# Why does it matter that a passive transformation can be "interpreted" as an active transformation?

In most mathematics textbooks I have read (see e.g. Axler), the change of basis (COB) matrix has been introduced as a useful computational tool for expressing components of vectors in different bases. In mathematical physics texts (see e.g. Szekeres Chapter 3.6), these are referred to as passive transformations. It is then further stated that we can "interpret" the aforementioned passive transformation as an active transformation if we redefine the COB matrix (which is from one basis $$E$$ to another $$E'$$) to actually be the matrix of some linear operator (with respect to the basis $$E$$ fixed). Of course, the action of that linear operator on $$E$$ can easily be shown to fully define the operator.

My question is whether there is any deep meaning or reason to this interpretation or not? For two fixed bases $$E$$ and $$E'$$ and using the processes mentioned above, there appears to be an isomorphism between the space of passive transformation and the space of linear operators. But so what? Why is this always brought up in the physics literature (even though it never seems to be exploited in any deep way subsequently). Please advise if this question is more appropriate for Math SE.

After further reading, I wonder if perhaps this question is related to the dichotomy mentioned in this question.

• Pardon as I kinda hijack this question a little. I am not so much interested in the core part of this question, but rather want to point out that, if you do an active rotation of a spin half particle by 2pi, then the spin half particle gets an extra minus sign. By the equivalence of passive v.s. active, this means that under a passive rotation of the same, the spin half particle also gets an extra minus sign. An electron thus cannot be seen as an ordinary object. It must at least exist as weirdly as a mobius strip, whereby the weirdness is inherent to its very existence. Commented Apr 14, 2023 at 6:55
• @naturallyInconsistent Note that your question is not directly related to the original since it does not concern active/passive rotations in $\mathbb{R}^3$ directly but through a representation of $\operatorname{SO}(3)$ on the Hilbert space of a spin particle. Perhaps it is best to start a new thread for this. Or to give a short answer: passive rotation should give the same minus sign as an active rotation however this is quite irrelevant since it is just a phase shift and the phase is noncanonical anyways. Commented Apr 17, 2023 at 14:16

The subtlety arises from the fact the association between matrices and linear maps is non-canonical and requires a choice of a preferred basis of the vector space.

An active transformation on $$V$$ is a linear ismorphism $$F: V\to V$$. Note that this is well-defined without reference to a basis on $$V$$, and it really maps an element of $$V$$ to another element.

Choosing a basis $$e=(e_1, \dots, e_n)$$ of $$V$$, we can express any vector $$v\in V$$ uniquely in the form $$v = \sum_{i=1}^n v_i e_i,$$ and the map $$v\to (v_1, \dots, v_n)$$ gives an isomorphism $$\phi_e V \to \mathbb{R}^n$$. With respect to this identification, we get a map $$\phi_e \circ F \circ \phi_e^{-1} : \mathbb{R}^n \to \mathbb{R}^n,$$ to which we may associate a square matrix in the usual way.

A passive transformation (COB) is something different. Suppose $$e'=(e_1', \dots, e_n')$$ is another basis of $$V$$, then we get a map (and matrix) by $$\phi_{e'} \circ \phi_e^{-1}: \mathbb{R}^n \to \mathbb{R}^n.$$ Essentially what this does is: it takes a representation of a vector with reprect to the basis $$e$$ as input and gives the representation of the same vector with respect to the other basis $$e'$$ as output.

So: active transformation really move around a vector in $$V$$ while passive transformations merely alter how we look at the same vector.

