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

[nonequilibrium]

Hello,

So I'm reading about isospin in Griffith's Introduction to Elementary Particles, but the concept seems rather fishy, and I'm not quite sure what to make out of it.

For example, if p and n (proton and neutron) are seen as different states of the same system, then what does \frac{1}{\sqrt{2}} \left( p + n \right) possibly mean? I suppose that expression makes sense if p and n really are different states of the same system, but not if they are kind of similar. Being the same or not is not really a continuous scale. So how serious should I take things like \frac{1}{\sqrt{2}} \left( p + n \right)? And what does it mean to you?

 

[naima]

Hello,

So I'm reading about isospin in Griffith's Introduction to Elementary Particles, but the concept seems rather fishy, and I'm not quite sure what to make out of it.

For example, if p and n (proton and neutron) are seen as different states of the same system, then what does \frac{1}{\sqrt{2}} \left( p + n \right) possibly mean? I suppose that expression makes sense if p and n really are different states of the same system, but not if they are kind of similar. Being the same or not is not really a continuous scale. So how serious should I take things like \frac{1}{\sqrt{2}} \left( p + n \right)? And what does it mean to you?

p and n differ because one of their quarks is up while it is down for the other.
Have you the same question for the spin of an electron which can be up or down?

 

[nonequilibrium]

p and n differ because one of their quarks is up while it is down for the other.
Have you the same question for the spin of an electron which can be up or down?

So the "upness" and the "downness" of a quark are analogous to the "upness" and the "downess" of the spin of an electron? This was not communicated to me.

Do you have a reference for that?

Thank you for the reply.

 

[naima]

have you read this page ?

 

[Vanadium 50]

Did Griffiths really write that? I'm surprised, because it's not in an eigenstate of T3, which means it doesn't commute with the Hamiltonian.

 

[nonequilibrium]

Thank you both for replying.

@ naima: I have now, but the confusion remains, it has merely shifted to the quark level: if u and d quarks can (approximately) be seen as two states of one system, then an expression like \frac{1}{\sqrt{2}}(u + d) should make sense. Does it? And if so, what does it mean?

@ Vanadium: I'm not following what you're saying. Are you asking me whether Griffiths wrote down the expression} \frac{1}{\sqrt{2}}(p+n)? If so: no he did not. But he did say that p and n can approximately be seen as two states of one system, so that one can define p as isospin up, also written down as \left( \begin{array}{c} 1 \\ 0 \end{array} \right); analogous for the neutron. But if so, an expression like \frac{1}{\sqrt{2}}(p+n) should make sense. Same question as above: does it? And if so, what does it signify?

 

[francesco85]

Hello! My opinion is the following: I think that there are two kinds of considerations that one could make (I will define the proton and the neutron as the so called mass eigenstates):

1- let us take QCD (without taking into account the electroweak symmetry): in the particular limit in which the "vector isospin" symmetry is exact, then the proton and the neutron are undistinguishable (as well as every other combination): it is just a matter of definition what you define the proton and what the neutron.

2- in the limit in which the "vector isospin" symmetry is broken (e.g. by different quark masses, electroweak corrections and possibly other effects), the proton and the neutron are physically different (for example they have different masses); in this case one can of course make all the linear combinations that can be made but, in my opinion, they do not correspond to physical observable states: they are not mass eigenstates.

Then I think that the only way to give meaning to \frac{1}{\sqrt{2}}(p+n) is in the QCD with the exact "vector isospin" symmetry; if the world was described by such a theory, I think that the way to identify p, n and every possible linear combination is just by experiments: one define experimentally p and n by giving a prescription of preparation of these states (and, in turn, definite results for the experiments, roughly speaking); then every linear combination can be seen by the result of the experiment (I actually don't know whether, given two experimental prescriptions for preparing two states, it is possible to give a prescription for preparing a linear combination..)

A question to Vanadium50: I don't understand your comment: why is T3 important? T1 and T2 commute with the hamiltonian (I suppose you are talking of the QCD hamiltonian): the combination \frac{1}{\sqrt{2}}(p+n) is an eigenstate of T1 (perhaps apart from a sign); why isn't it allowed, in your opinion?

