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As I understand it, a natural system of units is one in which the numerical values of $c$ and $\hbar$ are unity, i.e. $c=\hbar =1$.

What I find confusing is that they are still dimensionful, i.e. $[c]=LT^{-1}$ and $[\hbar]=ML^{2}T^{-1}$. So, how can the action be dimensionless, $[S]=1$, when it has the same dimensions as $\hbar$?

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    $\begingroup$ The combination $S/\hbar$ is dimensionless. If you happen to be working in a unit system where $\hbar=1$, then you would write $S/\hbar = S$, which is dimensionless. $\endgroup$
    – Mark Mitchison
    Dec 15, 2015 at 15:26
  • $\begingroup$ Related: physics.stackexchange.com/q/28957/2451 $\endgroup$
    – Qmechanic ♦
    Dec 15, 2015 at 15:29
  • $\begingroup$ @MarkMitchison Is that all there is to it then? I feel like such a fool now :-\ ... So is the point that a quantum mechanical action contains factors of $\hbar$ and so we can simply divide through such that we end up with a dimensionless action $S/\hbar$ which is simply $S$ in natural units. Is there any particular reason for wanting the action to be dimensionless? Is it simply that we want its numerical value to be independent of any system of units? $\endgroup$
    – Will
    Dec 15, 2015 at 15:45
  • $\begingroup$ @Will The quantum path integral contains the factor $\exp(iS/\hbar)$. The argument of the exponential must, as always, be dimensionless. You are correct that the numerical value of the quantity $S/\hbar$ is independent of the choice of units, so that it is meaningful to discuss "large" and "small" contributions to the phase $S/\hbar$ from different field configurations. $\endgroup$
    – Mark Mitchison
    Dec 15, 2015 at 16:13
  • $\begingroup$ Can you do an experiment to verify that the dimension of some physical quantity is indeed what the textbook says it is? $\endgroup$
    – Count Iblis
    Dec 19, 2015 at 18:01

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It is dimensionless in the sense of mass dimension.

Setting $\hbar = c = 1$ means we only need to fix one base unit, which is usually taken to be the energy measured in $\mathrm{eV}$. Now, since $c=1$, this means that because of $E=mc^2$ becoming $E=m$ both $E$ and $m$ are measured in $\mathrm{eV}$. They might represent different dimensions (mass and energy), but they are measured in the same unit. Now, $E=h\nu$ means that inverse time is measured in $\mathrm{eV}$, so time is measured in $\mathrm{eV^{-1}}$. And so on.

Now, the "mass dimension" of a quantity is simply the power of $\mathrm{eV}$ it is measured in. Since the action is the integral of an energy against time, it has units of $\mathrm{eV}\cdot\mathrm{eV}^{-1} = \mathrm{eV}^0$, i.e. it has mass dimension zero.

You are right in that it is not "dimensionless". But having mass dimension zero means for any quantity $Q$ that there are powers of $\hbar$ and $c$ such that $\frac{Q}{\hbar^n c^m}$ is dimensionless, and since $\hbar = c = 1$, $\frac{Q}{\hbar^n c^m} = Q$, so there is no numerical difference between those quantities, and one sloppily says that $Q$ is dimensionless.

If you are somewhat worried that $\frac{Q}{\hbar^n c^m} = Q$ "looks wrong" from a dimensional analysis standpoint, then yes, that's right - the convenience in the formulae we get from $\hbar = c = 1$ comes inevitably with the loss of a large part of dimensional analysis, all that is left is the mass dimension for that.

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  • $\begingroup$ Ah ok, this makes a lot more sense now. So in the case of the action, is it simply that a quantum mechanical action will naturally contain factors of $\hbar$ in it, such that we can divide through to obtain a dimensionless quantity $S/\hbar$ which can be written as $S$ in natural units?! Is there any particular reason for wanting it to be dimensionless? Is it from physical intuition that it is a fundamental quantity in nature and thus its numerical value should be the same in all systems of units? $\endgroup$
    – Will
    Dec 15, 2015 at 15:56
  • $\begingroup$ @Will: Go e.g. through the derivation of the path integral in quantum mechanics. You get the term $\mathrm{e}^{-\mathrm{i}S/\hbar}$. Writing that $\hbar$ everytime is just annoying, especially when you expand the exponential in a power series, so we set it to one. $\hbar = c = 1$ is much more for convenience because you don't have all those constants floating around in your formulae than anything else. $\endgroup$
    – ACuriousMind ♦
    Dec 15, 2015 at 16:06
  • $\begingroup$ So the fact that $S$ is treated as a dimensionless quantity is more for mathematical convenience rather than any deep physical meaning then? $\endgroup$
    – Will
    Dec 15, 2015 at 16:10
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    $\begingroup$ @Will: For me, yes. I cannot exclude someone else sees a "deeper" meaning, though. $\endgroup$
    – ACuriousMind ♦
    Dec 15, 2015 at 16:37
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    $\begingroup$ "the convenience in the formulae we get from ℏ=c=1 comes inevitably with the loss of a large part of dimensional analysis" which is one of the reasons we don't start students off this way: they have to know how to use dimensional analysis to their advantage first. $\endgroup$
    – dmckee --- ex-moderator kitten
    Dec 15, 2015 at 18:07
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To awnser your second question :

