Superselection rule

**touqra** · Nov7-06, 07:11 PM

What does superselection rule mean?

**CarlB** · Nov7-06, 09:47 PM

Let me try. I hope that the herd will correct me if I am wrong.

In quantum mechanics, if you have two solutions to, for example, the Dirac equation, you can sometimes add those two solutions together to get a new solution to the Dirac equation. As long as those two equations are just mathematical objects and are not associated with any particles, this can always be done.

But if you think of your two solutions as being actual particles, there are sometimes "superselection rules" that prevent you from combining the solutions. One of the superselection rules is charge. You cannot make arbitrary linear combinations between particles with different charge. For example, if you have a Dirac wave function for an electron, you cannot combine this with a Dirac wave function for a neutrino.

You can, however, combine wave functions for a spin up electron with wave functions for a spin down electron. In this case there is no superselection rule. The way this is treated is that "states with different charges live in different superselection sectors".

It's interesting to note that particles from different generations (but with all other quantum numbers the same) are in the same superselection sector. Thus you can mix the electron with the muon and tau. And you can mix the various neutrinos. But you can't mix electrons with quarks.

I've thought about this for some time, but so have a lot of other, far more intelligent, people. I will express some opinions on this now.

If two particles have different spin, then they will use different wave functions as different spin particles use different wave functions. For example, spin-1/2 particles use the Dirac wave equation. It is sort of intuitive why such particles cannot live in the same superselection sector.

But with particles of the same spin, the superselection rules are not so intuitive. My opinion is that you can rewrite the Dirac equations so that spin-1/2 particles which do not live in the same superselection sector use "different" Dirac equations. It turns out that this is a very natural thing to do in Clifford algebra; when one replaces spinors with more general objects, one finds that one can divide the resulting "generalized Dirac" equation into a set of more or less identical Dirac equations. The superselection rule then prevents you from moving from one of those identical Dirac equations to the other.

On the other hand, the fact that one can mix electrons with muons and tau suggest that these particles must share the same Dirac equation in a way that, for example, neutrinos and electrons do not. It is my belief that this is evidence that the electron, muon and tau are composite particles, and furthermore that they share the same set of components. And it is these components that the natural Clifford algebra superselection rules come from.

Carl

**zbyszek** · Nov9-06, 09:55 PM

Physically meaningful solutions of an equation E must be invariant under transformation T. Solution s1 and solution s2 are invariant under T. Superposition s1+s2 is also a solution of E, but it is not invariant under T. Thus s1+s2 is not a miningful solution.

This is the superselection rule in action.

Cheers!

**vanesch** · Nov10-06, 12:41 AM

Originally Posted by zbyszek

Physically meaningful solutions of an equation E must be invariant under transformation T. Solution s1 and solution s2 are invariant under T. Superposition s1+s2 is also a solution of E, but it is not invariant under T. Thus s1+s2 is not a miningful solution.

If s1 and s2 are quantum states, and T is a linear operator representing the invariant transformation, I fail to see how s1 and s2 could possibly be invariant, but not their sum ?

I thought that a superselection rule was an (arbitrary) limitation of the Hilbert space of states into a union of subspaces, so that when one obtains, by linear evolution, a state which is a sum of elements of the subspaces (which belongs to the sum of subspaces, but not their union), one automatically has to apply projection.

A typical example is electric charge. It is assumed (= superselection rule) that one cannot have superpositions of states which correspond to different electric charges.

Decoherence tries to explain superselection as a dynamical phenomenon.
In the example of charge, decoherence theory tells you that states with different charge couple differently to the EM field, so very quickly a (forbidden by the superselection rule) superposition of states with different charges will entangle with the EM field, so that all interference effects on the particle alone will be suppressed, and it will look *as if* the superposition was forbidden and one had to apply projection.

**Careful** · Nov10-06, 12:53 AM

Originally Posted by vanesch

Decoherence tries to explain superselection as a dynamical phenomenon.
In the example of charge, decoherence theory tells you that states with different charge couple differently to the EM field, so very quickly a (forbidden by the superselection rule) superposition of states with different charges will entangle with the EM field, so that all interference effects on the particle alone will be suppressed, and it will look *as if* the superposition was forbidden and one had to apply projection.

In your notion of ``existance'' you are very willing to separate the kinematical from the dynamical aspect.

In my mind a state only ``exists'' when it is dynamically stable over some respectable period of time.

