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13

1) The axial vector current $j^{\mu 5}$ is a pseudovector $$j^{\mu 5}~:=~\overline{\psi}\gamma^{\mu}\gamma^5\psi~=~j^{\mu}_R-j^{\mu}_L,\qquad j^{\mu}_{R,L}~:=~ \overline{\psi}_{R,L}\gamma^{\mu}\psi_{R,L}, $$ $$\psi_{R,L}~:=~P_{R,L}\psi,\qquad P_{R,L} ~:=~\frac{1\pm\gamma^5}{2} . $$ The $4$-divergence $d_{\mu}j^{\mu 5}$ is a pseudoscalar. That the axial ...


10

As far as I understand, due to conservation of angular momentum, the resulting system of neutral pions would need to have angular momentum 1, therefore, the identical neutral pions would be in an anti-symmetric state, which does not seem possible as they are bosons. Note that a neutral rho meson can decay into two neutral pions and a $\gamma$, although this ...


9

[...] $\Delta^+ \rightarrow p + \pi^0$, [...] $\Delta^+ \rightarrow n + \pi^+$, which process is favored: the proton and neutral pion or neutron and charged pion [?] Since the kinematics (and corresponding "phase space" factors) for the two final states are presumably as good as equal, the evaluation of the branching ratio $$\text{BR} := \frac{\Gamma[ ...


9

In a travesty of overlapping historical notations, we have both strong isospin, under which the neutron and proton are the two projections of the nucleon and are raised and lowered by the pion, and weak isospin, in which the left-handed parts of the $(u,d)$, $(e,\nu_e)$, $(c,s)$, $(\mu, \nu_\mu)$, $(t,b)$, and $(\tau,\nu_\tau)$ doublets are raised and ...


9

It's actually the other way around. That axial rotation of the pions ensures that, given its non-vanishing v.e.v., given by the condensate (assumed to be produced by QCD: a fact!), they therefore must be the Goldstone modes of the SSB of the axial charges. But, first, the chiral condensate is required so as to unleash all this. Take L=2, for simplicity, and ...


9

The QCD and QED themselves conserve parity. The conclusion of this statement is that all corresponding effective vertices must conserve the parity. The only coupling of $\pi^{0}$ to $\gamma$ conserving the parity is $$ L_{\pi^{0}} \simeq \frac{\pi^{0}}{\Lambda}\epsilon^{\mu\nu\alpha\beta}F_{\mu\nu}F_{\alpha\beta}, $$ which doesn't allow your decay process $\...


9

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7

They have 2 main differences. The first of them is very straightforward: They have different spins: As you pointed out, both are bound states of 2 spin 1/2 particles, therefore you can find the possible spins of such a bound state using the usual rules of angular momentum addition in Quantum Mechanics. 1/2 + 1/2 can give you either 3 spin 1 states (the ...


7

To toot my own horn a little1 may I suggest the series of papers published by the $f_\pi$ collaboration from Hall C at Jefferson Lab: Determination of the pion charge form factor for $Q^2=0.60-1.60 \,\mathrm{GeV^2}$, 2007 DOI:10.1103/PhysRevC.75.055205 and arXiv:nucl-ex/0607007 Charged pion form factor between $Q^2=0.60$ and $2.45 \,\mathrm{GeV^2}$. I. ...


7

Within a few MeV, from isospin symmetry of the strong interaction, it is similar to the antiproton proton annihilation which happens even at rest: Role of Delta exchange for proton-antiproton annihilation into two-pion and three-pion channels . No problem of energy conservation due to the much smaller masses of the pions to the two nucleons. This is an ...


7

You seem to be misunderstanding what the effective theory with the residual nuclear force between hadrons mediated by pions is: It is an effective theory which is a good approximation of the underlying nuclear force between quarks mediated by gluons. It is not that one of these is correct and the other false, or that they represent two different forces - ...


6

Just summing together all the comments and providing some more explicit calculations, we have conservation of four-momentum (which is the amalgamation of the conservation of energy and conservation of momentum), we have: $$p^{\mu}_{1}+p_{2}^{\mu}=p_{1}'^{\mu}+p_{2}'^{\mu}+p_{\pi}^{\mu}$$ Taking the inner product of each side with itself, we get: $$\left\...


6

The main point is that the quark and the antiquark inside a charged meson must necessarily have different flavors so the decay cannot be via the flavor-preserving electromagnetic interaction, but must wait for the weak interaction. In contrast, the decay of a neutral pion needs no flavor change and can use the much faster electromagnetic interaction.


6

You might consider 241162 and 102575. This is actually a history of science question, on a half-century old landmark in the intellectual history of 20th century physics, whose importance cannot be overstated. Perhaps it belongs to another site. For a quartic potential, following Schwinger, the σ mass may be sent to infinity, thereby also introducing the ...


