For the Lorentz transformation $x \to x'=\Lambda x$,

the active transformation is $\phi(x) \to \phi'(x)=\phi(\Lambda^{-1}x)$

and the passive transformation is $\phi(x) \to \phi'(x)=\phi(\Lambda x)$.

I know that the active transformation changes the field and the passive transformation changes the coordinate. But how can I understand the difference between these two equations, especially the second equation for a passive transformation?


2 Answers 2


Passive point of view:

Alice observes some field $\phi$ at location $x = x_0$ in her lab at Princeton USA, and finds field value $\phi(x_0)=\phi_0$. Bob observes Alice's measurement from his lab in Cambridge, UK. In his frame he sees the Princeton location as $x' = \Lambda x_0$ and confirms the same field value as Alice, which to him reads $\phi'(x') = \phi(x_0)=\phi_0$, hence $$ \phi'(\Lambda x_0) = \phi(x_0) $$

Active point of view:

Alice again observes field $\phi$ at location $x = x_0$ in her lab and finds field value $\phi(x_0)=\phi_0$. But this time Bob decides to reproduce her experiment identically in his lab and measure the same field at the exact same location relative to his frame, $x' = x_0$. All goes well and Bob finds the same value as Alice, meaning $$ \phi'(x_0) = \phi(x_0) $$ If the location of Alice's observation as seen from Bob's frame is ${\bar x}_0 = \Lambda x_0$, then conversely $x_0 = \Lambda^{-1}{\bar x_0}$ and Bob can say that $$ \phi'(x_0) = \phi(\Lambda^{-1} {\bar x_0}) $$

See for instance these notes on QFT on manifolds, particularly the following table after Eq.(8):

enter image description here

  • $\begingroup$ I get an "access denied" message when clicking the link you posted, can we get those notes from another link? $\endgroup$
    – Kirov
    Mar 23, 2017 at 18:45
  • $\begingroup$ @Kirov Sorry about the link, I think the notes disappeared when the author left Stanford. Perhaps it's worth contacting him directly about this? There is a recent contact in his last paper on arXiv, lanl.arxiv.org/pdf/1701.08777v1. $\endgroup$
    – udrv
    Mar 24, 2017 at 7:45

There is a lot of confusion in the literature regarding the so-called active and passive interpretation of transformations when it comes to scalar fields. However, this terminology and the corresponding dichotomy has its origins in the applications of linear algebra (e.g. computer vision) where it is more relevant and the concepts are more clear. The Wikipedia article on this subject makes this point very clear.

Transformation of vector spaces:

Consider a spatial transformation $T:\mathbb R^3 \to \mathbb R^3$. This can be interpreted to either transform a vector $v = v_1 e_x + v_2 e_y + v_3 e_z \in \mathbb R^3$ keeping the basis fixed or to transform the initial basis $\{e_x,e_y,e_z\}$ of $\mathbb R^3$ keeping the vector $v$ fixed. These two lines of interpretation of $T$ go by two names.

\begin{aligned} &\textit{Active (alibi) transformation}: && \text{vector } v \text{ rotates} \left(T: v \mapsto v'=Tv \equiv v_1' e_x + v_2' e_y + v_3' e_z\right), \\ & && \text{basis }\{e_x,e_y,e_z\} \text{ remains unchanged}. \\ \\ &\textit{Passive (alias) transformation}: && \text{vector } v \text{ stays put}, \\ & && \text{basis rotates } \left(\{e_x,e_y,e_z\}\mapsto \{T^{-1}e_x, T^{-1}e_y, T^{-1}e_z\}\right). \end{aligned}

From the first interpretation $v = T^{-1}v'$, it follows that $v = v_1' \tilde e_x + v_2' \tilde e_y + v_3' \tilde e_z$ where $\tilde e_x := T^{-1} e_x$, $\tilde e_y := T^{-1} e_y$ and $\tilde e_z := T^{-1} e_z$ are the transformed basis vectors from the second interpretation. Thus, the original vector $v$ in the rotated basis $\{\tilde e_x, \tilde e_y, \tilde e_z\}$ (in the passive point of view) has exactly the same coordinates $(v_1',v_2',v_3')$ as the rotated vector $v'$ in the original basis (the active point of view).

Transformation of scalar fields

This dichotomy is not very useful when it comes to scalar fields and, therefore, the literature lacks a canonical definition for these concepts. One way of thinking about them could be as user @udrv has written about. Here is another way which is equally popular. A scalar field is a real-valued map $\phi : \Omega \subset \mathfrak{M}_4 \to \mathbb R$. Consider a transformation $T:\Omega \to \Omega' \subseteq \Omega$ of the underlying spacetime domain. Now, one can imagine either a rotated field $\phi_A := \phi \circ T^{-1}: \Omega' \to \mathbb R$ or an oppositely-rotated field $\phi_P := \phi \circ T: \Omega \to \mathbb R$ to visualize this transformation. The two new fields can be interpreted in the following manner.

\begin{aligned} &\textit{Active (alibi) transformation}: && \text{field configuration } \phi\big|_{\Omega'} : \Omega' \to \mathbb R \text{ has morphed into } \phi_A : \Omega' \to \mathbb R,\\ & && \text{leaving the spacetime domain } \Omega' \text{untouched}. \\ \\ &\textit{Passive (alias) transformation}: && \text{field configuration } \phi_P \text{ is simply } \phi \text{ acting on a rotated domain},\\ & && \text{which is to say, } \phi_P(x) = \phi(T(x)) \text{ where } T:\Omega \to \Omega', x \mapsto x'. \end{aligned}

Contrast this with @urdv's answer where (s)he has casted $\phi_A\, (=\phi')$ according to the passive interpretation. This should tell you that any field redefinition obtained from a spacetime transformation can be seen in both active and passive interpretations and such vacuous names/interpretations hold no physical or mathematical value.

  • $\begingroup$ Dear Nanashi No Gombe. It is usually frown upon to directly copy-paste identical answers. (The problem is if everybody start to copy-paste identical answers en mass.) $\endgroup$
    – Qmechanic
    Jan 20, 2020 at 16:55
  • $\begingroup$ @Qmechanic It's not exactly identical, but yeah, I get your point. Thanks for pointing that out. :) $\endgroup$ Jan 20, 2020 at 16:56

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