## What's the meaning of the commutator? Not satisfied with usual answer

The usual answer to this question is that if the commutator between two observables A and B is zero, then there are states that have a definite value for each observable. If [A,B] isn't zero, then this isn't true.

Now, in general [A,B] = iC, where C is Hermitian. I'd like to know if there's an intuitive interpretation of the operator C. Evidently it's some sort of 'measure' of how much A and B don't commute, but is there a more concrete interpretation?
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 This is a good question I'm interested in... I think the "measure of non commutivity" might be the best explanation -- for example, the classic is with [x,p] = ihbar, which leads directly to the most famous manifestation of the HUP, because even in a particle's ground state it has nonzero fluctuations.
 There is a concept in Classical Mechanics called a Poisson bracket. Dirac noticed the following transition from Classical to Quantum Mechanics: $$\left\lbrace f, g \right\rbrace_{\mathrm{Poisson}} \rightarrow \frac{1}{i \, \hbar} \, \left[ \hat{f}, \hat{g} \right]$$ So, if you know the physical meaning of the observables corresponding to the (Hermitian) operators $\hat{A}, \hat{B}$, I would say the observable corresponding to what you wrote as the operator $\hat{C}$ would be the the Poisson bracket of A, and B, times the reduced Planck constant. $$C = \hbar \, \left\lbrace A, B \right\rbrace$$

## What's the meaning of the commutator? Not satisfied with usual answer

 Quote by Dickfore There is a concept in Classical Mechanics called a Poisson bracket. Dirac noticed the following transition from Classical to Quantum Mechanics: $$\left\lbrace f, g \right\rbrace_{\mathrm{Poisson}} \rightarrow \frac{1}{i \, \hbar} \, \left[ \hat{f}, \hat{g} \right]$$ So, if you know the physical meaning of the observables corresponding to the (Hermitian) operators $\hat{A}, \hat{B}$, I would say the observable corresponding to what you wrote as the operator $\hat{C}$ would be the the Poisson bracket of A, and B, times the reduced Planck constant. $$C = \hbar \, \left\lbrace A, B \right\rbrace$$
What's the intuition behind the Poisson bracket?
 Oh, so you just wanna be a wise guy. What do you mean by "intuition" as far as a physical equation is concerned?

 Quote by dEdt What's the intuition behind the Poisson bracket?
I can't say what the "intuition" is, but it's another way of formulating classical mechanics that, according to my book, "provides the most direct transition between CM and QM".

Recognitions:
 Hello. Please allow me to present intuitive straightforward physical interpretation. Thank You. Let's introduce two observables. Say, impulse-observable $\mathcal{P}$ and position-observable $\mathcal{X}$. What do these observables do? They act on particle, let's call it $\psi$. So, when impulse-observable $\mathcal{P}$ acts on particle $\psi$, the result is that we observe particle's impulse $p$: $$\mathcal{P} \psi = p \psi$$ Similarly, when position-observable $\mathcal{X}$ acts on particle $\psi$, the result is that we observe particle's position $x$: $$\mathcal{X} \psi = x \psi$$ Suppose we first observe impulse and right afterwards we also observe position. We expect this result: $$\mathcal{X}\mathcal{P} \psi = xp \psi$$ Now suppose we first observe position and right afterwards we also observe impulse. We expect this result: $$\mathcal{P}\mathcal{X} \psi = px \psi$$ So, if we subtract last 2 equations, we find $$\left(\mathcal{P}\mathcal{X}-\mathcal{X}\mathcal{P} \right) \psi = 0$$ In other words, $\left[\mathcal{P},\mathcal{X}\right] = 0$. If so, then we get unique result no matter how we observe the particle. Suppose now we don't get the unique result as we observe particle in different manners. Suppose we get $$\mathcal{X}\mathcal{P} \psi = ab \alpha$$ and $$\mathcal{P}\mathcal{X} \psi = cd \beta$$ Suppose this time $\alpha \ne \beta$. Let's subtract equations now: $$\left(\mathcal{X}\mathcal{P}-\mathcal{P}\mathcal{X}\right) \psi = ab \alpha -cd \beta$$ In other words, $\left[\mathcal{X},\mathcal{P}\right] \ne 0$. What does it mean? It means we can perform infinitude of measurements and get a different result every time we perform another and yet another measurement. So in other words: There are no states that have a definite value for each observable if $\left[\mathcal{X},\mathcal{P}\right] \ne 0$. So if we denote $\left[\mathcal{X},\mathcal{P}\right] = \mathcal{C}$, we find that $\mathcal{C}$ boosts and dislocates particle during measurements. So if successive measurements $\mathcal{C}$ keep pushing particle away from its original position, boosting its velocity, then observables at hand are not quite measurable simultaneously. I hope this explained it a bit. Cheers.