• Thank you for your very nice answer. Question though -- I think your $\phi_e$ as defined is a map $\phi_e: \mathbb{R}^n \to V$ (or otherwise you want to write instead $\phi_e \circ F \circ \phi_e^{-1}: \mathbb{R}^n \to \mathbb{R}^n$)? Also, I feel like something is missing in terms of associating a particular passive transformation to a particular active transformation. Does one perhaps need $F$ in your last equation/function composition somewhere?
– EE18
Commented Apr 14, 2023 at 0:06
• I did mess up the direction of the isomorphisms, thanks for spotting it, I've now fixed it. Commented Apr 14, 2023 at 0:20
• Thank you. But still, and to confirm, do you also need to specify that the basis $e'$ is obtained from the basis $e$ via $F$; that is, $F(e) = e'$?
– EE18
Commented Apr 14, 2023 at 0:25
• To answer your question: sort of. You may prefer to insert the identity map: $\phi_{e'} \circ I \circ \phi_e^{-1}$. Of you want to insert another ismorphism $F$ here, that is possible, but it is probably more useful to see that as a passive transformation after an active transformation, since $\phi_{e'} \circ F \circ \phi_e^{-1}$ is the same as $( \phi_{e'} \circ \phi_e^{-1}) \circ ( \phi_e \circ F \circ \phi_e^{-1})$ (we get this by inserting the identity map $I = \phi_e^{-1}\circ \phi_e$) Commented Apr 14, 2023 at 0:29

I would argue that it can be very convenient to switch between the two perspectives and thus is actually exploited in physics literature precisely because the mathematics is easier to handle, rather than there being any "deep significance". But being mathematically convenient to handle is a big win and pretty significant in my book :-).

See this paper by Cotler et. al: https://arxiv.org/abs/1702.06142. One of the points of the paper is to start with an abstract Hilbert space $$\mathcal{H}$$ and Hamiltonian $$\hat{H}$$ and see if there is a unique, preferred tensor product factorization of $$\mathcal{H}$$ in which $$\hat{H}$$ is $$k$$-local. This criteria and the precise details of the paper are not important for this answer.

What is a tensor product factorization? Well, it is an isomorphism $$T: \mathcal{H} \rightarrow \otimes_i\mathcal{H}_i$$. So, we are sending our utterly abstract Hilbert space $$\mathcal{H}$$ into a tensor product factorization of said space. For example, a two spin-1/2 particle system is dimension 4. Thus, the only non-trivial tensor product factorization is $$T: \mathcal{H} \rightarrow (\mathcal{H}_1 \cong \mathcal{C}^2) \otimes (\mathcal{H}_2 \cong \mathcal{C}^2)$$.

Looks familiar? Well it should. This is basically how we represent choosing a basis for a vector space, i.e., by defining an isomorphism. Thus, changing the tensor product factorization of an abstract vector space can be thought of as a "change of basis" and will sometimes be referred as such. More intuitively, your choice of basis and choice of tensor product factorization both contribute to how your vectors are labelled.

Now, to answer the questions the paper asks, Cotler et. al is interested in looking at these mathematical structures determined by the 3-tuple $$(\mathcal{H}, \hat{H}, \mathcal{T})$$ where $$\mathcal{T}$$ is an equivalence class of tensor product factorizations $$T$$ that are equivalent on common sensical or physical grounds. They want to keep $$\mathcal{H}$$ and $$\hat{H}$$ fixed, vary $$\mathcal{T}$$ and check for each $$\mathcal{T}$$ if $$\hat{H}$$ is $$k$$-local.

Cotler et. al could if they really wanted to stick with this perspective to answer the question. However, it is really confusing to do so. What if we could instead keep $$\mathcal{H}$$ and $$\mathcal{T}$$ constant while varying $$\hat{H}$$? This is much simpler because we can actually fix a tensor product factorization and simply change the Hamiltonian (instead of changing between equivalence classes of isomorphisms)! Well, it turns out that these two perspectives are equivalent and very analogous to the active vs. passive transformation perspectives mentioned in your post.

Passive transformation refers to the description of the vector in two different coordinate systems.

Passive transformations describe the point of reference changing.

Active transformation is a transformation of the vector with respect to the same coordinate system.

Active transformations describe the vector changing.

https://en.wikipedia.org/wiki/Active_and_passive_transformation

Active transformation is used when you have your frame of reference for your vector, passive transformation is when your vector is your frame of reference.

I hope that helps.