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\begin{document}
\title{Two hard-core anyons in harmonic trap}
%\author{Li Yang$^1$, and Han Pu$^{1,2}$}

%\affiliation{$^{1}$Department of Physics and Astronomy, and Rice Center for Quantum Materials,
%Rice University, Houston, TX 77251, USA \\
%$^2$Center for Cold Atom Physics, Chinese Academy of Sciences, Wuhan 430071, P. R. China}

\maketitle

\section{general formulation}
A two spinless particle in 1D are described by the two-body wavefunction $\Psi(x_1,x_2)=\langle x_1,x_2|\Psi \rangle$. The reduced one-body density operator is \[ \hat{\rho}_1 = \int dx_2\, \langle x_2| \Psi \rangle \langle \Psi |x_2 \rangle = \int dx_1 \int dx'_1 \int dx_2\, |x'_1\rangle \langle x'_1| \langle x_2| \Psi \rangle \langle \Psi |x_2 \rangle  |x_1\rangle \langle x_1| = \int dx_1 \int dx'_1 \, |x'_1\rangle \rho(x_1,x_1') \langle x_1|   \]  where the one-body density matrix (OBDM) is defined as:
\[ \rho(x,x')= \langle x'|\hat{\rho}_1| x \rangle = \int dx_2 \, \Psi^*(x,x_2) \Psi(x',x_2) \]
The real space and momentum space density profiles can be easily obtained from OBDM as
\begin{eqnarray}
	n(x) &=& N \langle x| \hat{\rho}_1 |x \rangle = N \rho(x,x) = N \int dx_2 \, \Psi^*(x,x_2) \Psi(x,x_2) = N \int dx_2 \, |\Psi(x,x_2)|^2   \\
	n(p) &=& N \langle p| \hat{\rho}_1 |p \rangle =\frac{N}{2\pi} \int dx \int dx' \,e^{ip(x-x')}\, \rho(x,x')
	\end{eqnarray}
with $N=2$. Furthermore, given a single-particle state $\varphi(x)$, the probability to find the particle in this state is \[  P_\varphi = \langle \varphi| \hat{\rho}_1 |\varphi \rangle= \int dx \int dx' \,\varphi(x) \varphi^*(x') \rho(x,x') \]

We will focus on the ground state of such a two-particle system in a harmonic trap with hardcore interaction. The Hamiltonian of the system reads:
\[ H =H_0+H_{\rm int}= -\frac{\hbar^2}{2m} \left( \partial_1^2 + \partial_2^2 \right) + \frac{1}{2} m\omega^2 (x_1^2+x_2^2) + g\delta(x_1-x_2)  \] where the interaction strength takes the limit $g \longrightarrow \infty$.

\section{free fermions in harmonic trap}
Now consider two free fermions in a harmonic trap, whose single-particle eigenstates are labeled as $\varphi_i(x)$ where $i =0,$ 1, 2, ..., and satisfy the orthonormal condition $ \int dx \, \varphi_i^*(x) \varphi_j(x) = \delta_{ij}$. The ground state wavefunction of the two-fermion system is given by \[ \Psi_F(x_1,x_2) = \frac{1}{\sqrt{2}} \,\left[ \varphi_0(x_1) \varphi_1(x_2) -\varphi_0(x_2) \varphi_1(x_1) \right]  \]
Using the expression above, one can readily find the following:
\begin{eqnarray}
\rho_F(x,x') &=& \frac{1}{2} \left[ \varphi_0^*(x) \varphi_0(x') +  \varphi_1^*(x) \varphi_1(x') \right] \\
n_F(x) &=& |\varphi_0(x)|^2 + |\varphi_1(x)|^2 \\
n_F(p) &=& |\tilde{\varphi}_0(p)|^2 + |\tilde{\varphi}_1(p)|^2
\end{eqnarray}
where \[ \tilde{\varphi}_i(p) = \frac{1}{\sqrt{2\pi}} \int dx \, e^{-ipx}\,\varphi_i(x) \] is the single-particle momentum space wavefunction.

Finally, the probability of finding the particle in the $i^{\rm th}$ harmonic oscillator eigenstate $\varphi_i(x)$ is given by \[ P_0=P_1=1/2\,,\;\;\;P_{i\neq 0,1}=0 \]
The corresponding reduced density operator can be written as
\[ \hat{\rho}_F = \frac{1}{2} \left( |\varphi_0 \rangle \langle \varphi_0| + |\varphi_1 \rangle \langle \varphi_1| \right) \] with the associated von Neumann entropy \[ S = {\rm Tr}[-\hat{\rho}_F \ln \hat{\rho}_F ] =\ln 2 \]

\section{hardcore anyons in harmonic trap}
Now consider two hardcore anyons in a harmonic trap. Using the anyon-Fermi mapping, the corresponding wavefunction of the anyons is given by
\[ \Psi_\kappa (x_1,x_2) = A_\kappa(x_2-x_1) \,\Psi_F(x_1,x_2) \]
where $\kappa$ is the anyon statistical parameter and
\[ A_\kappa (x_2-x_1) = e^{i\pi (1-\kappa) \theta(x_2-x_1)}= \left\{ \begin{array}{ll} e^{i\pi(1-\kappa)}\,, & x_2>x_1 \\ 1 \,, & x_2 <x_1 \end{array}  \right.  \] where $\theta(x_2-x_1)$ is the Heaviside step function.
The anyon wavefunction satisfies the following exchange properties:
\[ \Psi_\kappa(x_1,x_2) = e^{i\pi \kappa \epsilon(x_1-x_2)} \,\Psi_\kappa(x_2,x_1) \]
where \[ \epsilon(x) =  \left\{  \begin{array}{cl} 1\,, & x>0 \\ 0 \,, & x=0 \\ -1\,, & x<0 \end{array} \right.\]
Two special cases are: (1) For $\kappa=0$, $ A_\kappa (x_2-x_1) = {\rm sgn}(x_1-x_2)$, and the anyons correspond to hardcore bosons; (2) for $\kappa=1$, $A_\kappa (x_2-x_1) = 1$, and the anyons correspond to hardcore, i.e., free, fermions. Note that $A_\kappa = A^{\kappa+2}$, hence we can restrict the values of $\kappa$ to be $\kappa \in [0, 2)$.

Since $|\Psi_\kappa| = |\Psi_F|$ independent of $\kappa$, the real space density profile $n(x)$ is independent of $\kappa$. In other words, all hardcore anyons have the same real space density profile independent of their statistical parameter.

From the two-body wave function for two hard-core anyons in a harmonic trap, the OBDM can be calculated using the formula given in the first section combined with the anyon-fermi mapping:
\begin{align*}
\rho_\kappa(x,x')= \langle x'|\hat{\rho}_1| x \rangle &= \int dx_2 \, \Psi_{\kappa}^*(x,x_2) \Psi_\kappa(x',x_2)\\
& = \int dx_2 \,  e^{-i\pi (1-\kappa) \theta(x - x_2)} e^{i\pi (1-\kappa) \theta(x' - x_2)} \Psi_{F}^*(x,x_2) \Psi_{F}(x',x_2)\\
& = \rho_{F}(x,x') + \epsilon(x' - x)(e^{\epsilon(x' - x) i\pi (1-\kappa)} - 1)\int_{x}^{x'} dx_2 \, \Psi_{F}^*(x,x_2) \Psi_{F}(x',x_2)
\end{align*}

The latter integral can be calculated ``analytically" using the error function.

Using the projection formula given in the first section, the anyon OBDM can be used to numerically calculate the anyon's projection values. For one of the fermions in the two-body ground state, the probability of finding the particle in either the ground state or the first excited state was $P_{0} = P_{1} = \frac{1}{2}$ and all other projections were 0. An anyon in the two-body ground state can be found in any of the excited states (assuming $\kappa \neq 1$), however, with diminishing probability for excited states with more energy.

For instance, if two hard-core bosons are in the two-body ground state of a harmonic trap, then the probability of finding one of those particles in the $\phi_0$ state is .718, the probability for the $\phi_1$ state is .1667, and so on, as is illustrated in the top left plot of Fig. \ref{fig:projections} ($\kappa = 0$).
\begin{figure}[h]
	\includegraphics[scale=.5]{"../Plots/Anyon Projection Values"}
	\caption{Projections}
	\label{fig:projections}
\end{figure}
\begin{figure}[h]
\includegraphics[scale=.5]{"../Plots/Momentum"}
\caption{Momentum Distribution}
\label{fig:momentum}
\end{figure}

The OBDM can also be used to numerically find the momentum distribution of an anyon for various values of $\kappa$ (see Fig. \ref{fig:momentum}). The y-axis indicates the probability of finding the anyon with a particular momentum. Only the hard-core boson case ($\kappa = 0, 2$) and the fermion case ($\kappa = 1$) have momentum distributions that are symmetric about $p = 0$.

The von Neumann entropy can be determined from the OBDM by transforming from the position basis to the harmonic potential eigenbasis. The resulting matrix will be infinite in size, but for the purposes of calculations only the first five terms in each dimension need to be considered -- projections after $P_5$ are on the order of $10^{-3}$ and are thus negligible. If $\lambda_1, \lambda_2,\ldots$ are the eigenvalues of this matrix, the von Neumann entropy is given by
\[ S =  -\sum_i^\infty \lambda_i \log(\lambda_i) \]
Thus, the von Neumann entropy for hard-core anyons in the two-body ground state of the harmonic oscillator are given in Fig. \ref{fig:entropy} for various values of $\kappa$. The fermion case ($\kappa = 1$) obtains the analytic result $S = \log(2) = 0.6931$.

\begin{figure}[h]
\includegraphics[scale=.5]{"../Plots/EntropyPlot"}
\caption{Plot of von Neumann Entropy $(S)$}
\label{fig:entropy}
\end{figure}

Evidently, the entropy is greatest for $\kappa$ near 1 but not 1 and obtains its minimum value for hard-core bosons ($\kappa = 0$). It is also reassuring that the von Neumann entropy is symmetric about $\kappa = 1$, so $S_{\kappa = 1/2} = S_{\kappa = 3/2}$ and so on.

A. Minguzzi et al.\footnotemark\,claim that for large $p$ the momentum distribution $n(p)$ of hardcore bosons decays like $1/p^4$.
\footnotetext{A. Minguzzi, P. Vignolo, and M. Tosi, Physics Letters A \textbf{294}, 222 (2002).}
I have attempted to reproduce this calculation below in Section IV. Their analytic result is that
\[ \lim_{p \rightarrow \infty} n(p) = 2\sqrt{\frac{2}{\pi }} \frac{(\hbar m \omega)^{3/2}}{p^4}  \]
In fig. \ref{fig:tails} I have plotted on a log-log scale the anyonic momentum distribution $n_\kappa (p)$ for various $\kappa$. The orange line depicts the value $\log(n_\kappa(p))$ for $p \in [-25, -2] \cup [2,25]$. The blue line is the log-log plot of $1/p^4$. These plots confirm that hardcore bosons as well as anyons follow a $1/p^4$ momentum distribution decay. Even though anyons do not have a symmetric momentum distribution for $\kappa \in (0,1) \cup (1,2)$ (see fig. \ref{fig:momentum}), it is still true that the anyonic momentum tails follow a $1/p^4$ decay, regardless of the sign of $p$ (see fig. \ref{fig:tails}).

However, these numerical results are not in complete accord with the analytic coefficient presented by Minguzzi et al. For instance, if $N(p)$ is the numerically calculated value for the momentum distribution at momentum $p$, then
\[ N(p) = \dfrac{C}{p^4} \]
should hold for some constant $C$ assuming $p$ is sufficiently large. This constant, therefore, can be approximated at each point as $C = N(p) * p^4$. If $N(p)$ has been calculated at values $\{ p_{1},p_{2}, \ldots, p_{k} \}$, then $C \approx \text{Mean}(N(p_{1}) * p_{1}^4, N(p_{2}) * p_{2}^4, \ldots, N(p_{k}) * p_{k}^4) $. Using this fact, one can infer from the data above that $C \approx 0.513959 $.

If the $\frac{1}{2\pi}$ prefactor is included in Minguzzi's coefficient, my numerical results should be in agreement with Minguzzi's coefficient, yet they still differ by a factor of 2. According to Minguzzi's paper, when the prefactor is included we have $C = \frac{1}{\pi} \sqrt{\frac{2}{\pi}} \approx 0.253974\ldots$, which is about half what I calculated from my numerical data.

