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This problem is from Srednicki page 19. Why $U^{-1}(\Lambda)\phi(x)U(\Lambda) = \phi(\Lambda^{-1}x)$?

Can anyone derive this?

$\phi$ is a scalar and $\Lambda$ Lorentz transformation.

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  • $\begingroup$ Title change suggestion for future searchability: "Why does a Lorentz scalar field transform as $U(\Lambda)^{-1}\phi(x)U(\Lambda) = \phi(\Lambda^{-1}x)$?" $\endgroup$ – joshphysics Jan 29 '14 at 19:21
  • $\begingroup$ Yes this post was rushed but blame the flue. Thanks for your answers, I'll go through them carefully when the fever goes down. Cheers $\endgroup$ – Your Majesty Jan 29 '14 at 20:23
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That equation is, in fact, the definition of a Lorentz-scalar, but perhaps a few words will convince you that it is a well-motivated definition.

A helpful starting analogy.

Forget about relativistic field theory for a moment. Let's consider, instead, someone who wants to measure the temperature everywhere in a room. The temperature can be represented by a scalar field, namely a function $T:\mathrm{room}\to \mathbb R$ where $\mathrm{room}$ is some subset of three-dimensional Euclidean space $\mathbb R^3$. Now, suppose that someone takes the temperature distribution and rotates it by a rotation $R\in\mathrm{SO}(3)$, by drawing a picture, you should be able to convince yourself that the new temperature distribution $T_R$ that he would measure would be related to the old temperature distribution as follows: \begin{align} T_R(R\mathbf x) = T(\mathbf x), \end{align} In other words, the value of the transformed (rotated) temperature distribution at the transformed point is the same as the value of the un-transformed temperature distribution at the un-transformed point.

Classical field theory.

Now, let's go to classical relativistic field theory. Consider some scalar field on four-dimensional Minkowski space $\phi:\mathbb R^{1,3}\to \mathbb R$. By analogy with the temperature distribution, we define a Lorentz transformed field $\phi_\Lambda$ (often denote $\phi'$ in physics) by \begin{align} \phi_\Lambda(\Lambda x) = \phi(x). \end{align} for all $x\in \mathbb R^{3,1}$ and for all $\Lambda\in\mathrm{SO}(1,3)^+$. Notice that this can be re-written as follows: \begin{align} \phi_\Lambda(x) = \phi(\Lambda^{-1} x). \tag{$\star$} \end{align} The transformed field evaluated at a spacetime point $x$ agrees with the un-transformed field at the spacetime point $\Lambda^{-1} x$.

QFT.

But now, let's consider QFT. In this case, $\phi$ assign an operator (well really an operator distribution) to each spacetime point. Now in relativistic QFT, there exists a unitary representation $U:\mathrm{SO}(3,1)^+\to U(\mathcal H)$ of the Lorentz group acting on the Hilbert space $\mathcal H$ of the theory which transforms states $|\psi\rangle\in \mathcal H$ as follows: \begin{align} |\psi\rangle \to U(\Lambda)|\psi\rangle \end{align} Now suppose that $A:\mathcal H\to\mathcal H$ is a linear operator, is there some natural way that such an operator transforms under $U(\mathcal H)$? Yes there is, recall that when we make a change of basis in a vector space, this induces a change in the matrix representations of operators by similarity transformation. If we think of the Lorentz transformation as a change of basis, then it is natural to define a transformed operator by \begin{align} A_\Lambda = U(\Lambda)^{-1} A U(\Lambda). \end{align} If we apply this to the operator $\phi(x)$ at a given spacetime point $x$, then we have \begin{align} \phi(x)_\Lambda = U(\Lambda)^{-1} \phi(x) U(\Lambda) \tag{$\star\star$} \end{align} The transformation law used in QFT then follows by demanding that $(\star\star)$ which derives from the notion of transforming a linear operator on $\mathcal H$ agrees with the notion $(\star)$ of transforming a field in classical field theory. Explicitly, in this notation if we demand that \begin{align} \phi(x)_\Lambda = \phi_\Lambda(x), \end{align} then we obtain the desired definition of a Lorentz scalar field; \begin{align} U(\Lambda)^{-1} \phi(x) U(\Lambda) = \phi(\Lambda^{-1}x). \end{align}

Note.

The notion of scalar, vector, and tensor fields used in QFT might remind you of the notions of scalar, vector, and tensor operators used in the non-relativistic quantum mechanics of, for example, particles with angular momentum. This is not an accident; they are closely related concepts.

The additional complication we get in QFT is that fields are operator-valued functions of spacetime, not just operators, so we have to decide what to do with the spacetime argument of the field when we transform. We dealt with this complication above by essentially combining the notion of tensor operator in quantum mechanics, with the notion of field transformation in classical field theory.

For more mathematical remarks on tensor operators on Hilbert spaces, see

Tensor Operators

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  • $\begingroup$ I like your pedagogical skills. Awesome! (Unpossible to understand my English at the moment) $\endgroup$ – Your Majesty Jan 29 '14 at 20:55
  • $\begingroup$ @LoveLearning Thanks, much appreciated! Glad it helped. $\endgroup$ – joshphysics Jan 29 '14 at 21:22
  • $\begingroup$ Love your answer! I just wanted to point out that for something as simple as temperature (well it's not that simple when you dig into it) there's no consensus on how it in fact transforms!: Example: physics.stackexchange.com/questions/83488/… $\endgroup$ – R. Rankin Dec 4 '18 at 3:52
  • $\begingroup$ @R.Rankin Thanks, and that's interesting. It seems to me though that the lack of consensus in the post you linked refers to Lorentz boosts. I'd be surprised if there is a lack of consensus for spatial rotations which is what I referred to in the analogy. $\endgroup$ – joshphysics Dec 4 '18 at 8:06
  • $\begingroup$ there isn't of course (: I've been trying to think of boosts as just a spatial-time rotation (albeit a hyperbolic one). Thanks again for the great answer I enjoyed it. $\endgroup$ – R. Rankin Dec 4 '18 at 9:44
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It's a definition of a scalar field. You look at the same field from another reference frame and you only see it with a "shifted" argument. To prove, you expand it formally in powers of $x$, I guess.

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  • $\begingroup$ I agree that it is a definition, so why simultaneously claim that it can be proven? It's a well-motivated definition, but motivation and proof are different. $\endgroup$ – joshphysics Jan 29 '14 at 18:15
  • $\begingroup$ That's how PS motivates it... but why motivate if it can be derived? Can it be derived? Srednicki doesn't define it, he says we should expect... $\endgroup$ – Your Majesty Jan 29 '14 at 18:22
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    $\begingroup$ One can say that in a scalar field there is nothing to transform but the argument. The proof shows that it is transformed as expected, not differently. $\endgroup$ – Vladimir Kalitvianski Jan 29 '14 at 18:40

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