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Isospin: how serious must I take it? Superposition of proton and neutron?

by nonequilibrium
Tags: isospin, neutron, proton, superposition
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DrDu
#37
Apr3-12, 07:40 AM
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Quote Quote by DrDu View Post
Exactly!
So probably it would be better to consider a superposition of p+e and n., or, more generally,
[itex]p +e \leftrightarrow \left( \begin{array}{c} 1 \\ 0 \end{array} \right)[/itex] and [itex]n \leftrightarrow \left( \begin{array}{c} 0\\1\end{array} \right)[/itex]
I just realized that p+e has the same charge as n but would not the same ordinary spin, so that such a superposition would violate univalence superselection rule. Hence one has to add a chargeless fermion e.g. the antineutrino.
Hence in the course of beta decay a neutron (isospin down) would continuously transform into a superposition of an isospin up (p+e + anti-nu) and down (n) state.
francesco85
#38
Apr3-12, 07:57 AM
P: 62
Quote Quote by Vanadium 50 View Post
Of course it is, as your example points out. In that example, Q = T_3 + 1/2.
This does not mean that T3 is the electric charge up to a constant (up to a constant means T3->k*T3, where k is a number); what you have written means that you are considering a U(1) factor under which the doublet is charged with charge 1/2 (the operation of you have called "up to a constant" is not for free, you have to introduce another group, namely this U(1)); this is well-known to everybody who studied the basics of the standard model and group theory (even if this is a slightly different case); what is this U(1) you have introduce? Moreover, it can be seen as a definition, within the SM, nothing to do with T3 (it is just a convention, which can be redefined). And if I put the U(1) charge not equal to 1/2 but to 1/3 or 0, what happens? Why isn't that quantity "absolutely conserved"? Moreover you are still talking about eigenstates of T3 in a situation in which you are considering the electromagnetism, which explicitly breaks the isospin invariance. Do you mean that the neutron and the proton have the same mass, in your opinion?


@mr. vodka: sorry once again for my notation; [itex]P^2=P^{\mu}P_{\mu}=h^2-\vec{p}^2[/itex] (as operators), as you have written.
In my opinion, the reason why we assume only mass eigenstates is that, when the time goes to "infinity", a good approximation of the situation we are considering is that of free particles; now, if I understood correctly, one-particle irreducible representations of the Poincarč group are "classified" by mass and spin; so this clarifies, in my opinion, why we see, in scattering processes, only mass eigenstates: in few words, because irreducible representations of the Poincarč group are "defined" by mass and spin; you may ask why we build irreducible representations of the Poincarč group; well, this is related to Poicarč invariance; a good reference for this is Weinberg. Moreover, one may ask why one considers also energy eigenstates ("delocalized"); in my opinion, this has to do with formal scattering theory, which allows some kind of manipulations.
And, finally, when you ask if you interpreted correctly my point of view, the answer is yes, you interpreted correctly my point of view.


@tom.stoer

Quote Quote by tom.stoer View Post
of course you can; look at neutrino oscillations or at the Kaon system
Mmm.. are you sure about that? Can you see my answer(s) to this post (in particular my first answer) and can you give me your opinion about those?

http://www.physicsforums.com/showthread.php?t=591256

Thank you very much!
Francesco
DrDu
#39
Apr3-12, 09:55 AM
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Quote Quote by francesco85 View Post
so this clarifies, in my opinion, why we see, in scattering processes, only mass eigenstates:
I doubt this. Neutrino oscillations are a good counter example.
DrDu
#40
Apr3-12, 10:04 AM
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Quote Quote by mr. vodka View Post
tom.stoer: I'm not sure if I get what you're aiming at. Are you using charge as an argument for why the p+n state is not possible? However, that's not really an argument, but more of a restating of the fact that we do not see a p+n state. But more likely I misinterpreted the aim of your post, so I'd appreciate any clarification.

As to francesco, wouldn't your argument "prove" that there can only be energy eigenstates in nature? Yet this is of course not true.
Consider ordinary spin: An electron with spin up and a proton with spin down. Would you conclude from the fact that spin points up on one particle and down on the other that superpositions of electrons and protons exist?
francesco85
#41
Apr3-12, 10:05 AM
P: 62
Quote Quote by DrDu View Post
I doubt this. Neutrino oscillations are a good counter example.
Hello! I'm not a neutrino expertise, but I know the "usual" paradigm; (and indeed I made reference to it in this thread: http://www.physicsforums.com/showthread.php?t=591256, where I made also reference to an alternative possible interpretation); after all, neutrino osccillations are a consequence of the fact that we do not observe neutrino mass eigenstates experimentally, aren't they? In my opinion, this is the reason why one usually assumes the "usual" framework: as long as we neglect neutrino masses, we have three degenerate mass eigenstates; then linear combinations are possible; by definition, the electronic neutrino is the neutrino produced in e.g. beta decay in which an electron is produced (and,in my opinion, it is possible to produce it since it is a mass eigenstate, at this level); let's now see the fate of the neutrino: it oscillates: the point is that at this stage we are not neglecting neutrino masses no more. My point of view in the intepretation is to consider the "real" mass eigenstates also at the stage of the production through e.g. beta decay. I ask you the same question I have asked tom.stoer: what do you think about this interpretation? Thank also to you :)
Francesco

