Fermi and Gamow-Teller Transition

February 29, 2016

GoLuckyRyan Basic , , , , , , , , , , Leave a comment

The beta decay is caused by the weak interaction. The weak interaction is very short range, because the mediate particles, the W^{\pm} and Z^0 bosons are 80 GeV and 91 GeV respectively. The effective range is like 10^{-3} fm. So, the interaction can assumed to be a delta function and only the coupling constant matter. The Fermi coupling constant is

1.17 \times 10^{-11} (\hbar c)^2~ \mathrm{MeV^2}

The fundamental process of beta decay is the decay of quark.

\displaystyle u \xrightarrow{W^+} d + e^+ + \nu_e

Since a pion is made from up and down quark, the decay of pion into position and electron neutrino is also due to weak interaction.

The Hamilton of the beta decay is

\displaystyle H_w(\beta^{\pm})=G_V \tau_{\mp} + G_A \sigma \tau_{\mp}

where G_V is the vector coupling constant, the term is called Fermi transition. The \tau_{\pm} is the isospin ladder operator. The beta+ decay changes the isospin from +1/2 (neutron) to -1/2 (proton). The G_A is the axial coupling constant, the term is called Gamow-Teller transition. \sigma is spin operator. Because of this operator, the Gamow-Teller transition did not preserve parity.

The G_A is different from G_V, which is caused by the effect of strong interaction. The Goldberger-Trieman relation

\displaystyle g_A = \frac{G_A}{G_V} = \frac{f_\pi g_{\pi N}}{M_N c^2} = -1.3

where f_\pi \sim 93~\textrm{MeV} is the pion decay constant. g_{\pi N} \sim 14 \times 4\pi is the coupling constant between pion and nucleon.  This, we can see the effect of the strong interaction, in which pion is the meson for strong nuclear force.

 

The transition probability can be estimated by Fermi-Golden rule

\displaystyle W(p_e)=\frac{2\pi}{\hbar}|\left< \psi_f|H|\psi_0\right> |^2 \rho(E_f)

the final state wavefunction

\displaystyle \left|\psi_f\right> = \frac{1}{\sqrt{V}} e^{ik_e r} \frac{1}{\sqrt{V}} e^{ik_{\nu}r} \left|j_f m_f\right>

\displaystyle e^{ikr} = \sum \limits_{L}\sqrt{4\pi (2L+1)} i^L j_L(kr) Y_{L0}(\theta)

using long wavelength approximation, the spherical Bessel function can be approximated by the first term.

\displaystyle j_L(kr) \sim \frac{(kr)^L}{(2L+1)!!}

\displaystyle \left| \psi_f\right>=\frac{1}{V}(1 + i \sqrt{\frac{4\pi}{3}} Y_{10} + ...) \left|j_f m_f\right>

The first term 1, or L=0 is called allowed decay, so that the orbital angular momentum of the decayed nucleus unchanged. The higher order term, in which the weak interaction have longer range has very small probability and called L-th forbidden decay.

The density of state is

\displaystyle \rho(E_f) = \frac{V}{2\pi^2 \hbar^7 c^3} F(Z,E_e)p_e^2 (E_0-E_e) ( (E_0-E_e)^2-(m_{\nu} c^2)^2)^2

where the F(Z, E_e) is the Fermi function.

The total transition probability is the integration with respect to the electron momentum.

\displaystyle W = \int W(p_e) dp_e = \frac{m_e^5 c^4}{2 \pi^3 \hbar^7} f(Z,E_0) |M|^2

where f(Z,E_0) is the Fermi integral. The half-life

\displaystyle T_{1/2} = \frac{\ln{2}}{W}

To focus on the beta decay from the interference of the density of state, the ft-value is

\displaystyle ft = f(Z,E_0) T_{1/2} =\frac{2\pi^3\hbar^7}{m_e^5 c^4} \frac{\ln{2}}{|M|^2}

The ft-value could be difference by several order.

There is a super-allowed decay from 0^{+} \rightarrow 0^{0} with same isospin, which the GT does not involve. an example is

\displaystyle ^{14}\mathrm{O} \rightarrow ^{14}\mathrm{N} + e^+ + \nu_e

The ft-value is 3037.7s, the smallest of known.

Fermi Gamow-Teller
\Delta S=0 \Delta S=1
J_f=J_i + L J_f=J_i + L+1
T_f=T_i + 1

 

transition L \log_{10} ft_{1/2} \Delta J \Delta T \Delta \pi
Fermi GT
Super allowed 3.1 ~ 3.6 0^+ \rightarrow 0^+ not exist 0 no
allowed 0 2.9 ~ 10 0 (0), 1 0, 1 ; T_i=0\rightarrow T_f=0 forbidden no
1st forbidden 1 5 ~ 19 (0),1 0, 1, 2 0,1 yes
2nd forbidden 2 10 ~18 (1), 2 2, 3 no
3rd forbidden 3 17 ~ 22 (2), 3 3, 4 yes
4th forbidden 4 22 ~ 24 (3), 4 4, 5 no

The () means not possible if either initial or final state is zero. i.e 1^{-} \rightarrow 0^+ is not possible for 1st forbidden.

