4
$\begingroup$

eI am a mathematician trying to understand the role of representation theory in physics.

There have been countless questions on this site like this, but they sadly don't seem to answer me.

My perspective

Suppose we have a Hilbert space $\mathcal{H}$ with Hamiltonian $H$. We LOVE when there is a Lie group $G$ acting unitarily on $\mathcal{H}$ for three reasons (in diminishing order of importance):

  1. Since $G$ commutes with $H$, we get conservation laws (since it means $\mathfrak{g}$ (up to some multiplication by $i$) is made of Hermitian matrices which commute with $H$, and a short computation that their average value (nothing more I think) is conserved.

  2. We understand better the energy levels; since they're $G$ representations. We even have an easier time finding them.

  3. They conserve stationary states.

An example of those is found in the Hydrogen atom, where the $SO(3)$ action helps us find the energy levels via acting on $L^2(\mathbb{R^3})$.

Things I do not understand

  1. Why should I care about projective unitary representations? They only help with 3, which seem the least interesting. I know that physical states are the projective Hilbert space, but I don't see how that helps or is related. A partial answer is that given a projective representation that commutes with the projectivization of the hamiltonian, I can lift it to a unitary representation of a universal cover, and HOPE that it will commute with $H$ (but I don't see why it should).

  2. I was told about spin, which has that the Hilbert space we're working in is $\mathcal{H} = \mathbb{C}^2$. I wasn't told what the Hamiltonian is, but that $SU(2)$ is acting (via Pauli matrices on the lie algebra). I feel like I'm missing a piece of the story (in Hydrogen, we started with the Hamiltonian and noted it had symmetries. Here I wasn't even given a natural $SO(3)$ projective representation, just a 'hey check out this $SU(2)$ representation, those operators measure spin good luck).

  3. I don't understand Wigner's classification because it sounds beautiful; I don't understand why types of particles correspond to representations of $SO(3,1)$ (Is there a natural Hamiltonian invariant to that? So that energy levels are representations?). Maybe I'm trying to learn relativistic QM in a question and if that's the case tell me and I'll try reading elsewhere.

Final remarks

Maybe I'm approaching QM the wrong way (i.e Proj Hilbert space are states, we have hamiltonian and Schroedinger, and so we're interested in stuff that commute), maybe we should somehow start with a group. If so please tell me but also explain why.

$\endgroup$
8
  • 3
    $\begingroup$ This is a lot of questions to tackle with one post, but since you ask about (4), I can recommend Weinberg's QFT book. The first (non-historical) chapter is working through the Wigner classification start to finish. Understanding that might help answer some of your other questions. $\endgroup$ Apr 26, 2021 at 0:27
  • $\begingroup$ @Richard Myers I hoped that the spin question has a simple answer :(. Thanks for the recommendation anyway., I'll check it out $\endgroup$
    – Andy
    Apr 26, 2021 at 8:03
  • 1
    $\begingroup$ The key thing to remember is that a Hilbert space is NOT a vector space. Rather, it is a projective space. For this reason, we are interested in projective representations, not just faithful ones. $\endgroup$
    – Prahar
    Apr 26, 2021 at 8:09
  • $\begingroup$ @Prahar Mitra I agree states are in the projective hilbert space, but I don't see how projective representations interact with this. If I understand correctly your only hope of using them is lifting to a rep of the universal cover and hoping it commutes with some Hamiltonian, is this true? $\endgroup$
    – Andy
    Apr 26, 2021 at 10:20
  • 1
    $\begingroup$ @Andy I don’t understand this comment. Which representations appear in a Hilbert space have literally nothing to do with the Hamiltonian of the system. You can fix the Hamiltonian and construct quantum theories with ANY representation(s) of the group that you want. So, it is clear that the Hamiltonian by itself can say absolutely nothing about the representations that appear in the Hilbert space. What you would need is some additional input separate from the Hamiltonian. $\endgroup$
    – Prahar
    Apr 27, 2021 at 0:28

2 Answers 2

6
$\begingroup$

There is indeed a lot to tackle here, and as per site policy this question would be best split up into several. However, I will address your first question, which may shed some light on the remaining ones.