Is there any particular reason for wanting the action to be dimensionless? Is it simply that we want its numerical value to be independent of any system of units?

there are several reasons to it. One of them is that we don't want to bother about keeping track of the $\hbar$ and $c$'s everywhere, as they can add up pretty fast. One other, more fundamental reason is that many physical phenomenon are scale-dependent. In the natural system of units, lenght is measured in the same units as inverse energy, so looking at a phenomenon at different scales is the same as looking at it at different energies. Therefore, measuring everything in terms of energy (or mass) enables to clearly see how a theory behaves at different energy scales : this idea is one of the basic principles of the Renormalization Group method. In quantum electrodynamics for example, the mass of the electron varies according to the energy scale you're working at : this comes from the fact that the phenomenons involved in QED vary with the energy scale of the problem.

Also note that you cannot make any arbitrary choice of constants, it has to be consistent. For instance you cannot put $c=1$, $\hbar=1$ as well as the QED coupling constant $e=1$, because this would make the fine structure constant $\alpha = \frac{e^2}{\hbar c} = 1$, thus making the electromagnetic interaction much stronger than it really is ($\alpha = \frac{1}{137}$ at the atomic scale).

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The convention called natural units can be separated into two parts:

  1. Using a different set of base units. For example, instead of using the SI units $\text{kg}, \text{m}$ and $\text{s}$ as base units for a certain subspace of the unit space, the most common flavour of natural units uses $\hbar, c$ and $\text{eV}$ as base units for the same subspace. For example $1\,\text{fm}$ written in these base units equals $\frac{\hbar c}{197\,\text{MeV}}$.

  2. Not writing down some or all of these base units. E.g., in the most common flavour of natural units, $\hbar$ and $c$ are not written down. This is usually denoted as $\hbar=1; c=1$. You could as well not write down $\text{eV}$.

Convention 1 is practical as you have frequently used constants as base units. Convention 2 spares you the trouble of writing down some of these base units and builds upon the fact that you can always reconstruct the unit from the dimension of the quantity you are considering. Note that even when using natural units, writing $\hbar = c$ is technically wrong as the unwritten units do not match.

Taking this point of view, action is not dimensionless in natural units – you just do not write down the units that indicate the dimension.

Further reading and blatant self-advertising: I wrote a didactical paper (preprint) on the topic of natural units.

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  • $\begingroup$ One does not write $\hbar=c$, one writes $\hbar=c=1$ where it does not denote that $\hbar$ and $c$ have the same units, but that one is using a unit system such that both values are 1. I also think your view of Convention 2 is wrong, it is that quantities are scaled by the appropriate constants. $\endgroup$
    – Kyle Kanos
    Dec 19, 2015 at 17:49
  • $\begingroup$ @KyleKanos: I know why one writes $\hbar=c=1$. Still, $\hbar=1;c=1$ is technically more correct and less confusing to students as it does not contain $\hbar=c$. — quantities are scaled by the appropriate constants. – If you use $\hbar, c$ and $\text{eV}$ as base units, you do not need to scale quantities anymore. Convention 2 does nothing more than changing $\frac{\hbar c}{197\,\text{MeV}}$ to $\frac{1}{197\,\text{MeV}}$. $\endgroup$
    – Wrzlprmft
    Dec 19, 2015 at 17:59
  • $\begingroup$ When one writes $i\hbar\psi_{,t}=\hat{H}\psi\to i\psi_{,t}=\hat{H}\psi$, one is technically dividing both sides by $\hbar$ and is scaling $\hat{H}$ by $\hbar$ (so formally one should write $\hat{H}'$ to denote the scaling, but becomes cumbersome so one doesn't do it). I agree that $\hbar=1;c=1$ is more correct, but it's also more cumbersome and $\hbar=c=1$ is just as informative. $\endgroup$
    – Kyle Kanos
    Dec 19, 2015 at 18:05
  • $\begingroup$ @KyleKanos: There is no practical difference between scaling every dimensionful quantity appropriately and using a different base system of units (and not writing down some of them). $\endgroup$
    – Wrzlprmft
    Dec 19, 2015 at 18:11
  • $\begingroup$ There is no difference between appropriately choosing a base & scaling, yes. But there is a difference between scaling & simply not writing constants down. You are suggesting the latter (likely with the intention of meaning the former) which isn't functionally correct. $\endgroup$
    – Kyle Kanos
    Dec 19, 2015 at 18:13

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