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I read about how to solve the Schrödinger equation for the quantum harmonic oscillator in one dimension. It started with the Schrödinger equation, $$ \frac{p^2}{2m}\psi(x, t)+\frac{1}{2}m\omega^2x^2\psi(x, t)=\hat{E}\psi(x, t) $$ It makes sense to me; that looks like energy conservation. Putting differential equations for the momentum operator, $$ \frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x, t)+\frac{1}{2}m\omega^2x^2\psi(x, t)=\hat{E}\psi(x, t) $$ Then they try to make remove the dimensions from the equation, so they replace $\hat{E}$ with a dimensionless $\epsilon$ somehow, and divide the whole expression by $\hbar$. They then start a spree of confusing substitutions which mathematically are accurate, but appear to have no physical significance. I don't understand why we do any of that. What's wrong with just staying with those differential equations and waiting for numbers which you can plug in and solve/integrate?

That source said that they're trying to simplify this equation, because it's very messy and difficult to solve, but if you only substitute and cancel terms, you are not adding any new information. So if we aren't adding information to the equations, then what's the point of making these random-looking substitutions? I don't see any significance of expressing the energy with no dimensions.

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    $\begingroup$ WP due diligence. $\endgroup$ – Cosmas Zachos Nov 24 '18 at 11:47
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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.

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  • $\begingroup$ Is there a mistake in the math where you substitute $u$ into the equation? I don't know where the $2$ goes. My book says that we should take $E=\frac{1}{2}\epsilon\hbar\omega$ instead. $\endgroup$ – J. Doe Nov 24 '18 at 12:16
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    $\begingroup$ @J.Doe Nah, it's fine; I just checked it again. You'll notice that the RHS is now $2\epsilon\psi(x, t)$. A lot of resources use the substitution $\epsilon=\frac{2E}{\hbar\omega}$, but we don't need to care too much about those constants: it's enough to simply make sure that they don't disappear and cause actual mistakes. If you want, you can get rid of the $2$s without harming much. $\endgroup$ – user191954 Nov 24 '18 at 12:22
  • $\begingroup$ In the Note that the LHS is the Hamiltonian equation you've inserted an extraneous $\psi(x,t)$. $\endgroup$ – Ruslan Nov 24 '18 at 19:11
  • $\begingroup$ @Ruslan That's debatable: I didn't indicate that I'm specifically talking about the Hamiltonian operator. One usually uses a hat over the $\mathcal{H}$ for that. But since there does seem to be a strange mismatch between the two sides of the equation, I've removed that. $\endgroup$ – user191954 Nov 25 '18 at 2:51
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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 \begin{equation} a_{0}=\frac{4\pi^{2}\hbar^{2}\epsilon_0}{\pi me^{2}}=\frac{4\pi\hbar ^{2}\epsilon_{0}}{me^{2}}, \end{equation} 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.

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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.

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  • $\begingroup$ This seems to contradict Chair's answer: that says we should take $\xi=\sqrt{m\omega}{\hbar}$, $\epsilon=E/\hbar\omega$. $\endgroup$ – J. Doe Nov 25 '18 at 3:30
  • $\begingroup$ No - Chair's answer is correct and consistent with this one. You're misreading Chair's handling of the position. You've correctly reported the energy, but it is fully consistent with this answer. $\endgroup$ – Emilio Pisanty Nov 25 '18 at 8:28
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There are already very good answers addressing your concrete differential equation. I would like to address the statement

but if you only substitute and cancel terms, you are not adding any new information.

It's true that you don't "add new information" but you may expose information that is not visible at first sight. Here is a very simple example, at a highschool level. Consider the typical problem of shooting a ball up and describing the trajectory. Say we shoot the ball from height $0$, at a speed of $10\, $m/s. Then the height of the ball is represented, in meters and with $t$ in seconds, by $$ h(t)=-\tfrac12\,gt^2+10t. $$ Suppose you are asked to find at what time the height of the ball is maximum, and what is such maximum. If you know calculus, you may look for the $t$ such that $x'(t)=0$, and then evaluate $x$ at that $t$. But, without knowing calculus, and doing a bit of "substituting and cancelling terms", we may get $$ h(t)=-\tfrac12\,g\,(t^2-\tfrac{20}g\,t) =-\tfrac12\,g\,(t^2-\tfrac{20}g\,t+\tfrac{400}{g^2}-\tfrac{400}{g^2}) =-\tfrac12\,g\,(t-\tfrac{20}{g})^2+\tfrac{200}g. $$ Now $h$ is expressed in such a way that, since the first term is never positive, we can immediately see that the maximum height is $200/g$ meters, and that occurs precisely at $20/g$ seconds.

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We the physicists work with dimensions. However, mathematicians work with dimensionless parameters, like $x$ and $y$. For us, $x$ would be meters, but for a mathematician, $x\in\mathbb{R}$.

So, it turns out that there was a differential equation, called "Hermite's differential equation", which was well known before QM. Since it was a mathematical issue, it was stated in terms of $x$ and $y$, not "dimensionful" quantities.

So, what we found was that, after all those substitutions, the quantum harmonic oscillator became that HErmite's equation. And that was great, because we already knew the solution of that.

If we didn't know that equation, we would hardly have found the solution. Keep in mind that most QM problems do not have analytical solutions.

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  • $\begingroup$ So it was purely historical? Would it not be easier to just put all the numbers into an equation solver? It would be less confusing, certainly. $\endgroup$ – J. Doe Nov 24 '18 at 11:48
  • $\begingroup$ actually I don't think this is quite accurate. It is entirely possible to solve the ODE without knowing the specific "name" of the differential equation: the method of power series will do just this and this is often how solutions to the radial part of the SE of the 3D HO is presented. It just so happens that many ODEs that commonly occur in solving the SE for simple potentials were studied before, but there are plenty of examples - Poeshl-Teller, Morse etc - where the solution is expressible in terms of hypergeometric functions, with no "special names" to the particular solutions. $\endgroup$ – ZeroTheHero Nov 24 '18 at 17:42

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