}\frac{\partial^2}{\partial x^2}\psi(x, t)=\epsilon\psi(x, t)$$ Right now, there's no obvious difference between this equation and the original with $$E$$ in it: although it looks a bit cleaner, there's nothing trivially obvious when you look at it. So let's try another pair of substitutions, $$x=\alpha u$$ and $$\alpha=\sqrt{\frac{\hbar}{m\omega}}$$. Most books show these substitutions made one after the other, with expansions of the equation at each stage, but since the question says that they're already mathematically self-explanatory, I'll skip through to the result of applying those substitutions and simplifying stuff. You get this magically clean expression, $$u^2\psi(x, t)-\frac{\partial^2}{\partial u^2}\psi(x, t)=2\epsilon\psi(x, t)$$ Or, if I can be a bit looser with the notation, $$-\psi''+u^2\psi=2\epsilon\psi$$ and $$\psi''=(u^2-2\epsilon)\psi$$

This is obviously quite easy to solve in certain conditions. We can guess what happens if $$u\rightarrow\infty$$: the $$\epsilon$$-related terms become negligibly small, so we solve $$\psi''=u^2\psi$$. And that's easy enough to guess; the results are along the lines of $$\psi=Au^ke^{u^2/2}$$. Clearly that's not something you could achieve if you didn't have $$u$$ in the picture, and if you hadn't introduced $$\epsilon$$, it would have been sticky to figure out exactly what terms were being approximated to zero. A similar process can be followed to approximate solutions for $$u\rightarrow 0$$.

The next obvious advantage of removing the units is cleanness and generality of computations for different orders of magnitude. It's a common technique to assume all other units in a system to take values such that $$\hbar=1$$ and $$c=1$$. It makes a lot of calculations easier. In this case, we're setting the units of energy ($$E$$) as $$\hbar\omega$$ (just to verify, $$\hbar$$ has units of action, $$[E\ T]$$ (I'm using a nonstandard $$[E]$$ for energy), and $$\omega$$ is frequency, $$[T^{-1}]$$). We're also scaling the length to $$\sqrt{\frac{\hbar}{m\omega}}$$. Additionally, by setting these values, it's easy to make intuitive guesses about the scale of the problem (this is a more general advantage of removing dimensions, and isn't very specific to the quantum harmonic oscillator). This makes it easy to use the same equation for systems with different orders of magnitude (for example situations where the amplitude $$u$$ is really tiny and situations where it's huge), without loosing an understanding and 'feeling' for what's going on.

In your specific problem, and given the ubiquity of the harmonic oscillator, going to dimensionless variables means you can use the same basic solution for a huge number of problems, and recover the specific solution you need by simply adjusting the various scales.

More generally, there are a number of good reasons for this. First, finding "natural" units usually provide insight into the various scales of a problem. Second, using these natural units usually cleans up the resulting equations. As a third but less important reason, using a system of "natural" units where numbers of not small or large is computationally advantageous.

Consider the radial part $$\chi(r)=r R(r)$$ of the Schrodinger equation for the hydrogen atom. It is the solution to the differential equation

$$-\frac{\hbar^2}{2m} \frac{d^{2}}{dr ^{2}}\chi (r )+ \left(-\frac{e^{2}}{4\pi \epsilon _{0}r}+ \frac{\hbar^2}{2m}\frac{\ell(\ell+1)}{r ^{2}}\right)\chi (r ) =E\chi(r) \tag{1}$$ where $$m$$ is the electron mass, $$\hbar$$ is the reduced Planck constant, $$E$$ is the associated energy to $$\chi(r)$$, and $$\ell$$ is an integer.

Introduce the Bohr radius as a unit of length, defined as $$$$a_{0}=\frac{4\pi^{2}\hbar^{2}\epsilon_0}{\pi me^{2}}=\frac{4\pi\hbar ^{2}\epsilon_{0}}{me^{2}},$$$$ and the dimensionless quantity $$\rho = r/a_{0}$$.