 

[Vanadium 50]

You can't pick and choose which parts of an idea like isospin to accept. There are two elements: the total isospin of the system and the third component, T3. You can write down whatever combination that you like, but only states of definite T3 (unlike p+n) commute with the Hamiltonian and thus are realized in nature.

 

[francesco85]

You can't pick and choose which parts of an idea like isospin to accept. There are two elements: the total isospin of the system and the third component, T3. You can write down whatever combination that you like, but only states of definite T3 (unlike p+n) commute with the Hamiltonian and thus are realized in nature.

Why T3 and not T1? Suppose what you have said is true: then I build a theory in which I label my states not with eigenstates of T3 but as eigenstates of T1; of course the two ways of labelling the "base vectors" are possible; there is also a relationship among the two set of states: in the case in which we label the states with some quantum numbers + T3 the "base vectors" are something like |qn,t3=±1/2>, where qn are some other quantum numbers (energy, momentum, etc.) necessary to identify the state, while in the case in which we label the states with some quantum numbers+T1 the "base vectors" are something like |qn,t1=±1/2>; moreover the relation among the two sets of states is something like
|qn,t1=±1/2>=(|qn,t3=+1/2>±|qn,t3=-1/2>)(1/sqrt(2));
and both this sets are eigenstates of the hamiltonian.
So, if I chose T1 in order to label the states, why does nature realize only the T3 eigenstates? In my opinion, what you have said is false, in pure QCD.

ps (edit) : states that commute with the hamiltonian? What does it mean?

 

[naima]

Hello,

So I'm reading about isospin in Griffith's Introduction to Elementary Particles, but the concept seems rather fishy, and I'm not quite sure what to make out of it.

I would answer to this point.
Isospin is kept when you have strong interaction (you unplug the electromagnetic force).
you can use it to compute cross section.
look at
(1) p + n -> d + pi⁰ and
(2) p + p -> d + pi⁺
the system p + p is in a pure state with I = 1 while the system p + n is in a statistical superposition (with equal weight) of I = 1 and I = 0. So half of the mixture may interact to keep I equal to 1.
the experience give a partial cross section = 3,15 mb for (2) and 1,5 mb for (1)
You can see that the ratio is close to 2 as it would be if the symmetry was exact.

 

[Vanadium 50]

Why T3 and not T1?

Because T3 commutes with Q, and Q commutes with H, so T3 commutes with H.

In my opinion, what you have said is false, in pure QCD.

And we don't live in a pure QCD universe.

Why are you needlessly complicating this?

 

[francesco85]

Because T3 commutes with Q, and Q commutes with H, so T3 commutes with H.

You didn't answer my question: also T1 commutes with H (what is Q?), otherwise the hamiltonian would not be symmetric under the vector isospin symmetry. T3 or T1 are just ways to label the states, it does not have nothing to do with physics and to what nature should do.

And we don't live in a pure QCD universe.

Why are you needlessly complicating this?

So you are saying that even if we don't live in a pure QCD universe, T3 is a symmetry of the nature (and the masses of the quarks?) and moreover states are classified according to T3 (edit: not T1 or T2) and nature acts in such a way that we observe only such states?

I recall what is my point of view, from the first post from mine, in this thread:
in the case in which we take into account the "real and complete" theory, the isospin symmetry is broken and so we cannot classify the states according to the isospin. So neutron and proton are different particles. The fact that they have nearly equal masses, etc. is a signal that the isospin symmetry is almost exact. In this case we don't observe p+n and other combinations simply because they are not mass eigenstates of the theory. Nothing to do, in my opinion, with T3, T2 or T1.

What I have remarked in the post you quoted is valid only in pure QCD, as I have written: it is only in that case that it's "meaningful" to speak of isospin, in my opinion.Francesco

 

[Vanadium 50]

Q is charge.

Your postings are adding a lot of confusion to the mix. The OP's initial problem has a well-defined answer: you need to consider total isospin and it's third component together for the idea to be useful.

