A superselection rule is a rule that forbids superposition of quantum states. As stated by Lubos here, one cannot superpose states with different charges because of the conservation of charge:

An example in the initial comments involved the decomposition of the Hilbert space to superselection sectors ${\mathcal H}_Q$ corresponding to states with different electric charges $Q$. They don't talk to each other. A state with $Q=-7e$ may evolve to states with $Q=-7e$ only. In general, these conservation laws must be generalized to a broader concept, "superselection rules". Each superselection rule may decompose the Hilbert space into finer sectors.

Alternatively, one might say we don't have to consider such superpositions; it doesn't matter if there are other branches of the wavefunction with different $Q$ because they'll never interfere, so we might as well toss them out. This doesn't affect the results of experiments.

I'm confused why this same logic doesn't forbid almost any superposition whatsoever. For example, we often talk about a superposition of spin states $$|\psi \rangle = \frac{|\uparrow \rangle + |\downarrow \rangle}{\sqrt{2}}$$ or a superposition of momentum states $$|\psi \rangle = \frac{|p = p_0\rangle + |p = - p_0\rangle}{\sqrt{2}}$$ despite the conservation of angular momentum and momentum. Why exactly does superselection not forbid these kinds of superpositions as well?

Superselection has been discussed on this site a few times, but I haven't been able to find an argument that applies to conservation of charge but doesn't apply to conservation of momentum; this isn't a duplicate! One thought I had was that when we prepare a superposition of momentum states, we aren't really breaking superselection because there is a backreaction on the preparation apparatus, so we really have $$|\psi \rangle = \frac{|p = p_0, \text{app. recoils back}\rangle + |p = - p_0, \text{app. recoils forward} \rangle}{\sqrt{2}}$$ and the two states do have the same momentum. Then the original state we proposed is attained by just tracing out the apparatus; this doesn't decohere the superposition as explained here. This sounds plausible to me but then I don't understand why the same couldn't be said for charge conservation, leaving us with no superselection rules at all. (It's true that charge is discrete, but there are supposed to be superselection rules for continuous conserved quantities too.)

  • 2
    $\begingroup$ There is a reasonable discussion of super-selection rules in section 1-1 in Wightman's CPT, and in sections 2.2 and 2.7 in Weinberg's QFT. $\endgroup$ Commented Mar 27, 2018 at 21:49
  • 1
    $\begingroup$ Maybe the difference is that if you know you are in a Fock state, or a state connected to a Fock state by processes like pair production, this guarantees that you are in a charge eigenstate too, but the same is not true for $p$ or $S_z$. So I can imagine why in some contexts it might be a very practical consideration. But I think I agree with you that the difference between these two types of conservation is more of a difference of degrees than something that seems very fundamental. $\endgroup$
    – Rococo
    Commented Mar 29, 2018 at 23:21
  • 1
    $\begingroup$ Forbidding every superposition is a bizarre thing to say, given that (i) the statement "this state is (or is not) a superposition" is utterly meaningless, and (ii) a system might be constrained to follow multiple incompatible conservation laws, such as linear and angular momentum or different components of the latter; in those cases, the presumably allowed "no-superposition" states of one conservation law are superpositions of eigenstates of a separate conserved quantity. $\endgroup$ Commented Mar 29, 2018 at 23:22
  • 1
    $\begingroup$ @EmilioPisanty I guess I don’t really see that for charge. In an alternate universe QCD were somehow always weakly coupled, we could do scattering experiments with gluons, which do have color charge. Electric charge is almost like that, it just so happens the photon is neutral. $\endgroup$
    – knzhou
    Commented Mar 29, 2018 at 23:45
  • 1
    $\begingroup$ @EmilioPisanty Besides detecting charge through photons I would imagine one could make some weird measurement device involving, I don’t know, passing your state through a plasma might change the charge. I’m just spitballing here, the point is I don’t buy your reason charge is different. Under your reasoning superselectipn rules are just functions of our current technology? $\endgroup$
    – knzhou
    Commented Mar 29, 2018 at 23:46

3 Answers 3


Superselection makes sense not abstractly in an arbitrary Hilbert space but only in Hilbert spaces structured by the introduction of an algebra of distinguished observables of interest (i.e., that can be combined to prepare states), with prescribed commutation rules. These define the physics that is possible in the class of models considered, and superselection is a concept defined relative to these. (In particular, if the class of relevant observables changes, the concept of superselection changes with it. Enlarging the observable algebra may merge some sectors but typically creates others.)

Typical examples are the universal enveloping algebras of Lie algebras, or the $C^*$ algebras generated by the corresponding Lie groups. For example Heisenberg algebras and Heisenberg (or Weyl) groups correspond to canonical commutation rules, which are the basis of much of quantum physics.

