# Momentum operator generator of translation classical limit

Classical limit in quantum mechanics proof this question is based on my previous closed question but it is a more specific part and hopefully I will get help.

The classical limit of quantum mechanics is $$\hbar\to 0$$. In this case, the momentum operator becomes 0. But in classical physics momentum is still the generator of translation. how is this possible if the momentum operator becomes trivial?

The generator of $$x$$-translations is more precisely the $$x$$-derivative $$\frac{\partial}{\partial x}~=~i\frac{\hat{p}}{\hbar}~=~i\hat{k},$$ the wavenumber operator, which is independent of $$\hbar$$.

You are willfully misunderstanding the $$\hbar/S \to 0$$ classical limit. ℏ is dimensionful, so choosing enormous units to measure it with, like MKSA units to measure moving trains, makes it look small. The "proper", enormously subtle, classical limit runs on the above dimensionless ratio comparing the characteristic action quantity S of the system to ℏ.

The way ℏ appears to enter translations in the x-representation, $$e^{a{i\over \hbar} \hat p} f(x)= e^{a\partial_x} f(x)= f(x+a),$$ is as a rescaling/normalization of operators in the exponent to make it dimensionless, a habit of measuring normalization of phase-space operators in the Born commutation relation. You appear to be worrying about a non-issue. The gradient of a function is as huge as its local variation over a given scale.

I answered your earlier question. The answer here is similar. If you want to compare quantum to classical in a limit, you cannot do it piece by piece. You need to look at it as part of a coherent and self-consistent limit. In this case, trying to separately look at the limit of $$\hat{p}$$ in the $$\hbar \rightarrow 0$$ limit (see related comment in my answer to your earlier question about whether doing this as a limit is really the correct conceptual approach - but we'll go with it here) makes no sense because momentum is not an operator as viewed in the classical theory.

What follows is not a full, formal proof, but it outlines some relevant considerations.

The observables in quantum theory will apply the operator to a wavefunction and work from there, commonly to take an expectation value. The wavefunction also depends on $$\hbar$$ so, if you're going to take a limit, you need to know how the combined entity behaves not just one of part. As noted in the answer to your earlier question, you can always write a solution to the Schrodinger equation as $$\psi = R e^{iS/\hbar}$$ for real $$R$$ and $$S$$. It turns out that $$S$$ is then closely related to the action such that $$\nabla S$$ looks like the classical momentum. At the same time, $$\hat{p} \psi = \left( \nabla S \right) \psi$$. The right-hand side only depends on $$\hbar$$ through the argument to the exponential, and that goes away in, say, an expectation-value expression where you multiply by $$\psi^*$$.

The correct answer was given by Cosmas Zachos, however, I want to add a remark concerning the assertion that "$$\hat p$$ converges to $$0$$ in the classical limit"

One can actually look directly at the limit of the impulsion operator in a more mathematically precise way: the momentum operator is defined by $$\hat p = -i\hbar\nabla$$. Notice first that it is an unbounded operator on $$L^2$$, so one should be careful about what "convergence to $$0$$" means (in operator norm, $$\hbar\cdot\infty = ∞$$ does not converge to $$0$$).A way to look at the convergence of unbounded operator is by using weak convergence. An example is to look at the limit of its mean value $$\lim_{\hbar\to 0}\mathrm{Tr}(\hat p\,\rho)$$ for any nice density matrix $$\rho$$.

However, when doing the classical limit of quantum mechanics, one should forget that the size of operators also depend on $$\hbar$$.

A classical way to perform the limit $$\hbar\to 0$$ is to introduce the Wigner transform $$W_{\rho}(x,p) = \frac{1}{\hbar^d}\int_{\mathbb R^3} e^{-i\,y\cdot p/\hbar}\,\rho(x+y/2,x-y/2)\,\mathrm d y$$ which is indeed an object that converges to a classical distribution of the phase space. A quick computation then shows that $$\mathrm{Tr}(\hat p\,\rho) = \int_{\mathbb R^6} p\, W_{\rho}(x,p)\,\mathrm d x\,\mathrm d p.$$ In particular, if $$W_{\rho}(x,p)$$ converges to a classical function of the phase space $$f$$ in a suitable topology (which is the assumption that indicates that $$\rho$$ is $$\hbar$$ dependent), then $$\mathrm{Tr}(\hat p\,\rho) \to \int_{\mathbb R^6} p\, f(x,p)\,\mathrm d x\,\mathrm d p$$ which is not $$0$$.