Isospin: how serious must I take it? Superposition of proton and neutron?by nonequilibrium Tags: isospin, neutron, proton, superposition 

#1
Mar3112, 06:59 PM

P: 1,406

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 [itex]\frac{1}{\sqrt{2}} \left( p + n \right)[/itex] 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 [itex]\frac{1}{\sqrt{2}} \left( p + n \right)[/itex]? And what does it mean to you? 



#2
Apr112, 02:57 AM

PF Gold
P: 305

Have you the same question for the spin of an electron which can be up or down? 



#3
Apr112, 07:12 AM

P: 1,406

Do you have a reference for that? Thank you for the reply. 



#4
Apr112, 07:27 AM

PF Gold
P: 305

Isospin: how serious must I take it? Superposition of proton and neutron?
have you read this page ?




#5
Apr112, 08:04 AM

Mentor
P: 15,569

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
Apr112, 09:21 AM

P: 1,406

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 [itex]\frac{1}{\sqrt{2}}(u + d)[/itex] 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} [itex]\frac{1}{\sqrt{2}}(p+n)[/itex]? 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 [itex]\left( \begin{array}{c} 1 \\ 0 \end{array} \right)[/itex]; analogous for the neutron. But if so, an expression like [itex]\frac{1}{\sqrt{2}}(p+n)[/itex] should make sense. Same question as above: does it? And if so, what does it signify? 



#7
Apr112, 09:28 AM

P: 62

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 [itex]\frac{1}{\sqrt{2}}(p+n)[/itex] 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 [itex]p[/itex], [itex]n[/itex] and every possible linear combination is just by experiments: one define experimentally [itex]p[/itex] and [itex]n[/itex] 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 [itex]\frac{1}{\sqrt{2}}(p+n)[/itex] is an eigenstate of T1 (perhaps apart from a sign); why isn't it allowed, in your opinion? 



#8
Apr112, 10:14 AM

Mentor
P: 15,569

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
Apr112, 10:29 AM

P: 62

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? 



#10
Apr112, 10:43 AM

PF Gold
P: 305

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 + pi^{0} 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
Apr112, 10:48 AM

Mentor
P: 15,569

Why are you needlessly complicating this? 



#12
Apr112, 11:08 AM

P: 62

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 



#13
Apr112, 12:10 PM

Mentor
P: 15,569

Q is charge.
Your postings are adding a lot of confusion to the mix. The OP's initial problem has a welldefined 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
Apr112, 12:27 PM

P: 1,406

Hm a lot of replies (for which my thanks), but let me for the moment focus on the one that caught my attention:
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
Apr112, 12:35 PM

P: 62

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 



#16
Apr112, 01:34 PM

Sci Advisor
P: 5,307

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
Apr112, 02:22 PM

P: 1,406

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
Apr112, 03:10 PM

Sci Advisor
P: 5,307




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