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nrqed

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The short version is : when we use representations of the charges as differential operators to calculate their commutators, we always get -1 times the correct result. So, doe sthat mean that we cannot use the diferential operators to calculate the algebra??

The long version is:

The commutator of [itex] P_\mu [/itex] with a scalar field is

[tex] [P_\mu, \phi] = -i \partial_\mu \phi [/tex]

whereas the differential operator form is

[tex] P_\mu = i \partial_\mu [/tex]

This agrees with Baer and Tata (Weak Scale Supersymmetry) and with Aitchison (Intro to Supersymmetry), as well as with Greiner (Field Quantization). I have seen results with different signs in other books, though)

The Lorentz generators have commutation relation with a scalar field equal to

[tex] [M_{\mu \nu} , \phi ] = i (x_\nu \partial_\mu - x_\mu \partial_\mu) \phi [/tex]

this agrees with Greiner but disagrees with Srednicki.

Whereas the differential form representation is

[tex] M_{\mu \nu} = i ( x_\mu \partial_\nu - x_\nu \partial_\mu) [/tex]

If I compute the Lorentz algebra using the operators, I get commutators with the signs opposite to what I get when I use the differential operator form. The commutators based on the operators agree with my references. So, is there some reason preventing us to use the differential form to compute the algebra??