-5
$\begingroup$

The widely circulated folklore surrounding Planck’s constant $\hbar$ lends it an aura of importance. But could $\hbar$ be a constant of human convention which is dispensable? Does the unorthodox view in the paper https://arxiv.org/abs/1203.5557 make sense?

A disclaimer:

The question here is asked in the context of first and second quantization as a whole (i.e. quantum field thery, especially in the path integral framework). Anyone who is concerned with Planck's constant for first quantization in isolation (e.g. Schrodinger equation only) can skip this question.

And the quantization itself is NOT questioned here. Rather, the necessity of Planck's constant in the process of second quantization is called into question.

Exhibit 1, the quantization condition: $$ [x, p] = \left[x, -i\hbar\frac{\partial}{\partial x}\right] = i\hbar. $$ "Introducing $\hbar$ made the first time in history where multiplying a math identity by the same constant on both sides was reported to make a new physical principle". It comes from $$ \left[x, -i\frac{\partial}{\partial x}\right] = i, $$ which is the trivial identity it appears to be.

Exhibit 2, the path integrand of massless Dirac spinor: $$ \exp\left[\frac{i}{\hbar}S_\text{Dirac}\right] = \exp\left[\frac{i}{\hbar}\int \bar{\psi}(i\hbar\not{\partial}\psi) \right]= \exp\left[i\int i\bar{\psi}\not{\partial}\psi\right]. $$ Do you remember the $\hbar$ in $p = -i\hbar\partial/\partial x$ of Exhibit 1? It resurfaces here in the Dirac (first quantization) action $\int {\bar{\psi}(i\hbar\not{\partial}\psi)}$ in its relativistic incarnation. But lo and behold, it is canceled out by the $\frac{1}{\hbar}$ factor (for second quantization) in the path integrand. If the $\hbar$ from first quantization and the $\hbar$ from second quantization net out, why do we bother to introduce $\hbar$ in the first place?

Exhibit 3, the fine-structure constant: $$ \alpha = \frac{e^2}{\hbar c}. $$ Measurements of $\alpha$, $c$, and $e$ seem to be tantamount to measuring $\hbar$. The catch is that the whole schema hinges on the convention of unit of electron charge $e$ and measurement thereof. If you do proper rescaling of gauge field $A$ in the QED path integrand, only the fine-structure constant $\alpha$ remains. Electron charge $e$ drops out completely and you don't need $e$ anywhere in the Lagrangian. If we only invoke $\alpha$ in theory and in experiment and forego the notion of $\hbar$ and $e$, we don't sustain any loss of information. One might argue that when you do rescaling of certain field it’s simply a change of physics unit. However, if two parameters in a theory collapse into one parameter after rescaling ($\hbar$, $e$ -> $\alpha$), you might suspect there must be something redundant, which is nothing but the Planck’s constant $\hbar$.

So shall we regard Planck's constant $\hbar$ as only an arbitrary intermediate step which is subject to human convention? What is your take?


Added note 1:

One may argue that "$\hbar$ is dimensionful: it demarcates the physical scales separating classical from QM". The point is that the "dimensionful" $\hbar$ has to be compared with another "dimensionful" quantity, which is the action $S$ in the path integrand $$ \exp[i\frac{S}{\hbar}]. $$ Now if we rewrite it as $$ \exp[i\frac{10^{30} S}{10^{30}\hbar}], $$ both classical ($S$ or $10^{30} S$ begets the same classical equation) and quantum (path integrand not changed) physics are the same. Can we claim now that $10^{30}\hbar$ (instead of $\hbar$) demarcates the physical scales separating classical from QM? The absurdity with above argument stems from the fact we look at $\hbar$ and $S$ as two standalone quantities. We shouldn't. Rather, we should talk about $\hat{S} = \frac{S}{\hbar}$ only, never separating $S$ from $\hbar$. The classical limit is the tree/saddle point approximation of QFT path integrand $$\exp[i\hat{S}]. $$ Quantum loop corrections are high order (sort of Laylor expansion) effects beyond saddle point. The conventional $\hbar$ happens to be a proxy for bookkeeping loops. For massless Dirac spinor the $\hat{S}$ is (see Exhibit 2 above) $$ \hat{S} = \int i\bar{\psi}\not{\partial}\psi. $$ You don't see $\hbar$ in $\hat{S}$ at all.

