Photon-induced low-energy nuclear reactions

Abstract

We propose a new mechanism for inducing low-energy nuclear reactions (LENRs). The process is initiated by a perturbation which we assumed to be caused by absorption or emission of a photon. Due to the electromagnetic perturbation, the initial two-body nuclear state forms an intermediate state to make a transition into the final nuclear state through the action of another perturbation. In the present paper, we take the second perturbation to be also electromagnetic. We need to sum over all energies of the intermediate state. Since the upper limit on this sum is infinity it is possible to get contributions from very high energies for which the barrier penetration factor is not too small. By considering a specific reaction, we determine the conditions under which this mechanism may lead to significantly enhanced reaction rates. We find that the mechanism leads to very small cross-sections in free space. However, in a condensed medium, there exist several possibilities leading to enhanced cross-sections, which may lead to observable reaction rates even at relatively low energies. Hence we argue that LENRs are possible and provide a theoretical set-up which may explain some of the experimental claims in this field.

Appendix

In this appendix, we provide details of the angular integrals in the sum over energies in the intermediate state. We consider the initial state with $S=1/2$, $S_z=1/2$ and $l=0$. The intermediate state has $l=1$. With $S=1/2$ and $l=1$ we can form two states with $j=3/2$ and $j=1/2$. The states $|j,j_z\rangle$ that are relevant for us are

$$\begin{aligned}&|3/2,3/2\rangle = |1/2,1/2;1,1\rangle ,\nonumber \\&|3/2,-1/2\rangle = \sqrt{1/3}\ |1/2,1/2;1,-1\rangle \nonumber \\&\qquad \qquad \qquad \qquad + \sqrt{2/3}\ |1/2,-1/2;1,0\rangle ,\nonumber \\&|1/2,-1/2\rangle = -\sqrt{2/3}\ |1/2,1/2;1,-1\rangle \nonumber \\&\qquad \qquad \qquad \qquad + \sqrt{1/3}\ |1/2,-1/2;1,0\rangle , \end{aligned}$$

(47)

where on the right-hand side the states are shown in notation $|S,S_z;l,l_z\rangle$. Let us take the final photon direction to be along the z-axis, i.e. ${\hat{k}}_f = {\hat{z}}$. The corresponding polarisation vectors are along the ${\hat{x}}$- and ${\hat{y}}$-axes. Let us consider any one of the amplitudes appearing in eq. (32),

$$\begin{aligned} {{\mathcal {M}}}_1 = \langle n|\mathrm {e}^{i\vec k_1\cdot \vec {r}} \vec \epsilon \,'_{\beta '}\cdot \vec {p}|i\rangle . \end{aligned}$$

(48)

Since the initial state has $l=0$, we can express this as

$$\begin{aligned} {{\mathcal {M}}}_1 \equiv \int \mathrm {d}^3 r \psi ^*_n \mathrm {e}^{i\vec k_1\cdot \vec {r}} \vec {\epsilon }\,'_{\beta '}\cdot {\hat{r}} p_r\psi _i. \end{aligned}$$

(49)

We have $\vec {\epsilon }\,'_1\cdot {\hat{r}} = \sin \theta \cos \phi$ and $\vec {\epsilon }\,'_2\cdot {\hat{r}} = \sin \theta \sin \phi$. The intermediate wave function $\psi _n$ corresponds to $l=1$. We get non-zero contributions only for $\psi _n\sim Y_1^1$ and $\psi _n\sim Y_1^{-1}$. Hence, we have four relevant combinations involving the two polarisation vectors and the two spherical harmonics. Let us consider one of these angular integrals, given by

$$\begin{aligned} I_1\equiv & {} \int \mathrm {d}\Omega Y_1^{1*}({\hat{r}}) \mathrm {e}^{i\vec k_1\cdot \vec {r}} \vec {\epsilon }\,'_1\cdot {\hat{r}}\nonumber \\= & {} -{1\over 2} \sqrt{3\over 2\pi } \int \mathrm {d}\Omega \sin \theta \mathrm {e}^{-i\phi } \mathrm {e}^{ik_1r\cos \theta }\sin \theta \cos \phi .\nonumber \\ \end{aligned}$$

(50)

