# Planck Constant Significance

1. Jul 1, 2013

### msumm21

Is there an intuitive explanation of why E/w = 2*L, where E/w is the ratio of the energy of any photon to its angular frequency and L is the magnitude of the spin of any electron? Evidentially there is a mathematical derivation of L=hbar/2 from some base assumptions, and evidentially E=hbar*w was initially derived by Planck from experiments (but probably also now can be derived from the same base assumptions as L?), but is there some clearer way to see why this should be the case without going through all the math (or is the math simple and based on intuitive base assumptions)?

2. Jul 1, 2013

### The_Duck

In quantum mechanics, the energy of a wave is the number of times it oscillates per second, the momentum of a wave is how many times it oscillates per meter, and the angular momentum of a wave is how many times it oscillates as it goes around a circle. However, these three quantities don't seem to have the right units:

Energy: oscillations per second has units of Hz, not J as is usual for energy
Momentum: oscillations per meter has units of 1/m, not kg m/s as is usual for momentum
Angular momentum: oscillations per circle is dimensionless, but angular momentum has units of kg m^2 / s

$\hbar$ is the unit conversion factor that makes everything come out right. It has units of J s = kg m^2 / s

Energy: $\hbar \times$ (something with units of Hz) = something with units of J

Momentum: $\hbar \times$ (something with units of 1/m) = something with units of kg m/s

Angular momentum: $\hbar \times$ (something dimensionless) = something with units of kg m^2 / s

So $\hbar$ is, among other things, the proportionality constant between the usual units of frequency and the usual units of energy. So for any quantum mechanical wave, $E / \omega = \hbar$. This is just telling us that if we want, we can measure energy in units of Hz. Similarly, we measure momentum in units of 1/m and angular momentum in dimensionless units.

Now let's ask ourselves what are the possible angular momenta in quantum mechanics. As stated above, the angular momentum of a wave in QM is the number of times it oscillates as it winds around a circle. This number has to be an integer, since the wave has to connect back up with itself. So the possible angular momenta are thus 0, 1, 2, 3... . Or including the factor $\hbar$ to make the units more familiar, the possible angular momenta are $0, \hbar, 2 \hbar, 3\hbar ...$. For technical reasons the half-integer angular momenta 1/2, 3/2, 5/2... are also possible; unfortunately this fact is not straightforward to derive.

We can't know a priori which angular momentum the particles in our universe will have; we just know that it will be an integer or a half-integer (that is, an integer or a half-integer multiple of $\hbar$). The electron ends up having angular momentum $1/2 = \hbar/2$, while the photon has angular momentum $1 = \hbar$.

So we get your equation $E / \omega = \hbar = 2 \times$ (electron angular momentum).

Or, perhaps a more enlightening way to write this is in units where we measure E in units of Hz, so that $E = \omega$. Then

$E/\omega = 1 = 2 \times (1/2)$.

Last edited: Jul 1, 2013
3. Jul 1, 2013

### wotanub

There are some good ways to see how planck's constant arises (usually involving uncertainty relations and commutators), but in my opinion, none of them are intuitive to a layperson since it is usually revealed through the mathematics. The idea of the planck constant isn't intuitive at all since based on everyday observation it isn't obvious that energy or angular momentum should be quantized.

Planck's constant is a tiny number so the "levels" are so close together, they look continuous in the classical limit.

I don't think the fact that the energy, angular momentum, and spin are quantized in multiples of Planck's constant implies any special relationship between the observables, especially when you associate them with different systems (energy of photon and spin of electron, for example). Rather, I think it says that we chose our units in an unfortunate way, leading us to this "weird" looking number for the spacing instead of a "clean" integer.

4. Jul 1, 2013

### vanhees71

$\hbar$, the modified Planck constant, is a conversion factor from usual units to natural units, no more no less. It's common to set $\hbar=1$ in quantum mechanics (in relativistic quantum field theory also to set $c=1$), which brings the whole theory in its natural beauty :-).

5. Jul 1, 2013

### Jolb

There are a bunch of ways to answer this question, since all the mysteries of quantum mechanics trace back to Planck's constant being a nonzero value. [The magnitude of h does actually matter, so Vanhees' answer is a bit misleading.]

I think there are three ways to interpret Planck's constant which I find useful:

First, consider what h=0 corresponds to--the classical limit. If h were zero, then p and x would commute just like in classical mechanics, there would be zero uncertainty in the Heisenberg uncertainty relations, etc. Basically, things would reduce to classical mechanics if h were zero. The bigger the h, the "more quantum" things are. If we had a problem in front of us and we had no idea about whether or not to attack it with quantum mechanics, we would compare the relevant energy scales to h to see whether or not we're in the "quantum regime", and if all relevant energies are much greater than h, chances are that classical mechanics is a good approximation.

Second, consider the historical development of quantum mechanics. Niels Bohr used essentially the same argument as The_Duck did; he said that an electron in an atom must have an angular momentum equal to some integer multiple of hbar, and he got a really good prediction for the energy levels of hydrogen and other atoms. Landau levels can also be derived in the same way. So you can think of hbar as some fundamental, indivisible unit of angular momentum.

Third, consider the path integral formulation. If we interpret the path integral as something akin to a partition function, then h gives us some weighting for the probability of physical processes to occur. We think of the classical trajectory as having the highest probability for any physical trajectory, and other possible trajectories are weighted according to an exponential where the exponential "decay" constant is hbar. So if we think of the path integral in terms of classical statistical mechanics, it serves sort of the same purpose as Boltzmann's constant in the assignment of probabilities.

Last edited: Jul 1, 2013
6. Jul 2, 2013

### msumm21

Thanks for the replies. The interesting thing to me is the fact that the same hbar also quantizes spin (intrinsic angular momentum), because I believe I'd heard/read many times elsewhere that spin is NOT attributed to something spinning or orbiting around something. However, the math works out exactly if you do consider spin to be something circling around the electron because then, using Duck's explanation, $L=p*r=(h/\lambda)*r$ and the smallest wavelength that will give a consistent amplitude around this circle of radius r is $4\pi r$ so then $L=p*r=(h/(4\pi r))r = \hbar/2$ which works out perfect. If spin is indeed NOT something orbiting around something else, then the fact that it happens to be quantized by the same hbar that appears in those other equations seems surprising at first, but maybe there is some other reason why this should be expected?

Last edited: Jul 2, 2013