From 379a9eb27a7046db45d42b3efb439c6989f8d793 Mon Sep 17 00:00:00 2001 From: Kees van Kempen Date: Tue, 18 Oct 2022 17:06:07 +0200 Subject: [PATCH] 06: This proof sucks --- Exercise sheet 6/exercise_sheet_06.ipynb | 9 +++++---- 1 file changed, 5 insertions(+), 4 deletions(-) diff --git a/Exercise sheet 6/exercise_sheet_06.ipynb b/Exercise sheet 6/exercise_sheet_06.ipynb index 4dc034e..be4a9ac 100644 --- a/Exercise sheet 6/exercise_sheet_06.ipynb +++ b/Exercise sheet 6/exercise_sheet_06.ipynb @@ -222,15 +222,16 @@ "The condition for detailed balance is $$\\frac{P(s\\rightarrow s')}{P(s'\\rightarrow s)}=\\frac{\\pi(s')}{\\pi(s)}.$$\n", "Let's work to that expression by following the idea of the proof for the Ising model proof.\n", "\n", - "Starting from the left-hand side, consider the same cluster from the same seed and unit vector $\\hat{n}$ but from different states $s$ and $s'$. Flipping between $s$ and $s'$ is done by the same cluster $C$. The probability of this happening is given by the boundaries of the clusters, namely that they are not added to the cluster, with probability $1 - p_{add}$ per boundary site. Again, like in the Ising model proof, we take the boundary of $C$ to be of $m + n$ length with $m$ the number of aligned sites and $n$ the number of anti-aligned sites. This gives a ratio $$\\frac{P(s\\rightarrow s')}{P(s'\\rightarrow s)} = (1-p_{add})^{m-n}.$$\n", + "Starting from the left-hand side, consider the same cluster from the same seed and unit vector $\\hat{n}$ but between different states $s$ and $s'$. Flipping between $s$ and $s'$ is done by the same cluster $C$, but from opposite sign. The probability of this happening is given by the boundaries of the clusters, namely that they are not added to the cluster, with probability $1 - p_{add}$ per boundary site. This gives a ratio $$\\frac{P(s\\rightarrow s')}{P(s'\\rightarrow s)} = \\frac{\\prod_{^+} (1-p_{add}(s_i, s_j))}{\\prod_{^-} (1-p_{add}(s_i, s_j))}$$ for pairs $^\\pm$ with $s_i \\in C$ and $s_j$ neighbouring $s_i$ just outside the cluster with $\\pm(s_i \\cdot \\hat{n})(s_j \\cdot \\hat{n}) > 0$. Substituting the formula for $p_{add}$ in, we find $$\\frac{P(s\\rightarrow s')}{P(s'\\rightarrow s)} = \\frac{\\prod_{^+} \\exp(-2\\beta(s_i \\cdot \\hat{n})(s_j \\cdot \\hat{n}))}{\\prod_{^-} \\exp(2\\beta(s_i \\cdot \\hat{n})(s_j \\cdot \\hat{n}))} = \\prod_{} \\exp(-2\\beta(s_i \\cdot \\hat{n})(s_j \\cdot \\hat{n})) = \\exp(-2\\beta\\sum_{} (s_i \\cdot \\hat{n})(s_j \\cdot \\hat{n}))$$ with $$ describing both types of pairs.\n", "\n", "For the right-hand side, $$\\frac{\\pi(s')}{\\pi(s)} = e^{\\beta \\left[H_{XY}(s)-H_{XY}(s')\\right]}$$ just from the definition of $\\pi$.\n", "This difference in energy we can write as\n", "$$\n", - "H_{XY}(s)-H_{XY}(s') = \\sum_{ \\in boundary} \\left[ s_i \\cdot s_j - s'_i \\cdot s_j \\right]\n", - " = \\sum_{ \\in boundary} 2(s_i \\cdot \\hat{n})(s_j \\cdot \\hat{n})\n", + "H_{XY}(s)-H_{XY}(s') = \\sum_{} \\left[ s_i \\cdot s_j - s'_i \\cdot s_j \\right]\n", + " = 2\\sum_{} (s_i \\cdot \\hat{n})(s_j \\cdot \\hat{n})\n", "$$\n", - "by summing over all pairs $i, j$ on the boundary of $C$, with $i \\in C$ and $j$ an outside neigbour, as pairs not across the boundary do not give rise to an energy change. The last step is due to" + "by summing over all pairs $$ as above, as pairs not across the boundary do not give rise to an energy change. The last step is due to the flips from both signs to be accounted for.\n", + "This shows $$\\frac{P(s\\rightarrow s')}{P(s'\\rightarrow s)}=\\frac{\\pi(s')}{\\pi(s)},$$ thus that detailed balance holds." ] }, {