\section{momentum tail coefficient for hcb}

In the following section, I will try to reproduce the result from Minguzzi's paper that
\[ \lim_{p \rightarrow \infty} n(p) = 2\sqrt{\frac{2}{\pi }} \frac{(\hbar m \omega)^{3/2}}{p^4}  \]

\begin{center}
\begin{figure}[h]
	\includegraphics[scale=.45]{"../Plots/FullMomentumTails"}
	\caption{Momentum Tails for Anyons}
	\label{fig:tails}
\end{figure}
\end{center}

Recall that two body ground state for two fermions in a harmonic trap is
\[ \Psi_F(x_1,x_2) = \frac{1}{\sqrt{2}} \,\left[ \varphi_0(x_1) \varphi_1(x_2) -\varphi_0(x_2) \varphi_1(x_1) \right]  \]
\[ \varphi_{n}(x) = \dfrac{1}{\pi^{1/4} \sqrt{\alpha 2^n n!}} H_{n}(x/\alpha) e^{-\frac{1}{2}(x/\alpha)^2} \]
where $\alpha = \sqrt{\frac{\hbar}{m \omega}}$. Simplifying,
\[ \Psi_{F}(x_{1},x_{2}) = \dfrac{1}{\sqrt{\pi} \alpha^2} e^{-\frac{1}{2\alpha^2} (x_{1}^2 + x_{2}^2)} (x_{2} - x_{1}) \]
By the Bose-Fermi mapping, the two-body ground state for two hard-core bosons must be
\[ \Psi_{B}(x_{1},x_{2}) = \dfrac{1}{\sqrt{\pi} \alpha^2} e^{-\frac{1}{2\alpha^2} (x_{1}^2 + x_{2}^2)} |x_{2} - x_{1}|  \]
The one-body density matrix is then
\begin{align*}
\rho_{B}(x, x') &= \int_{-\infty}^{\infty} \Psi_{B}^{*}(x ,x_{2}) \Psi_{B}(x' ,x_{2}) dx_{2}\\
& = \dfrac{1}{\alpha^4 \pi} e^{-\frac{1}{2\alpha^2} (x^2 + x'^2)} \int_{-\infty}^{\infty} e^{-x_{2}^{2}/\alpha^2} |x_{2} - x| |x_{2} - x'| dx_{2}
\end{align*}
Let $Q_{2} = x_{2}/\alpha$, $Q = x/\alpha$, and $Q' = x'/\alpha$ to obtain
\begin{equation}
\rho_{B}(x, x') = \dfrac{1}{\alpha \pi} e^{-\frac{1}{2} (Q^2 + Q'^2)} \int_{-\infty}^{\infty} e^{-Q_{2}^{2}} |Q_{2} - Q| |Q_{2} - Q'| dQ_{2}
\end{equation}
Please note that I have only scaled $Q_{2}$ via u-substitution. The other variables have just been rewritten to simplify the expression. I am not scaling the length. Variables $Q$ and $Q'$ are just shorthand for $ x/\alpha$ and $x'/\alpha$ respectively.

The momentum distribution is
\begin{align*}
n_{B}(p) &= \dfrac{N}{2 \pi \hbar} \int_{-\infty}^{\infty} dx \int_{-\infty}^{\infty} dx' e^{i \frac{p}{\hbar} (x - x')} \rho_{B}(x,x')\\
& = \dfrac{N}{2 \alpha \pi^2 \hbar} \int_{-\infty}^{\infty} \alpha dQ \int_{-\infty}^{\infty} \alpha dQ' e^{i \alpha \frac{p}{\hbar} (Q - Q')}  e^{-\frac{1}{2} (Q^2 + Q'^2)} \int_{-\infty}^{\infty} e^{-Q_{2}^{2}} |Q_{2} - Q| |Q_{2} - Q'| dQ_{2}\\
& = \dfrac{N\alpha }{2 \pi^2 \hbar} \int_{-\infty}^{\infty} dQ \int_{-\infty}^{\infty} dQ' e^{i \alpha \frac{p}{\hbar} (Q - Q')}  e^{-\frac{1}{2} (Q^2 + Q'^2)} \int_{-\infty}^{\infty} e^{-Q_{2}^{2}} |Q_{2} - Q| |Q_{2} - Q'| dQ_{2}\\
& =  \dfrac{N\alpha }{2 \pi^2 \hbar} \int_{-\infty}^{\infty}  dQ_{2} e^{-Q_{2}^2} \int_{-\infty}^{\infty} dQ  e^{i \alpha \frac{p}{\hbar} Q} e^{-\frac{1}{2}Q^2} |Q_{2} - Q|  \int_{-\infty}^{\infty} dQ'  e^{-i \alpha \frac{p}{\hbar} Q'} e^{-\frac{1}{2}Q'^2} |Q_{2} - Q'| \\
& = \dfrac{N\alpha }{2 \pi^2 \hbar} \int_{-\infty}^{\infty}  dQ_{2} e^{-Q_{2}^2} \left| \int_{-\infty}^{\infty} dQ  e^{i \alpha \frac{p}{\hbar} Q} e^{-\frac{1}{2}Q^2} |Q_{2} - Q| \right|^2
\end{align*}

The integral over $Q$ can be evaluated asymptotically as

\begin{equation}
\int_{-\infty}^{\infty} dQ  e^{i \alpha \frac{p}{\hbar} Q} e^{-\frac{1}{2}Q^2} |Q_{2} - Q| = \dfrac{-2 \hbar ^2 e^{-\frac{1}{2} Q_{2}^2}}{\alpha^2 p^2}
\end{equation}

for large p. Thus,
\begin{align}
\lim_{p \rightarrow \infty} n_{B}(p) & =
\dfrac{2}{\pi} \sqrt{\dfrac{2}{\pi}} \dfrac{(\hbar m \omega)^{3/2}}{p^4}
\end{align}
%
from which we can infer that the Tan contact for two harmonically trapped hard-core bosons is
\[
C_{B}= \dfrac{2}{\pi} \sqrt{\dfrac{2}{\pi}}
\]
This is consistent with the ground state energy of two bosons in the TG ($g \rightarrow \infty$) limit, which reads
\[
E_{B} = E_{0} - 2 \sqrt{\dfrac{2}{\pi}}\dfrac{1}{g}
\]
and from the adiabatic sweep theorem, we have (see Eur. Phys. J. Special Topics \textbf{226}, 1583 (2017))
\[
\dfrac{dE_{B}}{d(1/g)} = -2\sqrt{\dfrac{2}{\pi}} = - \pi C_{B}
\]

\section{momentum tail coefficient for hca}

To find the momentum tail coefficient for hard-core anyons, the calculation is very similar to the calculation for hard-core bosons. I will give the derivation for the result

\[
\lim_{p \rightarrow \infty} n_\kappa(p) = \cos^2\left(\frac{\pi \kappa}{2}\right) \dfrac{2}{\pi} \sqrt{\dfrac{2}{\pi}} \dfrac{(\hbar m \omega)^{3/2}}{p^4}
\]
where the $\kappa$ subscript denotes the anyon parameter of the anyon two-body system.

Like before the fermion and hard-core anyon wavefunctions are

\[ \Psi_{B}(x_{1},x_{2}) = \dfrac{1}{\sqrt{\pi} \alpha^2} e^{-\frac{1}{2\alpha^2} (x_{1}^2 + x_{2}^2)} |x_{2} - x_{1}| \]
\[ \Psi_\kappa(x_{1},x_{2}) = e^{-i \frac{\pi \kappa}{2}  \epsilon(x_{2} - x_{1})} \Psi_{B}(x_{1},x_{2})   \]

where the second relation holds up to a constant phase factor and follows from the anyon-fermi mapping in Section III of this note. So the anyon momentum distribution is

\begin{align*}
n_\kappa (p) & = \dfrac{N}{2 \pi \hbar} \int_{-\infty}^{\infty} dx \int_{-\infty}^{\infty} dx' e^{i \frac{p}{\hbar} (x - x')} \rho_\kappa(x,x')\\
& =  \dfrac{N}{2 \pi \hbar} \int_{-\infty}^{\infty} dx_{2}
\left|\int_{-\infty}^{\infty} dx  e^{i \frac{p}{\hbar} x} \Psi_\kappa^*(x,x_{2}) \right|^2
\end{align*}

Consider the integral
\begin{align*}
\int_{-\infty}^{\infty} dx \, e^{i \frac{p}{\hbar} x} \Psi_\kappa^*(x_{1},x_{2}) & =
\int_{-\infty}^{\infty} dx \, e^{i \frac{p}{\hbar} x}  e^{i \frac{\pi \kappa}{2}  \epsilon(x_{2} - x)} \Psi_{B}^*(x, x_{2})\\
& = \int_{-\infty}^{\infty} dx \, e^{i \frac{p}{\hbar} x} \left( \cos\left(\frac{\pi \kappa}{2}\right) + i \epsilon(x_{2} - x) \sin\left(\frac{\pi \kappa}{2} \right) \right) \Psi_{B}^*(x, x_{2})
\end{align*}
Because $\epsilon(x_{2} - x)\Psi_{B}^*(x, x_{2}) = \Psi_{F}(x, x_{2}) $, in the $p \rightarrow \infty$ limit, the $\sin$ term above results in an exponentially decaying term, which means it can be neglected because the $\cos$ term results in a power law $1/p^4$ decay. Thus,

\begin{align*}
\lim_{p \rightarrow \infty} \left[ \int_{-\infty}^{\infty} dx  e^{i \frac{p}{\hbar} x} \Psi_\kappa^*(x_{1},x_{2}) \right] & =
\cos\left(\frac{\pi \kappa}{2}\right) \int_{-\infty}^{\infty} dx  e^{i \frac{p}{\hbar} x} \Psi_{B}^*(x, x_{2})\\
& = -\cos\left(\frac{\pi \kappa}{2}\right) \dfrac{2 \hbar^2 e^{-(x_{2}/\alpha)^{2}}}{\sqrt{\pi}\alpha^2 p^2}
\end{align*}
where the last equality comes from the integrals evaluated in the previous section and $Q_{2}$ is shorthand for $x_{2}/\alpha$. So,
\begin{align*}
\lim_{p \rightarrow \infty} n_\kappa (p) & = \dfrac{2 N \hbar^3}{\alpha^4 \pi^2 p^4}
\cos^2\left(\frac{\pi \kappa}{2}\right)  \int_{-\infty}^{\infty} dx_{2}  e^{-2(x_{2}/\alpha)^{2}}\\
& = \dfrac{2}{\pi} \sqrt{\dfrac{2}{\pi}} \cos^2\left(\frac{\pi \kappa}{2}\right) \dfrac{(\hbar m \omega)^{3/2}}{p^4}
\end{align*}

Numerical simulation can confirm the analytic results I have just cited. We can infer the anyon momentum tail coefficient from numerical values for $n_{\kappa}(p)$.  The plot in fig. \ref{fig:AnyonCoeff} compares the analytic and numerical calculation for the tail coefficient divided by $ \frac{2}{\pi} \sqrt{\frac{2}{\pi}}$ (should just be $\cos^2\left(\frac{\pi \kappa}{2}\right)$). This confirms the calculations from Sections IV and V.

If we define the momentum tail coefficient as the Tan contact, then
\[
C_{\kappa} = \cos^2\left(\frac{\pi \kappa}{2}\right) C_{B}
\]
which is consistent with the result obtained in Eur. Phys. J. D \textbf{71}, 135 (2017). Note that in that paper, the anyon statistical parameter $\chi$ is equal to $(1 - \kappa)$ in our notation.

An anyon gas with interaction strength $g$ has the same energy as a bosonic gas with interaction strength $g' = \frac{g}{\cos\left(\frac{\pi \kappa}{2}\right)}$. Therefore, in the large $g$ limit, the two harmonically trapped anyons should have energy
\[
E_{\kappa} = E_{0} - 2\sqrt{\dfrac{2}{\pi}} \frac{1}{g'}
= E_{0} - 2\sqrt{\dfrac{2}{\pi}} \frac{\cos\left(\frac{\pi \kappa}{2}\right)}{g}
\]
Therefore, it seems that the adiabatic sweep theorem for anyon gas should read
\[
\frac{dE_{\kappa}}{d(1/g)} = - 2 \cos\left(\frac{\pi \kappa}{2}\right) \sqrt{\frac{2}{\pi}} = -\pi C_{\kappa} / \cos\left(\frac{\pi \kappa}{2}\right)
\]

\begin{center}
\begin{figure}[h]
	\includegraphics[scale=.5]{"../Plots/AnyonCoeff"}
	\caption{Anyon momentum tail coefficient plotted as a function of $\kappa$}
	\label{fig:AnyonCoeff}
\end{figure}
\end{center}

\section{Time Evolution of Momentum Tail for expanding hca}

Let's consider the case where two hard-core anyons in the ground state of a harmonic trap are released suddenly. That is, the trap frequency $\omega$ changes suddenly from $\omega_{0}$ to 0. The time evolution of a particle in a harmonic trap with a time dependent trap frequency can be found analytically.

