I'm studying QFT on Weinberg's book. And I have a question about its notation for Lorentz transformation property of the free fields (Chap.5).

In Sec. 5.1 of Vol. 1, annihilation fields $\psi_{\ell}^{+}(x)$ and creation fields $\psi_{\ell}^{-}(x)$ are given (as (5.1.4), (5.1.5)) so that they satisfy the transformation properties below: \begin{align} &U_{0}(\Lambda, a) \psi_{\ell}^{+}(x) U_{0}^{-1}(\Lambda, a)=\sum_{\bar{\ell}} D_{\ell \bar{\ell}}\left(\Lambda^{-1}\right) \psi_{\ell}^{+}(\Lambda x+a) \quad (5.1.6)\\ &U_{0}(\Lambda, a) \psi_{\ell}^{-}(x) U_{0}^{-1}(\Lambda, a)=\sum_{\bar{\ell}} D_{\ell \bar{\ell}}\left(\Lambda^{-1}\right) \psi_{\ell}^{-}(\Lambda x+a) \quad (5.1.7) \end{align} $U_0$ is the transformation operators, and $D$ is the representation matrices for the homogeneous Lorentz transformation.

My question is why we choose $\Lambda^{-1}$ rather than $\Lambda$, as the argument of $D$. Is it consistent with the transformation rule for one-particle states given in the scattering theory part (such as (3.1.1))?


2 Answers 2


Note that the reply is not very technical, but I hope my argumentation will explain the answer. We are always assuming active transformations below - acting with an operator on a state gives a different state, but in the same basis.

There seem to be two conventions in the literature on how one views, or introduces, symmetry transformation operators - either by their action on basis vectors, as in ref (1), or by examining their action on vector components, as in ref. (2). The difference then implies the way an operator changes.

In the first case, we get (ref. (1), eq. 1.5.13)

\begin{equation} A' = U^{-1} A\, U \label{first} \end{equation}

While in the second case we would get (ref. (2), eq. 4.41): \begin{equation} A' = UA\, U^{-1} \end{equation}

In ref. (3) the first convention is used, as can be inferred from eq. (2.5.3). We can use this transformation property to classify how field operators transform (I note again, under an active transformation - hence the same coordinate dependence on RHS and LHS below). To compare with OP's equations, let's examine vector fields under a homogeneous Lorentz transformation. The first convention gives us

$$ U^{-1}(\Lambda) A^\mu(x) \, U (\Lambda) = A'^\mu(x) $$

Knowing that we are dealing with a vector field we can express the RHS as follows \begin{align} A'^\mu(x') = \Lambda^\mu{}_\nu A^\nu(x) = \Lambda^\mu{}_\nu A^\nu(\Lambda^{-1}x') \end{align} By renaming $x' \to x$ we get $$A'^\mu(x) = \Lambda^\mu{}_\nu A^\nu(\Lambda^{-1}x)$$ inserting above $$ U^{-1}(\Lambda) A^\mu(x) \, U (\Lambda) =\Lambda^\mu{}_\nu A^\nu(\Lambda^{-1}x) $$

Using the fact that $U(\Lambda)^{-1} = U(\Lambda^{-1})$ we rewrite the equation above as

$$ U(\Lambda^{-1}) A^\mu(x) \, U (\Lambda^{-1})^{-1} =\Lambda^\mu{}_\nu A^\nu(\Lambda^{-1}x) $$

And finally, by renaming $\Lambda^{-1}\to\Lambda$ we reproduce the equation from ref. (3)

$$ U(\Lambda) A^\mu(x) \, U (\Lambda)^{-1} =(\Lambda^{-1})^\mu{}_\nu A^\nu(\Lambda x) $$ Where we can on this example explicitly see why the representation matrix depends on the inverse of the Lorentz transformation.

A complete proof on the field transformation properties can be found in ref. (4), eq. (7.6-17), and a physics based argumentation on the transformation properties can be found in ref. (5), eq. (5.75-5.76)


  1. Sakurai, Napolitano - Modern Quantum Mechanics, 2nd Edition
  2. Goldstein - Classical Mechanics, 3rd Edition
  3. Weinberg - Quantum Theory of Fields, vol. I
  4. W.K. Tung - Group Theory in Physics
  5. Duncan - Conceptual Framework of Quantum Field Theory

I don't think this has something to do with the passive/ active picture as the other answer pointed out. From section 4.2 we know that the necessary and sufficient condition for a state to transform appropriately under the Lorentz transformations, the creation operator must have the following transformation rule:

$$\begin{aligned} U_0(\Lambda, \alpha) a^{\dagger}(\mathbf{p} \sigma n) U_0^{-1}(\Lambda, \alpha)= & e^{-i(\Lambda p) \cdot \alpha} \sqrt{(\Lambda p)^0 / p^0} \\ & \times \sum_{\bar{\sigma}} D_{\bar{\sigma} \sigma}^{(j)}(W(\Lambda, p)) a^{\dagger}\left(\mathbf{p}_{\Lambda} \bar{\sigma} n\right) \end{aligned}\tag{4.2.12}$$

In section 2.5 we have deduced that the representation of the little group must be unitary:


In components this is

$$D_{\bar{\sigma} \sigma}^{(j)}(W(\Lambda, p))=D_{\sigma \bar{\sigma}}^{(j)*}(W^{-1}(\Lambda, p))$$

Here we have used $D^{-1}(W) = D(W^{-1})$.


$$\begin{gathered}U_0(\Lambda, b) a^{\dagger}(\mathbf{p}, \sigma, n) U_0^{-1}(\Lambda, b)=\exp (-i(\Lambda p) \cdot b) \sqrt{(\Lambda p)^0 / p^0} \\ \times \sum_{\bar{\sigma}} D_{\sigma \bar{\sigma}}^{\left(j_n\right)^*}\left(W^{-1}(\Lambda, p)\right) a^{\dagger}\left(\mathbf{p}_{\Lambda}, \bar{\sigma}, n\right)\end{gathered}\tag{5.1.12}$$

Now with

$$\psi_{\ell}^{-}(x)=\sum_{\sigma n} \int d^3 p v_{\ell}(x ; \mathbf{p}, \sigma, n) a^{\dagger}(\mathbf{p}, \sigma, n)\tag{5.1.5}$$

it's natural that

$$U_0(\Lambda, a) \psi_{\bar{\ell}}^{-}(x) U_0^{-1}(\Lambda, a)=\sum_{\bar{\ell}} D_{\bar{\ell}}\left(\Lambda^{-1}\right) \psi_{\bar{\ell}}^{-}(\Lambda x+a)\tag{5.1.7}$$

The argument $W^{-1}$ in $D$ just reflects our convention for denoting the creation operator with $a^\dagger$, so we would (somewhat inconveniently) convert $D(W)$ in $(4.2.12)$ to $D^\dagger(W^{-1})$ for notational consistency.


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