How is the physical meaning of an irreducible representation justified?

Question

This is maybe not an entirely mathematical question, but consider it a pedagogical question about representation theory if you want to avoid physics-y questions on MO.

I've been reading Singer's Linearity, Symmetry, and Prediction in the Hydrogen Atom and am trying to come to terms with the main physical (as opposed to mathematical) argument of the text. The argument posits, if I understand it correctly, that a quantum system described by a Hilbert space $H$ on which a group $G$ of symmetries acts by unitary transformations should have the property that its "elementary states" "are" irreducible subrepresentations of the representation of $G$ on $H$. She begins this argument in section 5.1:

Invariant subspaces are the only physically natural subspaces. Recall from Section 4.5 that in a quantum system with symmetry, there is a natural representation $(G, V, \rho)$. Any physically natural object must appear the same to all observers. In particular, if a subspace has physical significance, all equivalent observers must agree on the question of a particular state's membership in that subspace.

and continues it in section 6.3:

We know from numerous experiments that every quantum system has *elementary states*. An elementary state of a quantum system should be **observer-independent**. In other words, any observer should be able (in theory) to recognize that state experimentally, and the observations should all agree. Secondly, an elementary state should be indivisible. That is, one should not be able to think of the elementary state as a superposition of two or more "more elementary" states. If we accept the model that every recognizable state corresponds to a vector subspace of the state space of the system, then we can conclude that elementary states correspond to irreducible representations. The independence of the choice of observer compels the subspace to be invariant under the representation. The indivisible nature of the subspace requires the subspace to be irreducible. So elementary states correspond to irreducible representations. More specifically, if a vector $w$ represents an elementary state, then $w$ should lie in an *irreducible* invariant subspace $W$, that is, a subspace whose only invariant subspaces are itself and $0$. In fact, every vector in $W$ represents a state "indistinguishable" from $w$, as a consequence of Exercise 6.6.

(For people who actually know their quantum, Singer is ignoring the distinction between representations and projective representations until later in the book.)

My first problem with this argument is that Singer never gives a precise definition of "elementary state." My second problem is that I'm not sure what physical principle is at work when she posits that physically natural subspaces and elementary states should be observer-independent (i.e. invariant under the action of $G$). What underlying assumption of quantum mechanics, or whatever, is at work here? Why should a mathematician without significant training in physics find this reasonable? (I have the same question about the identification of elementary particles with irreducible representations of the "symmetry group of the universe," so any comments about this physical argument are also welcome.)

Singer goes on to use this assumption to deduce the number of electrons that fill various electron orbitals, and I won't be able to convince myself that this makes sense until I understand the physical assumption that allows us to use irreducible representations to do this.

Answer 1

Invariant states are not the only meaningful ones. Even in classical mechanics, a baseball traveling 90 mph toward my head is quite meaningful to me, even though it is of no consequence to my fellow mathematician a mile away.

The focus on invariant subspaces comes not from an assumption, but from the way physicists do their work. They want to predict behavior by making calculations. They want to find laws that are universal. They want equations and calculation rules that will be invariant under a change of observers.

Any particular calculation might require a choice of coordinates, but the rules must be invariant under that choice. Once we're talking about one particular baseball trajectory, that trajectory will look different in different coordinate systems; the rules governing baseball flight, however, must look the same in all equivalent coordinate systems.

The natural features of baseballs arise from the equivalence classes of trajectories of baseballs -- equivalence under the group action. Here, if we pretend the earth is flat, gravity is vertical, and air does not resist the baseball, the group is generated by translations and rotations of the plane. Any physically natural, intrinsic property of the baseball itself (such as its mass) or its trajectory (such as the speed of the baseball) must be invariant under the group action. If you don't know a priori what these properties will be, a good way to find them is to pass from individual instances (the baseball heading toward me at 90mph) to the equivalence class generated by individual instances under the group action (the set of all conceivable baseballs traveling at 90mph). Note that the equivalence class is invariant under the group action, and it is exactly this invariance that makes the equivalence class a useful object of the physicists' study.

More generally, if you are studying a physical system with symmetry, it's a good bet that the invariant objects will lead to physically relevant, important quantities. It's more a philosophy than an axiom, but it has worked for centuries.

Answer 2

I suppose an answer to your question, as simple as possible is this: You would probably be happier if not an irreducible representation, but rather a single function was declared the object of interest (indeed this function, upon normalization, would literally be the probability distribution at a given time of finding the particle there). However, since the group G acts on the phase space of the system and preserves the energy functional under study (because it's an electron orbiting a proton, the energy is a function of r alone), any function f you find will transform under rotations by the element g to the element g.f.

