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

1. Mar 31, 2012

### 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?

2. Apr 1, 2012

### naima

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?

3. Apr 1, 2012

### nonequilibrium

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?

4. Apr 1, 2012

### naima

5. Apr 1, 2012

Staff Emeritus
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.

6. Apr 1, 2012

### nonequilibrium

@ 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?

7. Apr 1, 2012

### 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?

8. Apr 1, 2012

Staff Emeritus
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.

9. Apr 1, 2012

### francesco85

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 someting 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?

Last edited: Apr 1, 2012
10. Apr 1, 2012

### naima

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.

11. Apr 1, 2012

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

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

Why are you needlessly complicating this?

12. Apr 1, 2012

### francesco85

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.

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.

Best,
Francesco

Last edited: Apr 1, 2012
13. Apr 1, 2012

Staff Emeritus
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.

14. Apr 1, 2012

### nonequilibrium

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

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).

15. Apr 1, 2012

### francesco85

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.

Best,
Francesco

Last edited: Apr 1, 2012
16. Apr 1, 2012

### 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.

17. Apr 1, 2012

### 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.

18. Apr 1, 2012

### tom.stoer

Yes, that's my intention.

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.

19. Apr 1, 2012

Staff Emeritus
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!

20. Apr 1, 2012

### francesco85

Q

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.

Best,
Francesco

Last edited: Apr 1, 2012
21. Apr 1, 2012

Staff Emeritus
Re: Q

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

22. Apr 1, 2012

### samalkhaiat

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

23. Apr 1, 2012

### 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.

24. Apr 1, 2012

### 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>)

25. Apr 1, 2012

### nonequilibrium

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