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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

  1. A Modern Introduction to Quanrum Field Theory Eq. 3.19, 3.20 and 3.21.

  2. Field Quantization-W. Greiner Eq. 2.38, 2.39, and 2.45.

  3. An introduction to Quantum Field theory- Peskin and Schroeder page 18.

  4. Lectures on Classical Field Theory by Suresh Govindrajan

  5. Lectures on Quantum Field Theory by Ashok Das page 212, Eq. 6.4

  6. Relativistic Quantum Physics-Tommy Ohlsson Eq. 5.66, page 119.

  7. A first book on quantum field theory by P. B. Pal Page 22, Eq. 2.38.

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    $\begingroup$ You might be interested in this question. $\endgroup$
    – knzhou
    May 1, 2018 at 18:18
  • $\begingroup$ en.wikipedia.org/wiki/Active_and_passive_transformation $\endgroup$ May 4, 2018 at 8:09
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    $\begingroup$ I'm not sure the transformation in your edit makes sense. $S$ is a functional of $\phi$; any symmetry must be expressed as a change in $\phi$, since that is $S$'s only argument. Translation invariance means using $\phi(x-a)$ instead of $\phi(x)$, not replacing all the instances of $x$ in the integral; $x$ is after all a dummy variable. $\endgroup$
    – Javier
    May 5, 2018 at 16:10
  • $\begingroup$ @AccidentalFourierTransform Do you disagree with (a) and (b)? See the Peskin and Schroeder, and other links I've added. $\endgroup$
    – SRS
    May 5, 2018 at 16:55
  • $\begingroup$ @AccidentalFourierTransform But that's exactly every reference that I cited did. $\endgroup$
    – SRS
    May 5, 2018 at 16:58

4 Answers 4

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Perhaps the most clear way to see what's going on is to compare the action $$ S_1[\phi]=\int_{\mathbb R^d} \phi(x)^2\mathrm dx\tag1 $$ to the action $$ S_2[\phi]=\int_{\mathbb R^d} x^2\phi(x)^2\mathrm dx\tag2 $$

The first one should be invariant under translations, while the second one should not.

We define the translation operator $T_a\in\mathrm{End}(\mathscr C(\mathbb R^d))$, with $a\in\mathbb R^d$, as (ref.1) $$ (T_a\phi)(x):=\phi(x-a)\tag3 $$

With this, an action is invariant under translations if and only if $$ S[T_a\phi]\equiv S[\phi],\quad\forall a\in\mathbb R^d\tag4 $$

What follows is a standard textbook exercise: $$ S_1[T_a\phi]\overset{(1)}=\int_{\mathbb R^d} (T_a\phi)(x)^2\mathrm dx\overset{(3)}=\int_{\mathbb R^d} \phi(x-a)^2\mathrm dx=\int_{\mathbb R^d} \phi(y)^2\mathrm dy=S_1[\phi]\tag5 $$ and $$ S_2[T_a\phi]\overset{(2)}=\int_{\mathbb R^d} x^2(T_a\phi)(x)^2\mathrm dx\overset{(3)}=\int_{\mathbb R^d} x^2\phi(x-a)^2\mathrm dx=\int_{\mathbb R^d} (y+a)^2\phi(y)^2\mathrm dy\neq S[\phi]\tag6 $$ where in both cases we defined $y:=x-a$. This is the expected result: $S_1$ is translation invariant, and $S_2$ is not.

Note that a very similar computation shows that both $S_1$ and $S_2$ are invariant under Lorentz transformations $T_\Lambda\in\mathrm{End}(\mathscr C(\mathbb R^d))$, with $\Lambda\in\mathrm{SO}(1,d-1)$, defined in the obvious way: $(T_\Lambda\phi)(x):=\phi(\Lambda x)$. Here, and as before, it proves essential that $\mathrm dx$ is Poincaré invariant (indeed, it's the Haar measure of $\mathrm{ISO}(1,d-1)$), and that $x\mapsto \Lambda x+a$ is one-to-one over the integration region, $\mathbb R^d$.

But this doesn't mean that $\mathrm dx$ transforms under $T_a,T_\Lambda$. Symmetry transformations are defined at the level of the field $\phi$, with no reference to $\mathrm dx$. Indeed, the definition of invariance $S[T\phi]=S[\phi]$ is quite general, and it is not limited to local functionals: $S[\phi]$ could be any functional, not necessarily given by an integral. The volume form $\mathrm dx$ doesn't transform under symmetry operations1, and neither does $x$. The "arrows and primes" notation $\phi\to\phi'$ can be ambiguous and imprecise. If you stick to well-defined operations, the picture is rather clear.

