17
$\begingroup$

I have seen books and papers mentioning "In the semiclassical limit, $\hbar$ tends to zero", "the scaled Planck's constant goes as $1/N$ where $N$ is the Hilbert space dimension" etc.

Could anyone explain these variable values taken by Planck's constant? What is the idea behind it? For all I know, $\hbar$ is a constant equal to $6.6 \times10^{-34} \ [\mathrm{m^2 \cdot kg/s}].$

$\endgroup$
5
  • 10
    $\begingroup$ I would recommend rewriting your question title. Pretty controversial in the current form. Maybe hint to the books and papers in the title. $\endgroup$
    – Newbie
    Jan 17 at 15:10
  • 2
    $\begingroup$ Related/possible duplicates: physics.stackexchange.com/q/56151/50583, physics.stackexchange.com/q/19770/50583 and their linked questions $\endgroup$
    – ACuriousMind
    Jan 17 at 15:17
  • 1
    $\begingroup$ According to the new SI Units changes, the value of the Plank's constant is now an exactly defined fixed value. $\endgroup$
    – Sam
    Jan 17 at 15:21
  • 2
    $\begingroup$ For another closely related question, see here. $\endgroup$
    – knzhou
    Jan 17 at 16:33
  • $\begingroup$ This is just a change of units. Imagine you are in units where the Planck constant is smaller and smaller. $\endgroup$
    – LL 3.14
    Feb 17 at 17:08

8 Answers 8

51
$\begingroup$

In a purely classical (Newtonian) universe, quantum effects would be absent, and the way to pretend this is true mathematically is to allow Planck's constant to approach zero, and see what the consequences are. Similarly, in a classical universe, relativistic effects would not exist, and this would be expressed by letting c approach infinity.

This does not mean that $\hbar$ or c are not constants: it means we can see the effects of removing quantum mechanical or relativistic considerations from our mathematical description of the world by assigning a value of zero to those constants, and seeing what happens.

$\endgroup$
2
  • 6
    $\begingroup$ Slightly more specific, we usually should take the limit as these values tend to zero, which could then be referred to as the "classical/non-relativistic limit". Just substituting zero may not be well defined. $\endgroup$
    – Kai
    Jan 18 at 21:28
  • $\begingroup$ Actually it is not the Planck constant that is converging to $0$, but the units in which it is expressed that make it be smaller and smaller. This is well explained in arxiv.org/abs/1502.06143 or arxiv.org/abs/2103.10946 $\endgroup$
    – LL 3.14
    Feb 17 at 17:14
8
$\begingroup$

Theories that are said to have real parameters really have a multidimensional parameter space, and the real parameters are coordinates in that space.

Often, multiple points in the parameter space produce isomorphic theories. E.g., take a toy theory with parameters $x$ and $y$, and the property that $(x,y)$ and $(ax,ay)$ are isomorphic for any $a>0$. This could be because $x$ and $y$ are both expressed in some unit, and since there is nothing else to fix the meaning of the unit, changing its size doesn't change the theory as long as both $x$ and $y$ are updated consistently.

You can avoid the redundancy in the parameterization by switching to $(r,θ)$ polar coordinates and then fixing $r$, perhaps to $1$, or perhaps to $6.6\times 10^{-34}$ for backward compatibility. This doesn't cover the whole $\mathbb R^2$ parameter space: it's missing $x=y=0$. But it's fine for most purposes if you know from experiment that $x$ and $y$ aren't both zero.

This is not the only possible choice. You could fix $x=1$, and take $y$ as your free parameter. This covers even less of the original $(x,y)$ space, but it's fine for most purposes if you know from experiment that $x>0$.

The points not covered by your new coordinates don't cease to exist just because they're invisible on your new parameter chart. But to "reach" them you have to switch coordinate systems, and that means, e.g., dropping your convention that $\hbar$ has a fixed value.

Note that "the $x=0$ limit" is usually not a well defined point in theory space. If you fixed $x=1$, then $y$ and $y/x$ denote the same theory, but the $x\to 0$ limit with $y$ fixed isn't the same as the $x\to 0$ limit with $y/x$ fixed. There is in fact more than one "$\hbar\to 0$ limit" of at least some quantum theories. If you take $\hbar\to 0$ and $n\to\infty$ while holding $ω$ and $E=n\hbar ω$ fixed, you get a classical wave theory, where quantum effects are unobservable because you can't isolate individual particles. If you take $\hbar\to 0$ and $ω\to\infty$ while holding $n$ and $E$ fixed, you get a classical particle theory, where quantum effects are unobservable because the interference fringes are infinitesimally thin. These theories live at different points on the boundary of the quantum theory space.

$\endgroup$
3
$\begingroup$

From a mathematical perspective, it is interesting to study quantum theories, where the (reduced) Planck's constant is not necessarily equal to its physical value. For instance,

  1. In the mathematical topic of deformation quantization, $\hbar$ is treated as a free parameter (=indeterminate).

  2. In the mathematical topic of geometric quantization, $\hbar$ is an arbitrary but fixed constant.

$\endgroup$
3
$\begingroup$

Strictly answering your question.

"In the semiclassical limit, ℏ tends to zero".