Careful

**vanesch** · Nov10-06, 03:01 AM

Originally Posted by Careful

In my mind a state only ``exists'' when it is dynamically stable over some respectable period of time.

Well, that's then a "dynamically stable state", no

**zbyszek** · Nov10-06, 06:35 AM

Originally Posted by vanesch

If s1 and s2 are quantum states, and T is a linear operator representing the invariant transformation, I fail to see how s1 and s2 could possibly be invariant, but not their sum ?

Example: T stands for Galilean transformations, s1-represent a state of N massive
particles, s2- a state of N+1 massive particles. All particles have the same mass.
Under the transformation s1 aquires an overall phase that depends on the total mass,
same with s2.
s1+s2 aquire overall phase and the relative phase so it transforms differently under T
than s1 and s2.

Cheers!

**vanesch** · Nov10-06, 06:51 AM

Originally Posted by zbyszek

Example: T stands for Galilean transformations, s1-represent a state of N massive
particles, s2- a state of N+1 massive particles. All particles have the same mass.
Under the transformation s1 aquires an overall phase that depends on the total mass,
same with s2.
s1+s2 aquire overall phase and the relative phase so it transforms differently under T
than s1 and s2.

Ah, you mean: s1 and s2 belong to different representations of the symmetry group! Yes, then you're right, but then there is no unique T acting on the entire hilbert space. There is then simply no symmetry representation for s1+s2 (that's probably the point you want to make), so there is no definition of T over the entire hilbert space. Just as well you can say that there *is* no such symmetry for the entire system, but only for certain subspaces.
But then, you've already split up the hilbert space in subspaces which are symmetric, and others which aren't (the ones "in between").

**Careful** · Nov10-06, 07:53 AM

Originally Posted by vanesch

Well, that's then a "dynamically stable state", no

No, it is one we know for sure to exist !

**LanceV** · Mar6-08, 02:08 PM

Hello Forum!

I am confused about this topic, too.
The post of @vanesch seems to suggest, that the superselection rules are additional postulates to the quantum theory, while in @zbyszek's post there seems to be a mathematical way to deduct them?
Or rephrased into a more direct question:
Why does the weak isospin "generate" (?) a superselection rule, while the usual spin does not? The transformation law is the same for a spin 1/2 particle?
What is the correct terminology anyways? The superselection rule selects the quantum nubers of a state, that a given state can interfere with?

**vanesch** · Mar6-08, 02:20 PM

Originally Posted by LanceV

Hello Forum!

I am confused about this topic, too.
The post of @vanesch seems to suggest, that the superselection rules are additional postulates to the quantum theory, while in @zbyszek's post there seems to be a mathematical way to deduct them?
Or rephrased into a more direct question:
Why does the weak isospin "generate" (?) a superselection rule, while the usual spin does not? The transformation law is the same for a spin 1/2 particle?
What is the correct terminology anyways? The superselection rule selects the quantum nubers of a state, that a given state can interfere with?

Superselection is an invention to avoid the superposition principle in some cases. The most notable is the superposition of different charge states, which is the case with isospin. Take the example of an isospin doublet: proton and neutron (and let us forget for the moment about the fact that they are not elementary constituents).

As you point out, in as much as isospin is mathematically of an identical form as normal spin, we can have superpositions of |z+> and |z-> in normal spin, which make up, say, |x+>. But we can't have that in isospin ; we can't have states which are the superposition of a proton and neutron state. This is what this particular superselection rule says: no superpositions of states with different charges. There are other superselection rules (I'm no expert).

But in decoherence theory, one simply explains the *apparent* validity of superselection rules by the extremely rapid decoherence of the "forbidden" superpositions. In other words, the superpositions are not "forbidden" by any law of nature, but the different components interact so quickly in different ways with the environment that they get almost instantaneously irreversibly entangled, which makes them *appear* almost always as mixtures, and which makes it impossible to observe any interference effects which would testify of their superposition.
The charge rule is simply explained: a charged state will interact (Coulomb...) so quickly with anything charged in the environment, that it will very quickly act differently than a non-charged state.

**LanceV** · Mar6-08, 03:05 PM

Thanks alot for the reply.

Originally Posted by vanesch

As you point out, in as much as isospin is mathematically of an identical form as normal spin, we can have superpositions of |z+> and |z-> in normal spin, which make up, say, |x+>. But we can't have that in isospin ; we can't have states which are the superposition of a proton and neutron state.

