Why do we obtain classical physics by taking the limit of Planck's constant to zero?

Why if we specifically set Planck's constant equal to zero (the limit of it) do we sometimes get classical physics? I mean, what does it mean physically to set the constant equal to zero? Or to say it in another way, what did the first human to do this think of in order to come up with it?
I have seen in various situations somebody taking h-->0, but what is making anybody who does this to take h-->0 in the first place? Why would somebody take h-->0 to take the classical limit in the first place?

• This claim is in the generality in which you present it wrong. See Classical limit of quantum mechanics and linked questions. Nov 14 '15 at 21:44
• @ACuriousMind Thank you, there are very enlightening answers there. I have edited my question in order to be more specific as to what I want. Nov 14 '15 at 21:48
• Chapter 1, section 6 of Landau Vol. 3 QM explains this nicely. Nov 14 '15 at 21:50
• Nov 14 '15 at 22:10
• Well, one obvious reason is that $h$ does not appear in the equations of classical physics. If $[x,p]=0$, every state is defined simultaneously by position and momentum, which differentiates between the classical and the quantum theory. This answer does not explain certain details, which appear in chapter VI, §1 of Quantum mechanics by Messiah. Nov 15 '15 at 6:16

What would somebody take h-->0 to take the classical limit in the first place?

Think for a minute about what the presence of Planck's constant in quantum theories does.

It creates a granularity to the amount of energy that can be stored in a mode (but not in general, to the amount of energy that can be stored in most systems).

Why does that matter?

Classically, a system in equilibrium disperses it's energy so that every available mode has (to a good approximation for macroscopic systems) an equal share of the energy. And it turns out that the value of that average is set entirely by the temperature of the system.

The classical problem is that an electromagnetic bath (as in cavity radiation) has an infinite number of modes, implying an infinite energy.

How does $h$ fix it?

Plank's ansatz was that each mode had a discrete energy-spectrum1 with granularity proportional to the frequency. This means that at high enough frequencies each mode could no longer have an energy close to the average, having to choose between zero and some value considerably larger than the classically assigned average. Add to that Boltzmann's rule for assigning the probability to occupation numbers and the mean energy of the (infinite number of) high frequency modes drops rapidly to zero, preventing the ultraviolet catastrophe.

So, why take $\lim_{h \to 0}$?

Classically there is no granularity, in Plank's approach there is, but it only matters when $hf \gtrapprox E$ for $E$ the classical average energy per mode. If you allow $h$ to get small, that condition is lost for a bunch of modes. Let $h$ go all the way to zero and it is lost for all modes. So you expect to recover the classical result in the limit of negligible $h$.

Is the procedure general?

Do note that it is not trivial to show that this limit gives the expected behavior in most cases. I am given to understand that the general problem is mostly solved, but only a few cases are accessible at a introductory level.

1 Watch out for the two meanings of "spectrum" here. In this case I'm talking about the allowed energies in a single frequency (or wavelength), not about a group of frequencies (or wavelengths)!

• Just a comment on the last part; the procedure is pretty well understood in great generality and rigour by now...but mostly by the mathematical/mathematical physics community. Almost every QM system you can think of is proved to converge (in a precise sense) to the classical counterpart, and also a good number of mathematically well-defined bosonic quantum field theories (with fermions it becomes less interesting to take the classical limit, however something can be done also in that case but surely it is more involved). Nov 17 '15 at 9:12
• Oh, boy. When a mathematical physicist says "more involved" I think about running for the hills. I'm glad to hear that this problem is more completely understood than I knew. Would it make sense to change that to some version of "only a few cases are accessible at the introductory level"? Nov 17 '15 at 15:25
• Yes I think that would be rather correct ;-) Let's say that the mathematical context on which these things have been developed (for the most part the Hörmander school) do not help accessibility to physicists (and perhaps also to a non-small part of mathematicians). There is however a point that is interesting at the physical level, and that is often not known. That point is that not only we have to check (prove) that the dynamics of the system reduces to the classical one in the limit, but also what happens to states (i.e. what they become in the limit). Nov 17 '15 at 15:58
• ((the answer to the last question being classical states, but in the statistical mechanics sense, i.e. probability distributions on the classical phase space)) Nov 17 '15 at 15:59

In classical physics there exists no h, and everything was going happily along until the data/measurements showed that classical electrodynamics + classical mechanics could not explain a number of data.

