Isospin breaking, charge symmetry and charge indepedence

In summary, isospin symmetry refers to the conservation of total isospin in strong interactions, while isospin breaking occurs when the full Hamiltonian is not invariant under isospin transformations. Charge symmetry means that the interaction remains the same when protons and neutrons are swapped, while charge independence means that the interaction is the same for all combinations of proton-proton, proton-neutron, and neutron-neutron. The nuclear force is isospin breaking and charge symmetric, as it does not conserve total isospin but is still symmetric under charge transformations.
  • #1
AntiElephant
25
0
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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  • #2
AntiElephant said:
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
 
  • #3
samalkhaiat said:
AntiElephant said:
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.

samalkhaiat said:
AntiElephant said:
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.

AntiElephant said:
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.

AntiElephant said:
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.
 
  • #4
Bill_K said:
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]
 

What is isospin breaking and why is it important in physics?

Isospin breaking is a phenomenon observed in particle physics where the interactions between particles with the same isospin but different charges are not exactly the same. This is important because it helps us understand the differences between particles and their interactions, and can provide insights into the fundamental forces of nature.

What is charge symmetry and how does it relate to isospin breaking?

Charge symmetry is a fundamental symmetry in physics where the properties of particles and their interactions are the same regardless of their charge. However, isospin breaking violates this symmetry, as particles with different charges have different interactions. This is because isospin is a property that is only conserved in strong interactions, but not in electromagnetic interactions.

What is charge independence and why is it important in nuclear physics?

Charge independence is a concept in nuclear physics where the properties of a nuclear system are independent of the charge of its constituents. This means that the interactions between protons and neutrons in a nucleus are the same regardless of their individual charges. It is important because it allows us to simplify calculations and understand the structure and behavior of nuclei.

How do we experimentally observe isospin breaking, charge symmetry, and charge independence?

There are several ways to observe these phenomena experimentally, such as measuring the mass differences between particles with the same isospin but different charges, studying the decay patterns of particles, or analyzing scattering experiments. These experiments can provide evidence for the breaking of symmetries and help us understand the underlying physical processes.

Can isospin breaking, charge symmetry, and charge independence be unified into one theory?

Currently, there is no unified theory that can fully explain the relationships between isospin breaking, charge symmetry, and charge independence. However, efforts are being made to develop a unified theory that can describe all of these phenomena and their connections to the fundamental forces of nature. This is an ongoing area of research in particle and nuclear physics.

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