Let the potential $V(x) = \frac{1}{2} m \omega^2(t) x^2$ where $\omega(t) = \omega_0$ for $t < 0$ and let $\phi_j(x; 0)$ be the $j^{th}$ excited state solution for the static case with trap frequency $\omega_0$. If $\phi_j(x; t)$ is the solution for the time dependent case, then
\[
\phi_j(x; t) = \frac{1}{\sqrt{b}}\phi_j(x/b; 0) e^{i (\frac{x^2}{2} \frac{\dot{b}}{b} - E_j \tau (t) ) }
\]

where $b(t)$ and $\tau(t)$ are determined by

\begin{align*}
\ddot{b} + \omega^2(t) b = b^{-3}\\
\tau(t) = \int^{t}_{0}dt' \, b^{-2}(t')
\end{align*}

In the sudden expansion case where $\omega(t) = \omega_0$ for $t < 0$ and $\omega(t) = 0$ for $t > 0$, solving these equations with the initial conditions that $b(0) = 1$ and $b'(0) = 0$ gives
\begin{eqnarray}
b(t) = \sqrt{1 + t^2}\\
\tau(t) = \arctan(t)
\end{eqnarray}

These equations can be used to obtain time dependent expressions for the anyonic wavefunction, one-body density matrix, and momentum distribution.

\begin{align*}
\Psi_{\kappa}(x_{1}, x_{2}; t) & =
\frac{1}{b} \Psi_{\kappa}(\frac{x_{1}}{b}, \frac{x_{2}}{b}; 0)
\exp \left[ \frac{i \dot{b}}{2 b} (x_{1}^2 + x_{2}^2) - i \tau(t) (E_0 + E_1) \right]\\
\rho_{\kappa}(x, x'; t) & = \frac{1}{b} \rho_{\kappa}(\frac{x}{b}, \frac{x'}{b}; 0)
\exp \left[ \frac{i \dot{b}}{2 b} (x'^2 - x^2) \right]\\
n_{\kappa}(p ; t) & = \dfrac{N}{2\pi b} \int dx \int dx'
\rho_{\kappa}(\frac{x}{b}, \frac{x'}{b}; 0)
\exp \left[ip(x - x') +	 \frac{i \dot{b}}{2 b} (x'^2 - x^2) \right]
\end{align*}
where
\begin{align*}
\Psi_{\kappa}(x_{1}, x_{2}; 0) & =
e^{i \pi (1 - \kappa) \theta(x_{2} - x_{1})} \Psi_{F}(x_{1}, x_{2}; 0)\\
\rho_{\kappa}(x, x'; 0) & = \int dx_2 \, \Psi_{\kappa}^*(x,x_2; 0) \Psi_\kappa(x',x_2; 0)
\end{align*}

\begin{center}
\begin{figure}[h!]
	\includegraphics[scale=.43]{"../Plots/MomDistExpandingPair"}
	\caption{Fermionization of HCB (left) and HCA (right $\kappa = \frac{1}{2}$) -- dashed red line corresponds to the fermion distribution}
	\label{fig:HCBFermionizationPair}
\end{figure}
\end{center}

Minguzzi et al. (Phys. Rev. Lett. 94, 240404) show using the stationary phase method that as $t \rightarrow \infty$ the momentum distribution of hard-core bosons approaches the fermion momentum distribution. Their numerical calculations also confirm this. I have performed numerical calculation to reproduce this fermionization for two hard-core bosons. These numerical results are depicted in fig. \ref{fig:HCBFermionizationPair}.

Using numerical simulation, I also confirmed that two expanding hard-core anyons will experience the same fermionization in their momentum distribution. These results are also in fig. \ref{fig:HCBFermionizationPair} in the plot on the right.

To calculate the time evolution of the momentum tail coefficient, we start by simplifying the expression for the momentum distribution to obtain
\begin{align*}
n_{\kappa}(p ; t) & = \dfrac{N b}{2\pi} \int du \int dv
\rho_{\kappa}(u, v; 0)
\exp \left[ibp(u - v) +	 \frac{i \dot{b}b}{2 } (v^2 - u^2) \right]\\
& = \dfrac{N b}{2\pi} \int dx_{2} \left| \int dx \, e^{ibpx} \Psi_{\kappa}(x, x_{2}; 0) e^{\frac{i \dot{b}b}{2} x^2}\right|^2
\end{align*}
%
Evaluating the integral in the modulus square we obtain
%
\begin{align*}
\int dx \, e^{ibpx} \Psi_{\kappa}(x, x_{2}; 0) e^{\frac{i \dot{b}b}{2} x^2}
= \cos\left(\frac{\pi \kappa}{2}\right)
\frac{1}{\sqrt{\pi}} e^{- x_{2}^2} \frac{-2}{b^2 p^2} e^{i \delta}
\text{	as } p \rightarrow \infty
\end{align*}
where $\delta$ is a constant phase shift that will be eliminated by the modulus square and can therefore be ignored. Thus,
\[
\lim_{p \rightarrow \infty} n_{\kappa}(p ; t) =
\frac{4N}{2 \pi^2 b^3 p^4} \cos^2\left(\frac{\pi \kappa}{2}\right) \sqrt{\frac{\pi}{2}}
= \frac{2}{\pi} \sqrt{\frac{2}{\pi}} \cos^2\left(\frac{\pi \kappa}{2}\right) \frac{1}{b^3 p^4}
\]
\begin{equation}
\lim_{p \rightarrow \infty} n_{\kappa}(p ; t) =
\frac{2}{\pi} \sqrt{\frac{2}{\pi}} \cos^2\left(\frac{\pi \kappa}{2}\right) \frac{1}{b^3 p^4}
\label{MomentumTail}
\end{equation}

I made two numerical calculations of the time evolution of the momentum tail coefficient for $\kappa = 0$ and $\kappa = \frac{1}{2}$. Both numerical calculations were in agreement with the analytic prediction given in equation 11. Plots of the time evolution are depicted in fig. \ref{fig:HCATimeDepPair}
%
\begin{center}
\begin{figure}[h]
	\includegraphics[scale=.44]{"../Plots/MomTailCoeffTimeDepRow"}
	\caption{Time evolution of momentum tail coefficient for $\kappa = 0$ (left) and similarly for $\kappa = 1/2$ (right)}
	\label{fig:HCATimeDepPair}
\end{figure}
\end{center}
%
\section{Time Evolution of oscillating hca}

In the last section, we considered the case where the trap frequency of the harmonic trap is suddenly changed from $\omega_0$ to 0. Now, we will consider the case where the trap frequency is changed from $\omega_0$ to a non-zero value $\omega_1$.

Again we can use the same equations as before to describe the time evolution of the system. It can be shown that if $\omega(t) = \omega_0$ for $t \leq 0$ and $\omega(t) = \omega_1$ for $t > 0$, then
\begin{align*}
b(t) = \sqrt{1 + \frac{\omega_0^2 - \omega_1^2}{\omega_1^2} \sin^2(\omega_1 t)}\\
\tau(t) = \frac{1}{\omega_0} \arctan\left(\frac{\omega_0}{\omega_1} \tan(\omega_1 t)\right)
\end{align*}
Under these conditions, the wave function and the momentum distribution will undergo oscillations with period $T = \pi/\omega_1$. Using the same expressions for the OBDM and momentum distribution from the previous section, we can plot the time evolution of the momentum distribution by solving the integrals numerically.

In fig. \ref{fig:OscHCATimeDepPair}, there are two plots depicting the time evolution of the oscillating anyon gas for the hard-core boson case ($\kappa = 0$) and the $\kappa = 1/2$ case. For the sake of clarity, only half the period is depicted (where full period $T = 3\pi$) because the time evolution of the latter half of the period is the same as the first half but reversed. So, the plots in fig. \ref{fig:OscHCATimeDepPair} only give time slices from $t = 0$ to $t = \frac{3\pi}{2}$

The expression for the momentum tail coefficient in equation \ref{MomentumTail} can be used to find the momentum tail coefficient in the oscillating case, as long as the appropriate function is used for $b(t)$. In fig. \ref{fig:MomTailCoeffComp} there is a plot comparing the time evolution of the momentum tail coefficient for the expansion case and the oscillating case. The time evolution of each has been plotted over one full period where $T = 3\pi$. The momentum tail coefficient decays to zero in the expanding gas case, but oscillates periodically in the case where the the trap frequency is changed to a non-zero value.

\begin{center}
\begin{figure}[h]
	\includegraphics[scale=.52]{"../Plots/MomDistOscillatingPair"}
	\caption{Time evolution of oscillating anyon gas $\kappa = 0$ (left) and similarly for $\kappa = 1/2$ (right)}
	\label{fig:OscHCATimeDepPair}
\end{figure}
\end{center}

\begin{center}
\begin{figure}[h]
	\includegraphics[scale=.4]{"../Plots/MomTailCoeffComparison"}
	\caption{Time evolution of momentum tail in expanding case and oscillating case}
	\label{fig:MomTailCoeffComp}
\end{figure}
\end{center}

\section{momentum tail coefficient for hcb (n-particle case)}

In this section I will show that for $N$ hard-core bosons in a harmonic trap, the momentum distribution has a momentum tail that decays like $1/p^4$. It can be shown that (see Girardeau et al. \footnotemark\,)
\footnotetext{M. D. Girardeau, E. M. Wright, and J. M. Triscari, Physical Review A \textbf{63} 033601 }
\begin{align*}
\Psi_B(x_{1}, x_{2}, \ldots, x_{N}) & = A_{N} \exp\left(-\frac{1}{2} \sum_{i = 1}^{N} x_{i}^2\right) \prod_{1 \leq j < k \leq N} |x_{k} - x_{j}|\\
\rho(x, x') & = \int dx_{2} \cdots \int dx_{N} \Psi_{B}^*(x, x_{2}, \ldots, x_{N}) \Psi_{B}(x', x_{2}, \ldots, x_{N})
\end{align*}
where $A_{N} = \dfrac{2^{(N - 1)N/4}}{\pi^{N/4}} \left( \prod_{n = 0}^{N} n! \right)^{-1/2}$. Thus, by letting $\tilde{A}_{N} = \frac{N}{2 \pi} A_{N}^2$
%
\begin{align*}
n_{B}(p) &= \dfrac{N}{2 \pi} \int dx \int dx' e^{i p (x - x')} \rho(x, x')\\
& = \dfrac{N}{2 \pi} \int dx_{2} \cdots \int dx_{N} \left| \int dx e^{i p x} \Psi_{B}^*(x, x_{2}, \ldots, x_{N}) \right|^2\\
& = \tilde{A}_{N} \int dx_{2} \cdots \int dx_{N} \exp\left(- \sum_{i = 2}^{N} x_{i}^2\right) \left( \prod_{2 \leq i < q \leq N} (x_{q} - x_{i})^2 \right)
	\left| \int dx e^{i p x} e^{-\frac{1}{2} x^2} \prod_{2 \leq j \leq N} |x_{j} - x| \right|^2
\end{align*}
%
For large $p$, we can approximate the integral inside the modulus as
\[
\int dx e^{i p x} e^{-\frac{1}{2} x^2} \prod_{2 \leq j \leq N} |x_{j} - x| = \left( \dfrac{-2}{p^2} \right) \sum_{j = 2}^{N} e^{-\frac{1}{2} x_{j}^2} e^{i p x_{j}} \prod_{k \neq j} |x_{k} - x_{j}|
\]
Plug this back into the original expression and abbreviating the condition under the product symbol
\[
n_{B}(p) = \tilde{A}_{N} \frac{4}{p^4} \int dx_{2} \cdots \int dx_{N} \exp\left(- \sum_{i = 2}^{N} x_{i}^2\right) \prod_{ i < q} (x_{q} - x_{i})^2
	\sum_{j = 2}^{N} \sum_{\ell = 2}^{N} e^{-\frac{1}{2} (x_{j}^2 + x_{\ell}^2)} e^{i p (x_{j} - x_{\ell})}
	\prod_{k \neq j} |x_{k} - x_{j}| \prod_{m \neq \ell} |x_{m} - x_{\ell}|
\]
Now, whenever $x_{j} \neq x_{\ell}$ in the double sum, there will be $e^{i p (x_{j} - x_{\ell})}$ term. Elements in the double sum with this term (after they are integrated over) will lead to higher order dependencies (e.g., $1/p^2$), and once these terms are multiplied by $1/p^4$ in the prefactor, they will no longer be leading order. Thus, we can neglect any terms in the double sum where $x_{j} \neq x_{\ell}$, giving
\begin{align*}
n_{B}(p) & = \tilde{A}_{N} \frac{4}{p^4} \int dx_{2} \cdots \int dx_{N} \exp\left(- \sum_{i = 2}^{N} x_{i}^2\right) \prod_{ i < q} (x_{q} - x_{i})^2
	\sum_{j = 2}^{N} e^{-x_{j}^2} \prod_{k \neq j} (x_{k} - x_{j})^2\\
	& = \tilde{A}_{N} \frac{4}{p^4} \sum_{j = 2}^{N} \int dx_{2} \cdots \int dx_{N}
	\exp\left(- \sum_{i = 2}^{N} x_{i}^2\right) \left( \prod_{ i < q} (x_{q} - x_{i})^2 \right)
	e^{-x_{j}^2} \prod_{k \neq j} (x_{k} - x_{j})^2\\
\end{align*}

