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I'm trying to make sense of Poisson bracket relation

$$\{L_i,A_k\}_{PB}~=~\epsilon_{ikl}A_l,\tag1$$

where $L_i$ is $i$th component of angular momentum, $A_k$ is $k$th component of an arbitrary vector, and $\epsilon_{ikl}$ is Levi-Civita tensor.

When I take $\vec A$ to be position, momentum or angular momentum, I can confirm that $(1)$ is true. But if I take $\vec A(\vec r)$ to be an arbitrary vector field, e.g. $$\vec A=\cos\left((x-3)^2y\right)\vec e_x+(y+2z)\vec e_y+2\cos\left((y-3)^2z\right)\vec e_z,\tag2$$ $(1)$ appears to be false. Moreover, somewhere I've read that arbitrary vector can be written as $$\vec A=\phi_1\vec r+\phi_2\vec p+\phi_3 (\vec r\times\vec p),\tag3$$ where $\phi_i$ are arbitrary scalar functions and $\vec r$ and $\vec p$ are position and momentum vectors. And $(2)$ can't be written in the form $(3)$.

Does this mean that an arbitrary vector field is not a vector quantity? What does the "vector" word mean in this term then?


EDIT: based on the answers and after some experimenting I seem to have found that $(1)$ should be true for such $\vec A(\vec r,\vec p)$, which for arbitrary rotation $R$ satisfies

$$R\vec A(R^{-1}\vec r,R^{-1}\vec p)=\vec A(\vec r,\vec p).\tag4$$

At least, $(4)$ is satisfied by $\vec r$, $\vec p$ and $\vec L$, while not satisfied by $(2)$. This seems to be what @Qmechanic calls dynamical active vector. Am I right? Where to read more about this notion? Googling this phrase doesn't seem to give anything related.

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  • $\begingroup$ $\uparrow$ Read where? $\endgroup$ – Qmechanic Feb 15 '15 at 15:44
  • $\begingroup$ @Qmechanic do you mean where I've read about $(3)$? See e.g. Landau, Lifshitz "Mechanics" $\S42$ problem 4. $\endgroup$ – Ruslan Feb 15 '15 at 16:41
  • $\begingroup$ @Ruslan Landau and Lifshitz write that any vector field that only depends on position and momentum can be decomposed in that manner. In general this is valid only in 3D (where the cross product is defined) and only if $x$ and $p$ are not parallel (i.e. if $x \times p \neq 0$). $\endgroup$ – Neuneck Feb 16 '15 at 14:38
  • $\begingroup$ @Neuneck how does $(2)$ not satisfy your italicized statement? It depends only on position $(x,y,z)$ and momentum (constant in $\vec p$). $\endgroup$ – Ruslan Feb 16 '15 at 15:13
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Comments to the question (v3):

I) The notions of vectors, tensors, scalars, etc, depend on contexts in physics, cf. e.g. this and this Phys.SE posts and links therein.

II) In OP's context, these notions refer to representations $\rho$ of the Lie group $SO(3)$ [and the corresponding Lie algebra $so(3)$] of 3D rotations, cf. e.g. Ref. 1. Let $\mathrm{i}L_k$, $k=1,2,3$, denote the $3$ antisymmetric real $3\times 3$ matrix generators of $so(3)$: $$\mathrm{i}(L_k)_{ij}~:=~\epsilon_{ijk}.\tag{A}$$

In this context, a vector$^1$ $V_k$ is an object that transforms in the 3-dimensional vector (or triplet) representation $\bf 3$ of $SO(3)$. Concretely this means that $$(\rho(\mathrm{i}L_k)V)_i~=~\epsilon_{ijk}V_j,\tag{B}$$ up to sign conventions. Similarly, a scalar $S$ is an object that transforms in the 1-dimensional trivial (or singlet) representation $\bf 1$ of $SO(3)$. Concretely this means that $$(\rho(\mathrm{i}L_k)S)_i~=~0.\tag{C}$$

Example: OP's vector field (2) can be viewed as an element in the vector representation of $SO(3)$. Not surprisingly, the group action of a rotation $R\in SO(3)$ on the vector field leads to rigid rotations of the vector field (2). However, the vector field (2) is not a dynamical active vector, see next section III.

