By Lorentz transformation I mean an element of the Poincaré group.

In Steven Weinberg's Quantum Theory of Fields vol 1., the Poincaré group is constructed in terms of its action on coordinates. I understand that in the context of acting on coordinates the set of Lorentz transformations form a group.

However, I do not understand why we can then just talk about the Poincaré group as an abstract group which has a physically significant action in other contexts. Emphasis on the qualifying clause of the aforementioned sentence. In particular, using Wigner's theorem to furnish a unitary representation of the Poincaré group over Hilbert space.

The only way that I can currently conceive of remedying this situation is that I misunderstand the definition of a Lorentz transformation. My current understanding is that a Lorentz transformation is defined to be an isometry of Minkowski space (or some equivalent definition). I proceed to give a different understanding of what a Lorentz transformation is w.r.t. to the understanding I currently hold.

Definitionally, a Lorentz transformation preserves predicted experimental results. This means two things.

  1. The speed of light in all inertial reference frames must be $c$ $\implies$ Lorentz transformations are isometries of Minkowski space.
  2. Born's rule, which (I think) essentially gives rise to all experimental predictions of Quantum Mechanics, must be preserved by Lorentz transformations $\implies$ Lorentz transformations are Wigner symmetries (physical state ray symmetries) $\implies$ by Wigner's theorem Lorentz transformations have a unitary representation on a Hilbert space of physical states.

Thus, a Lorentz transformation is both an isometry of Minkownski space AND a Wigner symmetry and potentially more. This understanding, if accurate, would solve my confusion because it unifies the two actions of a Lorentz transformation with a single definition of the Lorentz transform. But, in this case, it would be erroneous to merely call Lorentz transformations coordinate transformations.

In other words, it would be a necessary but not sufficient condition for a Lorentz transformation to be an isometry of Minkowski space. The truly necessary and sufficient condition would be that a Lorentz transformation is an isometry of Minkowski space and a Wigner symmetry (on projective Hilbert space).


2 Answers 2


You bring up a good point. Often when people talk about Poincaré transformations, it sounds like they are referring specifically to transformations of coordinates. But they are of course a lot more than that. Ultimately, what we are trying to capture when talking about, say, a Lorentz boost, is the very physical act of observing from a frame that is in motion with respect to another. Do we expect the universe to look different from this new vantage point? Of course we do! A car originally moving at a velocity $u$ will now be moving at a new velocity $v$, and similarly for a particle zipping around in a bubble chamber. The state—the ray in Hilbert space that encodes all the accessible information of the way the world is—is different.

This line of thinking can be made more rigorous, and you can find a nice discussion of this in Section 13.1 of Wald (1984), but to paraphrase:

Consider a generic set of states $\mathcal{S}$ of an unspecified form and a family of observers $\mathcal{O}$ equipped with measuring devices who live in Minkowski spacetime $(\mathbb{R}^4, \eta_{\mu\nu})$. We can associate with every point along the worldline of an observer an orthonormal basis $(e_a)^\mu$ with $(e_0)^\mu$ tangent to the worldline and $(e_i)^\mu$ (for $i\in\{1,2,3\}$) being the spatial axes with which the orientation of the measuring device is specified. Now, the results of measurements are always numbers, so for each state $s\in\mathcal{S}$ there is an associated set of numbers measurable at a point $x\in\mathbb{R}^4$ that completely specifies the state (to our satisfaction, at least). We can therefore assign a map $f_\mathcal{O}:\mathbb{R}^4\times\mathcal{S}\to\mathbb{R}^k$ to each family of observers that gives all possible outcomes from measurements made on these states by the observers. For different families, this map will in general be different, $f_{\mathcal{O}^\prime}\neq f_\mathcal{O}$, depending on how the observers move and orient their measuring device.

Now consider Poincaré transformations, which have an action on Minkowski space given by $\Pi:\mathbb{R}^4\to\mathbb{R}^4$ (note the wording here). Under the action of $\Pi$, the basis vectors at each point become a new orthonormal set $\Pi_*(e_a)^\mu$, where $\Pi_*$ is the pushforward of $\Pi$ and is the action of the Poincaré transformation on vectors/tensors. We can take this as the basis for a new family of observers $\mathcal{O}^\prime$. Now, if we accept that the laws of physics are the same in every frame related by Poincaré transformations, then any physically possible result of a set of measurements made in one frame must be a physically possible result for a set of measurements made in another. In other words, given any $s\in\mathcal{S}$, there must exist an $s^\prime\in\mathcal{S}$ such that $f_\mathcal{O}(x,s)=f_{\mathcal{O}^\prime}(\Pi(x),s^\prime)$ for each $x\in\mathbb{R}^4$. We therefore obtain a map $\tilde{\Pi}:\mathcal{S}\to\mathcal{S}$ defined by this condition, and it is the action of the Poincaré transformation corresponding to $\Pi$ on our physical states.