Added note 2:

What about the mass term and gauge interaction term in $\hat{S}$? Do we need $\hbar$ there? Take QED for example. The conventional parameter list of QED is $c$,$\hbar$, $m_e$, and $e$. If we adopt $c$,$\hat{m_e} = \frac{m_e}{\hbar}$, and $\alpha = \frac{e^2}{\hbar c}$, we don't need standalone $\hbar$ (see Exhibit 3). The beauty of using $\hat{m} = \frac{m}{\hbar}$ is that we should be talking about $\hat{p} = \frac{p}{\hbar}$ instead of $p$. Now we have the quantization condition: $$ [x, \hat{p}] = \left[x, -i\frac{\partial}{\partial x}\right] = i, $$ and Heisenberg's uncertainty principle: $$ \Delta x \Delta \hat{p} = \frac{1}{2}. $$ Again, it has to be emphasized that it's not merely a change of unit, since the exact number of parameters has been reduced (from 4 to 3 in QED, the same picture applies to any standard model interactions).

$\endgroup$
10
  • 6
    $\begingroup$ Didn't we already go over this? The only thing that's being done here is a change of units, from non-natural to natural with $\hbar=1$. We need $\hbar$ because we like to use a particular system of units, in which, for example, momentum is not measured in the same units as inverse distance. $\endgroup$ Commented Jun 20, 2018 at 20:03
  • $\begingroup$ When you set some quantity to 1, it's called natural unit. It's a totally different story when you don't have to set a quantity to any value and still have a workable theory in its absence. $\endgroup$
    – MadMax
    Commented Jun 20, 2018 at 20:09
  • 6
    $\begingroup$ My point is that by "not including $\hbar$" you are, in fact, setting its value to $\hbar=1$. The two statements are equivalent. You'll of course still have a workable theory, because the only thing that you've changed is the units in which you have to measure its predictions. $\endgroup$ Commented Jun 20, 2018 at 20:12
  • 6
    $\begingroup$ This rant by OP was previously posted as an answer. $\endgroup$ Commented Jun 20, 2018 at 20:17
  • 2
    $\begingroup$ For some thoughtful discussion about this, you might read the literature surrounding the upcoming redefinition of the SI units, which will fix $h$ (and equivalently $\hbar$) but which cannot eliminate it: $h$ is the conversion factor between frequency and energy. $\endgroup$
    – rob
    Commented Jun 20, 2018 at 20:18

3 Answers 3

3
+50
$\begingroup$

Crucially, ℏ is dimensionful: it demarcates the physical scales separating classical from QM. Its role is clearest at that interface.

A high-school lab harmonic oscillator for a macroscopic system with ω=2Hz, m=10g, and maximum amp $x_0$=10cm, has $S≈E/ω≈mω x_0^2/2$ ≈ Kg m²/(10⁴s) =ℏn, so that n ≃ 10³⁰. Such fantastic numbers dramatize how the dimensionfull ℏ sets the scale of quantum actions S (with the same dimension). In systems with $S/\hbar\gg 1$ like the above force, the functional integrand ( $\sim e^{iS/\hbar}$) Dirac invented in 1932, Dirac, Paul A. M. (1933), "The Lagrangian in Quantum Mechanics" Physikalische Zeitschrift der Sowjetunion 3 64–72, leads to stationarity of its enormous imaginary exponent, through destructive interference, and thus classical extremality. (The exact oscillator above is the most rewarding problem of the Feynman and Hibbs book.) Again: ℏ/S comparable to a non-small number indicates QM should be relevant, and the classical path extremizing the action virtually meaningless. So, we may work with ℏ-natural units in QM, but it would be dysfunctional to use them in classical mechanics, and we don't.

