# How big/small is Planck constant?

Gold Member

## Main Question or Discussion Point

I was thinking, given than h, c and G have different units, I can not compare a unit to other. But when we say that the quantum path integral recovers classical mechanics for very small h, there it seems we have a sense of how small h is.

So, h is small... respect to what? Is it respect to E? It can not be, because it has different units. Without G (thus quantum gravity), we can build a linear momentum p=E/c, but again we can not compare it with h. Is it to be compared respective to some angular momentum? Agaist the coupling constant? What if the coupling constant is of order unity, or even a lot bigger than unity? Do we still have quantum mechanics?

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.Scott
Homework Helper
Tiny - beyond imagination.

Planck's length (a combination of Planck's constant, the Gravitational constant, and the speed of light)
is just over 10^-35 meters, much, much smaller that a Proton.

Planck's time () is less than 10^-43 seconds.

There are similar bizarrely tiny values for Plancks temperature and energy.

meBigGuy
Gold Member
• 1 person
h has units of action or angular momentum, so when we say that it is "small" we have to have some typical action or angular momentum in mind. The fact that h has a numerical value of 6.6 * 10^-34 kg m^2/s in the SI units that we use for macroscopic objects indicates that it is going to be much smaller than the action or angular momentum of objects we encounter in our daily life, for which the SI system was devised. For example a hoop of mass 1 kg and radius 1 meter spinning about its axis of symmetry at 1 radian/s has an angular momentum of 1 kg m^2/s, which is more than 10^33 times larger than h. This is the sense in which h is small.

Gold Member
The best thing that I have come by myself, and with the help of a Mr Sommerfeld, is to consider the minimum angular momentum of a central force in "relativistic" classical mechanics.

I mean, I set the equilibrium of forces for a circular orbit,

M V^2/R = K / R^2

And using L=MVR

L V/R^2 = K/R^2

and thus

L= K /V

now if the maximum possible speed is C, the minimum possible angular momentum for a stable circular orbit is

L = K/c

So it seems that at least I could compare h with this L. But when I consider, for instance, electromagnetism as a central force, I find that L is actually 137 times smaller than h.

Staff Emeritus
2019 Award
You are right in that the numeric value of h doesn't tell you anything. (Indeed, it's 200 MeV-fm; doesn't that make it "large"?)

One way to see that its small is to ask what would be the typical range of quantum numbers for, say, a child on a swing. Since those are something like 1035, I think it's reasonable to conclude h is small.

atyy
Classical mechanics is recovered in the limit of Planck's constant going to zero. Since zero is small, that is also paraphrased as classical mechanics is recovered when Planck's constant is small.

ftr
The best thing that I have come by myself, and with the help of a Mr Sommerfeld, is to consider the minimum angular momentum of a central force in "relativistic" classical mechanics.

I mean, I set the equilibrium of forces for a circular orbit,

M V^2/R = K / R^2

And using L=MVR

L V/R^2 = K/R^2

and thus

L= K /V

now if the maximum possible speed is C, the minimum possible angular momentum for a stable circular orbit is

L = K/c

So it seems that at least I could compare h with this L. But when I consider, for instance, electromagnetism as a central force, I find that L is actually 137 times smaller than h.
You have a mistake in your calculation. This is the Bohr model, and the whole point of the theory is that L is multiple of h_bar. For the lowest orbit n=1 hence L=h_bar.

http://en.wikipedia.org/wiki/Bohr_model.

However, your question is good and I will try to come back to it when I have the time.

Here is a fantastic comparison.

The size of the human body is closer to the size of the observable universe than it is to plancks constant. I know I didn't help your question but I felt like sharing :P

clem
'Large' and 'small' have no meaning for constants that have dimensions.
$\e^2/\hbar c$ is small because it =1/137, and is dimensionless.
$\hbar$ is small for people, but large for atoms.

ftr
$\hbar$ is small for people, but large for atoms.
I am not sure what you mean by that. h_bar has one measured value. the heuristic argument taking h_bar to zero to get classical is just for illustration purposes, just like when we assume we have a 100 g particles in QM to get the classical comparison which we actually do not have.

clem
In units with c=1, hbar=200 Mev-fm, which is 'large'. I should have said 'nuclei'.
hbar has many 'measured values'.

Considering that classical physics is recovered when h -> 0, and our every day physical experiences seem classical, one could argue h is infinitely large in comparison!

ftr
In units with c=1, hbar=200 Mev-fm, which is 'large'. I should have said 'nuclei'.
hbar has many 'measured values'.
h_bar,c, G are called fundamental CONSTANTS of nature, the units of measurement is a different issue.

http://en.wikipedia.org/wiki/Natural_units