Yes, you can also ask what isospin looks like in a universe without electromagnetism. But a) that's not the world we live in, and b) that's not what the OP asked.

 

[nonequilibrium]

Hm a lot of replies (for which my thanks), but let me for the moment focus on the one that caught my attention:

There are two elements: the total isospin of the system and the third component, T3. You can write down whatever combination that you like, but only states of definite T3 (unlike p+n) commute with the Hamiltonian and thus are realized in nature.

So only eigenstates of T3 are realized in nature? Why is this (focussing on the word "eigenstates", not so much "T3")? I see two options:
1) By "realizing in nature" you mean "being a result of a measurement" in which case I agree, but that doesn't mean the p+n state is not allowed as a state of the system, so my question remains unanswered;
2) The concept of isospin is a more formal notion than for example spin, and actually only eigenstates are defined, unlike for example the concept of spin, where "spin up + spin down" states make sense.

I'm hoping for (2).

 

[francesco85]

Q is charge.

Your postings are adding a lot of confusion to the mix. The OP's initial problem has a well-defined answer: you need to consider total isospin and it's third component together for the idea to be useful.

Yes, you can also ask what isospin looks like in a universe without electromagnetism. But a) that's not the world we live in, and b) that's not what the OP asked.

Dear Vanadium 50,
first, please, read what one asks; second, answer the questions posed without repeating ad libitum the same thing you have written in your first post; third, you should be more precise, in my opinion.Q is the charge of what symmetry? T3? The total isospin? T1? Something else?

Then I recall some questions you didn't answer.

QCD case (in the exact isospin limit)- You have made a very precise statement: some states are not realized in nature because they are not T3 eigenstates: why T3 and not T1? Doesn't T1 commute with the QCD hamiltonian, in the limit considered? Why is T3 important and not T1? How does the real world described by that theory know something about T3, T1 or group theory?
(See my second post for the example)

Real world (isospin broken) - Is it meaningful to classify the states according to isospin in a theory which is not invariant under the isospin?

I repeat that my answer to the question (may it be right or wrong) is the following: the state p+n is not an eigenstate of the hamiltonian "of the real world" and so it cannot be prepared (for example through a scattering experiment,where only mass eigenstates can be preparaed); is this right or wrong? Let's discuss; tell me your opinion: is there a right/wrong part in the sentence I have made? In such a case, which parts are right and which are wrong?

(EDIT:in the case in which the original question is posed in the framework of pure QCD, I have given my interpretation in my first post in this thread; the same questions can arise: is it right/wrong? In this case, why?)This is why I have asked you why T3 is so important. This is why I have asked the other question.Francesco

 

[tom.stoer]

I think the resolution is not isospin but electric charge. Both up and down (or proton and neutron) have different electric charge for which no superposition of states with different charge are known.

 

[nonequilibrium]

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.

 

[tom.stoer]

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?

Yes, that's my intention.

However, that's not really an argument, but more of a restating of the fact that we do not see a p+n state.

No, one can go further; I think it has something to do with electric charge as a conserved quantity due to a continuous symmetry and the induced superselection sectors. But I am not sure about that.

 

[Vanadium 50]

T3 is electric charge. (Up to a constant)

The reason this is useful is that everything one learned about spin angular momentum l in eigenstates of m can be applied to isospin, where you have isospin T in eigenstates of T3. Or charge. (If you're in an eigenstate of one, you're not in an eigenstate of another)

As an example, this allows you to calculate the branching fraction rho -> pi0 + pi0. In isospin space, {T,T3} is {1,0} = {1,0} + {1,0}. For spin, the C-G coefficient for this is zero, and so it must be for isospin. So this decay is forbidden. Voila!

 

[francesco85]

Q

T3 is electric charge. (Up to a constant)

The reason this is useful is that everything one learned about spin angular momentum l in eigenstates of m can be applied to isospin, where you have isospin T in eigenstates of T3. Or charge. (If you're in an eigenstate of one, you're not in an eigenstate of another)

As an example, this allows you to calculate the branching fraction rho -> pi0 + pi0. In isospin space, {T,T3} is {1,0} = {1,0} + {1,0}. For spin, the C-G coefficient for this is zero, and so it must be for isospin. So this decay is forbidden. Voila!