In fact, this works for isometries $\phi$ of any general spacetime $(M, g_{\mu\nu})$ and it's easy to see that the group structure of $\phi$ is inherited by the map $\tilde{\phi}$. Since we weren't required to specify the nature of the elements of $\mathcal{S}$, the above is as much applicable to rays in Hilbert space as to classical states or anything else.


This is not a direct answer to your question, but an analogous example in a simpler setting that might help clear up some confusion about the relationship between group actions and group representations.

Consider the symmetric group $S_3$, the group of permutations of the set $X =\{1, 2, 3\}$. One way to represent an element $\sigma$ of $S_3$ is by listing the elements of $X$ with their image under the action of $\sigma$. For example, one such element is \begin{align} \sigma_1 = \begin{pmatrix}1 & 2& 3\\ 2&1&3\end{pmatrix} \end{align} The group product is defined by composition of permutations. For example let \begin{align} \sigma_2 = \begin{pmatrix}1 & 2& 3\\ 3&2&1\end{pmatrix} \end{align} Then \begin{align} \sigma_1 \cdot\sigma_2 = \begin{pmatrix}1 & 2& 3\\ 3&1&2\end{pmatrix} \end{align} Note that we have defined this group in terms of its action on the space $X$, much like we can define the Poincare group in terms of its action on spacetime coordinates.

Now let us consider linear representations of $S_3$. A linear representation of a group $G$ gives us a linear operator $M(g)$ acting on a vector space $V$ for each element $g$ such that $M(g_1)M(g_2)\vec{v} = M(g_1\cdot g_2)\vec{v}$, where $v$ is a vector in $V$.

It turns out that there are non-trivial representation of $S_3$ over the vector space $\mathbb{R}^2$. In one such representation (called the standard representation), we have, for example, \begin{align} M(\sigma_1) &= \begin{pmatrix}-\cos\frac{\pi}{3}& \sin\frac{\pi}{3}\\ \sin\frac{\pi}{3}&\cos\frac{\pi}{3}\end{pmatrix}\\ M(\sigma_2) &= \begin{pmatrix}-\cos\frac{\pi}{3}& -\sin\frac{\pi}{3}\\ -\sin\frac{\pi}{3}&\cos\frac{\pi}{3}\end{pmatrix}\\ M(\sigma_1 \cdot \sigma_2) &= \begin{pmatrix}-\cos\frac{\pi}{3}& \sin\frac{\pi}{3}\\ -\sin\frac{\pi}{3}&-\cos\frac{\pi}{3}\end{pmatrix} \end{align} You can verify that multiplying $M(\sigma_1)$ and $M(\sigma_2)$ yields $M(\sigma_1 \cdot \sigma_2)$

So we defined $S_3$ by its action on the set $X$, and we can map $S_3$ into the space of matrices acting on $\mathbb{R}^2$ in a way that preserves the group structure, even though $\mathbb{R}^2$ has nothing to do with $X$. Similarly, we can define the Poincare group by its action on spacetime coordinates, then map the group into the space of operators acting on a Hilbert space and study the action of these operators.

Admittedly, this is probably not the best example for clearing up the confusion, because there is a simple geometric explanation for why it works that doesn't necessarily work when we look at representations of a symmetry group on Hilbert space. We can map the points in $X$ to the vertices of an equilateral triangle in $\mathbb{R}^2$. The matrices $M(\sigma)$ are either reflections about the altitudes of the triangle or rotations by $\pi/3$, so we permute the elements of $X$ by transforming the triangle onto itself. Nevertheless, even when we don't have a simple geometric picture like this, can still study representations of a group on a Hilbert space.

  • $\begingroup$ I do think closer to the heart of my confusion is what you mention in the last paragraph. Because why should a transformation that was originally defined to act on coordinates have anything, physically (I get mathematically that this is all possible and interesting), to do with transforming physical state rays. $\endgroup$ May 22, 2023 at 4:18
  • 1
    $\begingroup$ @SillyGoose At least in non-relativistic QM, you can associate states $\left| \psi \right>$ with spatial wavefunctions $\psi(x)$. You want to know what happens to the states/wavefunctions when you apply a symmetry transformation to the system, so you want to ask about the linear action of symmetry transformations on the Hilbert space. In relativistic QM, you can think about states formed by applying field operators at different spacetime points to the vacuum, and again you want to know how such states transform when you transform the coordinates. $\endgroup$
    – d_b
    May 22, 2023 at 4:58

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