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\exnumber{4554}
\heading{Fermions in magnetic field, quantum phase transition} 
\auname{Itay Raveh}


{\bf The problem:}

\Dn


A $d$ dimensional container ($d=1,2,3$)  contains fermions of density $n$, temperature $T=0$, mass $m$ and spin $\frac{1}{2}$, having a magnetic moment ${\bar m}$. The container is placed in a magnetic field $H/{\bar m}$ so that the fermion spectra is $\epsilon_{{\bf p}}=\frac{{\bf p}^2}{2m}\pm H$ where ${\bf p}$ is the momentum. (Note that orbital effects are neglected, possible e.g. at d=2 with the field parallel to the layer).
\begin{itemize}
\item[(a)]  Evaluate the chemical  potential $\mu(H)$, for small $H$:
Consider first an expansion to lowest order in $H$ and then evaluate $d\mu/dH$ to note the change at finite $H$.
\item[(b)] Beyond which $H_c$ does  the consideration in (a) fail? Find $\mu(H)$ at $H>H_c$ and plot qualitatively $\mu(H)/\mu_0$ as function of $H/\mu_0$ (where $\mu_0=\mu(H=0)$) for $d=1,2,3$, indicating the values of $\mu(H)/\mu_0$ at $H_c$.
\item[(c)] Of what order is the  phase transition at $H_c$, at either $d=1,2,3$? Does the phase transition survive at finite $T$? (no need for finite $T$ calculations -- just note analytic properties of thermodynamic functions).
\item[(d)] The container above, called A, with $H\ne 0$ is now attached to an identical container B (same fermions at density $n$, $T=0$), but with $H=0$. In which direction will the fermions flow initially? Specify your answer for $d=1,2,3$ at relevant ranges of $H$.

\end{itemize}

\Dn\Dn


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\bf The solution:}

\Dn

\begin{itemize}
\item[(a)]
To find $\mu$, first we find the density of states $g(\epsilon)$, then the total number of particles $N(\mu,H)$, and invert the equation:

\Dn
$g(\epsilon)$ gets a separate contribution from each spin:
\[
g(\epsilon)=g_+(\epsilon)+g_-(\epsilon)=g_f(\epsilon-H)+g_f(\epsilon+H)
\]
Where $g_f(\epsilon)=Vc\epsilon^{\frac{d}{2}-1}$ is standard free-particle energy density. Here $c$ is the volume inside the unit $d$-sphere times $\frac{d}{2}\left(\frac{\sqrt{2m}}{2\pi}\right)^{d/2}$. 

\Dn
$N$ is given by:
\[
N=\int g(\epsilon)f(\epsilon-\mu)d\epsilon
\]
Where $f$ at $T=0$ is just a step function limiting integration up to $\epsilon_F=\mu$. So
\[
N=\int_{-\infty}^\mu g_f(\epsilon-H)+g_f(\epsilon+H)d\epsilon
\]
\[
=\frac{2Vc}{d} \left[ 
\epsilon^{d/2}|_{0}^{\mu+H}+\epsilon^{d/2}|_{0}^{\mu-H}
\right]
\]
\[
=\frac{2Vc}{d} \left[ 
(\mu+H)^{d/2}+(\mu-H)^{d/2}
\right]
\]
This gives $\mu_0=\mu(H=0)=\left(\frac{Nd}{4Vc}\right)^{2/d}=\left(\frac{nd}{4c}\right)^{2/d}$. 