T3 is not the electric charge up to a constant:
T_3|p>=\frac{1}{2}|p>
T_3|n>=-\frac{1}{2}|n>

(they are eigenstates of T3 just because we "label" our states with T3).

Moreover, what does the example you have given mean? That in processes mediated by QCD in the limit in which isospin is conserved, isospin is really conserved ?

To mr. vodka: sorry for having been unclear, I will try to explain better my thoughts here: as far as I know, in scattering processes one usually takes mass eigenstates as incident and outgoing free particles; with mass eigenstates I mean eigenstates of the operator P^2 (à la Weinberg, as far as I have understood) (sorry for my confusing notation; I often use energy and mass eigenstates as synonimous, since usually one takes as "base states" the state parametrized by P^{\mu}); for example, if we do not take mass eigenstates as final product, how can we define the phase space? As far as I know, I think it's not possible. Notice that I am talking about scattering processes (it's the only way in which I know particles can be produced). If you ask how the state p+n can be seen or if it can be produced in other methods or simply how it looks, actually I don't know how to answer this question, since this is not a mass eigenstate and I don't know how to produce this kind of particles (I'm sorry for that :). If, however, we restrict to scattering theory in which in the end of a process we would like to see only definite particles (in the sense meant by Weinberg: "mass,energy, momentum,spin and spin along z"-eigenstates, if I remember correctly ), I am quite sure that we ought to see only mass eigenstates.
What I am quite safe to say is the following: in the limit in which we consider only QCD with exact isospin, I think that labelling the states is just a matter of convention (at least in scattering processess) : this is the same kind of thoughts I have exposed in the first post I have sent in this thread.

I basically agree with tom.stoer (maybe with a slightly different motivation, if I have understood what he has said), in the sense that electromagnetism can be one of the keys: even if I don't take into account the superselection rules (which I don't know and so I don't want to talk about), what is sure is that the proton and the neutron have different charges and so they interact differently with the photon (so, this can make us distinguish the neutron from the proton, measurable thing); moreover electromagnetism is the (or one of the) responsible for the neutron and the proton to have different mass and so make them distinguishable. Of course, also in this case, I am just talking about production of free particles in a scattering process. Maybe there is something deeper in seeing that these states are actually electric charge eigenstates, but actually I cannot see what; maybe this can be a point of contact with what tom said (still in the hypothesis that I have understood what he has said) ;hope this discussion will clarify my point of view.


Francesco

 

[Vanadium 50]

T3 is not the electric charge up to a constant:
T_3|p>=\frac{1}{2}|p>
T_3|n>=-\frac{1}{2}|n>

Of course it is, as your example points out. In that example, Q = T_3 + 1/2.

 

[samalkhaiat]

Hello,

So I'm reading about isospin in Griffith's Introduction to Elementary Particles, but the concept seems rather fishy, and I'm not quite sure what to make out of it.

For example, if p and n (proton and neutron) are seen as different states of the same system, then what does \frac{1}{\sqrt{2}} \left( p + n \right) possibly mean? I suppose that expression makes sense if p and n really are different states of the same system, but not if they are kind of similar. Being the same or not is not really a continuous scale. So how serious should I take things like \frac{1}{\sqrt{2}} \left( p + n \right)? And what does it mean to you?

To good approximation, the strong (nuclear) forces are independent of the electric charge carried by NUCLEONS. The strong interaction is invariant under transformations which interchange proton and neutron. That is, it has an SU(2) isospin symmetry in which the p and n states form an iso-doublet. The group structure here is very similar to that of the usual spin. That p and n form a doublet means that
T_{3}p = \frac{1}{2}p, \ \ T_{3}n = -\frac{1}{2}n
and
T_{-}p = n, \ \ T_{+}n = p
That the strong force does not distinguish p from n means that the strong Hamiltonian commutes with the three generators of the SU(2) isospin symmetry.

Sam

 

[nonequilibrium]

@ Sam: Thanks for trying to help, but you're just telling me things that I already know. How does it relate to my question pertaining to the p+n superposition?

