There is a comparison between classic mechanics and quantum mechanics, although they work fine in different areas, it makes me wonder how does an operator degenerate into a function/symbol, when quantum mechanics degenerate into the form of classic mechanics? Compared with the relativity theory, when the speed decrease, I could see the correction of momentum due to the relativity theory effect is also decreasing gradually. But regarding quantum mechanics, this seems not the same situation, I couldn't imagine how an operator degenerates into a function/symbol gradually.
My question is:
How does an operator degenerate into a function as the quantum mechanics effect decreasing? Any examples?

$$\tag{0} \text{Quantum Mechanics}\quad\longleftrightarrow\quad\text{Classical Mechanics}$$

$$\tag{1} \text{Operator}\quad\hat{f}\quad\longleftrightarrow\quad\text{Function/Symbol}\quad f,$$

$$\tag{2} \text{Commutator}\quad \frac{1}{i\hbar}[\hat{f},\hat{g}] \quad\longleftrightarrow\quad\text{Poisson bracket}\quad \{f,g\}_{PB} , $$

$$ \text{Heisenberg's EOMs}\quad\quad\quad\quad\quad\quad \text{Hamilton's EOMs} $$ $$\tag{3} \frac{d\hat{f}}{dt}~=~\frac{1}{i\hbar}[\hat{f},\hat{H}]+ \frac{\partial\hat{f}}{\partial t}\quad\longleftrightarrow\quad \frac{df}{dt}~=~\{f,H\}_{PB}+ \frac{\partial f}{\partial t} .$$

The formula above is from https://physics.stackexchange.com/a/245340/287975

  • $\begingroup$ Related/possible duplicates: physics.stackexchange.com/q/32112/50583 and its linked questions $\endgroup$
    – ACuriousMind
    Commented Jun 27, 2021 at 16:45
  • 2
    $\begingroup$ Canonical quantisation isn't a spectrum from classical to quantum mechanics, it is a rough prescription for obtaining the corresponding quantum system from a classical one. $\endgroup$
    – Charlie
    Commented Jun 27, 2021 at 16:48
  • $\begingroup$ @Charlie, but classic mechanics is the approximation when h decrease to 0, so doesn't it mean there should be a transition to classic mechanics? $\endgroup$
    – Jack
    Commented Jun 27, 2021 at 16:53
  • $\begingroup$ Are you expecting an "in-between object" between Hilbert space operators and functions on phase space? I'm not completely sure what you mean. $\endgroup$
    – Charlie
    Commented Jun 27, 2021 at 16:54
  • $\begingroup$ in the classical limit you work with the expectation values, rather than the operators. These expectation values follow the classical equations (assuming the system has a classical limit). $\endgroup$
    – user65081
    Commented Jun 27, 2021 at 16:56

2 Answers 2


Much of what you are seeking may be in the PSE questions linked. Your question is actually malformed. The proper formulation to take your limit in is the phase-space formulation, but you are making a conceptual slip in classifying Weyl symbols, its working components, as classical functions in (1).

They are c-number functions of ordinary phase-space, but may (and often do) depend on ℏ. They are full quantum objects, Wigner maps of operators, and completely equivalent, as the Wigner map is invertible. (The inverse is the Weyl map, and the "symbols" you mention are precisely the Wigner images of operators.)

It is only the small-ℏ limits of these Wigner images that are classical: so a "symbol" degenerates to a classical observable, in your language. All books on phase space QM cover them. The best thing here is an example.

My example below is from ours, but follows Shewell's Thesis (1959) and the "pedagogical dilemma" discussion of J Dahl and M Springborg, Mol Phys 47 (1982) 1001-1019, and especially their appendix.

So the equivalence in your equation (1) is operator-to-symbol (Wigner map); while "classical" is a limit of the symbol: a focus on the ℏ-independent part of the Wigner image.

In general, the Wigner image of an operator $\hat f$, $$ f(x,p; \hbar) =\frac{\hbar}{2\pi} \int d\tau d\sigma ~ e^{i(\tau p + \sigma x)} \hbox{Tr}~\left ( e^{-i(\tau {\hat p} + \sigma {\hat x})} {\hat f} \right ) \\ =\bbox[yellow]{ \hbar \int dy~ e^{-iyp} \left \langle x +\frac{\hbar}{2}y \right | {\hat f}({\hat x},{\hat p}) \left | x-\frac{\hbar}{2}y \right \rangle } $$ contains ℏ corrections when there are QM ordering ambiguities in the observables, such as in the Wigner map of the square of the angular momentum, $\hat {\mathbf L} \cdot \hat {\mathbf L}$.

This one contains an additional term $−3 \hbar^2/2$ introduced beyond the mere classical expression ${\mathbf L} \cdot {\mathbf L}$. Moreover, upon expectation-averaging (RMS), this quantum offset accounts for the paradoxically nontrivial angular momentum ${\mathbf L} \cdot {\mathbf L}=\hbar^2$ of the ground-state Bohr orbit, whereas the standard Hydrogen quantum ground state has vanishing $\langle \hat {\mathbf L} \cdot \hat {\mathbf L}\rangle =0$. Dahl and Springborg resolve this "pedagogical dilemma" (conundrum) by studying the expectation values in phase space and demonstrating these extra terms exactly patch up the (semi-)classical Bohr values.

