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We often see a limit $\hbar \rightarrow 0$ in quantum mechanics and sometimes its related with Symmetry breaking. Can someone briefly write the story behind this limit. Thanks in advance

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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

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up vote 3 down vote accepted

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.

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$\hbar\rightarrow 0$, 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

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.

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