Now suppose you and a friend are both conducting the experiment simultaneously, and such that your points of view of the proton are related by a rotation g about the proton. When you observe a distribution f, your friend will simultaneously observe the distribution g.f, just by computing change of coordinates, not by any physical assumption. Now, add the physical assumption that you and your friend should be able to reconcile your results when you meet to discuss them, and you'll find that one cannot make a distinction between f and g.f. So you shouldn't study single distributions f, but rather orbits g.f of a given function. This is my attempt at answering your second question. It is not a mathematical but physical answer, since you asked a physical question.

To your first question, I would simply agree with José Figueroa-O'Farrill that the term elementary state doesn't mean so much, except that you can formally express any state as a linear combination of its "elementary" states, and that this is useful for computing various quantities (like the energy attributed to the state, using the Casimir element, as she does later in the book).

I suppose yet another perspective on the first question, which is (slightly) more mathematical than my original answer is that you want to study functions on R^3/SO(3). Since SO(3) action on R^3 is extremely non-free, the naive approach of just taking invariant functions, would yield the wrong thing. Since R^3 is contractible, homotopy theory would tell us that R^3/SO(3) is homotopy equivalent to "pt/SO(3)", which in homotopy world is precisely the category RepSO(3). Maybe someone with a better grasp of homotopy theory can flesh this out?

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I think at a later point in the book than you referred to (correct me if my memory is incorrect), Singer explains that you want to seek states which minimize the energy, which is either defined or derived, depending on your conventions, to be a simple function of the eigenvalues of a certain differential operator, which in this case coincides with the Casimir element EF + FE + H^2/2 in the enveloping algebra of sl_2(C), which is the complexification of the lie algebra so_3. So the physical problem of finding states with minimal energy is reduced to finding eigenvalues of a certain operator, which by the above discussion commutes with the SO(3) action. Since the operators in SO(3) commute with the Casimir, they preserve its eigenspaces, which means that f and g.f get assigned the same energy. So, the unique up to scaling "s" distribution has minimal energy, the three "d" orbits have the next highest energy, and so on. One could certainly ask why don't all the electrons just adopt the s orbital, but this is forbidden by an observational law (Pauli's exclusion principle) stating that distinct electrons can't have the same distribution, or more precisely that the span of the distributions of n electrons has dimension n. Well actually n/2 because of "spin", but nevermind for now.

By the way, at the risk of being accused of distracting with a pretty picture, I recommend

https://www.orbitals.com/orb/orbtable.htm

which depicts the distributions we're discussing by drawing the crests where they are maximal in length as complex numbers. It's amazing how they match with the grayscale observed data, and how complicated they can get for large numbers of electrons. Note that they are definitely NOT G-invariant! (although a cool feature you can see is that precisely one in each level is H-invariant, where H is the rotation through a fixed axis. This happens because you have odd dimensional SO(3) reps, and the element H in so_3 has weights -k, -k+2, ..., 0 , ... k-2, k on each such V, and so since H has a one-dimensional zero weight space, it must fix one of the vectors and this means one function should be invariant about the axis defining H).

Answer 3

I'm not sure that there is here a physical assumption, as much as a convenient way in which to talk about the physical system. Quantum mechanical representations are unitary and, albeit typically infinite-dimensional, they are also typically fully reducible. Hence in a way, if you understand the irreducible representations you can understand any representation.

For instance, Wigner argues this way in his 1939 paper on the unitary representations of the Poincaré group. This might address the second version of your question about elementary particles corresponding to irreducible representations of the Poincaré group.

I guess that what I am trying to say is that a state corresponding to a reducible representation $V\oplus W$ is physically indistinguishable from one made up of two states corresponding to $V$ and $W$, respectively. Hence nothing is gained by considering such states as "elementary".

Answer 4

Edit: OK, I think I figured out what she (Singer) is saying and what QY is asking. The restriction of physically real states to invariant subspaces is, as she points out, necessary such that all observers agree that it is in the same sub-space. For example, how do I absolutely prove that a chair is physically real? I ask an independent observer to verify the chair is there.

While observers may disagree on specific values to the result of some measurement due to differing reference frames, they must agree that they're measuring the same thing. This is equivalent to the relativistic notion of invariance. Certain quantities in relativity are frame-independent and will look the same to all observers (e.g. the magnitude of the four-momentum, the space-time interval, the proper time, etc.).