III) In the Hamiltonian formalism, for objects constructed out of e.g. $x_i$, $p_j$, $\delta_{ij}$, $\epsilon_{ijk}$, etc., and contractions thereof, there is also another notion. Let us for the purpose of this answer call it the notion of dynamical active vectors, tensors, scalars, etc. E.g. a dynamical active vector $V_k$ satisfies by definition $$ \{L_i,V_j\}_{PB}~=~\epsilon_{ijk}V_k\tag{D}$$ where $$L_i~:=~\epsilon_{ijk}x_jp_k\tag{E}$$ is the orbital angular momentum, and $$\{x_i,p_j\}_{PB}~=~\delta_{ij}~=~-\{p_i,x_j\}_{PB},\quad \{x_i,x_j\}_{PB}~=~0~=~\{p_i,p_j\}_{PB},\tag{F}$$ is the canonical Poisson brackets (PB). Similarly, a dynamical active scalar $S$ satisfies $$ \{L_i,S\}_{PB}~=~0.\tag{G}$$

Example 1: Consider the inner/dot product $S=a_kx_k$, where $a_k$ is a constant (dynamically passive) vector. Then $S$ is a scalar (C), but $S$ is not a dynamical active scalar (G).

Example 2: The inner/dot products $S=x_kx_k$, $S=x_kp_k$, $S=p_kp_k$, and $S=a_ka_k$ are all dynamical active scalars according to definition (G).

Example 3: A linear combination $$V_k~=~\sum_a S_{(a)} V_{(a)k},\tag{H}$$ of dynamical active vectors $V_{(a)k}$ with dynamical active scalar coefficients $S_{(a)}$ is again a dynamical active vector (D).

Example 4: The three objects $x_i$, $p_j$, and $L_k$ are all dynamical active vectors according to definition (D). In fact, these three objects constitute a basis for dynamical active vectors, in the sense of eq. (H), cf. Ref. 2 and OP's eq. (3).

References:

  1. G. 't Hooft, Introduction to Lie Groups in Physics, lecture notes; chapter 3 + chapter 9. The pdf file is available here.

  2. L.D. Landau & E.M. Lifshitz, Mechanics, vol. 1, 1976; $\S$42. Problem 3 & 4. Note that L&L have the opposite convention for the Poisson bracket (PB) as compared to eq. (F).

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$^1$ We assume here that the metric $g_{ij}=\delta_{ij}$, and hence will not distinguish between covariant and contravariant indices, cf. e.g. this Phys.SE post. Also we assume Einstein summation convention, i.e. summation over repeated indices.

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  • $\begingroup$ So, as I understand you, in 4. and 5. you contrast $A_k$, the function of a single phase space point, with $(2)$, a function of position $\vec r$. But what if we view $(2)$ as a function of phase space point $(\vec r,\vec p)$, which just doesn't depend on $\vec p$? What's the major difference between these two types of functions then? Or, going backwards, why isn't e.g. $\vec A(\vec r)=\vec r$ a vector field, which would also rigidly rotate under $R$? $\endgroup$ – Ruslan Feb 15 '15 at 7:32
  • $\begingroup$ So what is the definition of dynamical active vector — is it (D)? My vector $(2)$ is also constructed of $x_i$ (if we view it just as a function of current $\vec x$), but still doesn't satisfy (D). Should the dynamical active vector be linear in $x_i$ and $p_i$ or what? $\endgroup$ – Ruslan Feb 15 '15 at 15:09
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Feb 15 '15 at 16:53
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The Poisson bracket you wrote only works for position (which is not a vector in general, as coordinates do not transform as vectors or any other tensor quantity). For general vectors the correct Poisson bracket is $$\{L_i,A_m\} = - (\mathbf p\times\nabla_{\mathbf p} + \mathbf r\times\nabla)_i A_m,$$ which reduces to the relation you wrote if you take $A_m = r_m$.

As for the decomposition, that must work every time $\mathbf r$ is not parallel to $\mathbf p$, as in this case $\mathbf r,\mathbf p,\mathbf r\times\mathbf p$ is a set of independent vectors in $\mathbb R^3$ and therefore they constitute a basis.

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  • $\begingroup$ Hmm, does this hold for quantum commutator of the operators of corresponding quantities? I've taken $(1)$ from the commutator given as $(29.4)$ in this book and just checked that it works for $\vec r$. The authors of the book talk about radius vector as of a vector, while you say it's not a vector. What am I misunderstanding? $\endgroup$ – Ruslan Feb 14 '15 at 12:37
  • $\begingroup$ the angular momentum generates the rotations around the axis given by its direction, so the radius "vector" gets rotated and therefore it looks like it behaves like a vector. The same confusion occurs in special relativity, where the position vector transforms with Lorentz transformations, but of course one can consider coordinate changes which are not linear. $\endgroup$ – Phoenix87 Feb 14 '15 at 12:41
  • $\begingroup$ I still don't get in what cases position of particle doesn't transform as vector? Also, the Poisson bracket $(1)$ works not only for position but also for momentum, and even for angular momentum although it's a pseudovector. Do all three of these not transform as vectors? $\endgroup$ – Ruslan Feb 14 '15 at 16:09

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