5

Because your formulae for the pion wave functions are written in a confusing way. The quark composition of the positively charged pion is $$ |\pi^+\rangle = |u\bar d\rangle $$ while the neutral pion is $$ |\pi^0\rangle = \frac{|u\bar u\rangle - |d\bar d\rangle}{\sqrt{2}}. $$ These two composite formulae are completely analogous: these pions are mesons, i.e. ...


5

The reason the signs are flipped from what you expect has to do with the fact that the antiquark transforms in the opposite way under isospin rotations. If the ordinary quark doublet is a column vector $$q=(u, d)^T$$ and transforms under rotations as $$q\rightarrow U(R) q$$ the antiquark doublet is a row vector $$\bar{q}=(\bar{u}, \bar{d})\rightarrow \bar{q} ...


5

In the pion reference frame the two outgoing leptons are very boosted, hence helicity and chirality almost coincide. The angular momentum conservation forces them to have opposite spins, since the pion spin is zero. Therefore, they will have the same helicity, which is highly suppressed in this kinematic regime, because of the vector nature of the QED ...


5

You've concluded that, to conserve parity, $L$ must be odd. By the two pions are identical bosons, and so the wavefunction must by symmetric under exchange. If $L$ is odd, the wavefunction is anti-symmetric under exchange, and so this is forbidden.


4

The mass (a relativistic invariant quantity) of the $\pi^0$ is well measured and is not consistent with the mass of positronium (also well measured). By orders of magnitude, so no the neutral pion is not an electron positron pair.


4

None of them. All of them (and the rho and eta, too). It doesn't matter and doesn't have a unique answer as the meson exchange idea is an effective theory. If you insist on trying to write down rules, then charged pion exchanges can only occur between protons and neutrons resulting in the nucleons exchanging types. On the other hand neutral pions can be ...


4

I think I know where your misconception lies-- you appear to think that the individual quarks are the real thing, so you wonder why the $\pi^{+}$, made up of ''pure'' states of $|{\bar u}\rangle$ and $|d\rangle$, could decay into a ''mixed'' state. The problem here is that the physical thing is not the quarks, themselves, but the quark field. And this ...


4

How can we measure meson decay constants? I am not an experimental physicst, but I think that the best way to obtain the decay constant is to study processes like $\pi^+\to \mu^+ \nu$ and extract them from the branching ratio: $$\rm{Br}(\pi^+\to \mu^+\nu)=\dfrac{G_F^2 m_{\pi^+} m_\mu^2}{8 \pi}\left(1-\dfrac{m_\mu^2}{m_{\pi^+}^2} \right)^2 f_{\pi^+}^2 |V_{...


4

Since $\pi^0$ is a pseudoscalar particle, we have $$\langle 0|J^\mu_{em}|\pi^0 \rangle =0,$$ and the pion cannot decay into two leptons with a simple photon exchange. In the Standard Model, the leading-order contributions for this process are a box diagram and a $Z^0$ exchange, as you can see in fig. 1 of arXiv:0806.4782 (replacing a $c$ quark by a light ...


4

Here is the diagram you are discussing: It seems you are worried by the angular momentum carried by the W+. The W+ is a virtual particle in this reaction. In virtual paths the particle is off mass shell and its mass is unphysical, and angular momentum as a part of a four vector will be a complicated function also having unphysical measure, so ...


4

The 3 pions can be considered as 3 states of the same particle, the isospin being used to label the 3 states. Since pions are bosons, the total wave function must be symmetric (Pauli principle). The total wave function is the (tensorial) product of space-wave function, spin wave-function and isospin wave-function. Spin wave-function is symmetric since pions ...


4

The pion is a pseudoscalar, with spin-parity $J^\pi=0^-$. In decay of a pion at rest the two decay products have equal and opposite momentum. Choose a coordinate system so that the electron goes to the left, and the antineutrino goes to the right: e <--------- π -----------> ν The weak interaction, as you say, involves particles with left-handed ...


4

This is a cluster of questions, with a very good one in the end, which is, in fact, the very question that inspired Susskind to introduce technicolor for EW SSB in the late 70s. I'll be very schematic to get the ideas across, rather than catching every factor of $\sqrt 2$ and such. Both the strong and the EW interactions work on conserved currents and gauge ...


4

The short answer is that the force carriers are mostly pions, but the details get quite complicated. In physics we know about four fundamental interactions: the strong force, the electromagnetic force, the weak force, and gravitation. The neutron participates in all of them, but feebly. Its gravitational mass is small, it has no electric charge, and its ...


4

Why the pion lives in a representation of isospin SU(2) Sure, it is well described as a part of the meson octet. That is how the quarks were discovered. Do not forget that SU(2) is a subgroup of an SU(3), in this case weak SU(3). and is the mediator of the strong force It is a mediator in some of the models of the strong nuclear force generated by ...


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