# $\hbar \rightarrow 0$ in quantum mechanics

We often see a limit $\hbar \rightarrow 0$ in quantum mechanics and sometimes it's related with symmetry breaking. Can someone briefly write the story behind this limit?

• Usually $\hbar \rightarrow 0$ is used in a context of the limit where Quantum Mechanics turns into classical mechanics (when quantization goes away). – Brandon Enright Apr 19 '13 at 6:22
• Related to this question seems to be Landsman's paper "Between Classical and Quantum" arXiv version... – Alex Nelson Apr 19 '13 at 15:54

Often, the classical limit of a quantum system may be seen by simply taking the limit as $\hbar\rightarrow 0$.

For example, a quantum harmonic oscillator has energy levels which are multiples of $\hbar$.

$$E_n= \hbar\omega\left( n+\frac{1}{2} \right)$$

In the limit as $\hbar\rightarrow 0$, we can see that energy levels become continuous.

Another example is Heisenberg's uncertainty principle. The uncertainty in a particle's position ($\sigma_x$) times its uncertainty in momentum ($\sigma_p$), obeys the following relation: $$\sigma_x\sigma_p\geq\frac{\hbar}{2}$$

In the limit as $\hbar\rightarrow 0$, no such restriction between energy and momentum exists.

• $\hbar\rightarrow 0$, it Seems that we are changing the planks constant? How can we limit on a constant? – Unlimited Dreamer Apr 19 '13 at 6:35
• @UnlimitedDreamer That is exactly what we are doing. Of course it is just a trick, in the real world Planck's constant has a definite value. It just so happens that ignoring that value often gets rid of the "quantumness" of our model. – Alex L Apr 19 '13 at 6:42
• It is like focusing a microscope ( dimensions h_bar) and diminishing the magnification until it is zero. It is the relative sizes of h_bar and the size of uncertainty in momentum times space . When we are at even microns, let alone milimeters, and measurable everyday momenta the h_bar can be considered zero. – anna v Apr 19 '13 at 6:43
• It is worth noting that when a dimensionful quantity like $\hbar$ is taken to be small, it means that it is small compared to some quantity of the same dimension in the system. – leongz Apr 19 '13 at 7:22
• @UnlimitedDreamer We do the same sort of thing when we take $c \to \infty$ to get from relativistic equations to Newtonian equations. $c$ has a definite value, but by pretending it is infinite our equations become Newtonian. In the same way, when we pretend Planck's constant is zero our equations go from quantum to classical. Edit: Just realized this is an old question. Not sure why it popped up in my recent questions page. – Jold Sep 27 '15 at 22:39

I'll start by setting the tangential part of the question aside: In quantum field theory there are currents and charges which are conserved classically, but not quantum mechanically, that is, there are violations of these conservation laws to order $\hbar^2$, and hence the symmetries hold classically but not quantum mechanically. (See below for what that means.) The non-conservation pieces are called quantum anomalies and the corresponding symmetries anomalous. (Other symmetry breaking phenomena require subtleties of QFT:Goldstone's theorem.)

Now for your core question: Classical mechanics is, "basically" an effective theory of the full theory of the world, quantum mechanics, that is, its Classical limit.

Mathematically, that means that all quantities of interest are functions of the dimensionful $\hbar$, so, normally they entail a classical piece without $\hbar$ and a "quantum correction piece" involving powers of $\hbar$ normalized by characteristic quantities of the same dimension dominating the problem in question, so, actions, angular momenta, etc... so as to be dimensionless.

When these quantities assume macroscopic values, (so, for engineering angular momenta or actions), these powers, and so quantum corrections are infinitesimally small, and thus ignorable. The leading term in this notional expansion is called the classical limit, and many texts on the quantum oscillator remind the student how small such correction effects beyond that limit are, in practice, and when they step to the fore and control atoms and crystals, etc.

A fascinating technical feature happens in this limit---actually dramatized in the struggle of humanity to go beyond this limit, in the 20th century: The mathematical rules describing the full QM theory are very different than the math of the classical limit. QM was worked out in the 1920s, mostly, with a resolute insistence on keeping away from the confusing classical limit, and violating its mathematical structure with giddy revolutionary abandon!

The reformulation of the QM formulation in a way that made that limit plausible (beyond the early and reassuring Ehrenfest theorem) was only 69 years ago: the Phase space Formulation of QM. In that formulation, it is manifest at every step how QM is a "deformation" of classical mechanics (a mathematese and engineering term: basically means systematic theory of correction) and how, conversely, classical mechanics is an approximate summary effective description of QM. (For example, Dirac, in 1933, Ref 1, explained how the principle of least action follows from the $\hbar/S\rightarrow 0$ limit of amplitudes through constructive interference.) Observables are now described by $\hbar$-dependent Wigner transforms of operators, whose detailed behavior "morphs" into that of classical observables in phase space.

Your question is really a question on deformation quantization, but without conscious reference to it!

References

1 Paul A. M. Dirac, "The Lagrangian in Quantum Mechanics", Physikalische Zeitschrift der Sowjetunion, 3 (1933) 64–72}}

2 Also see J H Van Vleck, "The correspondence principle in the statistical interpretation of quantum mechanics" , Proceedings of the National Academy of Sciences of the United States of America 14 (1928) 178–188, (doi: 10.1073/pnas.14.2.178)

I think this limit can be best understood if one writes the probability amplitude for a particle to go from a point $X$ to a point $X'$ at a posterior time $T$. As you probably know, Feynman worked out that this amplitude reads:

$\langle x'|e^{-\frac{i\hat{H}T}{\hbar}} |x \rangle \sim \int \mathcal{D}[x]e^{\frac{iS[x](X,X',T)}{\hbar}}$

In this formula, $x(t)$ is any dynamical path that goes from $X$ to $X'$ in a time $T$ and $S[x](X,X',T)$ is the action of this path in the sense of Lagrangian mechanics.

If we forget for a moment about this path story and imagine the path integral as a simple wave packet with a non trivial phase $S$ then we see that the integrand is a sum of cosine and sine functions with a frequency that is inversly proportional to $\hbar$.

If you recall that e.g.

$\sin x + \sin y = 2 \sin \left(\frac{x+y}{2} \right)\cos \left(\frac{x-y}{2} \right)$

we see that idealy constructive summation of the harmonics only appear if $x = y$

In the limit $\hbar \rightarrow 0$, the effective frequency in the path integral tends to infinity and all the interferences are destructive except for any two paths $x_1(t)$ and $x_2(t)$ such that $S[x_1](X,X',T)\sim S[x_2](X,X',T)$. In general this is true if and only if $x_1$ and $x_2$ are neighbouring paths and if $S$ reaches an extremum at $x_1$.

At the end of the day in the limit $\hbar \rightarrow 0$ or rather $S/\hbar \rightarrow \infty$, the probability amplitude $\langle x'|e^{-\frac{i\hat{H}T}{\hbar}} |x \rangle$ is dominated by a most probable path $x_1(t)$ and a "tube of neighbouring paths to $x_1$".

Incindently, the path $x_1(t)$ satisfies the least action principle and this "tube of most probable paths" describes the classical trajectory for a particle going from $X$ to $X'$ in a given time $T$ with an uncertainty width whose typical size is typical size is characterized by the de Brooglie wavelength of the particle.