samalkhaiat
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Moderator's note: This is a subthread spun off from https://www.physicsforums.com/threads/isthegroundstateenergyofaquantumfieldactuallyzero.953766/.
But, you are absolutely right. As far as I know, the Belinfante expression [tex]J^{\mu\nu}_{Bel} = \int d^{3}x \left( x^{\mu} \theta^{0 \nu}(x)  x^{\nu} \theta^{0 \mu}(x) \right) , \ \ \ \ \theta^{\mu \nu} = \theta^{\nu \mu} \ ,[/tex] seems to fail to satisfy all the commutation relations of the Poincare’ algebra (to be more accurate, I should say that I have never been able to establish the correct bracket [itex]\big[i J^{0 j}_{Bel} , J^{0 k}_{Bel} \big][/itex] using the usual methods, and I don't know if somebody else had).
E. Leader, “Spin in Particle Physics”, Cambridge University Press (2001)
You may also find the attached PDF’s useful
I should have said that in certain cases in QFT, we can neglect “surface terms”. For example, the (onshell) difference between the Belinfante and the canonical expressions for the angular momentum can be written as [tex]J^{ij}_{Bel}  J^{ij}_{C} = \int d^{3}x \ \partial_{\rho}F^{\rho 0 i j} (x) .[/tex] Now, by careful analysis using wave packets, one can show that the forward matrix elements of the RHS do vanish. Also, the forward momentum space matrix elements (i.e., the physical quantities in collision processes) of the “surface terms” do vanish if these terms are divergences of local operators.... this statement is indeed "modulo surface terms". Particularly I don't see any necessity for the energymomentum tensor to be symmetric in the realm of special relativity.
But, you are absolutely right. As far as I know, the Belinfante expression [tex]J^{\mu\nu}_{Bel} = \int d^{3}x \left( x^{\mu} \theta^{0 \nu}(x)  x^{\nu} \theta^{0 \mu}(x) \right) , \ \ \ \ \theta^{\mu \nu} = \theta^{\nu \mu} \ ,[/tex] seems to fail to satisfy all the commutation relations of the Poincare’ algebra (to be more accurate, I should say that I have never been able to establish the correct bracket [itex]\big[i J^{0 j}_{Bel} , J^{0 k}_{Bel} \big][/itex] using the usual methods, and I don't know if somebody else had).
Again, one can show that the physical matrix elements of the generators are gauge invariant.For gauge fields, of course you can argue with gauge invariance,
You are right on the "uniqueness" part. I had the pleasure of knowing and working with Elliot Leader on the very same problem (in QCD) for many years. So, I suggest you have a look at his book:In my opinion, there's neither uniqueness in this split nor is there a really well defined treatment of spin in relativistic hydrodynamics,....
E. Leader, “Spin in Particle Physics”, Cambridge University Press (2001)
You may also find the attached PDF’s useful
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