In the above sentence, it means the contribution of this number to the overall result of an equation, becomes too small that can be ignored when compared to the contributions of the other terms in the same equation.

Imagine the following simple equation: $c=a^2+b^2$ if you evaluate this equation for $a=3$ and $b=1$ you most likely are not going to ignore $b$ since it will cause you some $10 \%$ error. Now try to evaluate this for $a=3 \times 10^9$ and $b=3 \times 10^3$. In this case, $b$ will contribute with approximately $10^{-7} \%$ of the overall result, and then you may drop $b$, and get $c=a^2$, especially if some uncertainties in your experiment are bigger than $10^{-7} \%$. This part of the experimental error is important, as it means that you cannot even detect the effects of $b$ meaning that, for all intents and purposes, it doesn’t exist. The mathematical way of saying drop $b$ is $b \to 0$. In your particular case, ℏ can be imagined as a sort of angular momentum. In an atom where angular momentums (orbital and spin) are of that order, it cannot be ignored in overall sum of angular momentums. In classical physics, however, where planets, stars and galaxies angular momentum are considered, you can definitely drop any factor of ℏ.

"the scaled Planck's constant goes as $1/N$ where $N$ is the Hilbert space dimension".

In this sentence it is self-obvious. You have a fraction, and the bigger the denominator the smaller the result.


Now, I guess that this kind of interrogations usually appear on arguments of “when (or why) quantum mechanics reduces itself to classical physics”. If this is the case, I don’t know the answer to this new question, but, I would say that, for those using arguments of the first kind, I don’t believe that there is a set of fundamental quantum mechanical equations that fully transforms into Newton’s laws when we set $ℏ \to 0$, in a similar way that relativity does when you set $c \to \infty$, for example. For relativity, just make a Taylor expansion of relativistic total energy, set $c \to \infty$ and, voila, you fully recover Newtonian mechanics especially if formulated in a Lagrangian or Hamiltonian fashion. Those using arguments of the second kind (using the Hilbert space dimension), may be in a better position, but that would be the subject of a different question and not strictly on “How can Planck's constant take different values?” I guess.

$\endgroup$
2
$\begingroup$

In perturbation theories one speaks about how to "perturb" classical Newtonian mechanics to include quantum mechanical effects (or in reverse- How from quantum view arrive at classical physics). For example, if you account for a momentum error in a second Newton law, you'll have :

$$ F = \frac d {dt} \left( p \pm\Delta p \right) $$

Substituting in place of momentum error Heisenberg uncertainty principle, gives :

$$ F= \frac {dp}{dt} \pm \frac{d}{dt} \left(\frac{n\hbar }{2 \Delta x}\right) $$

Now differentiating with respect to $t$, gives : $$ F = \frac {dp}{dt} \pm \frac{n\hbar }{2}~ \frac {\Delta v}{(\Delta x)^2}$$

Where $\Delta v$ is uncertainty in speed, and $\Delta x$ is uncertainty in position.

Given this equation, substitute $n=0$ (equivalent to resetting reduced Planck constant to zero) - and you will have ordinary classical Newton second law expression.

Substitute $n=1$ and you'll have semi-classical fix for a minimum error due to Heisenberg uncertainty.

Full quantum view is in case $n \ge 1,~n=1,2,3,\dots,\infty$, because there's no upper bound limit in error.

$\endgroup$
0
$\begingroup$

Niels' answer is correct, but the idea of the multiverse should be mentioned. This is a speculative theory that physics might not be the same everywhere. Constants might change from place to place.

There is no experimental evidence for this. As far as we know, the observable universe is uniform. But we can't see past the observable universe. Inflation and string theory together make it at least plausible that other regions have laws that could be very different.

For more, see The Multiverse, Science or Science Fiction? | Sean Carroll

$\endgroup$
1
  • 3
    $\begingroup$ The value of $\hbar$ is now fixed by definition in SI units, so a different constant would vary in this multiverse idea. $\endgroup$
    – J.G.
    Jan 17 at 21:12
0
$\begingroup$

The plank constant never disappears in Classical Mechanics since the observations are made with light ($E=\hbar\omega$). Classical Mechanics is inclusive on photons (some sort of averaging). The photon energies are small with respect to the kinetic energy of a classical body, so it all is withing experimental error bars, but it exists! This is the right physics.

$\endgroup$
0
$\begingroup$

From a mathematical perspective, it is equivalent to say

  • All relevant quantities (with the dimensions of an action) in the system are much greater than $\hbar$
  • The value of $\hbar$ is very small compared to the relevant scale

The latter limit is much easier to perform symbolically, by taking the limit $\hbar \to 0$. In reality, $\hbar$ stays constant, but the scales of lengths$\times$momenta or energies$\times$times in the system get much larger, so that we can neglect the influence of any terms involving $\hbar$.

Your second example involving the "scaled Planck constant" likely involves a similar idea. For numerical computations, all quantities need to be represented by a dimensionless floating point number in the computer, so you represent them as a multiple of some scale with the right dimensions, in a procedure called nondimensionalization. After this procedure, the scaled versions of all previously dimensionful constants depend on the scale that was used.

benrg's answer has a more theoretical explaination of this idea.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.