Because each integral in this sum is the same but with the indices rotated, we can write
\begin{align*}
n_{B}(p) & = \tilde{A}_{N} \frac{4 * (N - 1)}{p^4} \int dx_{2} \cdots \int dx_{N}
	\exp\left(- \sum_{i = 2}^{N} x_{i}^2\right) \left( \prod_{ i < q} (x_{q} - x_{i})^2 \right)
	e^{-x_{2}^2} \prod_{k \neq 2} (x_{k} - x_{2})^2\\
	& = \dfrac{C_{N}}{p^4}\\
C_{N} & = 4(N - 1)\tilde{A}_{N}  \int dx_{2} \cdots \int dx_{N}
	\exp\left(- \sum_{i = 2}^{N} x_{i}^2\right) \left( \prod_{ i < q} (x_{q} - x_{i})^2 \right)
	e^{-x_{2}^2} \prod_{k \neq 2} (x_{k} - x_{2})^2\\
\end{align*}
I have tried to simplify the expression for $A_{N}$ as much as I can, but I still have not been able to find an analytic expression for it. Nonetheless, this still shows that the momentum distribution has a $1/p^4$ momentum decay for arbitrary $N$. For numerical confirmation of this result, see next section

\section{momentum tail coefficient for hca (n-particle case)}

In this section, I will find the momentum tail for $N$ hard-core anyons. We will be able to use some of the results from the previous section to help us. The anyon-fermi mapping is given by
\[
A_{\kappa}(x_{1},x_{2}, \ldots, x_{N}) = \prod_{2 \leq j < k \leq N} e^{i\pi(1 - \kappa) \theta(x_{k} - x_{j})} = \exp\left[ i\pi(1-\kappa) \left( \sum_{j < k} \theta(x_{k} - x_{j}) \right) \right]
\]
Thus, the anyon wave function is
\[
\Psi_{\kappa}(x_{1}, \ldots, x_{N}) = A_{N} \times A_{\kappa}(x_{1},\ldots, x_{N}) \times \exp\left(-\frac{1}{2}
\sum_{i = 1}^{N} x_{i}^2\right) \prod_{1 \leq j < k \leq N} (x_{k} - x_{j})
\]
But as was mentioned in a previous section, we can also write this as
\[
\Psi_{\kappa}(x_{1}, \ldots, x_{N}) = A_{N}  \exp\left(-\frac{1}{2}
\sum_{i = 1}^{N} x_{i}^2\right) \exp\left( i\frac{\pi \kappa}{2} \Sigma_{j < k} \epsilon(x_{k} - x_{j}) \right) \prod_{1 \leq j < k \leq N} |x_{k} - x_{j}|
\]
Letting $\epsilon_{\kappa} \equiv \exp(i\frac{\pi \kappa}{2} \Sigma_{j=2}^N \epsilon(x_{j} - x))$ and $\tilde{A}_{N} = \frac{N}{2 \pi} A_{N}^2$, then the momentum distribution is
\begin{align*}
n_{\kappa}(p) & = \frac{N}{2\pi} \int dx_{2} \cdots \int dx_{N} \left| \int dx e^{ipx} \Psi_{\kappa}^*(x, x_{2}, \ldots, x_{N}) \right|^2\\
& = \frac{N}{2\pi} \int dx_{2} \cdots \int dx_{N} \left| \int dx e^{ipx} \exp(i\frac{\pi \kappa}{2} \Sigma_{j=2}^N \epsilon(x_{j} - x)) \Psi_{B}^*(x, x_{2}, \ldots, x_{N}) \right|^2\\
& = \tilde{A}_{N} \int dx_{2} \cdots \int dx_{N} \exp\left(- \sum_{i = 2}^{N} x_{i}^2\right) \left( \prod_{2 \leq i < q \leq N} (x_{q} - x_{i})^2 \right)
	\left| \int dx e^{i p x} e^{-\frac{1}{2} x^2} \epsilon_{\kappa} \prod_{2 \leq j \leq N} |x_{j} - x| \right|^2
\end{align*}
Expanding the $\epsilon_{\kappa}$ in the modulus square we obtain
\begin{align*}
\epsilon_{\kappa} & = e^{i\frac{\pi \kappa}{2}\epsilon(x_{2} - x)} \cdots e^{i\frac{\pi \kappa}{2}\epsilon(x_{N} - x)}\\
& = \left(\cos(\pi \kappa/2) + i\epsilon(x_{2} - x)\sin(\pi \kappa/2) \right)
\cdots \left(\cos(\pi \kappa/2) + i\epsilon(x_{N} - x)\sin(\pi \kappa/2)\right)
\end{align*}
To show how we should handle this product, it will be easier to look at the case for $N = 3$, where
\begin{align*}
 \int dx e^{ipx} \left(\cos(\pi \kappa/2) + i\epsilon(x_{2} - x)\sin(\pi \kappa/2)\right)
\left(\cos(\pi \kappa/2) + i\epsilon(x_{3} - x)\sin(\pi \kappa/2)\right) e^{-\frac{1}{2} x^2}  \prod_{2 \leq j \leq 3} |x_{j} - x| \\
= \int dx e^{ipx} \left( \cos^2(\pi \kappa/2)
+ i\frac{1}{2}\epsilon(x_{2} - x)\sin(\pi \kappa) + i\frac{1}{2}\epsilon(x_{3} - x)\sin(\pi \kappa)
 - \epsilon(x_{2} - x)\epsilon(x_{3} - x) \sin^2(\pi \kappa/2) \right) e^{-\frac{1}{2} x^2}  \prod_{2 \leq j \leq 3} |x_{j} - x|
\end{align*}
%
The integral over the first term can be found easily by using the calculations from the boson case. I will show that all other terms that have a $\epsilon$ function in them will not be leading order and can therefore be ignored. For instance, let's look at the integral over $x$ for the second term (ignoring prefactors)
%
\begin{align*}
\int dx e^{i p x} e^{-\frac{1}{2} x^2} \epsilon(x_{2} - x) |x_{2} - x| |x_{3} - x|
&= \int dx e^{i p x} e^{-\frac{1}{2} x_{2}^2} \epsilon(x_{2} - x) |x_{2} - x| |x_{3} - x_{2}| \\
& + \int dx e^{i p x} e^{-\frac{1}{2} x_{3}^2} \epsilon(x_{2} -x_{3}) |x_{2} - x_{3}| |x_{3} - x| + o (\frac{1}{p^2})\\
&= 0 - \frac{2}{p^2} e^{ipx_{3}}e^{-\frac{1}{2} x_{3}^2} \epsilon(x_{2} -x_{3}) |x_{2} - x_{3}|
\end{align*}

Using \textit{Applications of Fourier Transforms to Generalized Functions} by M. Rahman (see the table on p. 159) to evaluate these integrals. The first integral is zero because the fourier transform of $\epsilon(x_{2} - x) |x_{2} - x|$ with respect to $x$ is 0. Evaluating the integral with respect to $x$ over other terms in the sum yields
\begin{align*}
\int dx e^{i p x} e^{-\frac{1}{2} x^2} \epsilon_{\kappa} \prod_{2 \leq j \leq 3} |x_{j} - x| = \left( \frac{-2}{p^2} \right)
 \cos^2(\frac{\pi \kappa}{2}) \sum_{j=2}^{3} e^{-x_{j}^2/2 + ipx_{j}} |x_{3} - x_{2}|\\
 + \frac{i}{2} \sin(\pi \kappa) \left( \frac{-2}{p^2} \right) |x_{2} - x_{3}| \epsilon(x_{2} - x_{3})
 \sum_{j=2}^{3} (-1)^{j} e^{-x_{j}^{2}/2 + ipx_{j}}
\end{align*}
where the last term in the sum with two $\epsilon$ functions will evaluate to zero. Now, to find the momentum distribution for the $N = 3$ case, we must find the modulus square of this expression. I will leave out the derivation for the following result, but if you use the same reasoning for the boson case you will find that

\begin{align*}
\left| \int dx e^{i p x} e^{-\frac{1}{2} x^2} \epsilon_{\kappa} \prod_{2 \leq j \leq 3} |x_{j} - x| \right|^2 & =
\frac{4}{p^4} |x_{2}-x_{3}|^2 \left[ \cos^4(\pi \kappa /2) \sum_{j} e^{-x_{j^2}} + \frac{\sin^2(\pi \kappa)}{4} \epsilon(x_{2} - x_{3})^2 \sum_{j} e^{-x_{j}^2} \right]\\
& = \frac{4}{p^4} |x_{2}-x_{3}|^2 \sum_{j} e^{-x_{j^2}} \left[ \cos^4(\pi \kappa /2) + \frac{\sin^2(\pi \kappa)}{4} \right]\\
& = \cos^2(\pi \kappa /2)\frac{4}{p^4} |x_{2}-x_{3}|^2 \sum_{j} e^{-x_{j^2}}
\end{align*}
Comparing this with the $N$-particle expression for bosons, we can see that for the $3$ particle case that for large $p$
\[
n_{\kappa}(p) = \cos^2(\pi \kappa /2) \frac{C_{3}}{p^4}
\]

It is not clear that for the $N$-particle case whether the $\kappa$ dependence can be expressed as the power of a cosine function. Letting $\epsilon_{ij} \equiv \epsilon(x_{i} - x_{j}) $ and similarly $|x_{ij}| \equiv |x_{i} - x_{j}|$ recall that
%
\begin{align*}
n_{\kappa}(p) & = \tilde{A}_{N} \int dx_{2} \cdots \int dx_{N} \exp\left(- \sum_{i = 2}^{N} x_{i}^2\right) \left( \prod_{2 \leq i < q \leq N} (x_{q} - x_{i})^2 \right)
	\left| \int dx e^{i p x} e^{-\frac{1}{2} x^2} \epsilon_{\kappa} \prod_{2 \leq j \leq N} |x_{j} - x| \right|^2\\
\epsilon_{\kappa} & =  \left(\cos(\pi \kappa/2) + i\epsilon(x_{2} - x)\sin(\pi \kappa/2) \right)
\cdots \left(\cos(\pi \kappa/2) + i\epsilon(x_{N} - x)\sin(\pi \kappa/2)\right)\\
& = \sum_{q = 1}^{N} \cos^{N-q}(\pi \kappa /2) (i \sin(\pi \kappa /2))^{q-1}
\sum_{2\leq i_{1} < \ldots < i_{q - 1} \leq N} \epsilon(x_{i_{1}} - x) \cdots \epsilon(x_{i_{q-1}} - x)
\end{align*}

So making extensive use of Theorem 4.4 from Rahman, and abbreviating the summation condition for $i_{1}, i_{2}, \ldots, i_{q - 1}$
\begin{align*}
\int & dx e^{i p x} e^{-\frac{1}{2} x^2} \epsilon_{\kappa} \prod_{2 \leq j \leq N} |x_{j} - x| =\\
-\frac{2}{p^2}& \left[ \sum_{q = 1}^{N} \cos^{N-q}(\pi \kappa /2) (i \sin(\pi \kappa /2))^{q-1}
 \sum_{i_{1} < \ldots < i_{q - 1}} \sum_{j \neq i_{1}, \ldots, i_{q-1}} e^{-x_{j}^2/2 + ipx_{j}} \epsilon_{i_{1}j} \cdots \epsilon_{i_{q-1}j} \prod_{k \neq j} |x_{kj}| \right]
\end{align*}

