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Many posts here turn around the question how exactly spacetime symmetries are represented on (projective) Hilbert spaces in quantum mechanics. The question here is why quantum states should live in (projective) representation spaces of these symmetry groups in the first place.

After all, spacetime symmetries merely reflect a redundant and arbitrary choice we make to describe a system in classical mechanics. Physical objects themselves are of course unaffected by a symmetry transformation, that's the whole point of a symmetry. The prominent role of such "artifacts" of our classical description in the construction of quantum mechanics seems surprising (see, e.g., L. E. Ballentine: Quantum Mechanics, chapter 3, and S. Weinberg, The Quantum Theory of Fields, chapter 2).

  1. Is there a good a priori reason to expect that quantum states should live in (projective) representation spaces of spacetime symmetries, besides the fact that quantum mechanics "works", as we see a posteriori?
  2. Is there a formulation of quantum mechanics, or QFT for that matter, which gets rid of the redundancy (spacetime symmetries), instead of carrying it over from classical mechanics, sticking to "bad habits"?

Edit: In what sense do spacetime symmetries merely reflect a redundant and arbitrary choice of how we describe a classical system? By spacetime symmetries, I mean elements of the Galilean group or of the Poincaré group in non-relativistic and relativistic physics, respectively, acting on vector spaces and objects that live on them (tensors). Now, for example, rotating a system actively (including everything with which it interacts) does not have observable consequences. Equivalently, a passive rotation acts on the basis vectors and also transforms tensor components, such that the tensor itself as a geometrical object is invariant.

In this sense, applying spacetime symmetries, whether active or passive, to a system produces different descriptions of the system, all of which are equivalent in that they describe the same physical situation.

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    $\begingroup$ I don't know why you claim that "spacetime symmetries merely reflect a redundant and arbitrary choice we make to describe a system in classical mechanics" - this is true for gauge symmetries, but false for global symmetries like Lorentz symmetry. For example, whether a system is rotationally symmetric or not is an objective property of the system, not a quirk of our description. $\endgroup$
    – ACuriousMind
    Mar 28 at 20:13
  • $\begingroup$ @ACuriousMind: I'm taking the view point of passive transformations: e.g., in special relativity, a rotation acts on the basis of a tangent space to spacetime. Then the components of a tensor change accordingly in our description, but the tensor (the geometric object) itself doesn't change. Put differently, whether a system is rotationally symmetric or not does not depend on our choice of basis vectors. Our choice does not affect the rotational symmetry of a physical system. $\endgroup$
    – Figaro
    Mar 28 at 20:29
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    $\begingroup$ Then I don't understand why you're taking the passive viewpoint here. Claims about symmetries of physical systems are usually phrased in active language - a rotationally symmetric system is symmetric if rotating all the variables doesn't change the action, and in particular rotated solutions to the equation of motion will still be solutions to the equation of motion. This is also the quantum definition of dynamical symmetry (commutator of the symmetry operator acting on the states with Hamiltonian and hence time evolution is zero). $\endgroup$
    – ACuriousMind
    Mar 28 at 20:40
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Your question is founded on a false premise. Spacetime symmetries are not redundancies. Picking up my tea cup and moving it across the room is a genuine transformation, not a gauge transformation of some kind, and claiming that the laws of physics are invariant under such transformations is making a genuine statement about the universe.

One could imagine a universe in which there was a special direction in space along which all objects naturally accelerate more readily than they would in any other direction, or a special point (in empty space) towards which all objects naturally feel a force. The claim that the universe we inhabit does not feature any special directions or points in space is a real, physical claim which may or may not end up being true.

If we did live in a universe with e.g. a special acceleration direction, then a rotation of all the matter in the universe would have a tangible effect on its behavior. The orbits of stars and planets would be different after the rotation, and the form of the physical laws governing their behavior would have to change.

I'm taking the view point of passive transformations: e.g., in special relativity, a rotation acts on the basis of a tangent space to spacetime. Then the components of a tensor change accordingly in our description, but the tensor (the geometric object) itself doesn't change.

Active and passive are not equivalent points of view that you are free to adopt at your leisure. An active transformation refers to a genuine, physical transformation of the physical matter or fields under consideration, while a passive transformation refers to a change of coordinates.