In quantum mechanics, states that lie in some invariant subspace should look the same to all observers. The reason we suppose that, in QM, physically real states must lie in invariant subspaces (which seems stricter than relativity which arguably allows for non-invariant physically real states) is because the act of measuring a quantum system disturbs it. So it is the only way to verify you actually have something as opposed to just a relic produced by the observation itself.

My previous answer (left here for reference purposes):

I'm pretty sure that the "elementary states" that she is referring to correspond (at least physically) to the basis states of the given Hilbert space which are not necessarily the same as the states that the particle is actually in. So, for example, for the spin of a spin-1/2 particle (like an electron), we could choose to measure along a variety of axes. Suppose we choose to measure the spin along the $z$-axis. The spin will either be up or down. The basis states corresponding to these outcomes are

$$|0\rangle = \left( \begin{array}c 1 \\ 0 \end{array}\right) \text{ and }|1\rangle= \left( \begin{array}c 0 \\ 1 \end{array}\right).$$

Note that it is possible for the particle, before you measure it, to be in a completely different state so, at least prior to a measurement being made, these states do not necessarily correspond to an actual physical state. They're simply the basis of the Hilbert space the particle is in. For example, prior to measurement the particle's state might be something like

$$|\psi\rangle = \frac{1}{\sqrt{2}}\left(|0\rangle + |1\rangle\right).$$

Now, it is possible that she's instead talking about the actual physical state of the system which is a little different. Consider the product state of two particles

$$|\psi\rangle = \frac{1}{\sqrt{2}}\left(|00\rangle + |10\rangle\right).$$

It is called a product state because it can be written

$$|\psi\rangle = \frac{1}{\sqrt{2}}\left(|0\rangle + |1\rangle\right) \otimes |0\rangle$$

where the state of one particle is $|a\rangle = \frac{1}{\sqrt{2}}\left(|0\rangle + |1\rangle\right)$ and the state of the other is $|b\rangle = |0\rangle$. If this is the case, it is important to note that in quantum mechanics there exist composite non-product states. These are the so-called entangled states. An example of one is the following Bell state,

$$|\psi\rangle = \frac{1}{\sqrt{2}}\left(|00\rangle + |11\rangle\right).$$

If this is the meaning Singer meant (as opposed to the basis states) then the elementary state would be the smallest factorable states one could reduce a system to, i.e. it might be the individual states of a product state or it might be an entangled state.

Answer 5

I have been struggling with very similar issues and was almost ready to ask here, but then found this question, so I'll instead post my ramblings as an "answer" to avoid duplication of questions (of course, it doesn't answer anything).

The official story seems to go something like this (for example in Weyl's classic): Look at something like $H=-\Delta + V(|x|)$ on $L^2(\mathbb R^3)$. Then the natural action of rotations $R\in SO(3)$ on $L^2$ by the unitary maps $(R\cdot f)(x) = f(R^{-1}x)$ is a symmetry of the system; this feels obvious, especially if I don't specify what I mean by symmetry. "Therefore" the irreducible representations of $SO(3)$ must be important.

I would really love to have this "therefore" explained in mathematical style, but after some reading around, my current impression is that there is nothing of the sort available. Rather, it seems to be a vaticinium ex eventu: For example, the spherical harmonics play a key role when analyzing $H$ above, and these do give us finite-dimensional representations. The representation on $\mathbb C^2$ (and for some reason we have now also found it convenient to slightly change $SO(3)$ to $SU(2)$) shows up in the description of spin. "Therefore," we conclude that representations must be relevant in general.

Since this answer is written from a position of ignorance, comments are very welcome.

Answer 6

I'll go against my better judgment and give an answer. I'll leave you with the caveat that my understanding of this area is very superficial, so there might be mistakes in what follows.

The first piece of the puzzle is that to each quantum mechanical system one can associate a Hilbert space $H$. So let's fix such a system and its associated Hilbert space $H$. The pure states of the system are, by definition, elements of the projectivization $\mathbb{P}H$ of $H$; and each observable is to be thought of as a selfadjoint operator $A$ on $H$. The spectral theorem then tells us that $A$ comes from integration against a spectral measure, and this spectral measure allows us to say something about the probability of the observable being in some pure state of the system.