Rewrite the Coulomb potential in terms of the dimensionless variable $$\rho$$, we get $$V(r)=-\frac{e^{2}}{4\pi \epsilon _{0}r}=-\frac{e^{2}}{4\pi \epsilon _{0}} \frac{me^{2}}{(4\pi \epsilon _{0})\hbar ^{2}}\frac{1}{\rho } =-\frac{me^{4}}{(4\pi\epsilon _{0})^{2}\hbar ^{4}} \frac{1}{\rho}=V(\rho).$$ Performing the substitution from $$r$$ to $$\rho$$,transforms the differential equation into $$\frac{-me^{4}}{2(4\pi \epsilon _{o})^{2}\hbar ^{2}} \frac{d^{2}}{d\rho ^{2}}\chi (\rho ) +\left[ -\frac{me^{4}}{(4\pi\epsilon _{o})^{2}\hbar ^{4}\rho} +\frac{me^{4}}{2(4\pi \epsilon_{o})^{2}\hbar ^{2}} \frac{\ell(\ell+1)}{\rho ^{2}}\right] \chi (\rho ) =E\chi(\rho ).$$ The Bohr energy $$\bar E=\frac{me^{4}}{2(4\pi \epsilon _{o})^{2}\hbar ^{2}}% \approx 13.6 eV ~\sim 2.2\times 10^{-18}J$$ is an obvious choice for an energy scale.

Dividing by this throughout yields the much cleaner expression $$-\frac{d^{2}}{d\rho ^{2}}\chi (\rho )+\left[-\frac{2}{\rho } +\frac{\ell(\ell+1)}{\rho ^{2}}\right] \chi (\rho )=\frac{E}{\bar E}\, \chi (\rho )$$ entirely in terms of the dimensionless variables $$\bar{V}(\rho )=\frac{V(\rho )}{\bar{E}}=-\frac{2}{\rho},\quad\nu =-\frac{E}{\bar{E}} .$$

This illustrates that, by simply going to dimensionless coordinates, we have a sense of the energies involved in atomic physics: not MeVs or GeVs, just eVs. Moreover, sizes in atomic physics are usually the size of the Bohr radius, i.e. $$\sim 10^{-11}m$$. We never have to manipulate small quantities, like $$10^{-18}J$$ or $$10^{-11}m$$.

In addition to being clean, this form is also amenable to computer solution: computers only work with dimensionless quantities, in the sense that it doesn't matter to them what the choice of units is.

There's two core reasons:

• It provides an important physical insight about the characteristic dimensions of the system, and how those depend on the base parameters.
• It removes notational clutter, making the equation easier to manage.

As you note, de-dimensionalizing a differential equation doesn't change it in any fundamental way, and it doesn't magically make it more solvable. All the changes are cosmetic, but cosmetic changes still matter; we're humans with limited monkey brains and simpler notation does make for an easier time.

The most important reason, though, is that you do get important physical insights from the process. The time-independent Schrödinger equation for the problem, $$\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x)+\frac{1}{2}m\omega^2x^2\psi(x) = E \psi(x),$$ has three relevant dimensionful parameters, $$m$$, $$\omega$$ and $$\hbar$$, and if you have three dimensionful parameters covering a three-dimensional space of quantities (i.e. $$[m]=[M]$$, $$[\omega]=[T^{-1}]$$ and $$[\hbar]=[M\:L^2\:T^{-1}]$$ are all algebraically independent) then you have a rigid system, in the sense of the Buckingham Pi theorem: any two copies of the problem will share the same behaviour, and they will be identical up to a re-scaling.

(On the other hand, if you add a fourth parameter within that three-dimensional space, like e.g. a quartic term $$\frac14\alpha x^4$$, then you will have one remaining 'shape' parameter, and not all copies of the system will have isomorphic behaviour. But I digress.)

Here, moreover, the fact that you have three parameters allows you to form uniquely-determined characteristic quantities for all physical dimensions, including in particular

• a characteristic length, $$\sqrt{\hbar/m\omega}$$,
• a characteristic momentum, $$\sqrt{\hbar m \omega}$$,
• a characteristic energy, $$\hbar\omega$$,

and through them any other dimension you care to name. This means that, when we do the variable substitutions \begin{align} x & = \sqrt{\hbar/m\omega} \ \xi \\ p & = \sqrt{\hbar m\omega} \ \pi \\ E & = \hbar\omega \ \epsilon, \end{align} what we're doing is identifying a single canonical copy of the problem, $$-\frac{1}{2}\frac{\partial^2}{\partial \xi^2}\psi(\xi)+\frac{1}{2}\xi^2\psi(\xi) = \epsilon \psi(x),$$ together with the canonical re-scaling that tells you what the relevant length and energy scales are for the problem.

And, once you do have the equation in a form without extraneous parameters and the only free handle is the de-dimensionalized energy $$\epsilon$$, it becomes much more clear exactly which parameters matter and which ones don't (or, rather, the parameters that don't matter have been whisked away). The resulting differential equation is mathematically equivalent to what you started with, but you've removed clutter and that makes it easier to work with, particularly when you go on to include this as part of a larger system.