@ francesco: Thanks for trying to clear that up, I appreciate it. You're using some notation I'm not yet familiar with, so you're probably presuming more knowledge on my side than I have. I'm a last year undergrad, currently taking a Griffiths level introductory course in particle physics. I don't know what you mean by P^2, unless you mean the square of the parity operator, but that seems pretty unrelated to mass, so I'm sure it's not that. I googled "mass eigenstates" and what I found was that the mass operator squared is \hat H^2 - \hat p^2 (in natural units) which is of course pretty logical come to think of it. And I see p+n is not an eigenstate. So I suppose the question that remains is: why don't we allow non-mass energy eigenstates to (freely) exist?

I'm not sure how this rhymes with what tom.stoer is saying though, despite you obviously seeing a parallel yourself. Tom.stoer's argument seems to be that a p+n state cannot exist (freely?) as it is not a charge eigenstate, and the latter can --if I understand him correctly-- be justified, having "something to do with electric charge as a conserved quantity due to a continuous symmetry and the induced superselection sectors. But I am not sure about that."

So according to one argument p+n cannot (freely?) exist because it's not a mass eigenstate, according to the other argument because it's not a charge eigenstate. Both still leave the question open as to why these are prerequisites, and maybe they have a common explanation, which is perhaps what francesco meant by "I basically agree with tom.stoer".

I hope I have interpreted everybody correctly.

 

[lpetrich]

Seems like you may mean something like
(1/sqrt(2))*(|pn> + |np>)
and
(1/sqrt(2))*(|pn> - |np>)
where the proton and neutron exchange their identities or flavors.

Mathematically, this works out much like 3D angular momentum, where the proton and neutron are analogous to the spin orientations +1/2 and -1/2 of a spin-1/2 particle. Thus, the term "isotopic spin" or "isospin" for that interrelation.

Isospin multiplets for a proton/neutron pair:

|1,1> = |pp>
|1,0> = (1/sqrt(2))*(|pn> + |np>)
|1,-1> = |nn>
|0,0> = (1/sqrt(2))*(|pn> - |np>)

 

[nonequilibrium]

No, I mean what I typed: \frac{1}{\sqrt{2}}(u+p).

 

[lpetrich]

So you are talking about an isolated particle that is a mixture of proton and neutron states?

X = (1/sqrt(2))*(|p> + |n>)

It would *not* have a well-defined electric charge, for starters. Consider the charge operator Q. To have a well-defined charge, a physical state must be an eigenstate of Q.
Q|p> = |p>
Q|n> = 0
But
Q.X = (1/sqrt(2))|p>
is not a multiple of X!
Thus, X is not an eigenstate of Q.

 

[samalkhaiat]

@ Sam: Thanks for trying to help, but you're just telling me things that I already know. How does it relate to my question pertaining to the p+n superposition?

What I told you means the following: If you observe the nucleons by means of electromagnetic interactions, they will not behave as superposition of n and p. This is because the em interactions break the isospin symmetry and give rise to the p-n mass difference; you cannot make a superposition of states with different masses. However, the strong interaction of the nucleons does not care about this mass difference and the nucleons behave like
|N\rangle = a|p\rangle + b|n\rangle

 

[nonequilibrium]

@lpetrich: Yes, we established that above, but why does it have to have a well-defined charge? After all we appear to live in a world where systems don't have to have, for example, a well-defined energy or momentum or etc. Why exclude the same possibility for charge? That requires an argument.

@sam: "you cannot make a superposition of states with different masses" but why? If it is not allowed from the get-go, then the whole p \leftrightarrow \left( \begin{array}{c} 1 \\ 0 \end{array} \right) and n \leftrightarrow \left( \begin{array}{c} 0 \\ 1 \end{array} \right) business seems to lose its footing, since if you don't allow superpositions, then how do you know you can still apply the whole spin formalism, which is crucial for the isospin concept?