ps [EDIT]: a question has just occured to my mind: if it is possible to produce in a scattering states which are not mass eigenstates, how do you compute the cross section or the decay rate for such a process? In the phase space factor there i an explicit mass, or am I wrong?
DrDu
#42
Apr3-12, 10:23 AM
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Francesco, in that thread you were citing you wrote:
"The Feynman diagram for every such process is weighted with factors of the PMNS matrix; so why are we speaking about electron neutrinos? The point is that we are not able to measure and observe the mass eigenstates of the neutrinos, our experimental apparatus are not so powerful. What we can say, in my opinion, is that we can treat the system as a statistical mixture of three kind of neutrinos. If we were able to see the mass eigenstates, then we would not observe oscillations."

The three mass eigenstates would still form a superposition and not a statistical mixture.
The situation is not much different from the emission of radiation by an excited atom. The radiation will not be an energy eigenstate but have a certain linewidth which is due to the exponential decay of the excitation. In the case of photons the measurement devices are so refined that you can detect the phase of the components of the different energy eigenstates.
francesco85
#43
Apr3-12, 10:27 AM
P: 62
Quote Quote by DrDu View Post
Francesco, in that thread you were citing you wrote:
"The Feynman diagram for every such process is weighted with factors of the PMNS matrix; so why are we speaking about electron neutrinos? The point is that we are not able to measure and observe the mass eigenstates of the neutrinos, our experimental apparatus are not so powerful. What we can say, in my opinion, is that we can treat the system as a statistical mixture of three kind of neutrinos. If we were able to see the mass eigenstates, then we would not observe oscillations."

The three mass eigenstates would still form a superposition and not a statistical mixture.
The situation is not much different from the emission of radiation by an excited atom. The radiation will not be an energy eigenstate but have a certain linewidth which is due to the exponential decay of the excitation. In the case of photons the measurement devices are so refined that you can detect the phase of the components of the different energy eigenstates.
Thanks for your answer; I'm also sorry beacuse I have also edited a question which might have not seen: if a state which is not a mass eigenstate is "really" produced, how can you compute the cross section or the decay rate for such a process? Isn' there a mass parameter in the phase space? For the moment, I'm going to reflect on your answer!
DrDu
#44
Apr3-12, 10:45 AM
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Quote Quote by francesco85 View Post
Thanks for your answer; I'm also sorry beacuse I have also edited a question which might have not seen: if a state which is not a mass eigenstate is "really" produced, how can you compute the cross section or the decay rate for such a process? Isn' there a mass parameter in the phase space? For the moment, I'm going to reflect on your answer!
That's not much different from calculation of the total cross section from partial wave amplitudes.
tom.stoer
#45
Apr3-12, 10:52 AM
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Hey guys, we have a thread for oscillations of neutrinos or mass-eigenstates which are allowed) and we have this thread for superpositions of "charge eigenstates" which are forbidden due to superselection rules. I think here we should focus on the latter one.
francesco85
#46
Apr3-12, 11:19 AM
P: 62
Quote Quote by DrDu View Post
That's not much different from calculation of the total cross section from partial wave amplitudes.
Eh?I don't understand. Can you write an explicit formula for the phase space factor in which a linear combination of two particles with different masses is produced, please (and also the scattering cross section in terms of the invariant amplitude)? Or can you give a reference?
Moreover I think I didn't understand your comparison with the emission of radiation: let me be precise and distinguish mass eigenstates from energy eigenstates: if I have understood your comparison, the role of the observable "energy" in your example is played by "mass" in my interpretation; but the two situations are different, from my point of view. In your case you mean that a superposition of energy eigenstates is possible (and I agree with this), but what I stress is that only mass eigenstates can be produced: notice that a linear superposition of energy eigenstates of photons is still a mass eigenstate. Is it wrong or did I misunderstand your comparison? In this case why?
One more final question, which might be helpful to clarify my point of view: suppose we have the standard model with a right handed neutrino and we add a Yukawa term analogous to that of the quark. After the electroweak symmetry breaking and diagonalization of the mass matrices, what is the difference between quarks and neutrinos? Nobody have doubts that in calulating effective low energy operators from high energy contribution (a very awful expression to indicate all contribution to hadronic state which are deduced by quark interactions, very very very roughly speaking; e.g. the mixing of the k kbar system already cited) mass eigenstates should be used. What is different in the case of neutrinos?
nonequilibrium
#47
Apr3-12, 01:11 PM
P: 1,412
Quote Quote by strangerep View Post
No, you don't need full QFT to understand this stuff. Just basic QM, some group theory, and some understanding of how the two fit together.

Let's hit the reset button and start again...

In fact, let's put the subject of isospin and the proton/neutron thing to one side for a little while and make sure you have the essential prerequisites...