Now one of the principles of special relativity is that the "laws of quantum mechanics" should be observer-independent, provided the observers are in (uniform) relative motion. Compare this to the classical notion of the "laws of mechanics" being the same in all (inertial) frames of reference. So whenever one has a group of symmetries $G$ of a quantum mechanical system, there ought to be an induced action of the group on the states of the system. The idea is that each $g \in G$ takes one frame of reference to another one. More specifically, $g$ ought to give us some kind of map $T_g$ from $\mathbb{P}H$ to itself, which takes a pure state $\hat{\psi}$ to another $T_g(\hat{\psi})$. (Think about the action of the Poincaré/Lorentz group on Minkowski space time.) We impose two extra conditions on the map $g \to T_g$. The first is that $T_{gh}(\hat{\psi}) = T_g \circ T_h(\hat{\psi})$ for all $g,h \in G$ and $\hat{\psi} \in \hat{H}$, the intuition being that changing frame of reference from A to B and then to C is the same as going from A to C, as usual. The second stipulation is that $(T_g(\hat{\psi_1}), T_g(\hat{\psi_2})) = (\hat{\psi_1}, \hat{\psi_2})$. We justify this by saying that if $G$ is really a symmetry group of our system, i.e. one that doesn't mess up the quantum-mechanical laws, then it shouldn't mess up probabilities. (I'm being purposefully vague here, because I don't want to say much about probability.)

A celebrated theorem of Wigner then tells us that such a map $g \to T_g$ comes from a unitary representation of $G$ on $H$, at least if $G$ is a connected semisimple Lie group (the semisimple condition may be weakened to a vanishing condition on the Lie algebra cohomology $H^\ast(\mathfrak{g})$, which I cannot recall). This means that for each $g \in G$ there is a unitary operator $U_g$ on $H$ such that $T_g(\hat{\psi}) = \widehat{U_g(\psi)}$ for all $\psi \in H$, and the map $g \to U_g$ is a homomorphism of $G$ into the unitary group $U(H)$ of $H$. If there are some continuity conditions imposed on the map $g \to T_g$, then there are analogous continuity conditions for $g \to U_g$.

So this explains how unitary representations show up when we're dealing with a symmetry group $G$ and a quantum system $H$. The final piece is now how this representation decomposes. The guiding physical intuition is that breaking up the representation into subrepresentations breaks up the quantum system into smaller, "elementary" pieces. The simplest of these pieces, the so-called elementary states (I like to think of these as elementary particles, although this is probably nonsense), therefore correspond to the "simplest" subrepresentations: the irreducible ones.

Hopefully this is at least somewhat helpful, and not entirely rubbish.

Answer 7

First, a bit of background. Consider a Hamiltonian $H$ on a finite-dimensional Hilbert space $\mathcal{H}$. Suppose that $H$ is gauge invariant, i.e. $G^{-1}HG = H$ for all $G$ belonging to a unitary representation of some group $\mathcal{G}$ on $\mathcal{H}$. It follows that $G^{-1}U(t)G = U(t)$ (here $U(t) = e^{-itH}$), and since the representation is unitary, it also follows that

$$\langle x' | U(t) | x \rangle = \langle x' | G^{-1}U(t)G | x \rangle = \langle x' | G^*U(t)G | x \rangle = \langle Gx' | U(t) | Gx \rangle$$

which implies that

$$\mathbb{P}(x \overset{t}{\longrightarrow} x') = \mathbb{P}(Gx \overset{t}{\longrightarrow} Gx').$$

That is, any gauge invariance of the Hamiltonian is automatically reflected in the transition probabilities: the eigenspaces of $H$ are invariant under the action of $G$, so that any unitary representation $D : G \rightarrow U(H)$ restricts to the eigenspaces. As José points out, typically the reps are reducible, etc. It is also worth mentioning that while the physically interesting states are gauge invariant, energy eigenstates need not be (and in the example below, are not).

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Now, an example of how this works in practice that requires no mathematically fancy machinery. Consider the 2D quantum Ising gauge theory

$$H = -\lambda \sum_{\square @} \prod_{\ell \in \square} \sigma_3(\ell) - \sum_{\ell @} \sigma_1(\ell)$$

where a link $\ell$ with time coordinate $t$ is understood to be spatial—and hence has a well-defined time coordinate $t$—if we write $\ell @ t$ (or suppressing the time argument). An elementary plaquette either has zero or two temporal links: in the former case we call the plaquette spatial and similarly write $\square @ t$.