\Dn
To find $\mu(H)$ for small $H$ we expand to second order of $H$ and apply$\left(\frac{d}{dH}\right)_{n,T}$
\[
N=\frac{2Vc}{d} 2\left[ 
\mu^{d/2}+\frac{d}{2}\left(\frac{d}{2}-1\right)\mu^{\frac{d}{2}-2} H^2
+O(H^3)
\right]
\]
\[
0= 
\frac{d}{2}\mu^{\frac{d}{2}-1}\frac{d\mu}{dH}+\frac{d}{2}\left(\frac{d}{2}-1\right)\left(\frac{d}{2}-2\right)\mu^{\frac{d}{2}-3} H^2
\]
Taking $H\rightarrow 0$, $d\mu/dH=0$. This can also be seen from the symmetry of the system to $H\rightarrow -H$. We can search for $\mu$ to second order in $H$: $\mu(H)=\mu_0+\mu_2 H^2$
\[
N
=\frac{2Vc}{d} \left[ 
(\mu_0+\mu_2 H^2+H)^{d/2}+(\mu_0+\mu_2 H^2-H)^{d/2}
\right]
\]
Expanding in the small parameters $\mu_2 H^2+H$ , $\mu_2 H^2-H$
\[=\frac{2Vc}{d} 2\left[ 
\mu_0^{d/2}+\frac{d}{2}\mu_0^{\frac{d}{2}-1} \mu_2H^2+\frac{d}{2}\left(\frac{d}{2}-1\right)\mu_0^{\frac{d}{2}-2} H^2
+O(H^3)
\right]
\]
From comparing $H^2$ terms we get
\[
\mu_2=-\mu_0^{-1}\left(\frac{d}{2}-1\right)
\]
So $\mu$ is increases with $H$ for $d=1$, is constant for $d=2$, and decreases for $d\geq 3$. Notice that for $d=2$, $\mu(H)$ is constant for $H<H_c$ (see  (b)) with no approximations.

\Dn 

\item[(b)]
At arbitrary $H$, we can no longer assume both spin states contribute: the negative energy spin ($-H$) states might include more then N states with energy lower than the ground state for positive energy spin ($+H$).


The critical $H$ is therefore when the energy levels up to $H$ in the $(-H)$ spins are equal to $N$:
\[
N=\int_{-\infty}^H g_f(\epsilon-H)d\epsilon
=\frac{2Vc}{d} \left[ 
\epsilon^{d/2}|_{0}^{2H}
\right]
=
\frac{2Vc}{d} 
(2H)^{d/2}
\]
\[
H_c=\frac{1}{2}\left(\frac{nd}{2c}\right)^{2/d}
=\frac{1}{2}\ 2^{2/d}\ \mu_0
\]
$\mu(H)$ above $H_c$ is given by
\[
N=\int_{-\infty}^\mu g_f(\epsilon-H)d\epsilon
=\frac{2Vc}{d} \left[ 
\epsilon^{d/2}|_{0}^{\mu+H}
\right]
=
\frac{2Vc}{d} 
(\mu+H)^{d/2}
\]
\[
\mu(H)=\left(\frac{nd}{2c}\right)^{2/d}
-H
=2^{2/d}\ \mu_0
-H
\]

Notice that $\mu$ must be continuous in $H_c$, as for $H$ infinitesimally smaller than $H_c$ the contribution of  $g_f(\epsilon+H)$ to the equation is also infinitesimal  ($O(H_c-H)$) for smooth $g_f(\epsilon)$. In fact, for $d\geq 3$ we have $g_f(0)=0$ and the contribution is $O\left((H_c-H)^2\right)$ thus giving a continuous derivative of $\mu$ at $H_c$.

Let us prove this formally.

$\mu$ is continuous at $H_c$:

Consider $\delta H > 0$. The density is constant and so is equal calculating at $H^-=H_c - \delta H$ and $H^+=H_c + \delta H$.
\[
n=\frac{2c}{d} \left[ 
(\mu^-+H^-)^{d/2}+(\mu^--H^-)^{d/2}
\right]
=\frac{2c}{d}  
(\mu^++H^+)^{d/2}
\]
We already know $\mu^+(H_c)=H_c$ so taking $\delta H $ to 0 gives
\[ 
(\mu^-+H_c)^{d/2}+(\mu^--H_c)^{d/2}
=  
(2H_c)^{d/2}
\]
Which is solved by $\mu^-=H_c$

\Dn
\Dn


$\mu$ is differentiable at $H_c$ for $d\geq 3$:

