- #1
soviet1100
- 50
- 16
Hello,
I'm reading Tong's lecture notes on QFT and I'm stuck on the following problem, found on p.11-12.
A scalar field [itex] \phi [/itex], under a Lorentz transformation, [itex] x \to
\Lambda x [/itex], transforms as
[itex] \phi(x) \to \phi'(x) = \phi(\Lambda^{-1} x) [/itex]
and the derivative of the scalar field transforms as a vector, meaning
[itex] (\partial_\mu \phi)(x) \to (\Lambda^{-1})^\nu{}_\mu (\partial_\nu \phi) (y). [/itex]
where [itex] y = \Lambda^{-1}x [/itex]
Could someone please explain the steps above? If the transformation is an active one, meaning that the field itself is rotated, then how does [itex] x \to \Lambda x [/itex] make sense. I don't get how he got the transformation property of the derivative either.
I'm reading Tong's lecture notes on QFT and I'm stuck on the following problem, found on p.11-12.
A scalar field [itex] \phi [/itex], under a Lorentz transformation, [itex] x \to
\Lambda x [/itex], transforms as
[itex] \phi(x) \to \phi'(x) = \phi(\Lambda^{-1} x) [/itex]
and the derivative of the scalar field transforms as a vector, meaning
[itex] (\partial_\mu \phi)(x) \to (\Lambda^{-1})^\nu{}_\mu (\partial_\nu \phi) (y). [/itex]
where [itex] y = \Lambda^{-1}x [/itex]
Could someone please explain the steps above? If the transformation is an active one, meaning that the field itself is rotated, then how does [itex] x \to \Lambda x [/itex] make sense. I don't get how he got the transformation property of the derivative either.