How much do you understand about intrinsic angular momentum (spin) in ordinary QM? E.g., do you know how to derive the result that total spin comes in integer or half-integer amounts, and that, given a specific total spin, the range of eigenvalues for a component of spin in a given direction depend on the former? E.g., for a particle of (total) spin 1/2, the possible eigenvalues for spin in the z direction are -1/2, +1/2. But for a particle of (total) spin 1, the possible eigenvalues for spin in the z direction are -1, 0, +1. Do you know how to derive these results starting from nothing more than an abstract Hilbert space and the rotation group SO(3)? If not, then it's essential to study (eg) Ballentine section 7.1. (Most QM textbooks cover this, but I'm most familiar with Ballentine's treatment.)

But if you think you do indeed know how to derive this, then compose a few paragraphs that sketch the essential ideas and steps so I can see what you're missing, if anything.
I'm familiar with the derivation yes. However, I don't recall SO(3) being mentioned (but I'm quite familiar with group theory, so no need to hold back). I took a quick look at Ballentine (happened to be my lying on my desk) and I don't see SO(3) being mentioned in the derivation either. It seems the derivation in Ballentine is similar to the one I saw in my QM class: we basically define the angular momentum operator J as something that satisfies the well-known commutation relations, and then it was stated that J˛ and J_z form a CSCO (that it's a SCO is clear, but we didn't see an argument for the "Complete" part [which, I suppose, depends on the context of the angular momentum]; maybe this is related to your SO(3) reference). One defines the classifying quantum numbers j and m such that [itex]\hat L^2 |j,m\rangle = j(j+1) \hbar^2 |j,m\rangle[/itex] and [itex]\hat J_z |j,m \rangle = m \hbar |j,m \rangle[/itex]. Since the size of the angular momentum should > 0, it follows that [itex]j \geq 0[/itex] 'although there is probably a direct proof for this). One can also prove that for a given j that m is restricted on both sides. One then defines the ladder operator [itex]\hat J_\pm := \hat J_x \pm i \hat J_y[/itex], which can be proven to decrease or increase m. Hence it follows that if we fix j, and take the maximal m for that j, that [itex]\hat J_+ |j,m \rangle = 0[/itex]. By using the commutation relations one can rewrite [itex]\hat J_- \hat J_+[/itex] in function of J˛ and J_z, such that the previous equation gives an equation in terms of a random j and its maximal m. One can repeat this for the minimal m. One can also give an argument that the distance between the maximal m and the minimal m is an integer (having to do with the ladder operators), and this fact combined with the previous two equations, leads to the conclusion (after limited arithmetic) that [itex]2j[/itex] is an integer. With some similar arguments one can also argue that [itex]-j \leq m \leq j[/itex].

But I don't think my confusion stems from me misunderstanding (regular) spin (?).
tom.stoer
#48
Apr3-12, 01:53 PM
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P: 5,366
As I said; I guess strangerep wants to discuss the states [itex]|l,l_3\rangle[/itex] and (forbidden) superpositions i.e. superselection rules for angular momentum; think about

[tex]|1,1\rangle\,+\,|1,-1\rangle[/tex]
[tex]|1,1\rangle\,+\,|2,2\rangle[/tex]
DrDu
#49
Apr3-12, 02:25 PM
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Quote Quote by tom.stoer View Post
As I said; I guess strangerep wants to discuss the states [itex]|l,l_3\rangle[/itex] and (forbidden) superpositions i.e. superselection rules for angular momentum; think about

[tex]|1,1\rangle\,+\,|1,-1\rangle[/tex]
[tex]|1,1\rangle\,+\,|2,2\rangle[/tex]
Neither of the two superpositions is forbidden by a superselection rule.
nonequilibrium
#50
Apr3-12, 02:30 PM
P: 1,412
I didn't even know there were any forbidden superpositions when it comes to regular spin...
tom.stoer
#51
Apr3-12, 02:40 PM
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Then how shall
[tex]|1,1\rangle\,+\,|2,2\rangle[/tex]
be realized in nature?

A simple definition of a superselection rule are states |1> and |2> for which <1|A|2>=0 holds for all observables A. I think this is satisfied for the above mention angular momentum eigenstates in QM.
nonequilibrium
#52
Apr3-12, 02:44 PM
P: 1,412
Not to be smart-***-y, but how would you realize e.g. [itex]|1,1 \rangle[/itex] in nature?
DrDu
#53
Apr3-12, 02:48 PM
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Quote Quote by tom.stoer View Post
Then how shall
[tex]|1,1\rangle\,+\,|2,2\rangle[/tex]
be realized in nature?

A simple definition of a superselection rule are states |1> and |2> for which <1|A|2>=0 holds for all observables A. I think this is satisfied for the above mention angular momentum eigenstates in QM.
As a chemist I would call this a pd hybrid. In a hydrogen atom you could prepare it by a pi/4 pulse of light starting from L=1, L_y=1. However, now I suspect that this is meant to be an artificial example and you only consider elements of the SO(3) algebra as observables?
tom.stoer
#54
Apr3-12, 02:50 PM
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Quote Quote by mr. vodka View Post
Not to be smart-***-y, but how would you realize e.g. [itex]|1,1 \rangle[/itex] in nature?
a massive spin-1 particle with z-component = +1, e.g. a rho meson


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