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Isospin breaking, charge symmetry and charge indepedence

by AntiElephant
Tags: charge, indepedence, isospin, symmetry
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AntiElephant
#1
May17-14, 04:43 PM
P: 20
Hi, I'm just getting a little confused with all the definitions here and I need some confirmation on what I say is correct or not;

Isospin symmetry: The property that an interaction is independent of the [itex] T_3 [/itex] value?
Isospin breaking: The property that it is dependent on [itex] T_3 [/itex]?
Charge symmetry: The property that the interaction is unchanged upon swapping protons with neutrons and neutrons with protons.
Charge independence: The interaction is the same for pp, pn and nn.

I've honestly looked but I haven't managed to get a full confirmation on what isospin symmetry actually means. Are these correct? I swear I also have two sources telling me that the nuclear force is and isn't charge symmetric. The latter case because;

1) The proton and neutron have different masses
2) They have different EM contributions.

But the nuclear force is also isospin breaking right? Especially for one pion exchange. Because the one pion exchanges for pp/nn are different for np (the former only have [itex] \pi^0 [/itex]). So is the nuclear force not charge symmetric, not charge independent, and isospin breaking?
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samalkhaiat
#2
May17-14, 07:29 PM
Sci Advisor
P: 910
Quote Quote by AntiElephant View Post
Isospin symmetry: The property that an interaction is independent of the [itex] T_3 [/itex] value?
It means that the total value of [itex]T_{3}[/itex] is conserved in the strong interaction. Or, that the strong interaction Hamiltonian commutes with the "Iso-spin charge", i.e. [itex][H_{st} , T_{ 3 } ] = 0[/itex].
Isospin breaking: The property that it is dependent on [itex] T_3 [/itex]?
This meant "in real life" the full Hamiltonian has a piece which does not commute with isospin charge. Normally it is the electromagnetic piece: [itex]H_{tot} = H_{0} + H_{em}[/itex], [itex][H_{0}, T_{3}]=0[/itex], but [itex][H_{em},T_{3}] \neq 0[/itex].

Charge symmetry: The property that the interaction is unchanged upon swapping protons with neutrons and neutrons with protons.
Which charge? [itex]T_{3}[/itex] is also called "charge".
Charge independence: The interaction is the same for pp, pn and nn.

I've honestly looked but I haven't managed to get a full confirmation on what isospin symmetry actually means. Are these correct? I swear I also have two sources telling me that the nuclear force is and isn't charge symmetric. The latter case because;

1) The proton and neutron have different masses
2) They have different EM contributions.

But the nuclear force is also isospin breaking right? Especially for one pion exchange. Because the one pion exchanges for pp/nn are different for np (the former only have [itex] \pi^0 [/itex]). So is the nuclear force not charge symmetric, not charge independent, and isospin breaking?
In real life Isospin is an approximate symmetry for the strong interaction, but good approximate symmetry, because the mass difference between n & p (or up and down quarks) is tiny and one can (as first approximation) neglect the electromagnetic interaction and treat it as exact symmetry. In the second (beter) approximation, we treat the iso-spin breaking electromagnetic Hamiltonian as a perturbing piece and do our purturbation theory.

Sam
Bill_K
#3
May18-14, 06:01 AM
Sci Advisor
Thanks
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P: 4,160
Quote Quote by samalkhaiat View Post
Quote Quote by AntiElephant View Post
Isospin symmetry: The property that an interaction is independent of the [itex] T_3 [/itex] value?
It means that the total value of [itex]T_{3}[/itex] is conserved in the strong interaction. Or, that the strong interaction Hamiltonian commutes with the "Iso-spin charge", i.e. [itex][H_{st} , T_{ 3 } ] = 0[/itex].
NO. The fact that the strong interaction is independent of T3 means that the strong interaction Hamiltonian commutes with T2, that is, total isospin.

Quote Quote by samalkhaiat View Post
Quote Quote by AntiElephant View Post
Isospin breaking: The property that it is dependent on [itex] T_3 [/itex]?
This meant "in real life" the full Hamiltonian has a piece which does not commute with isospin charge. Normally it is the electromagnetic piece: [itex]H_{tot} = H_{0} + H_{em}[/itex], [itex][H_{0}, T_{3}]=0[/itex], but [itex][H_{em},T_{3}] \neq 0[/itex].
NO. T3 is essentially electric charge. For p and n, T3 = Q - 1/2. And all of H commutes with it, including Hem. The fact that p and n have different masses, for example, is because [Hem, T] ≠ 0. The correct statement is that [Hem, T2] ≠ 0.

Quote Quote by AntiElephant View Post
I swear I also have two sources telling me that the nuclear force is and isn't charge symmetric. The latter case because;
1) The proton and neutron have different masses
2) They have different EM contributions.
The nuclear force is charge symmetric. These two effects are not about not the nuclear force, they're about the electromagnetic force.

Quote Quote by AntiElephant View Post
But the nuclear force is also isospin breaking right? Especially for one pion exchange. Because the one pion exchanges for pp/nn are different for np (the former only have [itex] \pi^0 [/itex]). So is the nuclear force not charge symmetric, not charge independent, and isospin breaking?
No. pp and nn are purely T = 1. Scattering involving np is different because it includes T = 0.

samalkhaiat
#4
May18-14, 02:27 PM
Sci Advisor
P: 910
Isospin breaking, charge symmetry and charge indepedence

Quote Quote by Bill_K View Post
NO. The fact that the strong interaction is independent of T3 means that the strong interaction Hamiltonian commutes with T2, that is, total isospin.


NO. T3 is essentially electric charge. For p and n, T3 = Q - 1/2. And all of H commutes with it, including Hem. The fact that p and n have different masses, for example, is because [Hem, T] ≠ 0. The correct statement is that [Hem, T2] ≠ 0.
To understand your "No" I want you write down the stronge interaction Hamiltonian, i.e. [itex]H_{N \pi}[/itex], and show me how it commutes with [itex]T^{2}[/itex] and NOT with Isospin charge [itex]T^{a}[/itex]?

[tex]T^{a}= \int d^{3} x N^{ \dagger } \frac{ \sigma^{ a } }{ 2 }N + \epsilon^{ a b c } \phi^{ b } \dot{ \phi }^{ c } \ \ [/tex]

The relation [itex]Q_{e} = B / 2 + T^{ 3 }[/itex] is correct in the lowest order, i.e. when you ignore the em interaction and it follows from the [itex]SU(2)[/itex] symmetric Lagrangian
[tex]
\mathcal{ L } = i \bar{ N } \gamma^{ \mu } \partial_{ \mu } N + (1 / 2 ) ( \partial_{ \mu } \phi^{ a } )^{ 2 } + i g \bar{N} \gamma^{5} \sigma . \phi N / 2
[/tex]
which is also [itex]U_{B}(1)[/itex] invariant with conserved B-charge given by
[tex]B = \int d^{3} x N^{\dagger} N[/tex]


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