The fact that (pure) physical states are elements of a projective Hilbert space is, in fact, the key to understanding why you should be interested in projective representations.

  1. Let $\mathcal H_1=\{\psi\in \mathcal H \ | \ \Vert\psi\Vert=1\}$ be the set of of unit vectors in $\mathcal H$. Define the equivalence relation $\sim$ such that $\psi\sim \phi \iff \exists\alpha\in\mathbb R : \psi= e^{i\alpha} \phi$. The set of (pure) physical states, which I'll denote by $\mathcal P$, is the set of equivalence classes of $\mathcal H_1$ under $\sim$. Define the ray product $\Phi \cdot \Psi$ between elements of $\mathcal P$ by choosing any two representatives $\phi,\psi\in \mathcal H_1$ of $ \Phi$ and $ \Psi$; then $\Phi \cdot \Psi := |\langle\phi,\psi\rangle|$.
  2. The probabilities of various measurement outcomes can ultimately be expressed in terms of ray products between physical states. With this in mind, a symmetry transformation is an automorphism $T:\mathcal P \rightarrow \mathcal P$ such that for all $\Phi,\Psi\in \mathcal P$, $T(\Phi)\cdot T(\Psi) = \Phi \cdot \Psi$.
  3. A (not necessarily linear) map $A:\mathcal H\rightarrow \mathcal H$ induces a corresponding map $\widetilde A:\mathcal P \rightarrow \mathcal P$ in the obvious way. Clearly unitary maps on $\mathcal H$ induce symmetry transformations on $\mathcal P$, as do antiunitary maps; the content of Wigner's theorem is that all symmetry transformations on $\mathcal P$ can be induced by either a unitary or antiunitary map on $\mathcal H$.
  4. This correspondence is not one-to-one; for any (anti)unitary operator $U:\mathcal H\rightarrow \mathcal H$ which induces a symmetry transformation $\widetilde U$, the operator $U' = e^{i\theta}U$ also induces $\widetilde U$ because the phase factor is lost when descending to the projective space $\mathcal P$.
  5. If we want to define the action of a symmetry group $G$ on $\mathcal P$, we should seek a unitary or antiunitary representation of $G$ on $\mathcal H$, which subsequently descends to a group action on $\mathcal P$ as above. However, requiring a genuine representation is too strong a requirement. A projective (anti)unitary representation $\rho$ of $G$ is also perfectly acceptable because even though $\rho(g_1) \circ \rho(g_2) = e^{i\theta}\rho(g_1g_2)\neq \rho(g_1g_2)$ at the level of maps on $\mathcal H$, it still descends to the desired group action on $\mathcal P$ as described in the previous point.

In summary, we would like to define the action of symmetry groups $G$ on physical states. Wigner's theorem tells us that the correct thing to do is to seek a projective unitary or antiunitary representation of $G$ on $\mathcal H$, which descends to a group action on $\mathcal P$ which leaves ray products - and therefore the probabilities of various measurement outcomes - invariant.

None of this has said anything at all about the Hamiltonian. Your statement that we are only interested in group representations which commute with the Hamiltonian is not correct. If the Hamiltonian does commute with some representation or another, then it is essentially as you say, but this need not be the case. For example, the (projective) unitary representations of e.g. $\mathrm{SO(3)}$ can be used to define the angular momentum observables via Stone's theorem, regardless of whether these representations commute with $H$.

Finally, the Galilean group is the symmetry group of Galilean spacetime. Given any nonrelativistic quantum theory in $3+1$ dimensions, we should be able to define the action of this symmetry group via a (projective) unitary representation on $\mathcal H$; if this representation is irreducible, then it can be labeled by the eigenvalues of the corresponding Casimir invariants, which can be roughly interpreted as the mass $m$ and spin $s$ (the linked article explains in more detail). It is in this sense that we say that a particle is defined (in terms of its mass and spin) in terms of an irreducible representation of the underlying symmetry group of the theory.