In any case, and not to be mislead to think that invariance of the measure is necessary and/or sufficient for invariance of $S$, the reader should consider other symmetry transformations, such as: internal symmetries (e.g., the $\mathbb Z_2$ transformation $(T\phi)(x):=-\phi(x)$), and non-isometric external symmetries (i.e., conformal transformations). The simplest example of a conformal transformation is the dilatation $$ (T_\lambda\phi)(x):=\lambda^\Delta\phi(\lambda x)\tag7 $$ with $\lambda\in\mathbb R$. The canonical example of a dilatation-invariant theory is $$ S_3[\phi]=\int_{\mathbb R^d} \frac12(\partial_x\phi(x))^2-\frac{1}{n!}\phi(x)^n\ \mathrm dx\tag8 $$ which satisfies $$ \begin{aligned} S_3[T_\lambda\phi]&\overset{(8)}=\int_{\mathbb R^d} \frac12(\partial_x(T_\lambda\phi)(x))^2-\frac{1}{n!}(T_\lambda\phi)(x)^n\ \mathrm dx\\ &\overset{(7)}=\int_{\mathbb R^d} \frac12\lambda^{2(\Delta+1)-d}(\partial_y\phi(y))^2-\lambda^{n\Delta-d}\frac{1}{n!}\phi(y)^n\ \mathrm dy \end{aligned}\tag9 $$ where I set $y:=\lambda x$. This equals $S_3[\phi]$ iff $2(\Delta+1)=n\Delta=d$, that is, $\Delta=(d-2)/2$ and $n=2d/(d-2)$ (which, for $d=4$, becomes $\Delta=1$ and $n=4$; i.e., $\phi^4$ theory in four space-time dimensions is scale invariant, as is well-known; more generally, $n$ is an integer only in $d=3,4,6$ space-time dimensions).

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More generally, let $\phi\colon\mathbb R^d\to V$, with $V$ a finite-dimensional vector space. We say $T\in\mathrm{End}(\mathscr C(\mathbb R^d,V))$ is a symmetry of a functional $S\colon\mathscr C(\mathbb R^d,V)\to\mathbb R$ if $$ S[T\phi]=S[\phi],\qquad\forall \phi\in\mathscr C(\mathbb R^d,V)\tag{10} $$

It is typically the case that symmetry transformations are coordinatised by some manifold $\mathcal M$, so that $T\colon\mathcal M\to \mathrm{End}(\mathscr C(\mathbb R^d,V))$. Moreover, we assume that $T_0=1$, with $0\in\mathcal M$ the origin of $\mathcal M$ and $1$ the identity transformation.

In this case, there is a weaker notion of symmetry: we say $S$ is quasi-invariant under $T$ if the directional derivative of $S[T_x\phi]$ vanishes at the origin of $\mathcal M$: $$ \lim_{t\to0}\frac{\mathrm d}{\mathrm dt}S[T_{vt}\phi]=0,\qquad\forall \phi\in\mathscr C(\mathbb R^d,V)\tag{11} $$ for some $v\in\mathrm T_0\mathcal M$. The pushforward (differential) of $T$ at the origin is usually denoted by $\delta$: $$ \delta_v=\lim_{t\to0}\frac{\mathrm d}{\mathrm dt}T_{vt}\tag{12} $$ with components $\delta_v=v^a\delta_a$, with $a=1,2,\dots,\dim\mathcal M$. Quasi-invariance is therefore equivalent to $$ 0=S[\phi+t\delta_v\phi]-S[\phi]=tS'[\phi]\cdot\delta_v\phi+\mathcal O(t^2)\tag{13} $$ where $\cdot$ denotes summation-integration, and $'$ denotes a functional derivative. Noether's first theorem is precisely the statement that there is a current for every quasi-symmetry of $S$, which is divergenceless whenever $\phi$ satisfies $S'[\phi]\equiv 0$.