The modulus square of this expression is then
\begin{align*}
 & \left| \int dx e^{i p x}  e^{-\frac{1}{2} x^2} \epsilon_{\kappa} \prod_{2 \leq j \leq N} |x_{j} - x| \right|^2 =\\
\frac{4}{p^4}  \sum_{q = 1}^{N} & \sum_{r = 1}^{N} \cos(\pi \kappa /2)^{2N-q-r} (i \sin(\pi \kappa /2))^{r+q-2} (-1)^{r-1}
\sum_{i_{1} < \ldots < i_{q - 1}} \sum_{j_{1} < \ldots < j_{r - 1}}\\
&\sum_{k \neq i}
\sum_{\ell \neq j}
e^{-(x_{k}^{2} + x_{\ell}^{2})/2} e^{ip(x_{k} - x_{\ell})}
\epsilon_{i_{1}k} \cdots \epsilon_{i_{q-1}k}
\epsilon_{j_{1}\ell} \cdots \epsilon_{j_{r-1}\ell}
\prod_{\alpha \neq k} |x_{\alpha k}|
\prod_{\alpha \neq \ell} |x_{\alpha \ell}|
\end{align*}
where the sums are all supposed to be in one line but I have split them into two to fit them on the page. Suppose that $k \neq \ell$. Then the integral with respect to $x_{k}$ will be a fourier transform. This can only lead to vanishing terms or non-leading order terms. Due to all the singularities, this fourier transform will be split into a sum over every singularity as per Theorem 4.4. If every fourier transform in this resultant sum is negligible, then the fourier transform overall is negligible. Here I have enumerated every possible fourier transform that could result

\begin{enumerate}
\item $\mathcal{F}[|x_{k} - x_{i}|] = \mathcal{O}(\frac{1}{p^2})$
\item $\mathcal{F}[\epsilon(x_{k} - x_{i})|x_{k} - x_{i}|] = 0$
\item $\mathcal{F}[\epsilon(x_{k} - x_{i})|x_{k} - x_{i}|^2] = \mathcal{O}(\frac{1}{p^3})$
\item $\mathcal{F}[\epsilon(x_{k} - x_{i})^2|x_{k} - x_{i}|^2] = \mathcal{O}(\frac{1}{p^3})$
\item $\mathcal{F}[|x_{k} - x_{i}|^2] = \mathcal{O}(\frac{1}{p^3})$
\end{enumerate}
When any of these terms is multiplied by the $\frac{4}{p^4}$ prefactor, that term will no longer be leading order, and it can therefore be ignored. So, we can ignore any terms with $k \neq \ell$ giving us

\begin{align*}
\frac{4}{p^4}  \sum_{q = 1}^{N} & \sum_{r = 1}^{N} \cos(\pi \kappa /2)^{2N-q-r} (i \sin(\pi \kappa /2))^{r+q-2} (-1)^{r-1}
\sum_{i_{1} < \ldots < i_{q - 1}} \sum_{j_{1} < \ldots < j_{r - 1}}\\
&\sum_{k \neq i, j}
e^{-x_{k}^{2}}
\epsilon_{i_{1}k} \cdots \epsilon_{i_{q-1}k}
\epsilon_{j_{1}k} \cdots \epsilon_{j_{r-1}k}
\prod_{\alpha \neq k} |x_{\alpha k}|^2
\end{align*}

%\section{two anyons in harmonic trap with finite contact interaction}
%Consider the Hamiltonian for two anyons:
%\[ H = -\frac{1}{2} \left( \partial^2_{x_1} +\partial^2_{x_2} \right) + \frac{1}{2}(x_1^2+x_2^2) + g\delta(x_1-x_2)  \] This can be decomposed to the COM and relative coordinates. The COM Hamiltonian is just the harmoinc oscillator, while the relative Hamiltonian takes the form ($x\equiv x_1-x_2$): \[ H_{\rm rel} = -\frac{1}{2} \partial^2_x + \frac{1}{2} x^2 + g\delta(x) \] The relative wavefunction has to satisfiy \[  \phi(x) = e^{i\pi \kappa \epsilon(x)} \,\phi(-x) \] This constraint, however, seems to indicate that the wavefunction is discontinuous at $x=0$. If this is the case, then the second order derivative term will result in a derivative in delta function, which cannot be cancelled out. The paper PRL {\bf 83}, 1275 (1998) seems to be relevant, but I am not sure how they got from Eq. (16) to (17).

\end{document}


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\begin{document}

\begin{center}
\Large \bf On two integrals appearing in the relativistic test particle operator.
\end{center}

We need to evaluate
\begin{align}
\psi_0 &= \int_0^p \frac{\exp{\Bigl[-(\sqrt{1+s^2}-1)/\Theta\Bigr]}}{\sqrt{1+s^2}} \, \rd s \nonumber \\
&= \int_0^\infty \frac{\exp{\Bigl[-(\sqrt{1+s^2}-1)/\Theta\Bigr]}}{\sqrt{1+s^2}} \, \rd s  - \int_p^\infty \frac{\exp{\Bigl[-(\sqrt{1+s^2}-1)/\Theta\Bigr]}}{\sqrt{1+s^2}} \, \rd s, \nonumber \\
\psi_1 &= \int_0^p \exp{\Bigl[-(\sqrt{1+s^2}-1)/\Theta\Bigr]} \, \rd s \nonumber \\
&= \int_0^\infty \exp{\Bigl[-(\sqrt{1+s^2}-1)/\Theta\Bigr]} \, \rd s  - \int_p^\infty \exp{\Bigl[-(\sqrt{1+s^2}-1)/\Theta\Bigr]} \, \rd s.
\end{align}
The two definite integrals can be expressed in terms of the modified Bessel functions $K$ as
\begin{align}
\int_0^\infty \frac{\exp{\Bigl[-(\sqrt{1+s^2}-1)/\Theta\Bigr]}}{\sqrt{1+s^2}} \, \rd s &=  e^{1/\Theta}K_0\left(\frac{1}{\Theta}\right) \nonumber \\
\int_0^\infty \exp{\Bigl[-(\sqrt{1+s^2}-1)/\Theta\Bigr]} \, \rd s   &= e^{1/\Theta}K_1\left(\frac{1}{\Theta}\right).
\end{align}
The remainder can be rewritten, using the change of variables
\begin{align}
x &= \sqrt{1+s^2}, \nonumber \\
\rd x &= \frac{s}{x}\rd s, \nonumber \\
s &= \sqrt{x^2-1},
\end{align}
so that
\begin{align}
\int_p^\infty \frac{\exp{\Bigl[-(\sqrt{1+s^2}-1)/\Theta\Bigr]}}{\sqrt{1+s^2}} \, \rd s &= \int_\gamma^\infty \frac{\exp{\Bigl[-(x-1)/\Theta\Bigr]}}{\sqrt{x^2-1}} \, \rd x, \nonumber \\
&= e^{-(\gamma-1)/\Theta}   \int_\gamma^\infty \frac{\exp{\Bigl[-(x-\gamma)/\Theta\Bigr]}}{\sqrt{x^2-1}} \nonumber \\
\int_p^\infty \exp{\Bigl[-(\sqrt{1+s^2}-1)/\Theta\Bigr]} \, \rd s &= \int_\gamma^\infty \frac{x}{\sqrt{x^2-1}}\exp{\Bigl[-(x-1)/\Theta\Bigr]} \, \rd x, \nonumber \\
&= e^{-(\gamma-1)/\Theta}\int_\gamma^\infty \frac{x}{\sqrt{x^2-1}}\exp{\Bigl[-(x-\gamma)/\Theta\Bigr]} \, \rd x.
\end{align}
We can therefore write them on the Laguerre integration form
\begin{align}
\psi_0 &= e^{1/\Theta}K_0(1/\Theta) - e^{-(\gamma-1)/\Theta} \int_\gamma^\infty \frac{1}{\sqrt{x^2-1}} w(x) \, \rd x , \nonumber \\
\psi_1 &= e^{1/\Theta}K_1(1/\Theta) - e^{-(\gamma-1)/\Theta} \int_\gamma^\infty \frac{x}{\sqrt{x^2-1}} w(x) \, \rd x , \nonumber \\
w &= \exp\Bigl[-(x-\gamma)/\Theta\Bigr].
\end{align}
To obtain an energy-independent quadrature we can translate $x-\gamma \mapsto x$.

\section{Partial temperature derivative}
For the evaluation of the Jacobian, we need to evaluate (with $\Theta = T/mc^2$)
\begin{align}
\frac{\partial \psi_0}{\partial T} &= \frac{1}{mc^2\Theta^2}\int_0^p \left(1-\frac{1}{\sqrt{1+s^2}}\right) \exp\Bigl[-(\sqrt{1+s^2}-1)/\Theta\Bigr] \nonumber \\
&= \frac{\psi_1 - \psi_0}{T\Theta}, \\
\frac{\partial \psi_1}{\partial T} &= \frac{1}{mc^2\Theta^2}\int_0^p \left(\sqrt{1+s^2}-1\right) \exp\Bigl[-(\sqrt{1+s^2}-1)/\Theta\Bigr]  \nonumber \\
&= \frac{\psi_2 - \psi_1}{T\theta},
\end{align}
where we introduce
\begin{align}
\psi_2 &= \int_0^p \sqrt{1+s^2} \exp{\Bigl[-(\sqrt{1+s^2}-1)/\Theta\Bigr]} \, \rd s \nonumber \\
&= e^{1/\Theta}[K_0(1/\Theta) + \Theta K_1(1/\Theta)] - e^{-(\gamma-1)/\Theta}\int_\gamma^\infty \frac{x^2}{\sqrt{x^2-1}}w(x)\,\rd x
\end{align}

\section{Asymptotic expansions}
In this section we give asymptotic expansions of the $\psi$ functions which can be useful in certain limits.

\subsection{Superthermal limit, $\gamma - 1 \gg \Theta$}

For the evaluation of the three special functions $\psi_n$, we require the evaluation of the integrals
\begin{align}
P_k = \int_{\gamma}^\infty \frac{x^k}{\sqrt{x^2-1}}e^{-(x-\gamma)/\Theta}\,\rd x,
\end{align}
for $k=0,\,1,\,2$. A useful asymptotic expansion valid for $\Theta \ll 1$ is obtained by repeated integration by parts
\begin{align}
P_k &= \Theta \frac{\gamma^k}{\sqrt{\gamma^2-1}} + \int_\gamma^\infty \frac{\rd}{\rd x}\left(\frac{x^k}{\sqrt{x^2-1}}\right)e^{-(x-\gamma)/\Theta} \rd x \nonumber \\
&= ... \nonumber \\
&= \Theta \sum_{n=0}^\infty \Theta^n  \left(\frac{\rd}{\rd \gamma}\right)^n\left(\frac{\gamma^k}{\sqrt{\gamma^2-1}}\right).
\end{align}
Carrying out this expansion to second-to-leading order yields
\begin{align}
P_k &\sim \Theta \frac{\gamma^k}{p} + \Theta^2 \frac{\gamma^{k-1}[(k-1)p^2 - 1]}{p^3} + \Ordo(\Theta^3/p^5).
\end{align}
This approximation is accurate in the superthermal limit: $p \gg \sqrt{\Theta}$.

This produces the following expressions for the special functions:
\begin{align}
\psi_0 &\sim e^{1/\Theta}K_0\left(\frac{1}{\Theta}\right) - \frac{\Theta}{p} e^{-(\gamma-1)/\Theta} + \Theta^2\frac{\gamma}{ p^3}e^{-(\gamma-1)/\Theta} \nonumber \\
\psi_1 &\sim e^{1/\Theta}K_1\left(\frac{1}{\Theta}\right) - \Theta \frac{\gamma}{p}e^{-(\gamma-1)/\Theta} +\frac{\Theta^2 }{p^3}e^{-(\gamma-1)/\Theta}  \\
\psi_2 &\sim e^{1/\Theta}K_0\left(\frac{1}{\Theta}\right) + \Theta\left[ e^{1/\Theta} K_1\left(\frac{1}{\Theta}\right)  - \frac{\gamma^2}{p}e^{-(\gamma-1)/\Theta}\right] - \Theta^2 \frac{\gamma(\gamma^2-2)}{p^3} e^{-(\gamma-1)/\Theta}\nonumber
\end{align}
At $\gamma-1 = 10\Theta$, each of these expanded forms have a relative error of approximately $10^{-8}$, compared to the exact expressions.