Likewise, $kT/\nu$ failing to be enormous can indicate the scale where the Planck radiation formula is accurate, instead of the divergent Rayleigh-Jeans one for such values.

In real life thermodynamic observables $f(x,p;\hbar)$ in physical chemistry, for instance, the so-called Wigner-Kirkwood expansion (in ℏ) separates classical results from quantum corrections, $f(x,p;\hbar)= g(x,p) + \hbar \partial_x\partial_p g(x,p) + ...$ Wigner all but discovered the phase space formulation of QM at around the same time his brother in law was achieving the above.

In this formulation, the variables are x and p and the parameter ℏ normalizes the phase-space cell xp and the momentum has no special association with it, so your confusions about its role in the Heisenberg commutation rule are plainly unwarranted. (In that formulation, $x\star p-p\star x=i\hbar$, so your trivial stunt won't work.) The corresponding uncertainty principle works in an interesting way, so $\Delta x \Delta p\geq \hbar/2 $, again reminding one that phase-space scales enormously larger than ℏ will never "see" that principle's effects.

On a notional planet in which this formulation were discovered first, their Schroedinger would derive his equation starting from the phase-space analog, $$ H(x+i\hbar\partial_p/2, p-i\hbar\partial_x/2)~ f(x,p)= E~f(x,p). $$ He would thus discover the momentum mapping to the standard Hilbert space $-i\hbar \partial_x$ that makes ℏ almost superfluous, in that limited setting. But, knowing where he started, he might not be as easily confused. On yet another planet, a matrix mechanics maven might choose to project x out in favor of momentum, and he would find something silly for coordinates, like $i\hbar\partial_p$, instead. (We never do this on our planet, since we already pegged to momentum, but recall the symmetry of the Fourier transform kernels.)

Takeaway: ℏ is a physical scale demarcating something meaningful, and cannot be argued away without nasty consequences. It is indispensable and lots of physical answers are crucially reliant on it, especially in the interstitial areas between CM and QM. Absorbing it into the units of the variables in physical observables, while tenable, tends to obscure the systematic separation between "small" and "large" variables which quantifies the separation between the quantum from the classical domain, both essential for best understanding our world. (Actually, in the active literature, numerical semiclassical pros routinely define a dimensionless ε in their expansions, having absorbed scales of ℏ in their units; but make no mistake: the expansion parameter is still very much there.) A sound and tractable illustration of this scale separation may well be Figalli et al, 2010.

N.B. on the hermeneutics of John's paper. It is fair to observe with him that ℏ is not there in classical mechanics, and can serve as a unit in QM and hide from view. However, John may have slipped into recalcitrant reductionism. He's chosen to live purely in QM, and has no interest in connecting to the classical world of trucks and hummingbirds ... it's all in the (QM) theory, for the proverbial notional somebody to work out. However, most of our lives and the engineering products we use actually work in CM, and the interface between the two pictures is central to physics, as I argue above, and not a mere technical detail for hacks to work out. To describe it, you'll need ℏ by some name or other.


Added note on "seeing" ℏ, but at risk of mission creep.

A similar dimensionful constant is c. One may simply consider it as a conversion factor between space and time, and use it as the immutable unit of velocities in which it appears to vanish; indeed, in HEP it is set to 1, in natural units, as well. Nevertheless, it is still ubiquitous in demarcating the boundary between relativistic and non relativistic effects. v close to 1 typifies relativistic laws and v less than 10000 is pre-Einstein physics. c is still there, in contrasting sizable (on the scale of 1) to minuscule velocities in laws, and similarly in electromagnetism. Relativistic corrections means small "velocity"-dependent effects.