Finally, it bears mentioning that usually $\mathcal M$ is a homogeneous space for some Lie group $G$; in which case we say that $S$ is invariant under $G$. If $G$ is connected and compact, then we can reconstruct it from its algebra $\mathfrak g=\mathrm T_1G$ by means of the exponential map. This algebra is generated by $\delta$, and therefore quasi-symmetry implies symmetry, and vice-versa. If $\exp$ is not surjective, then quasi-symmetry is truly weaker than symmetry.

References:

  1. Bogolubov, Logunov, Oksak, Todorov - General principles of quantum field theory, §7.1.C.

1: Here we are advocating for a passive point of view, as opposed to an active one. Internal transformations are always passive, so it is convenient to regard external ones as passive to, so as to have a uniform framework. It appears that the books OP is following introduce a mixed point of view, where both the fields and the spacetime point are allowed to vary, if somewhat redundantly. The passive point of view is, IMHO, the most general and clear one, and the one most people use nowadays.

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  • $\begingroup$ See also this old question by OP. $\endgroup$ May 1, 2018 at 21:21
  • $\begingroup$ In Ryder's QFT (page 83), the change in the action $\delta S[\phi]$ due to any spacetime transformation $x^\mu\to x^{\prime\mu}$ is computed as $\delta S=\int \mathscr{L}(\phi^\prime,\partial_\mu\phi^\prime)d^4x^\prime-\int \mathscr{L}(\phi,\partial_\mu\phi)d^4x$. But you do not change $dx$ to $dx^\prime$ in (5). @AccidentalFourierTransform $\endgroup$
    – SRS
    May 5, 2018 at 7:33
  • $\begingroup$ I did it and added to the body of the question itself. Even in the case when both the integrand and $dx$ are changed, the action is not translationally invariant contrary to your claim. @AccidentalFourierTransform $\endgroup$
    – SRS
    May 5, 2018 at 15:57
  • $\begingroup$ Note to self: "active" and "passive" should be switched, cf. physics.stackexchange.com/q/401629/84967 $\endgroup$ Jul 28, 2018 at 23:58
  • $\begingroup$ Note to self II: symmetries should be invertible, and therefore we may replace End by Aut. $\endgroup$ Aug 3, 2018 at 14:04
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If you would change $d^4x$ for $d^4x'$, you would just apply the substitution theorem, and you wouldn't say anything about translation invariance (you could see it as evaluating the same integral after reordering the terms in the Riemann sums). This is true also for non-translation invariant expressions.

$\phi$ is your physical state. $\phi'$ is a different state, not the same state written differently. It is related by a translation: the value of $\phi$ at $x$ is the same as the value of $\phi'$ at $x - a$. Now the action for this different state $\phi'$ is the same integral expression, but now evaluated in $\phi'$. Then you use the explicit relation between $\phi$ and $\phi'$ to express the same expression in terms of $\phi$ and now you do use the substitution theorem (which here essentially just is the translation invariance of the volume form $d^4x$) to conclude that the action of $\phi$ and $\phi'$ are the same if they are related by a translation.

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Adding a constant does not change the differential, so $d^4x'=d^4x$

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  • $\begingroup$ I know that. 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 $x$ to $x$. I'm suspicious whether it is $d^4x$ or $d^4x^\prime$ in step (1). Although it will not matter in the end. $\endgroup$
    – SRS
    Apr 28, 2018 at 6:43
  • $\begingroup$ You are right, I believe that it is only a matter of sloppy notation. $\endgroup$ Apr 28, 2018 at 8:04
  • $\begingroup$ @SRS: you have to change you integral bounds! This will compensate for $\phi(x-a)$. $\endgroup$
    – image357
    May 4, 2018 at 10:01
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It looks like you are getting too caught up in the notation and the formalism of it all. In this case the notation is not what is important, what is more important is the concept.

The concept may be easier to understand if you think about what the statement regarding the action means in one dimensional space rather than in four dimensional space-time.

What is meant is that the total area under a curve in the x-y plane doesn't change if you slide the curve back and forth along the x-axis. In order to show that this is true more formally, you want replace "under the integral sign" a curve $f(x)$ by a different curve $g(x)$, where the curve g looks exactly like the curve f was grabbed and slid over to the right (along the x-axis) by an amount $a$.

Well, obviously the two integrals are going to be equal if we are integrating from $x=-\infty$ to $x=+\infty$, right? This is all that is meant by the statement that the action is invariant under this transformation.

In the original problem the curve $f$ is called $\phi^2$. The fact that the function is squared does not matter for this proof. If could be any function.

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