%     CAN ``SOLID-STATE'' EFFECTS ENHANCE THE COLD FUSION RATE?
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%           J. Leggett and G. Baym (Urbana, Illinois)
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%           Submitted to Nature, April 24, 1989
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\c{\bf{CAN ``SOLID-STATE'' EFFECTS ENHANCE THE COLD FUSION RATE?}}
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\c{A. J. Leggett and G. Baym}
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\c{\it{Department of Physics, University of Illinois at Urbana-Champaign}}
\c{\it 1110 W. Green St., Urbana, Illinois 61801}
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\c{Submitted to {\it Nature}, April 24, 1989}
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    {\bf Recently two groups$^{1,2}$ have reported finding experimental
evidence for an unexpectedly high rate of nuclear fusion at room temperature
during the process of electrolytic deposition of deuterium on
palladium$^{1,2}$ and titanium.$^2$ To achieve the rate of neutron production
($\bf\sim$10$^{\bf -23}$/sec/pair) cited in ref. 2 (and {\bfit a
fortiori}$\,$ that
inferred in ref. 1) requires the solid-state environment to produce either an
unusual enhancement of the fusion reaction rate, or a large suppression of the
Coulomb barrier between deuterons -- the latter presumably arising from some
kind of sophisticated many-body screening effect in Pd and Ti, perhaps
associated with quasiparticles of large effective mass.$^2$ In this note we
point out that within the framework of the lowest-order Born-Oppenheimer
approximation a very severe constraint is imposed on {\bfit all}$\,$ such
enhanced
screening mechanisms in solids in equilibrium by observable behavior of a
$^4$He atom in the metal in question; and that unless the latter is quite
anomalous, no enhancement of the Coulomb barrier penetration anywhere near the
magnitude required to explain the fusion rates inferred from the experiments
is possible in a solid in equilibrium.}

    The theoretical rate, $\Gamma$, of neutron production by fusion per second
per deuteron pair is of the form $\Gamma=Aa^{-3}e^{-B},$ where $A$ is the rate
of the fusion reaction, d+d $\rightarrow$ $^3$He+n, per unit probability
density at ${\bf r} = 0$, ${\bf r}$ is the relative separation of the two
deuterons, $a^3$ is approximately the volume of classically accessible motion
for the pair, and $e^{-B}$ is a Coulomb barrier penetration factor.$^{3-5}$
The fusion rate is $\simeq 1.5\times10^{-16}$cm$^3$/sec; a generous lower
order-of-magnitude limit on $a$ in any solid-state environment is 0.1\AA, and
thus the experiments would require $B < 80$.  While for the D$_2$ molecule in
free space$^5$ $B \simeq$ 175, both theory$^4$ and experiment$^6$ suggest that
much smaller values can be obtained when the deuterons are bound together by a
muon or other heavy particle, which effectively screens the nuclear Coulomb
charges from one another down to very small distances.  A heavy electron
quasiparticle is unlikely to produce the required screening on sub-Angstrom
scales, since such quasiparticles have an extended structure in space; the
effective mass of the electrons on sub-Angstrom scales is the bare electron
mass.  On the other hand, one might suspect the presence of short-range
screening effects associated with the large electron densities in the metal.

    The essential point of our argument is that if the effective repulsion
of two deuterons at short distance is substantially weakened by solid-state
effects, then these effects should lead as well to greatly increased binding
of an $\alpha$ particle to the metal.  To make the argument precise, we
define, within the lowest-order Born-Oppenheimer approximation, the effective
``potential energy'' $V({\bf R}, {\bf r})$ of two deuterons placed in the
metal at ${\bf R} + {\bf r}/2$ and ${\bf R} - {\bf r}/2$ respectively.  [We
assume that two additional electrons are added to the metal as well, to
preserve overall charge neutrality; since these electron energies cancel out
in the argument we do not consider them explicitly.]
We split $V$ into three
contributions:  (a) the direct Coulomb interaction, $e^2/r$, of the two
deuterons with each other, (b) the Coulomb interaction, $V_c({\bf R}+{\bf
r}/2)+V_c({\bf R}-{\bf r}/2)$, of the two deuterons with the ``environment,''
{\it i.e.}, with all other charges, electronic and ionic, and (c) everything
else, {\it i.e.}, the kinetic energies and mutual interactions of the
environment, which we call K.  Here
$$V_c({\bf R}) = e\int d\,^3r{{\rho({\bf r})}\over{\vert {\bf r}-{\bf
R}\vert}},\eqno(1)$$
where $\rho({\bf r})$ is the charge density (expectation value) of the
environment.

