Proving the Last Term in the Poincaré Group Lie Algebra Identity

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SUMMARY

The discussion focuses on proving the last term in the Poincaré group Lie algebra identity, specifically addressing the expression involving the commutators of the operators \(L_{\mu \nu}\) and \(P_{\rho}\). The participant calculates the commutators and identifies the last term, \(i\hbar(x_{\mu}\partial_{\rho}P_{\nu}-x_{\nu}\partial_{\rho}P_{\mu})\), as unnecessary. The conclusion drawn is that this term should equal zero, and the correct interpretation involves recognizing that the derivative acts on the product of \(x_{\mu}\) and \(P_{\nu}\). This clarification resolves the confusion surrounding the calculation.

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malaspina
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Homework Statement



The problem statement is to prove the following identity (the following is the solution provided on the worksheet):

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Homework Equations



The definitions of [itex]L_{\mu \nu}[/itex] and [itex]P_{\rho}[/itex] are apparent from the first line of the solution.

The Attempt at a Solution



I get to the second line and calculate the commutators explicitly:

[itex]-i\hbar(\partial_{\rho}x_{\mu}-x_{\mu}\partial_{\rho})P_{\nu}+i\hbar(\partial_{\rho}x_{\nu}-x_{\nu}\partial_{\rho})P_{\mu}[/itex]

The derivatives of the coordinates give the metric tensor e.g. [itex]\partial_{\rho}x_{\mu}=g_{\rho \mu}[/itex]

Calculating the derivatives of the coordinates and rearranging I get:

[itex]-i\hbar g_{\rho \mu}P_{\nu}+i \hbar g_{\rho \nu}P_{\mu} +i\hbar(x_{\mu}\partial_{\rho}P_{\nu}-x_{\nu}\partial_{\rho}P_{\mu})[/itex]

The first two terms are the solution I'm looking for, so I'd deduce the last term should be equal to zero.

Is this correct? and if it is, how do I prove that the last term is in fact equal to zero?
 
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malaspina said:
Calculating the derivatives of the coordinates and rearranging I get:

[itex]-i\hbar g_{\rho \mu}P_{\nu}+i \hbar g_{\rho \nu}P_{\mu} +i\hbar(x_{\mu}\partial_{\rho}P_{\nu}-x_{\nu}\partial_{\rho}P_{\mu})[/itex]

The first two terms are the solution I'm looking for, so I'd deduce the last term should be equal to zero.

The last term should not be there. You need to be careful when evaluating [itex]-i\hbar(\partial_{\rho}x_{\mu}-x_{\mu}\partial_{\rho})P_{\nu}+i\hbar(\partial_{\rho}x_{\nu}-x_{\nu}\partial_{\rho})P_{\mu}[/itex]

For example, if you bring in the ##P_{\nu}## in the first term you get ##-i\hbar(\partial_{\rho}x_{\mu}P_{\nu}-x_{\mu}\partial_{\rho}P_{\nu})##

The first term in the parentheses should be interpreted as ##\partial_{\rho}(x_{\mu}P_{\nu})## where the derivative acts on the product of ##x_\mu## and ##P_{\nu}##.
 
TSny said:
The last term should not be there. You need to be careful when evaluating [itex]-i\hbar(\partial_{\rho}x_{\mu}-x_{\mu}\partial_{\rho})P_{\nu}+i\hbar(\partial_{\rho}x_{\nu}-x_{\nu}\partial_{\rho})P_{\mu}[/itex]

For example, if you bring in the ##P_{\nu}## in the first term you get ##-i\hbar(\partial_{\rho}x_{\mu}P_{\nu}-x_{\mu}\partial_{\rho}P_{\nu})##

The first term in the parentheses should be interpreted as ##\partial_{\rho}(x_{\mu}P_{\nu})## where the derivative acts on the product of ##x_\mu## and ##P_{\nu}##.
Jesus, I was completely dumbfounded and it was so obvious. Now that I've re-done it I don't even know how I was missing it. Thank you a lot!
 

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