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It is often stated that the nuclear $\beta$-decay is entirely described by the single V-A hamiltonian density: $$\mathscr H _{V-A} =\frac{G_F}{\sqrt 2} \overline p (g_V + g_A\gamma ^5 )\gamma ^{\mu}n\,\overline e(1-\gamma ^5 )\gamma _\mu \nu \,+\,\text{h.c.},$$ where $n,p,e,\nu$ are the Dirac fields of the particles involved.

I've learnt how to calculate, for example, the free neutron's lifetime from this hamiltonian, using the interaction picture formalism of QFT.

However, when it comes to computing transition amplitudes beetween bound nuclear states, it seems to me that $\mathscr H _{V-A}$ can't be the whole story, since it basically describes transitions beetween free particle states.

For example, suppose that I have to calculate the partial width of: $$^{14}\text O\to \,^{14}\text N ^* + e^{+}+\nu, $$ and that the wave-functions and energies of $\text O $ and $\text N^*$ are known, given by, say, shell-model wave-functions. What is the right formalism to treat such a problem or, in other words, to describe the transition of a proton bound in $\text O$ to a neutron bound in $\text N$?

For example, is it possible to treat $p$ and $n$ (or more conveniently $N$, the nucleon field) as Schrodinger fields with a Hamiltonian density such as: $$\mathscr H _N = \mathscr H _{\text Schr.}- N ^\dagger I _3 {\Delta m}N+N ^\dagger V N, $$ where $\Delta m$ is the $p$-$n$ mass difference and $V$ a phenomenological central potential (in the context of the shell model)?

Any help/reference is appreciated, thank you.

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The following will be correct to order $0$ in ${v_{\text {nucleon}}}/{c}$.

We consider two nucleon states $$\vert N_{i,f} \rangle = \sum _{\tau ,m}\intop \phi _{i,f}(\mathbf p)\vert \mathbf p,m,\tau\rangle$$ (where $\tau$ and $m$ are the eigenvalues of $I_3$ and $S_z$). Assuming that we restrict our attention to non-relativistic states: $$|\mathbf p |\gg m_N c \implies \vert \phi _{i,f}(\mathbf p)\vert ^2\approx 0$$

we observe that the $L^2({\mathbb R ^3})\otimes \mathbb C ^2_{\text {spin}}\otimes \mathbb C ^2_{\text {isospin}}$ operator:

$$H_\text{W}=\frac{G_F}{\sqrt 2}\tau _+[g_V j^0 _L(\mathbf x)-g_A \boldsymbol \sigma \cdot \mathbf j _L (\mathbf x)]+\text {h.c.}$$ $$j^\mu (\mathbf x)=\overline e (\mathbf x)\gamma ^\mu (1-\gamma^5)\nu (\mathbf x)$$

gives a good approximation for the relativistic matrix elements: $$\langle N_f , \text {leptons}\vert H\vert N_i,\text {leptons}\rangle,$$ if we take as the nucleon spatial wave-function $$\psi (\mathbf x )=\intop \frac{\text d ^3 \mathbf p}{(2\pi)^{\frac{3}{2}}}e^{i\mathbf p \cdot \mathbf x }\phi (\mathbf p ),$$which is just the non-relativistic prescription for the position-representation of the ket.

In the presence of $A$ nucleons, the generalization of $H_{\text W}$ is clearly: $$H_\text{W}=\sum _{k=1} ^A \frac{G_F}{\sqrt 2}\tau ^{(k)} _+[g_V j^0 _L(\mathbf x ^{(k)})-g_A \boldsymbol \sigma ^{(k)} \cdot \mathbf j _L (\mathbf x^{(k)})]+\text {h.c.}.$$

Finally, if we neglige the correction to the electron's wave function due to the Coulomb interaction with the nucleus (which is actually a raw approximation), we can compute the decay rate of $^{14}\text O $ to $^{14} \text N ^*$ by the following observations:

  1. The dipole approximation, that is $j^{\mu} (\mathbf x ) \approx j^{\mu}(0)$, will be very good since the leptons wavelenghts are hundreds of $\text fm$, while the nucleons are confined within a few $\text fm$.
  2. The $^{14}\text O $ and $^{14} \text N ^*$ are part of an isospin triplet with $I_{3}=1,0$ respectively. The spin of both nuclei is $0$.
  3. For a pure $0\to 0$ Fermi transition, the matrix element is proportional to $T_\pm$, the total isospin ladder operator.
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