The Hilbert spaces of interest are the (continuous unitary) irreducible representation spaces of these algebras, Lie algebras, or groups. These (or more precisely the classes of equivalent such representation spaces) are called the (superselection) sectors of the theory. Since they constitute different Hilbert spaces it makes no sense to superimpose vectors of the different sectors. One can define an inner product on the direct sum of these Hilbert spaces, but the algebra of operators still map each sector into itself, hence there is no way to create (in a physically relevant way) a superposition from pure states within different sectors.

For finite-dimensional Heisenberg algebras/groups, all continuous unitary irreducible representations are equivalent (Stone-von Neumann theorem); hence for nonrelativistic N-particle theories, there are no superselection rules (that would specify superselection sectors).

Once one also accounts for spin, the situation becomes more complicated: a mixture of a fermionic and a bosonic state makes no longer physical sense since the two state vectors behave differently under a rotation by 360 degree - though formally it is still defined. No amount of new physics will change that.

For infinite-dimensional Heisenberg algebras/groups as they occur in (relativistic or nonrelativistic) quantum field theory, the Stone-von Neumann theorem is no longer valid, and there are uncountably many inequivalent continuous unitary irreducible representations, hence there are uncountably many superselection sectors, distinguished by their essentially different behavior at spacelike infinity.

In more technical terms: The most interesting superselection rules, accounting for superconductivity, charge, baryon number, etc., arise due to unimplementable Bogoliubov transformations, involving limits that are so singular that they lead out of the Hilbert space representing the vacuum sector. In particular, charged states have a sufficiently different asymptotic structure from uncharged ones, since the Coulomb field is long range, and they belong to different superselection sectors. This is a general property of charges in gauge theories, see Strocchi, F., & Wightman, A. S. (1974). Proof of the charge superselection rule in local relativistic quantum field theory. Journal of Mathematical Physics, 15(12), 2198-2224. No amount of new physics will change that.

Under certain conditions, superselection sectors can be classified; see, e.g., the article DHR superselection theory from nLab. DHR stands for Doplicher, Haag, and Roberts; see, e.g.,

  • Doplicher, S., & Roberts, J. E. (1990). Why there is a field algebra with a compact gauge group describing the superselection structure in particle physics. Communications in Mathematical Physics, 131(1), 51-107.

Superselection rules have nothing to do with conservation laws. In spite of momentum conservation, states of different momentum can be superimposed since the Lorentz transformations that turn one momentum state into another are unitary and hence are defined on the same representation space.

  • $\begingroup$ +1 In that case, disjoint states which give rise to inequivalent representations through GNS, are to be thought as giving rise to superslection sectors? That is then the way to understand the distinct folia? By the way, considering the charge example in the OP, would this mean that disjoint states are actually describing different systems (for example, one describing a field of charge $q$ and the other a field of charge $2q$ and so forth)? $\endgroup$
    – Gold
    Commented Apr 3, 2018 at 20:52
  • $\begingroup$ @user1620696: Yes, exactly, for all but the last statement. For the latter, the states define the same system in states of different total charge. $\endgroup$ Commented Apr 3, 2018 at 20:53
  • $\begingroup$ indeed, when I wrote I didn't consider that in QFT we have particle creation/annihilation, so that even if we have a specific vacuum $\omega$ selected and work in a Fock space picture, operating with the creation/annihilation operators we get states of different charge on the same folium. Is that your point? $\endgroup$
    – Gold
    Commented Apr 5, 2018 at 2:57
  • $\begingroup$ @user1620696: no. The conventional language is that a system is defined by its algebra of observables, and that the states (whether given as wave functions, state vectors, density matrices, density operators, or linear functionals) describe its possible states. - Interactive QFTs in dimension >2 have no creation/annihilation operators in the traditional sense, by Haag's theorem. $\endgroup$ Commented Apr 6, 2018 at 13:26
  • 1
    $\begingroup$ @ungerade: As I said, you need to specify an algebra of observables of interest. If this algebra is the universal enveloping algebra of the Lie algebra of the Poincare group, you may consider UIR with fixed spin and different mass as superselection sectors. But in real physics, the algebra of physical observables is much bigger, and superpositions of mass states are possible (e.g. in neutrino oscillations). $\endgroup$ Commented Jun 11, 2023 at 15:49

I think the OP is getting close to answering his own question by suggesting that "superselection rules are just functions of our current technology." In my opinion, the answer is just a more precise statement of this suggestion: superselection rules capture beliefs we have about the nature of possible physical observables in the universe, limited by current experimental probes and theoretical models.