\footline={\hss\folio\rm\hss}

    Consider now the ``chemical energy'' $U({\bf R}+{\bf r}/2)$ of an $\alpha$
particle placed at {\bf R}+{\bf r}/2, {\it i.e.}, its potential energy
relative to that
of an $\alpha$ particle placed at infinite distance from the metal.  If we
use
the exact environment wavefunction for the ground state of the two deuteron
problem as a variational wavefunction for the $\alpha$-particle problem,
noting that the $\alpha$ has charge $2e$, we see that $U({\bf R}+{\bf r}/2)\le
K+2V_c({\bf R}+{\bf r}/2)$, with a similar inequality at ${\bf R}-{\bf r}/2$.
Adding the two inequalities we find, for all ${\bf R}$ and ${\bf r}$, that
$$V({\bf R},{\bf r})\ge{1\over2}[U({\bf R}+{{\bf r}\over2})+U({\bf R}-{{\bf
r}\over2})]
+{e^2\over r}.\eqno(2)$$

    In the limit of zero temperature, which we consider initially,  the
actual
energy eigenvalue $E_{2d}$ of the deuteron pair is bounded above by the energy
of two deuterons at infinite separation in the metal, which is $-2(E_d+K_d)$,
where $E_d$ = 1Ry is the binding energy of the deuteron in free space (Ry is
the Rydberg), and $K_d$ is its affinity to the metal.  Similarly, $U({\bf
R}\pm{\bf r}/2)$ is bounded below, for all {\bf R}, {\bf r}, by
$-(E_4+K_4+E^{\alpha}_{zp})$, where $E_4=5.802$Ry and $K_4$ are, respectively,
the free-space binding energy and affinity of the $^4$He atom, and
$E^{\alpha}_{zp}$ is the zero-point energy of the $\alpha$-particle in the
metal, {\it i.e.}, its energy relative to the minimum value of the
potential $U({\bf R})$.  Inserting these inequalities into (2) we
find
$$V({\bf R}, {\bf r})-E_{2d}\ge {{e^2}\over r} - \lambda{{e^2}\over
a_0},\eqno(3)$$
where $\lambda = {1\over 2}[E_4+K_4+E^{\alpha}_{zp}-2(E_d+K_d)]/{\rm Ry}$, and
$a_0$ is the usual Bohr radius. The inequality (3) is rigorous and applies to
arbitrary
position of the two deuterons, whether in the bulk or on the surface.

    Now clearly the exponent $B$ for penetration to ${\bf r}$ = 0 is bounded
below by its value for the potential on the right side of (3).  Further, it is
adequate for the present purpose to evaluate the latter by the simple WKB
approximation which gives a lower limit to the barrier penetration.$^4$
Thus, taking into account the reduced mass of the deuteron pair, we can write
    $$\eqalignno{B\ge B_0(\lambda)& \equiv 2\int^{a_0/\lambda}_0
dr\left({{M_de^2}\over{\hbar^2}}\right)^{1/2}\left({1\over r}
-{{\lambda}\over{a_0}}\right)^{1/2} \cr
&=\pi\left({{M_d}\over{m_e}}\right)^{1/2}\lambda^{-1/2}
\simeq 189 \lambda^{-1/2},&(4) \cr}$$
where $m_e$ is the mass of the electron and $M_d$ that of the deuteron.

    The deuterium affinity, $K_d$, is$^7$ 0.17 Ry in Pd and presumably of
similar size in Ti;
 to the best of our knowledge
there is no direct measurement of the helium atom affinity, $K_4$, for Pd or
Ti, but the fact that $^3$He desorbs from Ti at room temperature$^8$ suggests
that, as for other metals, $K_4$ is either positive or very small ($\ll$ 1
eV).  Similarly the zero-point energy of the $\alpha$ particle is not
measured, but it should be bounded above by
that of the deuteron; the latter can be estimated from the isotope-dependence
of K for the hydrogens$^7$ to be $\ll$ 1 eV.  If we neglect $K_4$ and
$E^{\alpha}_{zp}$ we find $\lambda\simeq 1.78$ and hence $B\ge 141$.  This
value corresponds to a fusion rate of order $10^{-50}$/sec per deuteron pair,
some 27 orders of magnitude below that inferred in ref. 2. To reduce $B$ to
the maximum value ($\sim$80) consistent with the fusion rates in ref. 2 would
require $K_4+E^{\alpha}_{zp}$ to be of order 100 eV, a quite incredible value,
which would have other easily observable consequences.  The bound given here
is likely orders of magnitude above the exact barrier penetration probability,
as one can see by applying the method to the free $D_2$ molecule, where it
gives a rate $\sim10^{-53}$/sec, some 10 orders of magnitude too fast.

    Unless the potential $V({\bf R}, {\bf r})$ is itself affected
on a scale of eV by room temperature, which seems most unlikely, a simple
extension of the above analysis to finite temperature with a
Maxwell-Boltzmann (or Bose) distribution for the deuterons shows that the
lower bound on $B$ at room temperature is negligibly different from that
quoted above.

    While the present argument cannot, strictly speaking, at present exclude
mechanisms which rely essentially on the presence of large numbers of
``third-party'' deuterons (since there is no measurement to date of helium
affinities under these conditions), nor those which rely on some unspecified
highly non-equilibrium energy distribution which cannot be mimicked by a
thermal distribution with a temperature of order room temperature, once such a
mechanism is specified in detail arguments similar to the above could be used
to estimate its maximum efficacy.  If the experimental results are confirmed,
it will be urgent to try to extend these arguments beyond the lowest
Born-Oppenheimer approximation.

    This research has been supported in part by NSF Grant DMR88-18713 and
through the MacArthur Professorship endowed by the John D. and Catherine T.
MacArthur Foundation at the University of Illinois.
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\frenchspacing
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\bigskip
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4) C. D. Van Siclen and S. E. Jones, {\it J. Phys. G} {\bf 12} (1986) 213\\
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7) R. L\"asser and G. L. Powell, {\it Phys. Rev. B}{\bf 34} (1986) 578.\\
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