# Interpreting the commutators of the Poincare generators

Suppose we have the usual commutators ($J$=Angular Momentum, $P$=Momentum, $K$=Boosts, $H$=Hamiltonian.) $$[J_i,J_j]=i\epsilon_{ijk}J_k\quad[J_i,K_j]=i\epsilon_{ijk}K_k\quad[J_i,P_j]=i\epsilon_{ijk}P_k.$$ and that $$[K_i,H]=iP_i.$$ A professor has said that the first three relations state that $\vec J,\vec K,\vec P$ are 3-vectors and that they rotate under spatial rotations. And that the interpretation of the fourth is that if we boost energy, we get momentum.

Can anybody give the chain of logical statements, beginning with these commutators, that leads to these interpretations? This has had me confused for over a year...

• You must simply exponentiate the commutators into finite transformations and you will materialise your professor's statements... – Valter Moretti Sep 28 '16 at 19:10

Under a linear transformation $T$ of the vector space, operators $O$ on it transform as $O\mapsto TOT^\dagger$. Since by definition the $J_i$ are the infinitesimal generators of rotation as $R(\phi) = \mathrm{e}^{\mathrm{i}J_i\phi}$, the finite rotation $\mathrm{e}^{\mathrm{i}J_i \phi}O\mathrm{e}^{-\mathrm{i}J_i\phi}$ implies that the infinitesimal change of any observable under rotation is $[J_i,O]$. (This follows by Taylor expanding the exponentials and keeping only the term to first order in $\phi$.)

So the commutation relations of the form $[J_i,O]$ tell you how $O$ changes under rotation.

The exact same reasoning goes through for the infinitesimal generators $K_i$ of the Lorentz boosts, so $[K_i,H]$ is the infinitesimal change of energy under a boost.

• Sheesh, it seems I don't know basic quantum mechanics... I'll go study Sakurai, I suppose. Anyway, thanks for this concise answer! – MaanDoabeDa Sep 28 '16 at 19:15
• I should ask to be sure, you are thinking in the Heisenberg picture, correct? That is where the $O\to TOT^\dagger$ is coming form, right? @ACuriousMind – MaanDoabeDa Sep 28 '16 at 19:46
• @MaanDoabeDa No, this isn't a time evolution, it's just in general how linear operators transform under linear transformations of the space they act on. There's nothing inherently quantum mechanical about this - think about how you would apply a change of basis (itself represented by a matrix) to a matrix in usual linear algebra. – ACuriousMind Sep 28 '16 at 19:49
• That was what I originally thought but then got confused. I just now realized because of your comment that, due to the Dirac notation, when we expand operators with terms like $\ket{m}\bra{n}$, such an object transforms like a mixed rank-2 tensor. Thank you for your comment, much appreciated! @ACuriousMind – MaanDoabeDa Sep 28 '16 at 23:18

For the sake of being explicit, I will demonstrate what @ACuriousMind and @Valter Moretti said to do. I won't include all the $i$'s though.

Supposing we induce a rotation of our coordinate system by an angle $\phi$ about some axis. Seeing as the operators $L_i$ are the generators of such rotations, any operator existing on our Hilbert space will now be transformed as $$\vec O\to e^{\vec\phi\cdot \vec L}\vec Oe^{-\vec\phi\cdot \vec L}\approx \vec{O}+[\vec\phi\cdot \vec L,\vec O]+\cdots,$$ where $\vec O=\hat{x_1} O_1+\hat{x_2} O_2+\hat{x_3} O_3$ and where we have used the Campbell Hausdorff Baker identity. Supposing we have the commutation relation $$[L_i,O_j]=\epsilon_{ijk}O_j,$$ we can rewrite the above as $$\vec{O}+[\vec\phi\cdot \vec L,\vec O]=\vec O+\phi_i \hat{x}_j[L_i,O_j]=\vec O-\vec\phi\times\vec O.$$ This says that the change induced in our operator is circulatory around $\phi$ in the direction opposite to the direction which the basis transformed. This makes sense, because the basis and components of a vector transform with transformations that are the inverses of one another. Hence, this operator is doing exactly what we would expect a vector to do under such a basis change.

Because $K,P,L$ all have the same commutation relations with $L$, they all behave as 3-vectors under rotation. From now on, you may remember this fact and not have to do it all over again.