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

dEdt
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?

VortexLattice
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.

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.
$$C = \hbar \, \left\lbrace A, B \right\rbrace$$

dEdt
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?

Dickfore
Oh, so you just want to be a wise guy. What do you mean by "intuition" as far as a physical equation is concerned?

VortexLattice
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".

What's the intuition behind the Poisson bracket?
There's no "intuition" behind the Poisson bracket; it's a reformulation of classical mechanics on phase space with new "coordinates": "generalized position" and "generalized momentum"; after Legendre transformation the Hamiltonian function, the Poisson brackets and the symplectic structure on phase space follows automatically. (You could also start with such a structure w/o using the Lagrangian as starting point, but this is difficult when it comes to identification of symmetries and conservation laws)

So the these concepts are mathematically identical in classical mechanics and there's not more intuition behind the Poisson brackets than behing the Lagrangian: you use a set of axioms, define a certain set of mathematical object (L, ..., x,p, H, ...) and derive the dynamics. The intuition is restricted to the moment where you assign these mathematical structures as a model to a specific physical system.

The_Duck
There is a straightforward physical interpretation when the operators A and B are the generators of some physical transformation.

Consider the case of the angular momentum commutation relation [Jx, Jy] = iJz.

The angular momentum operators are the generators of rotations. So the physical meaning of the commutation relation is: if you do a tiny rotation by an angle theta around the x axis, then a tiny rotation by an angle phi around the y axis, then a tiny rotation by -theta around the x axis, then a tiny rotation by -phi around the y axis, the result will be a tiny rotation by an angle theta*phi around the z axis.

(The same intepretation works for the corresponding classical Poisson bracket.)

Kraflyn
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.

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