\subsection{Non-relativistic or low-energy limit, $\mathrm{max}(\Theta,\,\gamma-1)\ll 1$}
In the non-relativistic limit, we may use the alternative form of the $\psi$ functions:
\begin{align}
\psi_n = \int_0^p (1+s^2)^{(n-1)/2} \exp\Big[ - (\sqrt{1+s^2} - 1)/\Theta \Big] \rd s.
\end{align}
By making a change of variables
\begin{align}
\frac{\sqrt{1+s^2}-1}{\Theta} &= x^2 , \nonumber \\
\frac{1}{\Theta}\frac{s\rd s}{\sqrt{1+s^2}} &= 2x\rd x, \nonumber \\
s &= \sqrt{ (1+\Theta x^2)^2 - 1} = x\sqrt{2\Theta}\sqrt{1+ \frac{1}{2}\Theta x^2} ,
\end{align}
the integral is recast into
\begin{align}
\psi_n &= \sqrt{2\Theta} \int_0^{x_m(p)} \frac{(1+\Theta x^2)^{n}}{\sqrt{1+\frac{1}{2}\Theta x^2}} e^{-x^2} \, \rd x, \nonumber \\
x_m &= \sqrt{\frac{\gamma-1}{\Theta}}.
\end{align}
The integrand can be Taylor expanded with
\begin{align}
\frac{(1+\Theta x^2)^{n}}{\sqrt{1+\frac{1}{2}\Theta x^2}} \sim 1 + \frac{4n-1}{4}\Theta x^2 + \frac{1}{4}\left(2n^2 - 3n + \frac{3}{8}\right)\Theta^2 x^4 + \Ordo\bigl(\Theta^3 x^6\bigr).
\end{align}
Using the integral identities
\begin{align}
\int_0^{x_m} e^{-x^2} \,\rd x &= \frac{\sqrt{\pi}}{2}\mathrm{erf}(x_m), \nonumber \\
\int_0^{x_m} x^2e^{-x^2} \,\rd x &= \frac{\sqrt{\pi}}{4}\mathrm{erf}(x_m) - \frac{1}{2} x_m e^{-x_m^2}, \nonumber \\
\int_0^{x_m} x^4e^{-x^2} \,\rd x &= \frac{3\sqrt{\pi}}{8}\mathrm{erf}(x_m) - \frac{3+2x_m^2}{4}x_m e^{-x_m^2},
\end{align}
we obtain
\begin{align}
\frac{1}{\sqrt{2\Theta}}\psi_n &\sim \left[ 1 + \frac{4n-1}{8}\Theta + \frac{3}{16}\left(2n^2- 3n + \frac{3}{8}\right)\Theta^2 \right] \frac{\sqrt{\pi}}{2}\mathrm{erf}(x_m) \nonumber \\
& - \frac{\Theta}{8} x_m e^{-x_m^2} \left[ 4n-1 + \Theta \frac{3+2x_m^2}{2}\left(2n^2-3n+\frac{3}{8}\right)  \right]
\end{align}
These expressions have a relative error $<10^{-8}$ when $\gamma-1 < 0.01$ (i.e. when $p<0.15$) or when $\Theta < 0.005$ (corresponding to temperatures $T < 2.5\,$keV).

%\section{Lorentz conductivity calculation}
%Consider the equation
%\begin{align}
%\frac{1}{\mathcal{V}'}\frac{\partial}{\partial p}\left(\mathcal{V}'\{A^p\} f\right) + \frac{1}{\mathcal{V}'}\frac{\partial}{\partial \xi}\left(\mathcal{V}' \{A^\xi\} f\right) +  = \frac{1}{\mathcal{V}'}\frac{\partial}{\partial \xi}\left(\mathcal{V}'\{D^{\xi\xi}\}\frac{\partial f}{\partial \xi} \right)
%\end{align}
%in the limit $D \gg A$ for which $f = f_0 + f_1+ ...$ where $f_0 = f_0(p)$. The equation then takes the form
%\begin{align}
%\frac{1}{\mathcal{V'}}\left(\frac{\partial \mathcal{V}'\{A^p\}}{\partial p} + \frac{\partial  \mathcal{V}'\{A^\xi\}}{\partial \xi}\right)f_0 + \{A^p\}\frac{\partial f_0}{\partial p} = \frac{1}{\mathcal{V}'}\frac{\partial}{\partial \xi}\left(\mathcal{V}'\{D^{\xi\xi}\}\frac{\partial f_1}{\partial \xi} \right)
%\end{align}
%Integrating over $\rd \xi \, \mathcal{V}'$ from $\xi = \xi$ to $1$ (where $\{A^\xi\}$ and $\{D^{\xi\xi}\}$ both vanish for $\xi=1$) yields
%\begin{align}
%\frac{\partial}{\partial p}\left( f_0\int_\xi^1\rd \xi \, \mathcal{V}'\{A^p\}\right) - \mathcal{V}'\{A^\xi\} f_0 = -\mathcal{V}'\{D^{\xi\xi}\}\frac{\partial f_1}{\partial \xi}.
%\end{align}
%Dividing by $\mathcal{V'}\{D^{\xi\xi}\}$ and again integrating over $\xi$, from $0$ to $\xi$, yields
%\begin{align}
%
%\end{align}

\end{document}


%!TEX root = ../thesis.tex
%*******************************************************************************
%****************************** Third Chapter **********************************
%*******************************************************************************
\chapter{Reduced Order Modeling of Comb-Drive in Electrolytes}

% **************************** Define Graphics Path **************************
\ifpdf
    \graphicspath{{Chapter3/Figs/Raster/}{Chapter3/Figs/PDF/}{Chapter3/Figs/}}
\else
    \graphicspath{{Chapter3/Figs/Vector/}{Chapter3/Figs/}}
\fi

\section{Summary}
The focus of this chapter will be the development of reduced order models for the comb-drive actuator in electrolytes. We will first review the Poisson-Nernst-Planck equations for describing ionic liquids between parallel plates. This will be followed by a discussion of how these equations facilitate the conceptualization of the system as a circuit, and how this conceptualization was used to develop the classic model. In particular, the assumptions behind the classic model will be enumerated, as will it's shortcomings when fit to data. Finally, we will present our newly developed models, the assumptions behind this work, and opportunities for further work with regard to these models.

\section{Poisson-Nernst-Planck (PNP) Equations}
We saw in Chapter 2 that modeling a comb-drive actuator in electrolytes reduces to modeling the overlapping regions of its fingers as parallel plates filled with a dilute ionic solution. This system is classically described by the Poisson-Nernst-Planck (PNP) equations. The one-dimensional PNP equations are

\begin{align}
    \frac{\partial c_\pm}{\partial t} = - D\frac{\partial}{\partial x}\bigg(-\frac{\partial c_\pm}{\partial x} \mp  \frac{ze}{k_B T}c_\pm \frac{\partial \phi}{\partial x}\bigg) \label{pnp_equations_1} \\
    -\epsilon \frac{\partial^2 \phi}{\partial x^2} = \rho_e = ze(c_+ - c_-) \label{pnp_equations_2}
\end{align}
where $c_+$ and $c_-$ are the concentrations of the negative and positive ions respectively, $\phi$ is the electric potential, D is the diffusivity of the electrolyte, which is assumed to be constant throughout the domain, $z$ is the valence number of the electrolytes, $k_B$ is the Boltzmann constant, $T$ is the temperature of the domain, $\epsilon$ is the relative permittivity of the aqueous solution, $\rho_e$ is the electric charge density, and $e$ is the charge of an electron.

Before performing further analysis on the PNP equations, it is important that we do two things. First, we non-dimensionalize the PNP equations, and then we communicate the physical intuition of the resulting equations. We non-dimensionalize the PNP equations as was done by Druzgalski et al \cite{Druzgalski2013}. The potential, time, concentration, and spatial coordinates are non-dimensionalized as

\begin{equation}\label{nondim_quants}
    \phi^* = \frac{\phi}{V_T}, \quad t^* = \frac{Dt}{g^2}, \quad  c^*_0 = \frac{c}{c_0}, \quad  x^* = \frac{x}{g},
\end{equation}
where $V_T = \frac{k_B T}{e}$ is the thermal voltage, $g$ is the length of the gap between comb-drive fingers, $c_0$ is the bulk concentration of an ion, and the rest of the terms in the equation are defined after \ref{pnp_equations_1} and \ref{pnp_equations_2}. Using these terms, the non-dimensional form of the PNP equations is
\begin{align}
    \frac{\partial c_\pm^*}{\partial t^*} = - \frac{\partial}{\partial x^*}\bigg(-\frac{\partial c_\pm^*}{\partial x^*} \mp  c_\pm^* \frac{\partial \phi^*}{\partial x^*}\bigg) \label{pnp_equations_nondim_1} \\
    -2 \gamma^2 \frac{\partial^2 \phi^*}{\partial x^{*2}} = \rho_e^* = c_+^* - c_-^* \label{pnp_equations_nondim_2}
\end{align}
where $\gamma = \frac{\lambda_d}{g}$. The summation in \ref{pnp_equations_nondim_1} is essentially the nondimensional current flux, $f_\pm^*$, of the positive and negative ions respectively.

\subsection{Circuit View of System using Linearized PNP Equations}
We have introduced and non-dimensionalized the PNP equations, as well as given physical intuition regarding its non-dimensional form. We will now use the non-dimensionalized PNP equations to explain the representation of electrolytes between parallel plates as a circuit.

We first make the critical assumption that the voltage applied to the parallel plate is small. A small voltage is defined as a voltage that is at most the same order of magnitude as the thermal voltage, $V_T$, which is about 25 $\mV$. When the applied voltage is small, the region between the parallel plates is described as three distinct parts, figure \ref{conc_profile}. The center of the domain is called the bulk, and is characterized by an equal concentration of positive and negative ions, low concentration gradients, and a linear potential profile. The two regions located close to the parallel plates, however, are characterized by high concentration gradients, unequal amounts of positive and negative ions, and a nonlinear potential profile. The concentration of positive ions is much higher than the negative at the parallel plate with a lower potential, and vice versa at that with one that is higher.

\begin{figure}[htpb]
    \begin{center}
    \includegraphics[width=0.7\linewidth]{Chapter3/figure/conc_profile.png}
    \caption{Schematic of concentration profile of negative and positive ions of electrolyte between parallel plates.}\label{conc_profile}
    \end{center}
\end{figure}

We can apply the non-dimensional PNP equations to these three regions separately, and use the assumption of a small applied voltage, to develop a circuit model for the system. We start with the bulk region in which the concentration gradient is zero, $\frac{\partial c_\pm^*}{\partial x^*} = 0$, and the potential profile is linear, $\frac{\partial \phi^*}{\partial x^*}=a \implies \frac{\partial^2 \phi^*}{\partial x^{*2}}=0$, where $a$ is some constant. Using these assumptions, \ref{pnp_equations_nondim_1} and \ref{pnp_equations_nondim_2} reduce to

\begin{align}
    \frac{\partial c_\pm^*}{\partial t^*} = - \frac{\partial}{\partial x^*}\bigg( \mp  c_\pm^* \frac{\partial \phi^*}{\partial x^*}\bigg) = 0 \label{pnp_equations_simple_nondim_1} \\
    -\gamma^2 \frac{\partial^2 \phi^*}{\partial x^{*2}} = \rho_e^* = c_+^* - c_-^* = 0\label{pnp_equations_simple_nondim_2}
\end{align}

So, in the bulk we can assume that the concentration of positive and negative ions are equal, and that the concentration does not change as a function of time. We can now write the net current flux in the system as
\begin{align}
    j = f_+^* - f_-^* & = (c_+^* + c_-^*)\frac{\partial \phi^*}{\partial x^*} \label{bulk_current_flux_1} \\
    & = 2c_0^*\frac{\partial \phi^*}{\partial x^*} \label{bulk_current_flux_2}
\end{align}
where \ref{bulk_current_flux_2} is obtained by using the fact that $c_+^* = c_-^*=c_0^*=1$ in the bulk. Equation \ref{bulk_current_flux_2} has the form $I=\frac{V}{R}$. We can, therefore, view the non-dimensional resistance and voltage of the bulk region, respectively, as

\begin{equation} \label{bulk_resist_volt}
    R_{Bulk}^* = \frac{1}{2c_0^*}, \quad V_{Bulk}^* = \int_{0}^{g} \frac{\partial \phi^*}{\partial x^*} dx^*.
\end{equation}

We can perform a similar analysis on the electric double layer regions between the parallel plates. In this region, the poisson equation is
\begin{align}
    -\gamma^2 \frac{\partial^2 \phi^*}{\partial x^{*2}} & = c_+^* - c_-^* = \rho_e^* \label{poisson_nondim_cap_1} \\
    &  = \text{sinh}(\frac{\phi^*}{2}) = \frac{dq^*}{dx^*}\nonumber
\end{align}
where \ref{poisson_nondim_cap_1} is obtained by assuming a boltzmann distribution.
\begin{equation} \label{boltzmann_distrib}
    c_+^* = c_0^* e^{-\frac{\phi z e}{k_B T}}, \quad c_-^* = c_0^* e^{\frac{\phi z e}{k_B T}}
\end{equation}
 Under the assumption of linearity, we rewrite \ref{poisson_nondim_cap_1} as
\begin{equation}\label{poisson_nondim_ode}
\gamma^2 \frac{d^2\phi^*}{dx^{*2}} - \phi^* =0.
\end{equation}.
Solving this ODE, and taking a first order derivative of it's solution, yields
\begin{equation} \label{poisson_nondim_odesol}
\frac{d\phi^*}{dx^*} = -\gamma^{-1} \phi^*,
\end{equation}
We now rewrite \ref{poisson_nondim_cap_1}, using the solution of \ref{poisson_nondim_ode}, as
\begin{equation} \label{poisson_nondim_odesol_charge}
-2 \bigg(\frac{\lambda_d}{g}\bigg)^2 \frac{d^2\phi^*}{dx^{*2}} = -2 \phi^* = \frac{dq^*}{dx^*}
\end{equation}
Equations \ref{poisson_nondim_odesol} and \ref{poisson_nondim_odesol_charge} indicate that the boundary regions between the electrolytes can be represented as a capacitor. We obtain the dimensionless linear differential capacitance of this region by taking the quotient of \ref{poisson_nondim_odesol} and \ref{poisson_nondim_odesol_charge}, yielding
\begin{equation} \label{lin_diff_cap}
\frac{dq^*}{dx^*}\frac{dx^*}{d\phi^*} = \frac{dq^*}{d\phi^*} = 2\frac{\lambda_d}{g} = 2\gamma = C_0
\end{equation}

Based on this analysis, it is clear that we can view electrolytes between parallel plates as a circuit composed of boundary oxide capacitors, and a bulk resistor, figure \ref{classic_circuit}.

