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\heading{E4550: Chemical potential: reaction in ideal quantum gases}
\auname{Shay Izbitski}
{\bf The problem:}
\Dn
${N}$ fermions of type ${F}$ that have spin ${\frac{1}{2}}$ are placed in a volume ${V}$. they can decay into a boson of type ${B}$ that has spin ${0}$ and a fermion of type ${A}$ that has spin ${\frac{1}{2}}$. the reaction is ${F\rightarrow A+B}$ and it has an energy gain of
${\epsilon_0}$ (i.e. ${A+B}$ have lower energy than ${F}$ ). the masses are $m_{F},m_{A},m_{B}$ respectively.
\Dn
(1) Assuming ideal gases at temperature ${T}$, write the equations
which determine the densities ${n_{F}, n_{A}, n_{B}}$ in
equilibrium.
\Dn
(2) Write the equations of (1) at ${T=0}$ and describe the
densities as functions of ${\epsilon_{ 0}}$. Find $\epsilon_c$
such that for $\epsilon_0>\epsilon_c$ the number of F fermions
vanishes.
\Dn
(3) Assume that the condensation of bosons B occurs at
$T_c$ such that $T_c\ll \frac{p_F^2}{2m_A}<\epsilon_0$
where $p_F$ is the Fermi momentum of fermions A. Evaluate $T_c$
and rewrite the condition $T_c\ll \frac{p_F^2}{2m_A}$ in terms
of the given parameters.
\Dn\Dn
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{\bf The solution:}
\Dn
(1)
The particles are fermions and bosons so clearly there densities are (assuming there is no degeneracy in the boson gas):
\[
n_F=2\sum_{p}\frac{1}{e^{\beta(\frac{p^2}{2m_F}-\mu_F)}+1}
\]
\[
n_A=2\sum_{p}\frac{1}{e^{\beta(\frac{p^2}{2m_A}-\mu_A)}+1}
\]
\[
n_B=\sum_{p}\frac{1}{e^{\beta(\frac{p^2}{2m_B}-\mu_B)}-1}+n^{0}_{B}
\]
Where ${n^{0}_{B}}$ is the bose condensation related term.
%%%%%%%%%%%%%%%
\Dn
(2)
At ${T=0}$ the bosons undergo a condensation meaning that
\[
n_B=n^{0}_{B}
\]
\[
\mu_B=0
\]
The chemical potential must obey the reaction by
\[
\mu_F+\epsilon_0=\mu_A+\mu_B
\]
and at ${T=0}$
\[
\mu_A=\mu_F+\epsilon_0
\]
in addition, the fermionic density must obey the reaction by
\[
n=n_A+n_F=const
\]
therefore at ${T=0}$ the fermionic density is
\[
n=\frac{1}{3\pi^2\hbar^3}(2m_F\mu_F)^{\frac{3}{2}}+\frac{1}{3\pi^2\hbar^3}(2m_A\mu_A)^{\frac{3}{2}}
\]
From all of the above it's clear that the condition for ${n_F=0}$ is ${\mu_F<0}$ leading to
\[
0=\mu_A-\epsilon_c \rightarrow n=\frac{1}{3\pi^2\hbar^3}(2m_A\epsilon_c)^{\frac{3}{2}}
\]
The density ${\frac{n_F}{n}}$ starts from the value ${\frac{n_F}{n}=\frac{m^{\frac{3}{2}}_{F}}{m^{\frac{3}{2}}_{F}+m^{\frac{3}{2}}_{A}}}$ at ${\epsilon_0=0}$ and decrease until it vanishes at ${\epsilon_0=\epsilon_c}$.
${n_A=n-n_F}$ and therefore the density ${\frac{n_A}{n}}$ starts from the value ${\frac{n_A}{n}=\frac{m^{\frac{3}{2}}_{A}}{m^{\frac{3}{2}}_{F}+m^{\frac{3}{2}}_{A}}}$ at ${\epsilon_0=0}$ and increase until it gets the constant value 1 at ${\epsilon_0=\epsilon_c}$.
%%%%%%%%%%%%%%
\Dn
(3)
Under the given assumption of $T_c\ll \frac{p_F^2}{2m_A}<\epsilon_0$ and using the fact that
${\frac{p_F^2}{2m_A}=\mu_F+\epsilon_0}$ at ${T=0}$ we get that
\[
\frac{p_F^2}{2m_A}=\mu_F+\epsilon_0<\epsilon_0 \rightarrow \mu_F<0
\]
${\mu_F<0}$ leads us to ${n_F=0}$, leading to ${n_A=n_B=n}$ which eventually leads us to
\[
\mu_F+\epsilon_0=\frac{(3\pi^2\hbar^3n)^{\frac{2}{3}}}{2m_A}
\]
putting that back in the condition we are given gives us
\[
T_c \ll \frac{(3\pi^2\hbar^3n)^{\frac{2}{3}}}{2m_A}
\]
From the other hand it's known that since ${n_B=n}$
\[
T_c=\frac{2\pi\hbar^2n^{\frac{2}{3}}}{m_B(2.612)^\frac{2}{3}}
\]
substituting this to the former condition we get the condition in the given parameters
\[
\frac{m_B}{m_A} \ll \frac{4\pi}{(3\pi^2\cdot2.612)^{\frac{2}{3}}}=0.69
\]
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\end{document}