Lorentz transformation of a scalar field

1. Aug 5, 2015

soviet1100

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 $\phi$, under a Lorentz transformation, $x \to \Lambda x$, transforms as

$\phi(x) \to \phi'(x) = \phi(\Lambda^{-1} x)$

and the derivative of the scalar field transforms as a vector, meaning

$(\partial_\mu \phi)(x) \to (\Lambda^{-1})^\nu{}_\mu (\partial_\nu \phi) (y).$

where $y = \Lambda^{-1}x$

Could someone please explain the steps above? If the transformation is an active one, meaning that the field itself is rotated, then how does $x \to \Lambda x$ make sense. I don't get how he got the transformation property of the derivative either.

2. Aug 5, 2015

Avodyne

A (classical) scalar field assigns a number $\Phi(P)$ to each spacetime point $P$. (I'll generalize to quantum fields at the end; it's easier to think about numbers for now.) Suppose that Alice uses coordinates $x$ to label a particular spacetime point $P$, and that Bob uses coordinates $x'$ to label the same spacetime point, and that their coordinates are related by a Lorentz transformation: $x'{}^\mu=\Lambda^\mu{}_\nu x^\nu$. Alice then uses a function $\phi$ of her coordinates for the field, with the property that $\phi(x)=\Phi(P)$ when $x$ is Alice's label for $P$. Bob uses a different function $\phi'$ of his coordinates for the field, with the property that $\phi'(x')=\Phi(P)$ when $x'$ is Bob's label for $P$. From this we have $\phi'(x')=\phi(x)$. Since $x'=\Lambda x$, we also have $x=\Lambda^{-1}x'$. Substituting for $x$ in $\phi'(x')=\phi(x)$, we get $\phi'(x')=\phi(\Lambda^{-1}x')$. We now change the dummy label $x'$ to $x$, and we get $\phi'(x)=\phi(\Lambda^{-1}x)$, which is Tong's first formula.

Tong's second formula then follows from the chain rule for derivatives. Let $\partial_\mu$ denote a derivative with respect to $x^\mu$, and $\bar\partial_\nu$ denote a derivative with respect to $y^\nu=(\Lambda^{-1})^\nu{}_\rho x^\rho$. We want to compute $\partial_\mu\phi'(x)$. Using $\phi'(x)=\phi(y)$ and the chain rule, we have $\partial_\mu\phi'(x)=\partial_\mu\phi(y)=\bar\partial_\nu\phi(y)\partial_\mu y^\nu$. Then we have $\partial_\mu y^\nu = \partial_\mu[(\Lambda^{-1})^\nu{}_\rho x^\rho] = (\Lambda^{-1})^\nu{}_\rho \partial_\mu x^\rho = (\Lambda^{-1})^\nu{}_\rho \delta_\mu{}^\rho = (\Lambda^{-1})^\nu{}_\mu$. Hence we get $\partial_\mu\phi'(x)=(\Lambda^{-1})^\nu{}_\mu\bar\partial_\nu\phi(y)$, which is Tong's second formula.

Note that it is a property of the Lorentz transformation matrices that $(\Lambda^{-1})^\nu{}_\mu=\Lambda_\mu{}^\nu$, so we can also write the second formula as $\partial_\mu\phi'(x)=\Lambda_\mu{}^\nu\bar\partial_\nu\phi(y)$, which looks nicer to me.

In QFT, $\Phi(P)$ is an operator instead of a number, but the key formula $\phi'(x')=\phi(x)=\Phi(P)$ still holds, so all of the above goes through.

Last edited: Aug 5, 2015
3. Aug 6, 2015

soviet1100

Ah, I see it now. Thanks a lot for the help.