Thought about this while I was looking at some stuff on quantum-classical correspondence and where precisely the difference between quantum and classical comes from.

Usually it's said that the key/fundamental difference is that in classical mechanics

$$[x, p] = 0$$

holds, and in quantum mechanics

$$[x, p] = i$$


However, thinking about what $p$ actually is from group theory,

$$p = k\frac{\partial}{\partial x},\ T(\Delta x) = e^{\frac{\Delta x}{k}p}$$

where $T$ is the translation operator and $k$ is some constant yields the result that

$$[x, p] = k$$

implying $k \neq 0$ if we want our canonical Hamiltonian momentum to be consistent with the group theoretic momentum.

Now, I know in classical mechanics there is the Poisson bracket, and $\{x, p\} = 1$ in that case, so does that mean the classical mechanics is simply using a different representation of the group (Poincare or Galilean), with a different Lie bracket (Poisson instead of commutator)?

If that's the case, is it simply the representation that makes QM different from CM? Or is it actually the $i$ vs $1$?

TL;DR: Group theory seems to imply that $x$ and $p$ cannot commute. How does that gel with classical mechanics, which is presumably also consistent with group theory in a way I'm missing?

Edit: My question isn't about how QM is different from CM, though that is the context. It's about how to reconcile commuting $x$ and $p$ with $p$ as the generator of translations.

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    $\begingroup$ Possible duplicate of What makes a theory "Quantum"? $\endgroup$
    – Hugo V
    Jan 29, 2019 at 18:01
  • $\begingroup$ That doesn't answer my question about classical mechanics and group theory $\endgroup$
    – gautampk
    Jan 29, 2019 at 18:03
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    $\begingroup$ If you really wished to talk group theory, the infinite-dimensional Lie Algebra $\{f,g\}$ of Hamiltonian mechanics (PB) is a very, very, very different Lie Algebra than the one ($[ f(x,p), g(x,p)]$ ) generating the Heisenberg group. This is the essence of Groenewold's theorem, ensuring quantization is not a functor. Classical limits are clever "cheats" marginalizing these deep differences. $\endgroup$ Jan 29, 2019 at 18:28
  • 1
    $\begingroup$ Right I didn't read your post properly... not enough coffee I guess. $\endgroup$ Jan 29, 2019 at 18:30
  • 3
    $\begingroup$ A bit of beating a dead horse, but the proper "translation-generating" expression you meant to write above is $f(x+\Delta)= \exp(\Delta\partial_x) ~~f(x)= \exp(-\Delta \{p,\cdot)~~ f(x)\equiv f(x)-\Delta \{p,f(x)\}+ \Delta^2 \{p,\{p,f(x)\}\}/2!+... $. This is the sense in which p generates translations in Hamiltonian mechanics. $\endgroup$ Jan 29, 2019 at 20:08

2 Answers 2


The saying you are asking about is not meant as a technical statement, but as a heuristic, much like saying that classical mechanics is quantum mechanics in the limit $\hbar\to 0$ (this is true in some sense, but simply taking that limit straightforwardly on standard QM does not yield classical mechanics).

In particular, the statement $[x,p] = 0$ is nonsensical in the standard classical Hamiltonian formalism. There is no "commutator" of classical Hamiltonian observables, the corresponding structure is the Poisson bracket which is always $\{x,p\} = 1$.

There is an operatorial formulation of classical mechanics, namely Koopman-von Neumann mechanics in which $x$ and $p$ are operators acting on a Hilbert space with $[x,p] = 0$.

Your claim that "group theory" forces $x$ and $p$ to have a non-zero commutator is not correct. What is correct is that the generator of spatial translations $t_x$ needs to have $[x,t_x] = \mathrm{i}$ and the generator of momentum translations $t_p$ needs to have $[p,t_p] = \mathrm{i}$. But if you're not coming from the Hamiltonian formalism, then it is not necessarily the case that $t_x = p$. In fact, Koopman-von Neumann mechanics indeed is free to introduce these generators as additional operators with no algebraic relation to $x$ or $p$, cf. e.g. the first chapter of Mauro's " Topics in Koopman-von Neumann Theory".