To explain that, we first give a mathematical definition of superselection sectors:

Definition 1: Consider a Hilbert space $H$ and a collection of subspaces $\{H_i\}$ such that: $H = \oplus_{i} H_i$. We say that $\{H_i\}$ are superselection sectors in $H$ if $\langle \psi_i| \mathcal{O}|\psi_j\rangle = 0$ $\forall$ $\psi_{i,j} \in H_{i,j}$, $i \neq j$, $\mathcal{O} \in O_{phy}$, where $O_{phy}$ is the set of all physical observables $\mathcal{O}$ acting on the Hilbert space $H$.

There are two important notions that have to exist a priori in this definition:

  • We have to specify a Hilbert space. To solve any quantum mechanical problem dogmatically, we would need a Hilbert space that contains states of the entire universe to ensure unitary evolution. However, practically, we often assume weak enough interactions between a test system $A$ and the rest of the universe $B$, so weak that we approximate $A$ as isolated (i.e. $H = H_A \otimes H_B$). So we can approximately obtain unitary evolution by restricting to the test system subspace $H_A$ for all interactions that act only on $H_A$. This freedom of choosing Hilbert spaces gives a bit of ambiguity to the meaning of superselection. Because it is possible that $\{H_i\}$ define superselection sectors in $H_A$, while $\{H_i \otimes H_B\}$ doesn't define superselection sectors in $H_A \otimes H_B = H$. We will see examples illustrating this later.

  • We have to define physical observables. Technically all self-adjoint operators on $H$ should be admissible candidates. But based on the current understanding of fundamental physics, we believe that certain self-adjoint operators are impossible to measure and certain transitions between states are impossible to engineer. Thus superselection rules today are practical tools that can be updated as we learn more about the world and find more physical observables that possibly break existing rules.

We now look at two examples to clarify why these two points above can cause confusion and to postulate a more complete definition of superselection rules.

Example 1: To illustrate the importance of the first bullet point, consider the OP's example of spins. Suppose we look at the Stern-Gerlach experiment and define $|\uparrow \rangle$ and $|\downarrow\rangle$ as eigenstates of the $\sigma_z$ operator. We take the Hilbert space $H_A$ spanned by these two states and postulate that the only physical observables on $H_A$ are $f(\sigma_z)$ where $f$ is some arbitrary analytic function. Now you may easily check that $|\uparrow \rangle$ and $|\downarrow\rangle$ define superselection sectors on $H_A$ alone!

But remember we have a freedom of choosing Hilbert space! Suppose now that we enlarge our Hilbert space to include a second particle $B$. Although we constrained physical observables on $H_A$ to be $f(\sigma_z)$, we don't have to constrain observables mixing $H_A$ and $H_B$. Even if we maintain conservation of total angular momentum $\sigma_z(A) + \sigma_z(B)$, we may well break conservation of $\sigma_z(A)$ or $\sigma_z(B)$ separately, thus mixing $\{|\uparrow\rangle \otimes H_B\}$ and $\{|\downarrow \rangle \otimes H_B\}$. This would imply $\{|\uparrow\rangle \otimes H_B\}$ and $\{|\downarrow \rangle \otimes H_B\}$ do not define superselection sectors on $H_A \otimes H_B$, as advertised in the first bullet point above. This is similar in spirit to OP's consideration of $p(particle) + p(recoil)$ on the Hilbert space $H_{\text{particle}} \otimes H_{\text{apparatus}}$.

Example 2: To illustrate the second bullet point, we consider OP's spin system without adding particle $B$. Instead we enlarge the set of physical observables by adding $\sigma_x$. In particular, we engineer a Hamiltonian $H= -\sigma_x$ by putting the spin in a unit of transverse magnetic field (as usual, we assume the interactions between the source of the magnetic field and the particle $A$ to be sufficiently weak that we can separate out $H_A$). Suppose we start the system in $|\downarrow \rangle$ at $t=0$, then at $t= \pi/2$, we flip the two sectors completely: $$ e^{-i Ht} |\downarrow\rangle = e^{i \pi \sigma_x/2} |\downarrow\rangle = i \sigma_x |\downarrow\rangle = i|\uparrow \rangle $$ Again, as advertised, the introduction of more possible physical observables (in this case the introduction of $\sigma_x$ into the Hamiltonian) breaks the conservation of angular momentum and therefore the superselection sectors.