\section{Classic Circuit Model of Comb-Drive Actuator in Electrolyte}
In this section, we will discuss the classic model of the comb-drive in electrolytes, which is based on the circuit model derived in the previous section, figure \ref{classic_circuit}.

\begin{figure}[htpb]
    \centering
    \begin{minipage}{\textwidth}
        \centering
        \includegraphics[width=0.6\textwidth]{Chapter3/figure/classic_circuit.png} % first figure itself
        \caption{Circuit schematic of the classic circuit model for describing a comb-drive actuator in electrolytes. It is composed of an oxide capacitor at the boundary, and a parallel resistor/capacitor in the bulk.}\label{classic_circuit}
    \end{minipage}\vfill
    \begin{minipage}{\textwidth}
        \centering
        \includegraphics[width=0.8\textwidth]{Chapter3/figure/leaky_ox_circuit.png} % second figure itself
        \caption{Circuit schematic of the leaky circuit model for describing a comb-drive actuator in electrolytes. In this case the boundary is composed of a parallel resistor and capacitor to represent the oxide.}\label{leaky_ox_circuit}
    \end{minipage} \vfill
    \begin{minipage}{\textwidth}
        \centering
        \includegraphics[width=0.8\textwidth]{Chapter3/figure/leaky_ox_stern_circuit.png} % second figure itself
        \caption{Circuit schematic of the leaky circuit model for describing a comb-drive actuator in electrolytes. It is similar to the leaky oxide schematic, except a stern layer is placed in series with the oxide resistor/capacitor.}\label{leaky_ox_stern_circuit}
    \end{minipage} \vfill
\end{figure}

\begin{comment}
\begin{figure}[htpb]
    \begin{center}
    \includegraphics[width=0.7\linewidth]{Chapter3/figure/classic_circuit.png}
    \caption{Circuit schematic of the classic circuit model for describing a comb-drive actuator in electrolytes. It is composed of an oxide capacitor at the boundary, and a parallel resistor/capacitor in the bulk.}\label{classic_circuit}
    \end{center}
\end{figure}
\end{comment}

The force, $F$, experienced by a comb-drive actuator in air can be calculated as
\begin{equation}\label{air_force_comb}
F = kx = \dfrac{Nb\epsilon_0}{g}V_{App}^2,
\end{equation}
where \textit{k}, the stiffness of the comb-drive, \textit{x}, its displacement, \textit{N}, the total number of comb-pair fingers, \textit{b}, the comb finger thickness (into the page), and \textit{g}, the distance between overlapping fingers, are properties of the comb-drive.  $\epsilon_0$ is the permittivity of free space, and $V_{App}$ is the voltage applied to an overlapping pair of fingers. In this scenario, fringe effects of the comb-drive tip can be ignored. This simplified model is most accurate with sufficient overlap between fingers\cite{Ye1998}. The classic model constructed by Mukundan et al. and Panchawagh et al. can be expressed by modifying \ref{air_force_comb} \cite{MukundanandPonce2009,Panchawagh2009},
\begin{equation}\label{classic_force_comb}
F = \dfrac{Nb\epsilon_0\epsilon_W}{g}f(\omega)V_{AppRMS}^2 = \dfrac{Nb\epsilon_0\epsilon_W}{g}f(\omega)\bigg(\dfrac{V_{App}}{\sqrt[]{2}}\bigg)^2,
\end{equation}
where $V_{AppRMS}$ is the root-mean-square of the applied voltage, $\epsilon_W$ is the relative permittivity of water, and $f(\omega)$ is a function of the applied frequency that is equal to \cite{MukundanandPonce2009,Panchawagh2009}
\begin{equation} \label{classic_frequency_eqn}
f(\omega) = \bigg|\dfrac{Z_{Bulk}}{Z_{Bulk}+2*Z_{Interface}(\omega)}\bigg| ^2, \quad Z_{Interface}=Z_{Ox}.
\end{equation}
Here $Z_{Bulk}$ is the impedance of the bulk resistor, and $Z_{Interface}$ the impedance of the interface between the bulk electrolyte and the electrode. In this case, $Z_{Interface}$ is equivalent to $Z_{Ox}$, the impedance of the native oxide capacitor. These impedances are expressed as
\begin{equation} \label{bulk_oxide_impedances}
Z_{Bulk} = R_{Bulk}, \quad Z_{Ox} = \dfrac{1}{j\omega C_{Ox}}.
\end{equation}
We note that $C_{Bulk}$ is ignored, because $C_{Ox}$ dominates the capacitive behavior in the range of frequency we consider. From equations \ref{classic_force_comb}-\ref{classic_frequency_eqn}, the time-constant that governs the frequency required to overcome ionic shielding, and facilitate actuation, is \cite{MukundanandPonce2009,Panchawagh2009}
\begin{equation} \label{classic_timeconst}
\tau = R_{Bulk}C_{Ox}.
\end{equation}
At higher concentrations the bulk resistance, $R_{Bulk}$, decreases so higher frequencies of applied voltages are required to overcome shielding, and facilitate actuation.

\subsection{Assumptions and Validity of Classic Circuit Model}
The classic model, while a major breakthrough in modeling electrostatic comb-drive actuators in electrolytes, makes three major assumptions: i) that the native oxide is a pure dielectric, ii) that the ion concentration of the bulk electrolyte is constant, and iii) that the Stern layer can be neglected compared to the oxide layer. Figure \ref{classic_circuit_fit} shows the displacement of a comb-drive actuator in KCl at concentrations of 0.1 mM, 0.5 mM, and 1 mM respectively. We see that, while the Classic model accurately captures the trend of displacement, it tends to significantly underestimate the displacement in the low frequency region, and overestimate it in the intermediate frequency regions.

In order to overcome these shortcomings, we develop a hierarchy of novel models that addresses assumptions i)-iii) both individually and in concert. Subsequently, we select the simplest model that explains the comb-drive displacement in electrolytes. The developed models are 1) Leaky Oxide, 2) Leaky Oxide + Stern, 3) Variable Resistor, and 4) Leaky Oxide + Stern + Variable Resistor models. These models, as well as the assumptions of the Classic Circuit model that they address, are listed in Table \ref{hiearchy_table}. We find that the model which removes assumptions i) and ii) is sufficient to accurately predict the displacement of a comb-drive actuator in electrolytes.

\begin{figure}[htpb]
    \begin{center}
    \includegraphics[width=0.7\linewidth]{Chapter3/figure/placeholder.png}
    \caption{Placeholder for classic model fit to comb-drive actuator displacement data at 0.1 mM, 0.5 mM, and 1 mM $KCl$ respectively}\label{classic_circuit_fit}
    \end{center}
\end{figure}

\begin{table}[!htb]
\begin{center}
{\begin{tabular}{|c|c|}
	\hline
	\textbf{Model (i)} & \textbf{Classic Model} \\
     &  \textbf{Assumptions Removed} \\
    \hline
    Leaky Oxide (1) & i) Pure Dielectric Oxide\\
    \hline
    Leaky Oxide  & i) Pure Dielectric Oxide \\
    + Stern (2)    & ii)  No Stern layer \\
    \hline
    Variable Resistor (3) & iii) Constant Bulk Resistor \\
    \hline
    Leaky Oxide+ & i) Pure Dielectric Oxide\\
    Stern+   & ii)  No Stern layer \\
     Variable Resistor (4) & iii) Constant Bulk Resistor  \\
    \hline
\end{tabular}}
\caption{Hierarchy of models developed in this paper, and the classic circuit model assumptions they address}\label{hiearchy_table}
\end{center}
\end{table}

\section{Novel Circuit Models of Comb-drive Actuator in Electrolytes}
In this section, we outline the development of the set of models summarized in table \ref{hiearchy_table}. In order to represent these models, we re-express the frequency dependence of the force on the comb-drive actuator, \ref{classic_force_comb}, as
\begin{equation}\label{general_red_force_comb}
F = \dfrac{Nb\epsilon_0\epsilon_W}{g}h_i(\omega)V_{AppRMS}^2
\end{equation}
where $h_i$ is the function of frequency that corresponds to the Leaky Oxide, Leaky Oxide + Stern, Variable Resistor, and Leaky Oxide + Stern + Variable Resistor models (i=1,2,3,4). Explicitly,

\begin{align}\label{gen_freq_eqn}
h_i(\omega) = \bigg|\dfrac{Z_{Bulk}}{Z_{Bulk}+2*Z_i(\omega)}\bigg|^2,  i=1,2,3,4,5
\end{align}

The Leaky Oxide and Leaky Oxide + Stern models yield analytical expressions for $h_i$, while the frequency dependence of the Variable Resistor and Leaky Oxide + Stern + Variable Resistor models are evaluated numerically, after solving systems of ordinary differential equations (ODEs).

The Leaky Oxide and Leaky Oxide + Stern models are created by modifying the interface impedance, $Z_i, i=2,3$. Figures \ref{classic_circuit}, \ref{leaky_ox_circuit}, and \ref{leaky_ox_stern_circuit} shows the circuit model for the Classic, Leaky Oxide, and Leaky Oxide + Stern models. For both the Leaky Oxide, and Leaky Oxide + Stern models, the bulk impedance is a constant value resistor, $Z_{Bulk}=R_{Bulk}$. However, their interface impedances vary. Noting that the leaky oxide and stern layer impedances are

\begin{equation} \label{oxide_stern_impedances}
Z_{Ox} = \frac{R_{Ox}}{1+j\omega R_{Ox}C_{Ox}}, \quad Z_{Stern} = \frac{1}{j\omega C_{Stern}},
\end{equation}

we can right the interface impedance of the Leaky Oxide + Stern model, and the Leaky Oxide models as

\begin{align}
Z_{LeakyOx}(\omega) = Z_{Ox}, \label{leaky_ox_imped} \\
Z_{LeakyOx+Stern}(\omega) = Z_{Ox}+ Z_{Stern}  \label{leaky_ox_stern_imped}.
\end{align}

The Variable Resistor and Leaky Oxide + Stern + Variable resistor models are modifications of the Classic and Leaky Oxide + Stern models respectively. Specifically, both of these models remove the assumption that the bulk resistance is constant, and instead model it as a function of bulk concentration. In addition, these models account add the stern layer, and the nonlinear electric double layer to the interface. We note that it is assumed that the concentration in the bulk changes in a manner that is spatially homogeneous. This yields a system of ODEs of the form,

\begin{align} \label{ode_states}
\dot{\mathbf{y}} = f(\mathbf{y},t,\omega), \nonumber\\ \mathbf{y} =[q^*,V_{Ox}^*,V_{St}^*,V_{EDL}^*,V_{Bulk}^*,i^*,c_0^*,R_{Bulk}^*]^T
\end{align}

where  $q$ is the surface charge on the Electric Double Layer (EDL) capacitor, $V_{Ox}^*$  is the voltage drop across the native oxide, $V_{Stern}^*$ is the voltage drop across the stern layer, $V_{EDL}^*$ is the voltage drop across the EDL, $V_{Bulk}^*$ is the voltage drop across the bulk, $i^*$ is the current, $c_0^*$ is the concentration of the bulk electrolyte, and $R_{Bulk}^*$ is the resistance of the bulk. All of these quantities are dimensionless,  and $V_{Bulk}^*$ is re-dimensionalized to $V_{Bulk}$ when evaluating displacement. The ODEs for the Leaky Oxide + Stern + Variable Resistor model are of the same form as equation (7).