  • $\begingroup$ Right, but if $t_x$ and $p$ don't coincide is there a guarantee that $p$ is conserved? $\endgroup$
    – gautampk
    Jan 29, 2019 at 18:19
  • $\begingroup$ @gautampk Yes, the KvN algebra has $[t_x,p] = 0$ and $[t_p,x] = 0$. $\endgroup$
    – ACuriousMind
    Jan 29, 2019 at 18:21
  • $\begingroup$ Ah right, thanks very much. I'll have a deeper look at KvN $\endgroup$
    – gautampk
    Jan 29, 2019 at 18:54

I'm not sure I totally understand your question. Quantum mechanics and classical mechanics are just mathematically completely different. The statement $[q, p] = 0$ in classical mechanics is pretty tongue-in-cheek, because a measurement is just a completely different thing in classical mechanics.

This leaves us with the question: what is the physical significance of the statement $\{q, p\} = 1$ in classical mechanics? And does $p$ "generate translations" in classical mechanics? If so, how?

To answer these questions, we must first discuss what a vector field is. As you may know, a vector field is often defined to be a first order differential operator which takes functions to functions.

Let's consider classical mechanics in one spatial dimension. Phase space would be $\mathbb{R}^2$, with a $q$ direction and a $p$ direction. Consider a test function $f : \mathbb{R}^2 \to \mathbb{R}$. A vector field $X$ will take the function $f$ to a new function $X(f)$. In coordinates, we can write $X$ as

$$ X = a\partial_q + b \partial_p $$ where $a, b : \mathbb{R}^2 \to \mathbb{R}$. This can be thought of as a directional derivative in the $(a, b)$ direction.

Now, for any function $h$ on phase space, we can define the hamiltonian vector field, denoted $X_h$, to be

$$ X_h(f) \equiv \{f, h\} $$ where $\{\cdot , \cdot \}$ is the Poisson bracket $$ \{f, h\} \equiv \frac{\partial f}{\partial q} \frac{\partial h}{\partial p} -\frac{\partial h}{\partial q} \frac{\partial f}{\partial p}. $$

We are now ready to discuss how a function $h$ can "generate" a transformation. Say you have a path in phase space $\gamma: [0,l] \to \mathbb{R}^2$. The first derivative $\dot \gamma$ should be regarded as a directional derivative as well, i.e. it takes a function defined on $Im(\gamma)$ to another function, defined by

$$ \dot \gamma(t) (f) \equiv \frac{d}{ds}f(\gamma(s))|_{s = t}. $$

A path is said to give the flow of the vector field $X_h$ is it satisfies $$ \dot \gamma = X_h $$ for all points $\gamma(t)$.

This is all to say that $h$ generates the transformation given by the flow of the vector field $X_h$. For our two dimensional example, this just amounts to solving the following differential equation:

$$ \dot q = \frac{\partial h}{\partial p} \\ \dot p = -\frac{\partial h}{\partial q} $$

We can see that for $h = p$, this just generates spatial translations: $\dot q = 1, \dot p = 0$.

It is interesting to gain a bit of intuition for what $X_h$ looks like. It is, in a way, the opposite of the gradient of $h$. Where as flow given by the gradient of $h$ would evolve points to places of highest $h$, while flow given by $X_h$ will evolve points to places of equal $h$. In other words, $X_h$ will move points along equipotential lines of $h$. So it is very reasonable that "$p$ generates spatial translations," because the lines of equipotential for the function $p$ are just lines that span all values of $q$.

The statement that $\{q, p\} = 1$ just means that if we evolve along the vector field given by $X_p$, then $q$ will change. That's all it means.