Example 3: Another example to illustrate the second bullet point. Let's escape from reality a bit more and imagine for fun that QM was discovered in 18th century, back when we believed in Galilean symmetry. To study states in QM, we found the projective representation of the Galilean algebra with $\{K_i\}$ generating Galilean boosts and $\{P_i\}$ generating translations. When we compose boosts and translations, we found that for a state with mass $M$: $$ e^{-i \vec K \cdot \vec v} e^{-i \vec P \cdot \vec a} = e^{i M \vec a \cdot \vec v/2} e^{-i (\vec K \cdot \vec v + \vec P \cdot \vec a)} $$ This means a superposition of states with different masses would break Galilean symmetry, a sacred law of physics back then. In order to avoid that problem, we would have guessed a superselection rule forbidding transitions between states with different masses. And that rule would have stood the test of experiments and theoretical models up until the introduction of Einstein's relativity (which effectively modified the Hamiltonian, one of the physical observables, and its fundamental symmetries)! This is a rather dramatic illustration but I think it makes the point that what we call physical is always a matter of contemporary belief.

To summarize the discussion above, we come back to the classic example of electric charge. Why have physicists established the convention that charges define superselection sectors? It is because the superselection of charges is in some sense universal. While the superselection rule of angular momentum depends on the choice of Hilbert space, and the superselection rule of mass depends on choice of physical observables, electric charge is somehow different. As long as we believe in the Standard Model, no matter what Hilbert space $H \subset H_{\text{universe}}$ we choose (it can be the Hilbert space of the lab, of the earth, or of the whole universe), the subspaces $\{H_Q \subset H\}$ indexed by the restricted total charge operator $Q|_H$, define superselection sectors in $H$ in the sense of definition 1. Colloquially, this is to say: states with different charges don't talk to each other, for all choices of Hilbert space and physical observables within the Standard Model.

This last summary makes it clear that superselection rules we have today are intimately connected to contemporary beliefs about the character of physical laws. Maybe one day, there will be a new physical observable that mixes charge sectors, or a larger Hilbert space in which charge in one universe can be traded for charge in another. By that time, we would have to repeal the principle of charge superselection and hopefully find new superselection rules that will make our lives easier.

  • 3
    $\begingroup$ This is an outstanding answer and it's worthy of a rather bigger one than the current run. For now, though, +1. $\endgroup$ Commented Mar 31, 2018 at 0:12
  • 1
    $\begingroup$ This is a valid answer only in as far all of physics is a matter of contemporary belief, and then it is a trivial statement. $\endgroup$ Commented Apr 1, 2018 at 12:19
  • $\begingroup$ All physics, like superselection rules, captures beliefs we have about the nature of possible physical observables in the universe, limited by current experimental probes and theoretical models. $\endgroup$ Commented Apr 1, 2018 at 16:51
  • $\begingroup$ @ArnoldNeumaier I agree with you that the statement about "beliefs" is trivial and I should probably delete it. Perhaps the more meaningful statement (meaningful to me but you can definitely disagree) appears in the last paragraph, that the criterion for deciding what we call "superselection" rests on a sense of universality, which distinguishes electric charge from examples that confuse the OP in the question statement. Clearing that confusion with some precise definitions and examples was the main objective of this answer. $\endgroup$ Commented Apr 1, 2018 at 18:45
  • $\begingroup$ As mentioned in my answer, the term superselection sector is a precise technical term. For the case of charge, superselection is a general property of gauge theories, see Strocchi, F., & Wightman, A. S. (1974). Proof of the charge superselection rule in local relativistic quantum field theory. Journal of Mathematical Physics, 15(12), 2198-2224. This has nothing to do with the standard model, as electrical charge is intrinsically linked to QED, which is a gauge theory. In particular, your last paragraph is void of meaning. $\endgroup$ Commented Apr 1, 2018 at 19:43

There's a talk by Robert Spekkens at PI on this exact topic.
Video: Are Superselection Rules Fundamental?

Apparently there is (was?) a long time debate among quantum optics people about whether coherent states are real or just a fiction.

I'm not well-versed enough to do a good summary of this talk. The short takeaway I got is that coherence is a convenient approximation to use in calculations, and this approximation becomes exact when your environment has infinitely many degrees of freedom.
If you want to describe everything as objectively as possible, including the environment, then coherence is just a fiction.

So the answer to your question, according to this talk, is... superposition does sort of forbid many superpositions?

Some references:
quant-ph/0507214 - Dialogue Concerning Two Views on Quantum Coherence: Factist and Fictionist
quant-ph/0610030 - Reference frames, superselection rules, and quantum information

  • $\begingroup$ This does not answer the original question, which is not about coherent states but about superselection. Note that for Fock spaces over infinitie-dimensional Hilbert spaces 9as needed for QFT), there are many coherent states that do not belong to the Fock space (the vacuum sector of a free theory) but to a nontrivial superselection sector. But the coherent states discussed in quantum optics are not of this kind. The talk by Spekkens is about photon number superselection which is an untenable minority position unrelated to the superselection rules enforced by quantum field theory. $\endgroup$ Commented Apr 1, 2018 at 15:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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