Taking the equation for $n$ for $H<H_c$
\[
\frac{d}{dH}n=\frac{2c}{d} \frac{d}{dH}\left[ 
(\mu+H)^{d/2}+(\mu-H)^{d/2}
\right]=0
\]
\[
\frac{d}{dH}(\mu+H)^{d/2}=-\frac{d}{dH}(\mu-H)^{d/2}
\]
\[
(\mu+H)^{d/2-1}(\frac{d\mu}{dH}+1)=-(\mu-H)^{d/2-1}(\frac{d\mu}{dH}-1)
\]
We know in the limit $H\rightarrow H_c^-$, $\mu\rightarrow H_c$ and with $d\geq 3$, $(\mu-H)^{d/2-1}\rightarrow 0$. So
\[
0=\lim_{H\rightarrow H_c^-}(\mu+H)^{d/2-1}(\frac{d\mu}{dH}+1)=(2H_c)^{d/2-1}\lim_{H\rightarrow H_c^-}(\frac{d\mu}{dH}+1)
\]
\[
\frac{d\mu}{dH}(H_c^-)=-1
\]
So $\frac{d\mu}{dH}(H_c^-)=\frac{d\mu}{dH}(H_c^+)=-1$ and $\mu$ is differentiable at $H_c$. This also shows that for $d=1$, $\lim_{H\rightarrow H_c^-}(\frac{d\mu}{dH}-1)=0$ to negate the "infinite" $\lim_{H\rightarrow H_c^-}(\mu-H)^{d/2-1}$.

Using the same method, for $d=3$ the second derivative is not continuous: 
\[
(\mu+H)^{d/2-2}\left(\frac{d\mu}{dH}+1\right)^2+(\mu+H)^{d/2-1}\frac{d^2\mu}{dH^2}=
-(\mu-H)^{d/2-2}\left(\frac{d\mu}{dH}-1\right)^2
-(\mu-H)^{d/2-1}\frac{d\mu}{dH}
\]
\[
\lim_{H\rightarrow H_c^-}
(2H_c)^{d/2-1}\frac{d^2\mu}{dH^2}=\lim_{H\rightarrow H_c^-}-4(\mu-H)^{d/2-2}=-\infty
\]

\Dn
\Dn


Graphs:


\includegraphics[scale=0.6]{4554graphs.png} 


\Dn

\item[(c)]
We can look at the magnetization at different values of $H$. For $H>H_c$, all spins are in one direction and magnetization is constant $M=N$. Otherwise
\[
M=\int_{-\infty}^\mu g_f(\epsilon+H) - g_f(\epsilon-H)d\epsilon
=\frac{2Vc}{d} \left[ 
(\mu-H)^{d/2}-(\mu+H)^{d/2}
\right]
\]
The susceptibility is 
\[
\chi=\frac{d}{dH}M=\frac{d}{dH}\frac{2Vc}{d} \left[ 
(\mu-H)^{d/2}-(\mu+H)^{d/2}
\right]
\]
So $\chi$ has the same continuity properties as $\frac{d}{dH}\mu$ at $H_c$. For $d=1,2$ the phase transition is of first order (discontinuity), and of second order for $d=3$ (derivative discontinuity).

At finite temperature, $f(\epsilon -\mu)$ is smoothed and  has a positive finite value at $\epsilon > H$ for all finite $H$, so the transition from one dominant spin to both spins dominant is smooth. Both "sites" always have non-zero occupation. The phase transition does not survive.  


\Dn

\item[(d)]
Particles will initially flow from higher chemical potential to lower
\Dn 

For $d=3$:


For all $H$, $\mu(0)>\mu(H)$ so particles will flow from B to A
\Dn 

For $d=2$:

If $H\leq H_c$, then the chemical potentials are the same and there will be no flow. Otherwise $\mu(0)>\mu(H)$ so particles will flow from B to A.
\Dn

For $d=1$:

If $H\leq H_c$, $\mu(0)<\mu(H)$ and particles will flow from A to B. Otherwise, there exists $H_1=3\mu_0$ where $\mu_0=\mu(H_1)$. For $H\in (H_c,H_1)$, $\mu(0)<\mu(H)$ and particles will flow from A to B. For $H\in (H_1,\infty)$, $\mu(0)>\mu(H)$ and particles will flow from B to A. 

\end{itemize}















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