In performing the integral over the polar angle $\theta$ we keep only the leading-order contribution. We obtain

$$\begin{aligned} I_1= \left( {1\over 2}\sqrt{3\over 2\pi }\, \right) 4\pi i\ {\sin k_1r\over k_1^2 r^2}. \end{aligned}$$

(51)

Here we have dropped terms of higher order in $1/k_1 r$, which will be small since $k_1$ takes rather large values. Similarly, we have the remaining integrals

$$\begin{aligned}&I_2 \equiv \int \mathrm {d}\Omega Y_1^{1*}({\hat{r}}) \mathrm {e}^{i\vec k_1\cdot \vec {r}} \vec {\epsilon }\,'_2\cdot {\hat{r}} = - iI_1 \end{aligned}$$

(52)

$$\begin{aligned}&I_3 \equiv \int \mathrm {d}\Omega Y_1^{-1*}({\hat{r}}) \mathrm {e}^{i\vec k_1\cdot \vec {r}} \vec {\epsilon }\,'_1\cdot {\hat{r}} = - I_1 \end{aligned}$$

(53)

and

$$\begin{aligned} I_4 \equiv \int \mathrm {d}\Omega Y_1^{-1*}({\hat{r}}) \mathrm {e}^{i\vec k_1\cdot \vec {r}} \vec {\epsilon }\,'_2\cdot {\hat{r}} = -iI_1. \end{aligned}$$

(54)

We next consider the final-state matrix element

$$\begin{aligned} {{\mathcal {M}}}_2 \equiv \langle f|\vec \epsilon \cdot {\hat{r}}|n\rangle . \end{aligned}$$

(55)

Let the initial photon momentum be

$$\begin{aligned} \vec k_i = k_i\left[ \sin \theta _i(\cos \phi _i{\hat{x}} +\sin \phi _i {\hat{y}}) + \cos \theta _i{\hat{z}}\right] . \end{aligned}$$

(56)

The corresponding polarisation vectors can then be taken as

$$\begin{aligned} \vec \epsilon _1= & {} \sin \phi _i{\hat{x}} - \cos \phi _i{\hat{y}}\nonumber \\ \vec \epsilon _2= & {} \cos \theta _i (\cos \phi _i{\hat{x}} + \sin \phi _i{\hat{y}}) - \sin \theta _i{\hat{z}}. \end{aligned}$$

(57)

By a suitable choice of coordinates, we can set $\phi _i=0$. In any case, it will cancel out in the final result. We again find four relevant angular integrals in ${{\mathcal {M}}}_2$. These are

$$\begin{aligned} I'_1\equiv & {} \int \mathrm {d}\Omega '\ \vec {\epsilon }_1\cdot {\hat{r}}'\ Y_1^1({\hat{r}}') = \left( {1\over 2}\sqrt{3\over 2\pi }\, \right) {4\pi \over 3} i\mathrm {e}^{i\phi _i}\nonumber \\ I'_2\equiv & {} \int \mathrm {d}\Omega '\ \vec {\epsilon }_1\cdot {\hat{r}}'\ Y_1^{-1}({\hat{r}}') = -I_1^{\prime *}\nonumber \\ I'_3\equiv & {} \int \mathrm {d}\Omega '\ \vec {\epsilon }_2\cdot {\hat{r}}'\ Y_1^{1}({\hat{r}}') = i\cos \theta _i I_1^{\prime }\nonumber \\ I'_4\equiv & {} \int \mathrm {d}\Omega '\ \vec {\epsilon }_2\cdot {\hat{r}}'\ Y_1^{-1}({\hat{r}}') = i\cos \theta _i I_1^{\prime *}. \end{aligned}$$

(58)

Let us now consider the case in which the initial and final polarizstion vectors are $\vec \epsilon _1$ and $\vec \epsilon \,'_1$. In this case, adding contributions from all the states in eq. (47), we obtain

$$\begin{aligned}&{{\mathcal {M}}}_2{{\mathcal {M}}}_1 = -4\pi i \sin \phi _i\int \mathrm {d}r' r'^2 R_f(r')r'R_n(r') \nonumber \\&\quad \int \mathrm {d}r\, r^2 R_n(r) {\sin k_1r\over k_1^2 r^2}p_rR_i(r), \end{aligned}$$

(59)

where $R_i$, $R_n$ and $R_f$ are the radial wave functions.