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\exnumber{3022}
\heading{Bosons with spin in harmonic trap}
\auname{Yuval Friedman}
{\bf The problem:}
$N$ Bosons that have spin1 are placed in a 3D harmonic trap.
The harmonic trap frequency is $\Omega$.
A magnetic field $B$ is applied, such that the interaction is $-\gamma B S_z$,
where $S_z=1,0,-1$, and $\gamma$ is the gyromagnetic ratio.
\Dn
(1) Write an expression for the density of one-particle states~$g(\epsilon)$.
\Dn
(2) Write an expression for the ${B=\infty}$ condensation temperature~$T_c$.
\Dn
(3) Write an equation for $T_c(B)$. It should be expressed in terms
of the appropriate polylogarithmic funtion.
\Dn
(4) Find the leading correction in ${T_c(B)/T_c \approx 1+\cdots}$
assuming that $B$ is very large.
It should be expressed in terms of an elementary function.
\Dn
(5) Find what is $T_c(B)/T_c$ for $B=0$, and what is the first-order
correction term if $B$ is very small.
\Dn
(6) Sketch a schematic plot of $T_c(B)/T_c$ versus $B$.
Indicate by solid line the exact dependence, and by dashed and dotted lines the approximations.
It should be clear from the figure whether the
approximation under-estimates or over-estimates
the true result, and what is the $B$ dependence of the slope.
\Dn
{\bf Tips:} The prefactors are important in this question. Do not use numerical substitutions. \\
Use the notation $L_{\alpha}(z)$ for the polylogarithmic function,
and recall that ${L_{\alpha}(1) = \zeta(\alpha)}$. \\
Note also that $L_{\alpha}'(z)=(1/z)L_{\alpha{-}1}(z)$, and that $\Gamma(n)=(n-1)!$ for integer $n$.
\Dn\Dn
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{\bf The solution:}
\Dn
\begin{itemize}
\item[(1)] For a 3D harmonic trap the energy is $E_n = (n_x+n_y+n_z)\Omega$. For a given enegy $\epsilon$ possible states should fulfill $n_x+n_y+n_z\leq \epsilon/\Omega$. The total amount of states possible:
\begin{equation}
N(\epsilon) = \int_{0}^{\epsilon/\Omega}\text{d}n_x\int_{0}^{\epsilon/\Omega-n_x}\text{d}n_y\int_{0}^{\epsilon/\Omega-n_x-n_y}\text{d}n_z=\frac{1}{6}\left(\frac{\epsilon}{\Omega}\right)^3
\end{equation}
\begin{equation}
g(\epsilon)=\frac{\text{d}N(\epsilon)}{\text{d}\epsilon} = \frac{1}{2}\frac{\epsilon^2}{\Omega^3}
\end{equation}
\item[(2)] The known result for $S_z=0$:
\begin{equation}
N=cT^{\alpha}F_{\alpha}(z)\label{eq:1}
\end{equation}
In our case: $c=\frac{1}{2\Omega^3}\text{ , }\alpha=3\text{ , } F_3(z)=\Gamma(3)\text{L}_3(z)=2\text{L}_3(z)$, so $\eqref{eq:1}$ becomes:
\begin{equation}
N=\big(\frac{T}{\Omega}\big)^3\text{L}_3(z)
\end{equation}
And the total amount of particles:
\begin{equation}
N=\big(\frac{T}{\Omega}\big)^3[\text{L}_3(z\eexp{\beta\gamma B})+\text{L}_3(z)+\text{L}_3(z\eexp{-\beta\gamma B})]\label{eq:2}
\end{equation}
Below condensation temprature $z\eexp{\beta\gamma B}=1 \implies z=\eexp{-\beta\gamma B}$. so $\eqref{eq:2}$ becomes:
\begin{equation}
N=\big(\frac{T}{\Omega}\big)^3[\text{L}_3(1)+\text{L}_3(\eexp{-\beta\gamma B})+\text{L}_3(\eexp{-2\beta\gamma B})]+\langle n_0\rangle
\end{equation}
For $B=\infty$ at $T=T_c$:
\begin{equation}
N=\left(\frac{T_c}{\Omega}\right)^3\text{L}_3(1) \implies T_c=\Omega\left(\frac{N}{\zeta(3)}\right)^{1/3}
\end{equation}
\item[(3)] The transcendental equation for $T_c(B)$:
\begin{equation}
N=\big(\frac{T}{\Omega}\big)^3[\zeta(3)+\text{L}_3(\eexp{-\gamma B/T})+\text{L}_3(\eexp{-2\gamma B/T})]\label{eq:3}
\end{equation}
\item[(4)] We want to solve $\eqref{eq:3}$ for $B\gg 1$ in leading order for $x=(T/T_c)-1$ where $T_c$ is the zero order solution.
Using $L_\alpha(z\ll 1)\approx z$ we get:
\begin{equation}
\text{L}_3(\eexp{-\gamma B/T})+\text{L}_3(\eexp{-2\gamma B/T})\approx \eexp{-\gamma B/T}+\eexp{-2\gamma B/T}\label{eq:4}
\end{equation}
$\eexp{-\gamma B/T}$ is the small parameter, so in order to simplify we will keep only first order terms (so we can neglect the last part in $\eqref{eq:4}$). Equation $\eqref{eq:3}$ becomes:
\begin{equation}
\frac{T}{T_c}\approx\left( 1+\frac{\eexp{-\gamma B/T}}{\zeta(3)}\right)^{-1/3}\implies x\approx -\frac{1}{3\zeta(3)}\eexp{-\gamma B/T_c}
\end{equation}
\item[(5)] For $B=0$:
\begin{equation}
N=\left(\frac{T_c(0)}{\Omega}\right)^3 3\text{L}_3(1) \implies T_c(0)=\Omega\left(\frac{N}{3\zeta(3)}\right)^{1/3}=3^{-1/3}T_c
\end{equation}
For $B\ll 1$ we will use the approximation:
\begin{equation}
L_\alpha(z\approx 1)\approx L_\alpha(1)+\frac{(z-1)}{z}L_{\alpha}'(z)\approx \zeta(\alpha)+(z-1)\zeta(\alpha-1)
\end{equation}
In order to simplify we take only first terms in $B$:
\begin{align}
\text{L}_3(\eexp{-\gamma B/T})+\text{L}_3(\eexp{-2\gamma B/T})&\approx 2\zeta(3)+\zeta(2)[\eexp{-\gamma B/T}+\eexp{-2\gamma B/T}-2]
\\ &\approx \nonumber 2\zeta(3)-\frac{3\gamma B}{T_c(0)}\zeta(2)
\end{align}
And for the leading order in $x=T/T_c(0)-1$ equation $\eqref{eq:3}$ becomes:
\begin{equation}
\frac{T}{T_c(0)}\approx\left( 1-\frac{\zeta(2)}{\zeta(3)}\frac{\gamma B}{T_c(0)}\right)^{-1/3}\implies x\approx\frac{1}{3}\frac{\zeta(2)}{\zeta(3)}\frac{\gamma B}{T_c(0)}
\end{equation}
\clearpage
\item[(6)]
\end{itemize}
\begin{center}
\includegraphics[scale=0.7]{C:/Users/User/Desktop/pic.png}
\end{center}
The large $B$ approximation cuts the $B=0$ axis at $1-\frac{1}{3\zeta(3)}\approx 0.723$, which is larger than $3^{-1/3}\approx 0.693$.
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\end{document}