There is one final thing to mention: what does it mean for two functions to "commute" in the sense of $\{g, h\} = 0$? To investigate this question, we use the Jacobi identity:

$$ \{ \{g, h\}, f\} + \{ \{h, f\}, g\} + \{ \{f, g\}, h\} = 0 $$ then rearrange a bit, also using the anti symmetry of the Poisson bracket $$ \{ \{g, h\}, f\} = \{ g, \{h, f\}\} - \{ h, \{g, f\}\} $$ and then using the definition of our Hamiltonian vector fields acting on a test function $f$ to write $$ X_{\{g, h\}} (f) = [X_g, X_h](f) $$ and finally $$ X_{\{g, h\}} = [X_g, X_h] $$ where $[\cdot, \cdot]$ is the vector field Lie Bracket defined by $$ [X_g, X_h] \equiv X_g X_h - X_h X_g. $$

It is a basic fact that the flows given by commuting vector fields also commute. So if $[X_g, X_h] = 0$, that means that evolving along $X_g$ for some time, and then evolving along $X_h$, gives the exact same result as evolving along $X_h$ and then $X_g$. So when $\{g, h\} = 0$, then $[X_g, X_h] = 0$, and the flows commute. This is why functions that commute with the Hamiltonian are said to produce "symmetries."

If you have been astute, you may have noted that if $\{g, h\}$ is a constant, then $X_{\{g, h\}} = 0$. In other words, $X_q$ and $X_p$ actually do commute, even though $\{q, p\} \neq 0$. This actually has an analogue in quantum mechanics. If you translate a wave function in position space by a distance $q$, and then by a momentum $p$, you will recieve the same exact wave function as if you translated by $p$ and then $q$ up to an unobservable phase $e^{iqp}$. So this failure to commute can not be observed by quantum measurements. This is interesting given the fact that in classical mechanics, $X_q$ and $X_p$ commute even though $\{q, p\} \neq 0$.

What does it all really mean though? I don't know. Classical mechanics and quantum mechanics are in many ways parallel stories with their own interesting twists and turns. They rhyme with each other, but certainly one is not the "limit" of the other in any precise sense, and they must both be understood on their own terms.

  • $\begingroup$ Thanks for your answer, it was an interesting read -- basically I think I just don't know enough about the modern formulation of classical mechanics (symplectic manifolds and the like). $\endgroup$
    – gautampk
    Jan 29, 2019 at 23:36
  • $\begingroup$ Specifically wrt quantum-classical 'correspondence' though, I guess my thought is that we have this collection of formalisms that we call 'classical' (Hamiltonian, Lagrangian, Newtonian, KvN, etc) and which are all equivalent to each other, and we have this other collection of formalisms that we call 'quantum' (Schrodinger, Heisenberg, Feynman) that are all also equivalent. I just wonder where specifically we fail when we try to draw an equivalence between the classical collection and the quantum collection $\endgroup$
    – gautampk
    Jan 29, 2019 at 23:38
  • $\begingroup$ Ack, I wasn't sure how much math you knew. Having said that, nothing in my answer is particularly advanced. The definition of vector fields as differential operators is a common thing to do in math, although it's not really crucial to understand any of the ideas. Should be in any normal intro manifolds class/book. $\endgroup$ Jan 29, 2019 at 23:43
  • $\begingroup$ With regards to your question "Now, I know in classical mechanics there is the Poisson bracket, and {x,p}=1 in that case, so does that mean the classical mechanics is simply using a different representation of the group with a different Lie bracket?" I would not say that what separates QM from CM is a choice of representation. While in the case of $\mathbb{R}^{2n}$ phase space is a vector space, it doesn't have to be in general. 1/2 $\endgroup$ Jan 29, 2019 at 23:47
  • $\begingroup$ No it was pitched fine, I'm good with manifolds and diff geom in general (learnt it from a proper mathematician even!) but just not in the specific context of phase spaces and classical mech (focus was more on Lorentzian stuff for GR) $\endgroup$
    – gautampk
    Jan 29, 2019 at 23:47

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