Fix expectation values
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@ -8,10 +8,10 @@
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Carlo sampling method. We determined the time of equilibrium to be approximately
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$5000$ Monte Carlo cycles, and used the following samples to find the probability
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distribution at temperature $T_{1} = 1.0 J / k_{B}$, and $T_{2} = 2.4 J / k_{B}$.
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For $T_{1}$ the mean energy per spin is $\langle \epsilon \rangle \approx -1.9969 J$,
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For $T_{1}$ the mean energy per spin is $\langle \epsilon \rangle \approx -1.9972 J$,
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with a variance $\text{Var} (\epsilon) = 0.0001$. And for $T_{2}$, close to the critical
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temperature, the mean energy per spin is $\langle \epsilon \rangle \approx -1.2370 J$,
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with a variance $\text{Var} (\epsilon) = 0.0203$. In addition, we estimated
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temperature, the mean energy per spin is $\langle \epsilon \rangle \approx -1.2367 J$,
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with a variance $\text{Var} (\epsilon) = 0.0202$. In addition, we estimated
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the expected energy and magnetization per spin, the heat capacity and magnetic
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susceptibility. We have estimated the critical temperatures of finite lattice sizes,
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and used these values to approximate the critical temperature of a lattice of
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@ -216,7 +216,7 @@ the magnetic susceptibility.
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\label{fig:phase_susceptibility_1M}
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\end{figure}
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Result of profiling using Score-P in Figure \ref{fig:scorep_assessment}.
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Assessment of profiling using Score-P in Figure \ref{fig:scorep_assessment}.
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{../images/profiling.pdf}
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@ -21,9 +21,9 @@ Continuing, we used the generated samples to compute energy per spin $\langle \e
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magnetization per spin $\langle |m| \rangle$, heat capacity $C_{V}$, and $\chi$.
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In addition, we estimated the probability distribution for temperatures $T_{1} = 1.0 J / k_{B}$,
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and $T_{2} = 2.4 J / k_{B}$. We found that for $T_{1}$ the expected mean energy
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per spin is $\langle \epsilon \rangle \approx -1.9969 J$, with a variance $\text{Var} (\epsilon) = 0.0001$.
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And for $T_{2}$, the mean energy per spin is $\langle \epsilon \rangle \approx -1.2370 J$,
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with a variance $\text{Var} (\epsilon) = 0.0203$.
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per spin is $\langle \epsilon \rangle \approx -1.9972 J$, with a variance $\text{Var} (\epsilon) = 0.0001$.
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And for $T_{2}$, the mean energy per spin is $\langle \epsilon \rangle \approx -1.2367 J$,
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with a variance $\text{Var} (\epsilon) = 0.0202$.
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We estimated the expected energy and magnetization per spin, in addition to the
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heat capacity and susceptibility for lattices of size $L = {20, 40, 60, 80, 100}$.
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@ -26,16 +26,16 @@ The lattice was initialized in an ordered and an unordered state, for both tempe
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no change in expectation value of energy or magnetization for $T_{1}$, when we
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initialized the lattice in an ordered state. As for the unordered initialized
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lattice, we first observed a change in expectation values, and a stabilization around
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$5000$ Monte Carlo cycles. The expected energy per spin is $\langle \epsilon \rangle = -2$
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and the expected magnetization per spin is $\langle |m| \rangle = 1.0$. % add something about what is expected for $T_{1}$ ?
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$5000$ Monte Carlo cycles. The expected energy per spin is $\langle \epsilon \rangle \approx -2$
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and the expected magnetization per spin is $\langle |m| \rangle \approx 1$. % add something about what is expected for $T_{1}$ ?
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For $T_{2}$ we observed a change in expectation values for both the ordered and the
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unordered lattice.
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% \begin{align*}
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% p(s|T=1.0) &= \frac{1}{e^{-\beta \sum E(s)}} e^{-\beta E(s)} \\
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% &= \frac{1}{e^{-(1/k_{B}) \sum E(s)}} e^{-(1/k_{B}) E(s)} \ .
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% \end{align*}
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For $T_{2}$ we observe an increase in expected energy per spin $\langle \epsilon \rangle \approx -1.23$,
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and a decrease in expected magnetization per spin $\langle |m| \rangle \approx 0.46$.
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For $T_{2}$ we observe an increase in expected energy per spin $\langle \epsilon \rangle \approx -1.25$,
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and a decrease in expected magnetization per spin $\langle |m| \rangle \approx 0.47$.
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% Burn-in figures
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\begin{figure}[H]
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\centering
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@ -69,7 +69,7 @@ We used the estimated burn-in time of $5000$ Monte Carlo cycles as starting time
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samples. To visualize the distribution of energy per spin $\epsilon$, we used histograms
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with a bin size $0.02$. In Figure \ref{fig:histogram_1_0} we show the distribution
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for $T_{1}$. Where the resulting expectation
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value of energy per spin is $\langle \epsilon \rangle = -1.9969$, with a low variance
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value of energy per spin is $\langle \epsilon \rangle = -1.9972$, with a low variance
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of Var$(\epsilon) = 0.0001$. %
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\begin{figure}[H]
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\centering
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@ -78,14 +78,14 @@ of Var$(\epsilon) = 0.0001$. %
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\label{fig:histogram_1_0}
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\end{figure} %
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In Figure \ref{fig:histogram_2_4}, for $T_{2}$, the samples of energy per spin is
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centered around the expectation value $\langle \epsilon \rangle = -1.2370$. %
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centered around the expectation value $\langle \epsilon \rangle = -1.2367$. %
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{../images/pd_estimate_2_4.pdf}
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\caption{Distribution of values of energy per spin, when temperature is $T = 2.4 J / k_{B}$}
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\label{fig:histogram_2_4}
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\end{figure} %
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However, we observed a higher variance of Var$(\epsilon) = 0.0203$. When the temperature
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However, we observed a higher variance of Var$(\epsilon) = 0.0202$. When the temperature
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increased, the system moved from an ordered to an unordered state. The change in
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system state, or phase transition, indicates the temperature is close to a critical
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point.
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