What does it mean if something Transforms Covariantly ? (Quantum Field Theory)

Elwin.Martin · Oct 31, 2011

What does it mean if something "Transforms Covariantly"? (Quantum Field Theory)

Referring to an isospinor, [itex]\psi[/itex] which transforms as [itex]\psi[/itex](x)→ [itex]\psi[/itex]'(x)=S(x) [itex]\psi[/itex](x) (S(x) being an n by n matrix)

I'm told that it is clear that [itex]∂_{μ}[/itex][itex]\psi[/itex] does not transform covariantly.

Now, correct me if I'm wrong, but it would appear that [itex]∂_{μ}[/itex][itex]\psi[/itex] ' can be found by the product rule to be S([itex]∂_{μ}[/itex][itex]\psi[/itex]) + ([itex]∂_{μ}[/itex]S)[itex]\psi[/itex].

What is meant by that it doesn't transform covariantly?

I know what covariant vs. contravariant indices are, but I don't know what it means for something to transform covariantly. I understand that we *want* a covariant derivative, but I don't understand why =|

Any and all help would be great!

**from Ryder, in case anyone was wondering

dextercioby · Oct 31, 2011

Re: What does it mean if something "Transforms Covariantly"? (Quantum Field Theory)

I don't a copy of Ryder handy, however I can tell you that [itex] \partial_{\mu}\psi [/itex] does not transform covariantly with respect S(x), because you have another term - [itex] (\partial_{\mu} S)\psi [/itex] - which spoils covariance. If the gradient of psi had transformed covariantly, this extra term would not have been there. Covariant comes from co-variant, in other words, the derivative of the spinor should have transformed like the spinor, i.e. with an identical matrix.

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What does it mean if something Transforms Covariantly ? (Quantum Field Theory)

Elwin.Martin

dextercioby