Re: Why is it rare for two deuterium nuclei to form helium-4 in fusion reaction

First of all, excellent question.  Not many students your age have read
enough about nuclear reactions or given them enough detailed thought to ask
such a question.  It turns out that this is one of the reactions I started
studying in graduate school, a project that was taken over by one of my
best friends.  It's difficult to measure the D + D -> 4He + gamma reaction
near stellar energies.  A 24 MeV gamma-ray is a very high-energy photon, so
it's easy to pick it out from other photons, but the huge flux of neutrons
swamps the large Sodium Iodide (NaI) scintillation spectrometers we use to
measure it.  You can google search any terms above for further information,
or refer to the textbook "Radiation Detection and Measurement" by Knoll.

Yes, generally nature favors the more energetically stable solution.  Given
that simple fact, one would indeed expect the reaction to proceed to the
most stable situation, 4He + gamma.  However, the details of the reaction
are governed by quantum mechanics.  In a full calculation of the reaction
rates, there are four "exit channels."  One is 3He + n, one which you
neglected above is 3H + p  (3H is tritium, which as you probably know is a
radioactive isotope of hydrogen commonly used in biological research and
hydrogen bombs), and a third is 4He + gamma.  The fourth "exit channel" is
actually by far the most common at low enegires, simple scattering (d + d
-> d + d).  In a quantum mechanical calculation, the probability of the
reaction ending in one of these four situations (assuming the collision
energy is low enough that the production of high-energy particles like
mesons and such is impossible) must sum to 1.  The incident particles are
described by wave functions, their interaction by a potential function, and
the outgoing particles and photons are described by different sets of wave
functions.  The potential function will create a mathematical operator
(here we slide into quantum mechanics) which describes the interaction
between the incoming and outgoing particles for each of the four
situations.  An integral (called an overlap integral) over the two wave
functions with the mathematical operator between them determines the
probability of the incoming particles ending up in one of the four outgoing
states.  When one actually does the math, the probability of particle
emission vs. gamma-ray emission (which is described by classical
electromagnetic potential operators rather than strong nuclear force
potential operators) is generally around 10,000 to 1.  That's just an order
of magnitude guesstimate, every reaction is different.  The difference
comes largely from the relative strengths of the strong nuclear force to
the electromagnetic force at that range when construction the reaction's
mathematical operator.  In the case of d + d -> 4He + gamma, however, we
have a special situation.  Normally, simple electric dipole radiation
dominates many reactions.  But it's quantum mechanically disallowed by
angular momentum conservation-like rules, so the system is forced to
radiate by electric quadrupole radiation instead.  This radiation is much
weaker than dipole radiation.  When electromagnetic radiation is broken up
into multipole components, the lowest orders (dipole, quadrupole, octupole)
dominate.  That's something you'll learn about in advanced E&M classes in
college.  So in this case, the gamma-emission is even more suppressed than
usual and the probability of proton or neutron emission is staggeringly
high relative to the gamma emission.

Some questions you might have:  What is a wave function?  A particle is
also a wave, in quantum mechanics.  It travels and diffracts and reacts
just like a wave.  The shape of that wave can be described by a
mathematical function.  How do we do that?  That's an entire first month's
worth of quantum mechanics courses.  The potentials and operators and
details of how the reactions proceed are further details best saved for a
formal course on quantum mechanics, though David Griffiths wrote and
excellent text on the subject from which you could pretty much teach
yourself (it's called, creatively enough, "Quantum Mechanics" and has a
picture of a live cat on the front and a dead one on the back).  The same
author wrote a good E&M textbook for undergraduates which explains
electromagnetic multipoles.

There's a lot of information here, and a lot of roads for you to follow to
a complete answer to your question.  And there are some strange mysteries
in this reaction, not usually talked about.  For example:  p-wave capture
strength at low energies, meaning with just one unit of angular momentum in
the incoming particles, is observed in the angular spread of the outgoing
radiation when polarized deuterons are used...but this incoming wave's
reaction probability is supposedly disallowed by simple quantum mechanical
rules.  We don't know the exact form of the strong nuclear force yet, one
of the last great frontiers of nuclear physics.  A complete answer would
take you all the way through my friend Konstantin's doctoral dissertation
(published just last year, but the paper isn't out yet), so don't expect to
learn it all overnight.  Good luck!