 

[samalkhaiat]

To put two objects in the same representation of some symmetry group Like SU(2), they need to have the same mass; only mass eigenstates can be put in superposition. The strong force DOES NOT CARE ABOUT THE MASS DIFFERENCE between n and p so we can put them in the same multiplet and consider their superposition.

 

[Vanadium 50]

@lpetrich: Yes, we established that above, but why does it have to have a well-defined charge?

Because charge is absolutely conserved.

The only way to have one nucleon in a superposition of different states of different charge is to have another nucleon in the orthogonal superposition, so you are still in an eigenstate of Q and T3.

The same thing can happen with ordinary spin. I can have two particles with spins S1 and S2, neither of which is in an eigenstate of Sz, but the total spin is in an eigenstate of Sz. This is another reason why isospin is used - there is an exact analogy between what is going on with spin and with isospin.

 

[tom.stoer]

you cannot make a superposition of states with different masses[/tex]

of course you can; look at neutrino oscillations or at the Kaon system

 

[tom.stoer]

So you are talking about an isolated particle that is a mixture of proton and neutron states?

X = (1/sqrt(2))*(|p> + |n>)

It would *not* have a well-defined electric charge

Yes, and I think this is forbidden by a superselection rule

 

[DrDu]

Yes, and I think this is forbidden by a superselection rule

Exactly!
So probably it would be better to consider a superposition of p+e and n., or, more generally,
p +e \leftrightarrow \left( \begin{array}{c} 1 \\ 0 \end{array} \right) and n \leftrightarrow \left( \begin{array}{c} 0\\1\end{array} \right)

 

[nonequilibrium]

For some reason the answers given aren't getting to me. I suppose I'll have to study a course in QFT before I'll be able to understand it(?)

 

[strangerep]

For some reason the answers given aren't getting to me. I suppose I'll have to study a course in QFT before I'll be able to understand it(?)

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.

 

[tom.stoer]

@strangerep, you should give mr vodka a hint what comes next; the states |l,l_3\rangle and superpositions like

|1,1\rangle\,+\,|1,-1\rangle
|1,1\rangle\,+\,|2,2\rangle

 

[DrDu]

Exactly!
So probably it would be better to consider a superposition of p+e and n., or, more generally,
p +e \leftrightarrow \left( \begin{array}{c} 1 \\ 0 \end{array} \right) and n \leftrightarrow \left( \begin{array}{c} 0\\1\end{array} \right)

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]

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; P^2=P^{\mu}P_{\mu}=h^2-\vec{p}^2 (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 Poincarè 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 Poincarè group are "defined" by mass and spin; you may ask why we build irreducible representations of the Poincarè group; well, this is related to Poicarè 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

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?

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

Thank you very much!
Francesco

 

[DrDu]

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]

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]

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: https://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 occurred 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]

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]

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]

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]

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]

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]

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 J² 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 \hat L^2 |j,m\rangle = j(j+1) \hbar^2 |j,m\rangle and \hat J_z |j,m \rangle = m \hbar |j,m \rangle. Since the size of the angular momentum should > 0, it follows that j \geq 0 '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 \hat J_\pm := \hat J_x \pm i \hat J_y, 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 \hat J_+ |j,m \rangle = 0. By using the commutation relations one can rewrite \hat J_- \hat J_+ in function of J² 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 2j is an integer. With some similar arguments one can also argue that -j \leq m \leq j.

But I don't think my confusion stems from me misunderstanding (regular) spin (?).

 

[tom.stoer]

As I said; I guess strangerep wants to discuss the states |l,l_3\rangle and (forbidden) superpositions i.e. superselection rules for angular momentum; think about

|1,1\rangle\,+\,|1,-1\rangle
|1,1\rangle\,+\,|2,2\rangle

 

[DrDu]

As I said; I guess strangerep wants to discuss the states |l,l_3\rangle and (forbidden) superpositions i.e. superselection rules for angular momentum; think about

|1,1\rangle\,+\,|1,-1\rangle
|1,1\rangle\,+\,|2,2\rangle

Neither of the two superpositions is forbidden by a superselection rule.

 

[nonequilibrium]

I didn't even know there were any forbidden superpositions when it comes to regular spin...