Performing a spacetime symmetry transformation - such as a rotation of the system - is an active transformation. Going back to the example referenced above, one could imagine that the universe is anisotropic, and that objects accelerate more readily in the direction normal to the plane of the Milky Way. Performing a rotation on the entire galaxy would then clearly change how it behaves. This is more than just a change of coordinates.

Put differently, the laws of physics are inherently invariant under passive transformations because we define it to be so. In Physics 101, we often change coordinates to describe motion along an inclined plane. We change the coordinates one way and the basis vectors the other such that the laws of physics remain invariant. This is a passive transformation.

On the other hand, we could actually pick up our experimental apparatus in our hands and rotate it. The coordinates of our objects would change, but the basis vectors would not. This is an active transformation (which incidentally doesn't leave the behavior of the system unchanged, because this model does not feature rotational invariance).

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    $\begingroup$ Thank you for this answer! I completely agree with its content, but I feel the example of a preferred direction normal to the plane of the Milky Way misses the point: rotating the Milky Way without rotating the preferred direction is not a spacetime symmetry (not an element of the Poincaré group) because we rotate some vectors (e.g. the normal) but not the one characterizing the preferred direction. So it's not surprising that this has observable consequences. $\endgroup$
    – Figaro
    Mar 28 at 21:42
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    $\begingroup$ On the other hand, a Poincaré transformation affects the entire system, including the preferred direction, and hence has no observable consequences. In this sense, I claim that Poincaré transformations (or Galilean transformations in non-relativistic physics), whether active or passive, merely reflect a choice of our viewpoint. $\endgroup$
    – Figaro
    Mar 28 at 21:42
  • $\begingroup$ @Figaro the whole point of the distinction active transformation vs passive transformation is to make a distinction between the two cases you illustrated here. $\endgroup$ Mar 28 at 21:54
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    $\begingroup$ @Quantumwhisp not sure to understand why there should be distinction. Citing Ballentine: Quantum Mechanics (1998), p. 68: "The transformation just described is in the active point of view, in which the object (in this case a function) is transformed relative to a fixed coordinate system. There is also the passive point of view, in which a fixed object is redescribed with respect to a transformed coordinate system. The two points of view are equivalent, and the choice between them is a matter of taste." $\endgroup$
    – Figaro
    Mar 28 at 22:08
  • $\begingroup$ @Figaro The points of view are equivalent only insofar as they have the same effect on the coordinates of a particle. That is not to say that passive and active transformations are equivalent. Consider that under your definition, every system is invariant under Poincare transformations, so the entire concept becomes meaningless. $\endgroup$
    – J. Murray
    Mar 28 at 22:24
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We should distinguish between two things:

  • A global symmetry transform (e.g., global rotation) clearly has no observable physical effect. If we rotate the whole universe, we haven't changed anything observable.

  • In contrast, the existence of a global symmetry (I mean the mathematical exisitence, in the model) typically does have observable consequences. That's because global symmetries tend to be splittable, as defined by Harlow and Ooguri. A splittable global symmetry is one that lets us transform part of the system without transforming the whole thing. That's clearly observable.

The spacetime symmetries in question are splittable, and that's why we care about their representations in quantum mechanics.

Here's another perspective: Most of the models we consider in quantum mechanics are models of only one part of a more complete physical system. In such a model, a global symmetry transform doesn't necessarily represent a global transform of the behind-the-scenes physical context in which the model is meant to apply.

Can we eliminate the unobservable global transforms from the formalism? With a splittable symmetry, we can transform an arbitrarily large part of the system, and that's observable, so disallowing a transform of the whole system seems like a difficult thing to enforce mathematically. I don't know how to do it, and I'm not sure what purpose it would serve.

Keep in mind that even local gauge symmetries are a ubiquitous feature of our current formulation of physics, even though everybody acknowledges that they are mere redundancies. We don't eliminate them from the formalism, we just acknowledge the redundancy.

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  • $\begingroup$ Thank you! Never heard of splittable symmetries before, so this answer is really helpful to dig further! $\endgroup$
    – Figaro
    Mar 28 at 23:01

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