Why does a Lorentz scalar field transform as $U^{-1}(\Lambda)\phi(x)U(\Lambda) = \phi(\Lambda^{-1}x)$?

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

The following answer of mine is inserted here even though I feel there is still some problems with it which I mention at the very end (I would appreciate any comments on it) :

We are viewing the Lorentz transformation as an active transformation, namely 'dragging' the point $x$ with value $\phi(x)$ to the new location $\Lambda x$ with value $\phi(x')$. Thus, Lorentz invariance of the theory tells us that an observable of the form $\langle \alpha | \phi(x) | \beta \rangle$ (for some basis states $|\alpha \rangle,|\beta \rangle$ of our Hilbert space) should remain invariant with respect to this transformation. In other words, the value of the observable above would be the same at $x$ and at $x^{\prime}$. Thus we should find the equality

\begin{equation} \langle \alpha^{\prime} | \phi(x^{\prime}) | \beta^{\prime} \rangle = \langle \alpha | \phi(x) | \beta \rangle . \end{equation}

Again, this equality arises because the value of the operator at $x$ has been 'dragged' to $x^{\prime}$ and the states have transformed under the unitary operator representing the Lorentz transform. Since this is a symmetry, the value of the observable should remain the same. Manipulating this a little we thus find the equality we are after, namely

\begin{equation} \langle \alpha^{\prime} | \phi(x^{\prime}) | \beta^{\prime} \rangle= \langle \alpha| U^{-1} \phi(x^{\prime}) U | \beta\rangle = \langle \alpha | \phi(x) | \beta \rangle . \end{equation}

Therefore, we find

\begin{equation} \phi(x) = U^{-1}(\Lambda) \phi(x^{\prime}) U(\Lambda). \end{equation} "

There are two main things I am feeling uncertain about with this answer:

1. I feel I may be confusing how to properly interpret 'active' transformations in this case: I know that classically the interpretation is $\phi'(x')=\phi(x)$, so it seems perhaps incorrect for me to say " We are viewing the Lorentz transformation as an active transformation, namely 'dragging' the point $x$ with value $\phi(x)$ to the new location $\Lambda x$ with value $\phi(x')$" .

2. In the same spirit as the above, I cannot convince myself as to why I should not instead expect that the correct equality is

\begin{equation} \langle \alpha^{\prime} | \phi^{\prime}(x^{\prime}) | \beta^{\prime} \rangle = \langle \alpha | \phi(x) | \beta \rangle , \end{equation}

(ie $\phi^{\prime}(x)$ on the left hand side is now \phi^{\prime}(x^{\prime}), which of course yields an incorrect relation.

Any comments or clarifications would be appreciated.