# Field transformation in Peskin-Schroeder (chapter 3)

1. Jun 27, 2012

### AnZa85

1. The problem statement, all variables and given/known data

There is something I don't understand about eq. 3.110 (there is no need of the complete equation actually) in Peskin Schroeder.
What I need to do is to use the unitary transformation law obtained for one-particle states to get the usual transformation law for the Dirac field (under Lorentz transformations).

2. Relevant equations

3. The attempt at a solution

I've been able to obtain the law stated in P.S.
I also checked the result with the similar law for scalar field transformation and still I don't understand.
I guess I might be wrong somewhere:

I started from Peskin's law for scalar fields:

$\Phi$(x) $\rightarrow$ $\Phi$'(x) = $\Phi$($\Lambda$-1x)

Here the book reads: the transformed field, evaluated at the boosted point, gives the same value as the original field evaluated at the point before boosting.
From this I understand that the previous relation - with explicit notation for coordinate systems - becomes:

$\Phi$(x(O)) $\rightarrow$ $\Phi$'(x(O')) = $\Phi$($\Lambda$-1x(O'))

which gives the correct law for scalars:

$\Phi$(x(O)) $\rightarrow$ $\Phi$'(x(O') = $\Phi$(x(O))

Now, in chapter 3.5, I find:

U($\Lambda$)$\Psi$(x)U-1($\Lambda$) = $\Lambda$1/2-1 $\Psi$($\Lambda$x)

Or the equivalent for scalar field (which is not in Peskin):

U($\Lambda$)$\Phi$(x)U-1($\Lambda$) = $\Phi$($\Lambda$x)

That looks good, provided that I understand the change in the tranformation action due to the fact that we are transforming the ladder operators in Dirac field.
But here comes my question: In deriving these equations, no change was made on coordinate system, so to me they read:

$\Phi$(x(O)) $\rightarrow$ $\Phi$'(x(O)) = $\Phi$($\Lambda$x(O))

Which is not the same - even accounting for the transformation change.

I apologize for the long post on such an inessential question but I could really use some help on this.
Thanks.