Why can't we have it? Interestingly I never thought about this, but now I am unsure. Am I missing something obvious? Or is it just an observation?

Originally Posted by vanesch

This is what this particular superselection rule says: no superpositions of states with different charges. There are other superselection rules (I'm no expert).

It seems to me, that most quantum numbers are connected to superselection rules, with only few exceptions (energy, angular momentum and spin). Once I thought it had something to do with internal and external symmetries, but there is also a ssrule connected to parity.

Originally Posted by vanesch

But in decoherence theory, one simply explains the *apparent* validity of superselection rules by the extremely rapid decoherence of the "forbidden" superpositions. In other words, the superpositions are not "forbidden" by any law of nature, but the different components interact so quickly in different ways with the environment that they get almost instantaneously irreversibly entangled, which makes them *appear* almost always as mixtures, and which makes it impossible to observe any interference effects which would testify of their superposition.

That's very intuitive. Thanks for the explanation.

**MaverickMenzies** · Mar6-08, 05:02 PM

When I've encountered super-selection rules (SSR) in the past (in the context of quantum mechanics) they have been presented as violations of the superposition principle. Thus, it requires picking a preferred set of commutating self-adjoint operators and declaring only these are observables for the system and denying the same interpretation to all others. As a consequence, the Hilbert space describing the states of the system can be view as possessing a unique decomposition as a direct sum of the eigen-subspaces (called super-sectors) of the preferred observables. The key word here is unique as any other alternative decomposition of the Hilbert space is interpreted as being non-physical.

Essentially, you artificial break the linear nature of quantum mechanics by assuming the existence of a unique decomposition of the Hilbert space. Consequently, one cannot prepare a system exhibiting the SSR in a superposition of eigen-states of the preserved observables.

An often quoted example of the super-selection rule is the one invoked to explain the absence of states which are superposition of charge eigenstates or nucleon eigenstates i.e. it is forbidden to prepare nucleons in superpositions like

a | proton > + b | neutron >.

I hope this helps.

**CarlB** · Mar6-08, 07:15 PM

I wonder if another place superselection rules don't apply is in the generations. That is, one forms superpositions of electron, muon and tau neutrinos when working out neutrino oscillation. The particles of the three generations carry the same quantum numbers, except perhaps for "generation", and so maybe there isn't any need for the absence or presence of a superselection rule.

**vanesch** · Mar7-08, 07:14 AM

Originally Posted by LanceV

Why can't we have it? Interestingly I never thought about this, but now I am unsure. Am I missing something obvious? Or is it just an observation?

I think superselection rules were simply introduced to "explain" the fact that we never saw a particle with was a superposition of a proton and a neutron, or that we never saw other interference effects of the kind. But I am not sure what is the real reason for the introduction of superselection rules.

**samalkhaiat** · Mar8-08, 10:30 PM

[quote=touqra;1154099]

What does superselection rule mean?

Read my selection rule: You can choose to do theoretical physics without knowing anything about superselection rules. Just ignor it! What you need to know is the following

Let G be a symmetry group and M(0), M(1/2), M(1), ... are its multiplets (representations). Then the combination

$LaTeX Code: a_{1}M(0) + a_{2}M(1/2) + a_{3}M(1) + ...$

with atleast two nonvanishing a's, is not allowed by the symmetry group G.

Examples

1) Lorentz group G = SO(1,3):

M(0) is the scalar (spin = 0) representation, M(1/2) is the spinor (spin = 1/2) rep. and M(1) is the vector (spin = 1) rep.
We know that Lorentz transformations do not mix, say, scalars with spinors (supersymmetry does).

2) Nuclear I-spin G = SU(2)

M(I=0) = trivial rep. is the isoscalar (singlet) $LaTeX Code: \\Lambda$ .
M(I=1/2) = fundamental rep. is the isospinor (doublet) (P,N).
M(I=1) = adjoint rep. is the isovector (triplet) $LaTeX Code: (\\Sigma^{+}, \\Sigma^{-}, \\Sigma^{0})$ .

SU(2) transformations mix particles occupying the same multiplet, for example, neutron with proton (in real life, we don't see such mixing because isospin is not an exact symmetry, it is broken by mass-difference and em-interaction), But particles belonging to different multiplets don't get mixed by SU(2) transformations.

regards

sam