Under a translation in spacetime i.e., $$x\mapsto x^\prime=x+a,\tag{a}$$ a scalar field $\phi(x)$ $$\phi(x)\mapsto\phi^\prime(x)=\phi(x-a).\tag{b}$$ My aim is to verify the invariance of an action of the form $S[\phi(x)]=\int d^4x~ \phi^2(x)$. This video by M. Luty shows the invariance (around time 17.40 minuties) as follows. $$\int d^4x~\phi^{2}(x)\mapsto\int d^4x~\phi^{\prime 2}(x)\tag{1}$$ $$=\int d^4x~\phi^{2}(x-a)\tag{2}.$$ Now by changing variables $x-a=y$, one has $$S[\phi(x)]\mapsto\int d^4y~\phi^2(y)=S[\phi(y)]\tag{3}$$ which readily establishes the invariance of the action.
Question I'm a bit confused about step (1). Since the coordinates also change shouldn't we also change $d^4x\to d^4x^\prime$ in step (1)? I know that differentials don't change by adding a constant to a variable. In fact, that's what is used in arriving at step (3) from step (2). But the step (1) looks like $\phi(x)$ is mapped to $\phi^\prime(x-a)$ and $x$ is mapped to $x$. I'm suspicious whether it is $d^4x$ or really $d^4x^\prime$ (which is in turn equal to $d^4x$) in step (1).
$S_2[\phi]$ of AFT's answer Using (a) and (b), (when both the intergrand and $dx$ are changed) we get, $$S_2[\phi]=\int x^2 \phi(x)^2 dx\mapsto \int (x+a)^2\phi(x-a)d(x+a)=\int (x+a)^2\phi(x-a)dx.$$ Changing variable $y=x-a$, I find $$S_2[\phi]\mapsto \int(y+2a)^2\phi^2(y)dy\neq S_2[\phi].$$ So it shows that even when $dx$ is changed to $dx^\prime$, the action is not translationally invariant.
References
A Modern Introduction to Quanrum Field Theory Eq. 3.19, 3.20 and 3.21.
Field Quantization-W. Greiner Eq. 2.38, 2.39, and 2.45.
An introduction to Quantum Field theory- Peskin and Schroeder page 18.
Lectures on Classical Field Theory by Suresh Govindrajan
Lectures on Quantum Field Theory by Ashok Das page 212, Eq. 6.4
Relativistic Quantum Physics-Tommy Ohlsson Eq. 5.66, page 119.
A first book on quantum field theory by P. B. Pal Page 22, Eq. 2.38.
No comments:
Post a Comment