 

decay time constant and line width

December 20, 2010

GoLuckyRyan advance, Basic, theory, Working on , , , , , , , , Leave a comment

the spectrum of energy always has a peak and a line width.

the reason for the line width is, this is decay.

i give 2 explanations, once is from classical point of view and i skipped the explanation for the imaginary part. so, i am not fully understand. the 2nd explanation is look better, but it is from QM. however, there is one hide question for that explanation is, why the imaginary energy is negative?

the simplest understanding of the relation is using fourier transform. (i think)

Fourier transform is changing the time-frame into the frequency frame. i.e, i have a wave, propagating with frequency w. we can see a wave shape when plot with time. and we only see a line, when we plot with frequency, since there is only 1 single frequency. however, for a general wave, it is composite of many different frequencies, using fourier transform can tell us which frequency are involved. And energy is proportional to frequency.

when the particle or state under decay. the function is like

f(t) = Exp(-R t) Exp ( i \omega_0 t)

where the R is decay constant, and ω0 is the wave frequency.

after fourier transform, assume there is nothing for t < 0

F(t) = \frac {1} { R + i ( \omega_0 - \omega )}

the real part is

Re(F(t)) = \frac {R} { R^2 + ( \omega_0 - \omega )^2}

which is a Lorentzian shape and have Full-Width-Half-Maximum (FWHM) is 2R. it comes from the cosine part of the fourier transform. thus, the real part.

and the imaginary part is

Im(F(t)) = \frac {\omega_0 - \omega}{R^2 + ( \omega_0 -\omega )^2 }

the imaginary part is corresponding to the since part, so, we can neglect it. (how exactly why we can neglect it? )

Thus, we can see, if there is no decay, R → 0, thus, there is no line width.

therefore, we can see the line width in atomic transition, say, 2p to 1s. but there are many other mechanism to the line width, like Doppler broadening, or power broadening. So, Decay will product line width, but not every line width is from decay.

**********************************

another view of this relation is from the quantum mechanics.

the solution of Schroedinger equation is

\Psi (x,t) = \phi(x) Exp \left( - i \frac {E}{\hbar} t \right)

so, the probability conserved with time, i.e.:

|\Psi(x,t)|^2 = |\Psi (x,0)|^2

if we assume the energy has small imaginary part

E = E_0 - \frac {i} {2} R \hbar

( why the imaginary energy is nagative?)

|\Psi(x,t)|^2 = |\Psi (x,0)|^2 Exp ( - R t)

that make the wavefunction be :

\Psi (x,t) = \phi (x) Exp( - i \frac {E}{\hbar} t ) Exp( - \frac {R}{2} t )

what is the meaning of the imaginary energy?

the wave function is on time-domain, but what is “physical”, or observable is in Energy -domain. so, we want Psi[x,E] rather then Psi[x,t], the way to do the transform is by fourier transform.

and after the transform, the probability of finding particle at energy E is given by

|\Psi(x,E)|^2 = \frac {Const.}{R^2 +(\omega_0 - \omega )^2}

which give out the line width in energy.

and the relation between the FWHM(line width) and the decay time is

mean life time ≥ hbar / FWHM

which once again verify the uncertainty principle.

decay

December 19, 2010

GoLuckyRyan Basic, Porperties, theory , , Leave a comment

the decay idea and mathematic is simple. so, i just state it.

Number of particle (time) = Initial # of particle × Exp( – time / T )

or in formula

N(t) = N(0) Exp \left( - \frac {t} {T} \right )

where T is time constant, which has a meaning that how long we should wait before it decay. T also has another name, “mean-lifetime“, coz when you find out the mean of their life by usually statistical method, integrate the whole area of the graph of decay time and make it equal to initial # of particle × “mean lifetime”. that is what you got. ( \int_0^\infty Exp( - \frac {t}{T}) dt= T )

some people like to write the equation is other way:

N(t) = N(0) Exp \left( - R t \right )

where R is the chance of decay in unit time. which is just the “invert” meaning of T.

we also have “Half-Life” t_\frac {1}{2} , which is the time that only half of the particle left. by the equation, we have:

t_\frac {1}{2} = ln(2) T

thus, a longer T, the particle live longer, as what is the T mean!

But above mathematics only tell us the statistic result of the decay, not about the mechanism, or physics of what cause the decay happen. why there is decay? why particles come out from nucleus? how many kind of decay ?

the easiest question is, there are 3 decay happen in nature and a lot more different decay happened in lab. the reason for only 3 decay is that, only these 3 live long enough to let us know. the other, they decay fast and all of them are done.

and the reason for nucleus decay is same as the reason for atomic decay. excited nucleus is unstable (why?) they will emit energy to become stable again.

and the physics behind decay, we will come back to it later.