$\endgroup$
6
  • $\begingroup$ Thanks for the answer. I feel like you got to the heart of my questions via the 2 points you raise. You claim that even if the group action doesn't commute with the Hamiltonian it is interesting. Why? Are you saying that operators that arise from lie algebras are interesting? But there are infinitely many embeddings of all Lie groups with no further requirement, so you must be assuming some compatibility. For 2, you're saying that experimentally we've found that in spin operators some representations arise and in mass operator's others do, so we label these by the casmir operator action. $\endgroup$
    – Andy
    Apr 26, 2021 at 17:18
  • $\begingroup$ @Andy Operators that implement symmetry transformations are interesting, and many (but certainly not all) symmetries, such as rotations, boosts, and translations, constitute Lie groups. The infinitesimal generators of those groups correspond to physical observables like momentum and angular momentum, which are interesting whether or not they are conserved. $\endgroup$
    – J. Murray
    Apr 26, 2021 at 18:19
  • $\begingroup$ @Andy If you have a particle trapped in place in e.g. an ion trap or a quantum dot, it's of great physical interest to know how the state of the particle changes under spatial rotations. This leads us to ask, what are the possible projective unitary representations of $\mathrm{SO}(3)$ on a quantum system? Each irreducible representation can be labeled by its spin, so now we know how to model particles with spin 0, spin 1/2, spin 1, ... $\endgroup$
    – J. Murray
    Apr 26, 2021 at 18:23
  • $\begingroup$ I came up with another motivation; Suppose we have a physical property (I'm thinking angular momentum) that we don't know the Hamiltonian to, but we know that it should respect a given group action of $G$, we can consider the center (i.e various casmir) operators of the universal lie algebra as a source of potential observables that measure our property (this happens with angular momentum). $\endgroup$
    – Andy
    Apr 26, 2021 at 21:37
  • $\begingroup$ I really liked your last comment; you're saying if I have a particle in a space $X$ and natural change of coordinates, of course I want to know when I change coordinates in this way. Thus I have both $G$ and $H$ acting on $\mathcal{H}$ and I want to understand how $G$ acts on $H$ eigenspaces. It thus obvs makes sense to 'diagonalize' each of $G, H$ by themselves and then study the interaction. Is this what you're getting at? How can this last step be done? $\endgroup$
    – Andy
    Apr 26, 2021 at 21:41
2
$\begingroup$

The departure point/motivation is the physics: the so-called statistical interpretation of the standard/textbook Quantum Mechanics, more precisely: that given a complex wavefunction $\psi (x)$, its squared modulus $\vert\psi (x)\vert^2$ has the interpretation of a probability density to find the particle at point $x\in\mathbb{R}$. This entails the fact that $\psi (x)$ and $e^{i\alpha}\psi (x), \alpha\in\mathbb{R}$ represent the same (pure) state. This has a consequence that the simplest states are elements of a projective Hilbert space. Physicists call them rays.

Then the whole symmetry Weyl-Wigner Lie group/Lie algebra machinery follows, as soon as symmetries are defined to conserve probability densities, i.e. have a particular action on rays.

The rest of your questions have answers on this website or in the first and second volumes of Weinberg's QFT book.

Let me add one comment about the spin (your point 2). Quantum spin is a consequence of the representation theory of the space-time symmetry group (Galilei group represented on a projective Hilbert space in non-relativistic QM, or Poincare group in case of special relativity). This is how spin is justified theoretically. It doesn't come out of the blue, the correct "wave-equation" for the H-atom in non-relativistic physics is the Pauli equation. It is derived from the representation theory of the Galilei group, just as Dirac's equation is derived from the representation theory of the Poincare group.

Additional comment: The Hamiltonian is fundamental, it properly defines the system by describing first if the system is made up of elementary subsystems (and if they interact with each other) and how the system as a whole (or various subsystems of it) interact(s) with other systems. The existence of a Hamiltonian is linked to the symmetry of time translation, i.e. to the possibility that states of the system change in time. If H "encodes the dynamics", it "drives the kinetics (evolution of states)". There is, however, a particular formulation of QM in which states are unimportant, but only observables are. It is said that in the Heisenberg picture, the dynamics doesn't influence the kinetics.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.