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

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SUMMARY

The discussion centers on the concept of covariant transformation in Quantum Field Theory, specifically regarding the isospinor \(\psi\) and its derivative \(\partial_{\mu}\psi\). It is established that while \(\psi\) transforms as \(\psi'(x) = S(x) \psi(x)\) using the transformation matrix \(S(x)\), the derivative \(\partial_{\mu}\psi\) does not transform covariantly due to the presence of an additional term \((\partial_{\mu} S)\psi\). This extra term indicates that the transformation of the derivative does not maintain the covariance expected in the transformation of the spinor itself.

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Elwin.Martin
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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
 
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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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