The main difference between these models is in the modification of $V_{Ox}^*$ to account for the parallel addition of $R_{Ox}$ to $C_{Ox}$, figures \ref{leaky_ox_circuit}. Unlike for the case of the Leaky Oxide and Leaky Oxide + Stern models, evaluating the functional dependence on force, $h_i(\omega)$, requires numerical integration. Specifically, we use the RMS integral of $V_{Bulk}$ to obtain

\begin{align} \label{numeric_vrms}
V_{RMS}(\omega) = \sqrt[]{\frac{1}{T} \int_{0}^{T}V_{Bulk}(\omega,t)^2dt} = h_i(\omega)V_{AppRMS},\nonumber\\ i = 3,4
\end{align}

\subsection{ODEs for Numeric Circuit Models}
We have introduced abstractly the Variable Resistor, and the Leaky Oxide + Stern + Variable Resistor models. In this section, we enumerate the non-dimensional ODEs that must be solved in order to evaluate these models. For the variable resistor model, the system of ODEs is

\begin{align}
\frac{dq^*}{dt^*} = i^* \label{charge_state} \\
\frac{dV^*_{EDL}}{dt^*} = \frac{i^*}{C_{EDL}^*} \label{vedl_state} \\%\frac{i}{dq/dV_{EDL}} \\
\frac{dV^*_{Stern}}{dt^*} = \frac{i^*}{C_{Stern}^*} \label{vstern_state} \\%\frac{i}{ \frac{C_{Stern} \lambda_{D}}{C_{Water} \lambda_{Stern}}dq/dV_{0}} \\
\frac{dV^*_{Ox}}{dt^*} = \frac{i^*}{C_{Ox}^*} \label{vox_state} \\
\frac{dc^*_0}{dt^*} = -2i^*sign(q^*) \label{concentration_state}\\
\frac{dR^*_{Bulk}}{dt^*} = \frac{-1}{2c_{0}^{*2}}\frac{dc^*_0}{dt^*} \label{resistor_state}\\
\frac{dV^*_{Bulk}}{dt^*} = \frac{di^*}{dt^*}R^*_{Bulk} + i^*\frac{dR^*_{Bulk}}{dt^*}\label{vbulk_state}  \\
\frac{di^*}{dt^*} = \frac{\bigg(\frac{dV^*_{Bulk}}{dt^*}-2\frac{d}{dt^*}V^*_{INT}-i^*\frac{dR^*_{Bulk}}{dt^*}\bigg)}{R^*_{Bulk}} \label{current_state}
\end{align}
where
\begin{equation}
    V^*_{INT}  = V^*_{EDL}+V^*_{Stern}+V^*_{Ox}
\end{equation}
 and each state has been defined in the paragraph following \ref{ode_states}. $C_0$ is the dimensionless linear differential capacitance of the electric double layer, \ref{lin_diff_cap}, and $C_{EDL}$ is the dimensionless nonlinear differential capacitance of the electric double layer which is expressed as
\begin{equation} \label{edl_nonlinear}
C_{EDL} = C_{0} cosh\bigg(\frac{V^*_{EDL}}{2}\bigg).
\end{equation}

We note four things. First, the differential equation for the bulk resistor, $R_{Bulk}^*$ is obtained by taking a time derivative of the bulk resistance expressed in \ref{bulk_resist_volt}. Second, with the exception of the $C_{EDL}^*$, the non-dimensional capacitance of circuit element $l$, $C_l^*$, is

\begin{equation}
    C_l^* = \bigg(\frac{C_{l} \lambda_{D}}{C_{Bulk} \lambda_{l}}C_0\bigg)
\end{equation}
where $C_{l}$ is the dimensional capacitance of the circuit element, $C_{Bulk}=\frac{\epsilon_W}{g}$ is the capacitance of the bulk which is assumed to be calculated based on the relative permitivitty of water, $\lambda_{l}$ is the thickness of the physical layer represented by circuit element $l$, and $\lambda_{D}$ is the debye lenght. Third, the voltages in this system are where non-dimensionalized using the thermal voltage, $V_T$. Finally, the voltage, $V_{l}$, bulk resistor, $R_{Bulk}^*$, capacitances, $C_l^*$, and time $t^*$ are the only parts of the ODE that are explicitly non-dimensionalized. The non-dimensional charge and current fall out of solving the ODEs with these quantities.

The system of ODEs for the Leaky Oxide + Stern + Variable Resistor model is identical to that of the Variable Resistor model, with the exception of a modification to \ref{vox_state} which is rewritten as
\begin{align} \label{vox_state_leakoxstern}
\frac{dV^*_{Ox}}{dt^*} = \frac{i^* - V^*_{Ox}/R^*_{Ox}}{C^*_{Ox}};
\end{align}

\section{Neglected Effects of Novel Models}
Up until this point, we have neglected two key effects in the development of our models. The first is the depletion layer that can develop in doped silicon, and the second is the formation of concentration gradients in the bulk electrolyte. We describe both of these effects in this section, and justify their explicit absence from our models.

\subsection{Depletion Layer Formation}
The silicon wafers used to fabricate the comb-drive actuators are p-type boron-doped to 0.008-0.01 \textOmega $\textrm{ }cm$. When a voltage is applied to the comb-drive, the resulting external electric field causes charge carriers to collect at the interface of the silicon and native oxide. This region is called the depletion layer, and can also be approximated as a capacitor. The relative importance of this capacitance, compared to other circuit elements in our models, is determined by the width of the depletion layer. We perform a scaling analysis in order to determine the importance of this depletion layer.

Figure \ref{depletion_layer} depicts the depletion layer in series with the native oxide and Stern layers, as well as the bulk electrolyte, for a single comb-drive finger. The analysis detailed in this section assumes that the native oxide is a pure dielectric, and makes use of dimensional quantities. We write Poisson's equation for the depletion layer as
\begin{equation}
\frac{d^2\phi}{dx^2} = -\frac{dE}{dx} = -\frac{e}{\varepsilon_{Si}}(n_+ - n_- + N_d), \quad -x_d \leq x \leq 0
\end{equation}
where $n_+$ and $n_-$ are positive and negative free charge carrier concentrations respectively, and $N_d$ is the concentration of the positive dopant. We solve for the electric field, assuming that there are no free charge carriers in the depletion layer.
\begin{equation}
E(x) = \frac{eN_d(x+x_d)}{\varepsilon_{Si}}.
\end{equation}
We next require that the electric field be continuous at the interface of the native oxide and the depletion layer, $x=0$. and solve for the depletion layer width, $x_d$.

\begin{figure}[htpb]
    \begin{center}
    \includegraphics[width=0.7\linewidth]{Chapter3/figure/depletion_layer.png}
    \caption{Illustration of one finger in a pair of comb-drive fingers. It depicts the Silicon substrate, the depletion layer, the native oxide, the Stern Layer, and the bulk electrolyte.}\label{depletion_layer}
    \end{center}
\end{figure}
% \begin{equation}
% E(x=0) = \frac{eN_dx_d}{\varepsilon_{Si}} = \frac{V_{Ox}}{t_{Ox}}.
% \end{equation}
% Re-writing equation (52) allows us to write the depletion layer width as
\begin{equation}
x_d = \frac{\varepsilon_{Si}}{t_{Ox}}\frac{V_{Ox}}{eN_{d}}.
\end{equation}
If we use the nominal values of the physical parameters in equation (51), shown in \ref{table_phys_param_nomvals}, then the depletion layer width is
\begin{equation}
x_d = 3 \times 10^{-8} V_{Ox}.
\end{equation}
We know that $V_{Ox}$ is at most O(1), so we can approximate the minimum capacitance of the depletion layer as
\begin{equation}
C_{D} \approx \frac{\varepsilon_0 \varepsilon_{Si}}{x_d} = 0.0034
\end{equation}

\begin{table}[!htb]
\begin{center}
{\begin{tabular}{|c|c|}
	\hline
	\textbf{Physical Parameters} & \textbf{Nominal Values} \\
    \hline
    $\varepsilon_{Si}$ & 11.68 \\
    \hline
    $t_{Ox}$ & 2 $nm$ \\
    \hline
    $N_d$  & $10^{25} m^{-3}$  \\
    \hline
\end{tabular}}
\caption{Nominal values of physical parameters used for approximated the width of the depletion layer.}\label{table_phys_param_nomvals}
%\begin{flushleft}
%Table III:
%\end{flushleft}
\end{center}
\end{table}

Since the nominal capacitance of the Stern and native oxide layers are O(0.1) and O(0.01) respectively, it is clear that the effects of the depletion layer capacitance is significant. While this is relevant for conceptualizing the physical system, explicitly modeling the depletion layer does not impact the ability of our reduced models to fit the data. The models presented in this chapter are lumped circuit models, rather than spatial models. As a result, the depletion layer capacitance, which is in series with the Stern or oxide layer capacitance depending on the model, can be lumped in with the capacitance of these layers. When these models are fit, the resulting value of the Stern and oxide capacitance  should be viewed as compensating for both these respective layers and the Depletion layer.

\subsection{Concentration Gradients}
The variable resistor models described in the previous section assumes that the concentration of the electrolytes changes uniformly across the bulk electrolyte, i.e. that the concentration has no gradients. While this is not generally true, the impact of concentration gradients is only relevant when large low-frequency voltages are applied to the bulk. In the case of the comb-drives fabricated for this paper, we postulated that the impedance of the native oxide at low frequencies was sufficiently high enough to prevent this. We confirm in the next chapter that, in the context of our current devices, the variable resistor models that account for concentration gradients fit the data identically to our models that assume uniform concentration. For the sake of completeness, we illustrate here how we account for concentration gradients in our variable resistor model. This derivation is derived based on the work of Bazant et al \cite{Bazant2004}.

Bazant et al. used matched asymptotics to derive expressions for the change in the concentration of a \textit{z:z} electrolyte. They first expressed the dimensionless concentration in the bulk as
\begin{equation} \label{concentration_asympt}
c^* = c_+^* + c_-^* = 2c_0^* + \gamma c_1^*
\end{equation}
where $c_1^*$ accounts for the variation in the bulk concentration, and $\gamma$ is as defined in \ref{lin_diff_cap}. The time derivative of the the bulk concentration is then
\begin{equation} \label{dcdt_asympt}
\frac{dc^*}{dt^*} = \gamma \frac{dc_1^*}{dt^*} = \gamma \frac{d^2c_1^*}{dx^{*2}},
\end{equation}
with the boundary condition
\begin{equation} \label{dcdt_asympt_bc}
\frac{dc_1^*}{dx^*}(x=0,t) = -2 sinh\bigg(\frac{V_{EDL}^*}{2}\bigg) \frac{dV_{EDL}^*}{dt^*}.
\end{equation}
In order to account for the concentration gradient, we replace \ref{concentration_state} with \ref{dcdt_asympt} and \ref{dcdt_asympt_bc}. We finally replace $R_{Bulk}^*$ and $\frac{dR_{Bulk}^*}{dt^*}$ with
\begin{equation}
R_{Bulk}^* = \frac{1}{c^*},
\end{equation}

and

\begin{equation}
\frac{dR_{Bulk}^*}{dt^*} = \int_{0}^{g^*} \frac{-1}{c^{*2}}\frac{dc^*}{dt^*}dx^*.
\end{equation}
respectively.

\section{Chapter Review and Further Work}
In this chapter, we justified the use of a circuit model to describe the behavior of electrolytes between parallel plates using the PNP equations, detailed the classic circuit model that falls out of this circuit model, as well as the assumptions behind this model, and presented a hierarchy of models that addressed the assumptions of the classic circuit model. Moreover, we noted important effects neglected by our reduced models, and stated that, as will be shown in chapter 4, that these had no impact on describing the comb-drive actuators in electrolytes that were fabricated for this dissertation.

In the future, it would be useful to numerically, and experimentally, explore how changes in the parallel plate system, especially the thickness of the native oxide, would impact the importance of effects neglected by our model.


\subsection{Feature Engineering}

Despite the variables provided by original data, we construct some new variables to improve model performance. As we are trying to forecast whether it would precipitate in the next day, the temperature difference in each day could be important. Rain is formatted by condensation of water in air, higher temperature could contribute to evaporation of water and lower temperature may lead to condensation. Therefore, the temperature differences could be useful in predicting precipitation.

With TMAX (Max temperature in the day) and TMIN in the original data, we construct TDIF = TMAX - TMIN to measure the temperature difference each day. By looking into TDIF in different days we find some interesting phenomena. The overall TDIF is significantly larger in summer than in winter, which can be seen from the range of TDIF in Figure \ref{tdif}. It is also interesting that TDIF at mountain areas is higher than downtown areas in summer, and right the opposite in winter. Those two phenomena indicate the existence of a seasonal pattern in TDIF, and this could be the same for PRCP.

\begin{figure}[h]
\centering
\begin{minipage}[t]{0.48\textwidth}
\centering
\includegraphics[width=6cm]{tdif1.png}
\end{minipage}
\begin{minipage}[t]{0.48\textwidth}
\centering
\includegraphics[width=6cm]{tdif2.png}
\end{minipage}
\caption{Stations with TDIF Records}
\label{tdif}
\end{figure}


\subsection{The Bellman equation for Markov Decision Processes}