$\endgroup$
6
  • $\begingroup$ Thank you @Cosmas Zachos for you interest in topic and detailed answer. Please see added note 1 and 2. And, what is you take on Exhibit 2? $\endgroup$
    – MadMax
    Commented Jun 22, 2018 at 16:43
  • $\begingroup$ Added note 2: yet again, if you stick to a pure QM system, you may choose to absorb ℏ in your units, and, indeed, many, including myself, do, on occasion. The nastiness begins when you wish to take the classical or semiclassical limit--when you want to connect to engineering, or the Rayleigh-Jeans formula, etc. Sticking to a limited quantum problem and optimizing your units hardly ever gives you new physics. $\endgroup$ Commented Jun 22, 2018 at 16:55
  • $\begingroup$ Added note 1: You might have missed my point. You may absorb any and all constants and numbers in definitions, but a huge $\hat S$ will lead to the principle of least action, so classical mechanics, while a small one won't, leading to QM propagators. That is, the numerical size of the dimensionless phase $\hat S$ in the integrand will determine whether you get a bona-fide quantum propagator or a δ-function one center around the classical trajectory. ℏ is the "secret sauce" that separates CM from QM behavior, whatever units, definitions, etc... you choose. $\endgroup$ Commented Jun 22, 2018 at 18:17
  • $\begingroup$ Exhibit 2: Needless to say, using a Dirac actions with dubious classical limits would not illustrate any of this point. $\endgroup$ Commented Jun 22, 2018 at 18:19
  • $\begingroup$ Thank you @Cosmas Zachos for the reply. "using a Dirac actions with dubious classical limits" - exactly, since $\hbar$ (tree/saddle point approximation) might not be a universal way of separating classical from QM. Zurek's decoherence picture makes more sense to me as for the transition from quantum to classical. $\endgroup$
    – MadMax
    Commented Jun 27, 2018 at 15:07
3
$\begingroup$

You're letting the math obscure the physics. Sure, you can cancel the $\hbar$ in the relation $$[x, - i \hbar \partial_x] = i \hbar$$ to get an equation without $\hbar$ in it, but that doesn’t matter. The point is that in quantum mechanics, the momentum is $$p = - i \hbar \partial_x$$ which, assuming a plane wave, means $$p = \hbar k.$$ Both sides of this relationship are measureable quantities. The momentum is how much 'oomph' the particle has when it hits you, while the wavenumber is related to how fast the phase changes in space. The mathematical identity $[x, - i \partial_x] = i$ is trivial, but the physical statement $p = \hbar k$ is profound.

You can't get rid of the $\hbar$ here because otherwise the two sides would have different dimensions. You could formally get rid of it by choosing natural units, just like setting $c = 1$, but that doesn't mean Planck's constant isn't doing anything. You need it to tell you what the natural units are.

$\endgroup$
3
  • $\begingroup$ Thank you @knzhou for your explaination. Please see added note 2. $\endgroup$
    – MadMax
    Commented Jun 22, 2018 at 16:40
  • $\begingroup$ @MadMax Yes, the $\hbar$ means something in the mass term. It means that a stationary mass $m$ electron has frequency $\omega = m c^2 / \hbar$. Here both $\omega$ and $m$ are perfectly measureable quantities that were defined well before quantum mechanics. So we need $\hbar$ to convert between their units. $\endgroup$
    – knzhou
    Commented Jun 22, 2018 at 17:20
  • $\begingroup$ @MadMax Again, you may switch to natural units so the identity is simply $\omega = m$, but then you're not doing anything different from everybody else. $\endgroup$
    – knzhou
    Commented Jun 22, 2018 at 17:21
-4
$\begingroup$

Short answer. One day, possibly, a purely theoretical formulation of the fine structure constant will be achieved, in such a way that its value will be calculated without using any empirical data, that is, without the intervention of the dimensioned physical constants. That day the Planck constant will appear as a derived constant, expressed as a combination of fundamental constants of electromagnetic nature. But while that advance does not arrive, we will have the Planck constant stuck in every corner of physics.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.