The irreps of a product of $\mathbb{Z}_2$'s can be used to construct a physical basis. A gauge invariant eigenstructure exists for $H$, and the physics are governed only by the parameter $\lambda$ and the gauge-induced unitary representation from $\mathbb{Z}^M_2$ to $U(\mathcal{H}_2^{\otimes L})$. Here $\mathcal{H}_2$ is the Hilbert space of a single spin variable, $L$ is the number of links in the lattice, and $M$ is the number of them that comprise a maximal tree (see also one of my answers here).

If you want a writeup that covers this in detail, including numerics via MATLAB, leave contact info in a comment and I'll send a PDF (lamentably I only have this in Word).

Answer 8

I assume that you've seen most of the wikipedia articles on Wigner's classification and the like, which from what I remember do not address the interesting question you raise.

A few years ago I thought about the closely related question: given a quantum system as a black box, how can one identify the elementary particles in this system? It is clear (in the sense described by e.g. Wigner's classification) how to do this when the system consists of a single particle, but in a multi-particle system (such as the universe) it is possible for a group of operators that commute with the Hamiltonian to act irreducibly. The answer I eventually settled upon is that identifying the elementary particles in a system is essentially as arbitrary as writing the Hamiltonian as a sum of a kinetic term and an interacting term. It would absolutely make my day if someone could correct or clarify my speculation here.

Once this is done, I am comfortable taking the answer to your second question (about the identification of elementary particles with irreducible representations of the "symmetry group of the universe") to be: elementary particles don't exist in any real sense (as in QFT), but in a classical quantum system one can usefully define them in terms of irreducible representations in this way.

Answer 9

Physics is usually performed by means of Hamiltonian or Feynman Path Integral approach. Then, for steady state (time independent, when energy is conserved by Noether theorem), the symmetry of the system is in fact symmetry of Hamiltonian or Lagrange functional. For such system Hilbert space should be simple sum of subspaces for which Hamiltonian acts in invariant way, because more or less U(t) = exp(-iHt) defines "evolution operator" in this Hilbert space. The full solution with time for Schroedinger equation comes from formula $\psi(t) = U(t)\psi_0$ where $\psi_0$ is the initial state. So every quantity which is invariant under group action also commutes with H, so with U(t) and then it is invariant under time evolution which means is invariant quantity.

The same case is for more complicated symmetries as for elementary particles in quantum field theory, namely Standard Model here where the most important symmetry is full Poincaré group, plus some internal field symmetries defining fermions, bosons etc...

Elementary state I presume is state which is steady state for such evolution but if I well remember it is not standard terminology for physicist.

For non-relativistic quantum mechanics try Schiff Quantum Mechanics

For elementary particles take a look: Weinberg The Quantum Theory of Fields, where You may find this stated more formally but with also physical meaning.

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Additional notes: in the very fundamental level of physics is the same as in the very fundamental level of mathematics: incomplete induction. From facts which we know we build construction based on logic and deduction. But "the real meaning of relations" is something hidden. So probably there is no better way to explain why eigenstates of Hamiltonian are so important beside: because it works. In every physical quantum system, as far as I know, explanation of behavior of system may be reduced to symmetry operations, and representations of it. Of course we are used to build simple formal systems based on axioms which explains this. And probably it is very best way, specially for educational purposes. But sometimes we forget, that in physics always there are some additional conditions, relations, influences, which we cannot state explicite, because we do not know them. We do not describe reality in physics, we only construct some models, and assume some objects which are purely theoretical. So if You ask: why do You care about invariant quantities, the answer is: because You may posses only a small amount of information about quantum system. Every macroscopic measurement takes finite amount of time whilst during that time system may change his state million times. Only thous states which are very slowly changing, or are steady states may be seen in experiment. We may of course see effects of such changes, by interference patterns for example, but in fact quantum mechanics is not about reality itself, but rather about state of our knowledge about reality. This is a matter of interpretation of course, so probably there are physicists which have different opinion. On the level of formal mathematical operations it may looks the same, but physics without explanation is not worth practice. And explanation is nothing more than incomplete induction. It is just the same as with Church-Turing thesis. Why do we ask: What is the most compelling reason to believe Church's thesis? which is in opinion of mathematicians not real question, and maybe philosophical one. Why? It clearly states that mathematic in this area uses Incomplete Induction ( because every model of computability we know...), and there is no additional justification. So physics do.

Answer 10

You can use group representation theory to reduce the dimensions of the problem. And you can explain energy degeneracies. That's all!