The datum that introduced the otherwise unknown and undefined h_bar was black body radiation. It was not possible to reconcile the fact that no excess ultraviolet radiation was observed from the experimental study of the radiation from bodies in the lab. Planck had the brilliant idea to assume that the electromagnetic radiation was not continuous but came in quanta of energy h*nu, where nu is the frequency, and could fit the data. Thus the constant took his name. It became obvious that a new framework for physical observations was emerging.

When the photoelectric effect, and then the spectra of atoms appeared in the labs the new framework became consistent by using this h*nu identification for photons that make up light. These data are not classical physics. Quantum mechanics was developed to explain it, and it soon became clear that all classical observations emerged from an underlying quantum mechanical framework, and this can be proven with advanced mathematical methods. For example, the way the classical electromagnetic radiation is built up by photons.

A second foundation stone for quantum mechanics is Heisenberg's Uncertainty Principle. , HUP . This is what sets the stage for deciding whether quantum mechanical methods are needed or whether classical theories are adequate. For example this inequality does not hold for classical physics, since h is a very small number. The constraints of the inequality appear at the level of particle physics, for micrometers and lower values. The statement that one gets classical physics when h is zero comes from the observation that classical momenta times classical dimensions multiplied have no lower bounds up to dimensions of micrometers and smaller . The handwaving statement "one gets classical physics if h is zero" is based on this relation.

The HUP is in the foundation of the mathematics of quantum mechanics, it appears in the commutator and anticommutator relations of operators representing measurable observables, like position and momentum. If h is zero, one has to fall back on the classical theories. , the consistency check also needs advanced mathematics to see the correspondence.

• "Planck had the brilliant idea to assume that the electromagnetic radiation was not continuous but came in quanta of energy hnu"* Planck never came up with this idea, he assumed EM field is described by the Maxwell equations. Where he introduced the quantization is in calculating the entropy of material oscillators (model of matter establishing the equilibrium radiation) and in quantifying energy exchange in a model of interaction of a single material oscillator with EM field. He never quantized EM radiation itself. Dec 28 '15 at 16:18

Why do we obtain Classical Physics by taking the limit of Planck's constant to zero?

If you set $h = 0$ in Plank's law you will get $B_{v}(v,T)=2v^2k_BT/c^2$ that is the Rayleigh–Jeans classical law which is valid up to one point and after that it comes in conflict with what was experimentally observed. see: Planck's law

Now if you do not make $h = 0$ from the beginning and you want to solve the equation

Plank's law = Rayleigh–Jeans law, for small frequencies where both are valid

you get $h = 0$ for a perfect fit.

There is no other explanation for your question than pure math.

• Thanks for the answer, but my question essentially was, why to set the constant equal to zero? Nov 14 '15 at 22:06
• As $\hbar \to 0$ you recover the result predicted from classical theory. So you do in fact recover classical physics in this case.
– user12029
Nov 14 '15 at 23:09
• But what made you think of making h-->0 in the first place? Nov 15 '15 at 0:48
• This time my answer is correct. The mistakes have been eliminated. The minuses that keep coming are just to discredit me. The first minus was justified. The others no. Nov 15 '15 at 1:08
• I think the statement "there is no other explanation than pure math" is unjustified. There is a deep physical justification for the classical limit being sometimes described as $\hbar \to 0$: in many cases (see dmckee's answer for example) quantum effects become irrelevant when the typical scale of action per particle/mode is much greater than $\hbar$. Nov 15 '15 at 18:38