This specific celebrated example helped (1959) convincingly dash Weyl's 1927 interpretation of his operator ordering prescription as a quantization recipe, by providing a dramatic "inconsistency", even though, of course, most in the late 40s (Groenewold's epochal Thesis) understood it (and any competitor) was no good as a consistent quantization recipe. (But this is far beyond your question, already).

The Weyl map mentioned above is $$ {\hat f}({\hat x},{\hat p}) =\frac{1}{(2\pi)^2}\int d\tau d\sigma dx dp ~f(x,p;\hbar) \exp \Bigl ( i\tau ({\hat p}-p)+i\sigma ({\hat x}-x) \Bigr ) . $$ It is subtly different than Weyl's original, in that the symbol may, in general, contain $\hbar$, whereas Weyl's never did: he chose its classical limit, as explained, believing his map was a quantization recipe, erroneously. So, for instance, to get the commutator on the left, you need the so called Moyal Bracket as your symbol, whereas Weyl (and, indirectly, Dirac!) might have used the classical limit of the MB, the Poisson Bracket!

I was elliptical about the classical limit of Wigner maps above, since ℏ is a dimensionful variable; the limit suppresses/truncates the dimensionless deformation parameter ℏ/S analogous to your v/c in the non-relativistic limit, where S is the characteristic action of the object. (Recall the inverse of it led to Dirac's understanding of the classical limit in his path integral.)

  • The takeaway is that operators are completely equivalent to c-number phase-space functions (Weyl symbols) and transitioning from quantum to classical mechanics involves suppression of ℏ.

  • Note in response to your edit attributing the list to a citation of an answer.

It is clear that answer uses "symbol" in the original (ℏ=0) Weyl sense, which is not how I would have placed it, myself. To the extent that $\longleftrightarrow$ indicates an invertible map, whereas $\longrightarrow$ an irreversible limit, on line (1) I would have placed, instead:

Operator $\hat f ~~~\longleftrightarrow~~~$ Symbol $f(x,p;\hbar) \qquad\longrightarrow~~$ Classical observable $f(x,p;0)$.

The keystone Symbol bridges Hilbert space to phase space, but, as illustrated above, it is thoroughgoingly quantum and equivalent to the operator.

In fact, as you may read up in books, symbols completely commute among themselves, but, to consistently work out quantum mechanics with them, you need to compose them noncommutatively through the star-product—which I tried to avoid for simplicity here. There are really two transitions on your list: one is the invertible translation of operators to their Wigner map images, the other being the limit/truncations from the latter to classical observables, as illustrated.


We need to extend your table of correspondencies by some more lines:

$$\text{Quantum Mechanics} \quad\longleftrightarrow\quad \text{Classical Mechanics} \tag{0}$$

$$\text{Schrödinger's Wave Function} \quad\longleftrightarrow\quad \text{Hamilton's Principal Function}$$ $$\Psi(q,t) \quad\longleftrightarrow\quad S(q,t) \tag{4}$$

$$\text{Schrödinger's equation} \quad\longleftrightarrow\quad \text{Hamilton-Jacobi equation}$$ $$i\hbar\frac{\partial\Psi(q,t)}{\partial t}=\hat{H}\left(q,\frac{\hbar}{i}\frac{\partial}{\partial q},t\right)\Psi(q,t) \quad\longleftrightarrow\quad -\frac{\partial S(q,t)}{\partial t}=H\left(q,\frac{\partial S(q,t)}{\partial q},t\right) \tag{5}$$

The correspondence in (4) and (5) is accomplished by $$\Psi(q,t)=A(q,t)e^{iS(q,t)/\hbar}$$ where the amplitude $A(q,t)$ is slowly varying in space and time, and the phase $S(q,t)/\hbar$ is rapidly varying in space and time (because of the smallness of $\hbar$). Such a $\Psi$ is called a quasi-classical wave function.

Now we are ready to derive how a quantum-mechanical operator comes about to correspond to a classical function.

As an example let's take the $x$-component of the momentum operator $\hat{p}_x=\frac{\hbar}{i}\frac{\partial}{\partial x}$. When we apply this operator to the wave function $\Psi$ from above we get $$\begin{align} \hat{p}_x\Psi &=\frac{\hbar}{i}\frac{\partial}{\partial x}\Psi =\frac{\hbar}{i}\frac{\partial}{\partial x}\left(Ae^{iS/\hbar}\right) \\ &=\left(\frac{\hbar}{i}\frac{\partial A}{\partial x}+\frac{\partial S}{\partial x}A\right)e^{iS/\hbar} \approx \frac{\partial S}{\partial x}Ae^{iS/\hbar} =\frac{\partial S}{\partial x}\Psi =p_x\Psi \end{align}$$

(At $\approx$ we used the smallness of $\hbar$ and the slow variation of $A$.)
So we have finally found: Applying the operator $\hat{p}_x$ to a wave function $\Psi$ is approximately equal to multiplying it by the function $\frac{\partial S}{\partial x}$, which is just the classical momentum $p_x$ in the Hamilton-Jacobi formalism.

You can repeat the same reasoning for any operators composed of the elementary operators $\hat{q}_i$ and $\hat{p}_i$, including the Hamiltonian operator $\hat{H}$ itself.


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